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\centerline{\bf The proof of the central limit theorem and Fourier
analysis I.}
\smallskip
\centerline{\it by P\'eter Major}
\centerline{The Mathematical Institute of the Hungarian Academy of
Sciences} \smallskip
\noindent
{\narrower {\narrower
{\it Summary:}\/ The central limit theorem belongs to the most
important results of probability theory. There is another classical
result of the probability theory, the law of large numbers which
states in a slightly informal language that under some mild
conditions the average of the partial sums of independent and
identically distributed random variables can be well
approximated by the expected value of these random variables.
The central limit theorem gives better, more substantial information
about the fluctuation of the average of these partial sums around
their expected value. It states that if we multiply this fluctuation
by $\sqrt n$, where $n$ denotes the number of random variables in the
average, then the distribution of the fluctuation with this
normalization is close to a (non-degenerated) distribution independent
of the number~$n$. Moreover, and this is a most remarkable fact, this
approximating distribution does not depend on the distribution of the
random variables whose average is taken. This distribution is a
``universal law" which is called the normal distribution in the
literature. In the series of problems described in this text we shall
formulate the above result in a more precise and more general form.
 
The method of the proofs also deserves special attention. This method,
which is called the characteristic function method, is actually the
application of the Fourier analysis in the proof of limit theorems.
One goal of this series of problems was to show that this
is a natural approach. The main difference between my text and the
discussion of the central limit theorem in usual text-books is that
I tried to supply a more definitive and detailed explanation about the
relation between the method of the proof of the central limit theorem
and some basic ideas of the Fourier analysis.
 
The theory of the Fourier series gives a natural method to prove
the local version of the central limit theorem, i.e.\ to describe
the asymptotic behaviour of the density and not of the distribution
function of the normalized partial sums of independent random
variables. Furthermore, if we understand what kind of additional
problems arise if we want to prove the central limit theorem in its
original (and not in its local) form and also understand the answers
supplied by the results of the Fourier analysis to these questions,
then we can prove the desired results.
 
Moreover, in such a way we get a natural method to prove some
refinements of the central limit theorem. We also can investigate the
question how good approximation is supplied by the central limit theorem
for the distribution of normalized partial sums of independent random
variables, and what kind of better approximation can be given. But
these problems are not discussed here. They will be the subject of the
second part of this series of problems.
 
Finally, I mention some parts of this work which may be of special
interest. They are the proof of the Stirling formula in a way suggested
by the proof of the local central limit theorem, (problem~2), the
necessary and sufficient condition of the relative compactness of a
sequence of probability measures expressed by means of their
characteristic functions (problem~22), a detailed discussion of
the question how the properties of a function or a measure are reflected
in the behaviour of their Fourier transform (problems~27--34 and the
discussion about these problems). \par} \par}
 
\bigskip\noindent
\centerline{\bf CONTENT}
\bigskip
\halign { \vbox{\hsize=10truecm \parindent=18pt
\vskip4pt # } \qquad & \hfill  # \quad & \hfill # \cr
& {problems} & {solutions}\cr
\noalign{\smallskip}
\item{A.} Local limit theorems. \dotfill &3--10 &42--50\cr
\item{B.} The definition of the normal distribution function and of
the characteristic function. Some important results related to these
notions. \dotfill  &10--13 &51--53\cr
\item{C.} The definition of the convolution and some of its important
properties. \dotfill &13--16 &53--56\cr
\item{D.} Convergence in distribution. \dotfill &16--22 &56--62\cr
\item{E.} The relation between the convergence of distribution
functions and the convergence of their characteristic functions.
\dotfill &22--24 &62--65\cr
\item{F.} The relation between the behaviour of a function or measure
and its Fourier transform. \dotfill &24--29 &65--70\cr
\item{G.} The central limit theorem. \dotfill  &29--35 &70--75\cr
\item{H.} The multi-dimensional central limit theorem. \dotfill  &35--38
&75--80\cr
\noalign{\medskip}
\item{}{\it Some additional remarks.} \dotfill  & 38--41 \cr
\noalign{\medskip}
\item{}{\it Appendix.} \dotfill   && 80--85\cr
\itemitem{a.)} The proof of the inversion formula for Fourier transforms
together with the proof of the Parseval formula. \dotfill && 80 \cr
\itemitem{b.)} The proof of Weierstrass second approximation theorem
together with the proof of Fej\'er's theorem. \dotfill  &&83 \cr}
 
\vfill\eject
 
\headline{\ifodd\pageno \hfill  {\it Central limit theorems and Fourier
analysis} \hfill  \else \hfill {\it P\'eter Major} \hfill \fi}
 
 
 
\noindent
{\script A.) Local limit theorems.} 

\medskip
Let us first consider the following problem: Let $\xi_j$,
$j=1,2,\dots$, be a sequence of independent and identically
distributed random variables which take integer values. We introduce
the notation $P(\xi_1=k)=p(k)$, $k=0,\pm1,\pm2,\dots$,
$\summ_{k=-\infty}^\infty p(k)=1$ and define the partial sums
$S_n=\summ_{j=1}^n \xi_j$, $n=1,2,\dots$. Let us consider the
probabilities $p_n(k)=P(S_n=k)$, $k=0,\pm1,\pm2,\dots$, $n=1,2,\dots$,
and try to give a good approximation of the probabilities
$p_n(k)$ in the case of large parameters $n$.
 
A good estimation can be given for these probabilities $p_n(k)$ by
means of the following method. Let us define the Fourier series
$$
P_n(t)=\sum_{k=-\infty}^\infty p_n(k)e^{ikt},\quad -\pi\le t\le \pi.
\tag1
$$
By a basic formula of the theory of Fourier series the coefficients of
the above Fourier series can be expressed by means of the formula
$$
p_n(k)=\frac1{2\pi}\int_{-\pi}^\pi e^{-ikt}P_n(t)\,dt,\quad
k=0,\pm1,\pm2,\dots. \tag2
$$
Hence, if we give a good asymptotic formula for the Fourier series
$P_n(t)$ and  estimate well the integral in formula~(2), then
we get a good estimate for the probabilities $p_n(k)$ we are
interested in. Let us observe that $P_n(t)=Ee^{itS_n}$, $-\pi\le
t\le\pi$. Besides, as $S_n$ is the sum of independent and
identically distributed random variables,
$$
P_n(t)=Ee^{it(\xi_1+\cdots+\xi_n)}=\(Ee^{it\xi_1}\)^n=\(P_1(t)\)^n,
$$
where $P_1(t)=\summ_{k=-\infty}^\infty P(\xi_1=k)e^{ikt}$. Furthermore,
$P_1(0)=1$, and as we shall see, $|P_1(t)|<1$ if $-\pi\le t\le\pi$ and
$t\neq0$ under some natural conditions. Hence in the estimation of the
probabilities $p_k(n)$ a so-called singular integral appears in
formula~(2), in which only a small neighbourhood of the origin gives an
essential contribution to the integral. Such expressions can be well
bounded by some standard methods of the analysis. Let us first show the
following identities which we later need. 

\medskip
\item{1.)} Let us show that
$$
\frac1{\sqrt {2a\pi}}\int_{-\infty}^\infty e^{-u^2/2a}\,du=1
$$
for all real positive numbers $a>0$. Furthermore,
$$
\frac1{\sqrt {2a\pi}}\int_{-\infty}^\infty e^{-(u-z)^2/2a}\,du=1
$$
for all real positive numbers $a>0$ and  {\it complex}\/ numbers~$z$.

\medskip
First we apply the above method in a special case. Let us consider the
Poisson distribution with parameter $\lambda=n$, that is the
distribution of a random variable $\eta$ for which
$P(\eta=k)=P_n(k)=\frac{n^k}{k!}e^{-n}$, $k=0,1,2,\dots$. The
distribution of a Poissonian random variable with parameter~$n$ agrees
with the distribution of the sum of $n$ independent Poissonian random
variables with parameter~1. Hence the above sketched method makes
possible to get a good estimate for the probabilities $P_n(k)$. We get
a really good estimate in the case when the number $k$ is close to the
expected value of the random variable $\eta$, i.e.\ if $n\sim k$. In
particular, by investigating the appropriate Fourier series we get a
good bound on the number $P_n(n)$. We do not exploit explicitly the
probabilistic content of the coefficients $P_n(k)$ of this Fourier
series, but the calculations indicate what kind of technical problems
has to be solved in the analogous problems of probability theory.
Besides, we get in such a way the proof of an important result of
the analysis, the proof of the Stirling formula. This is the content of
the next problem. 

\medskip
\item{2.)} Let us calculate the value of the Poissonian distribution
with parameter $n$ in the point $n$ with the help of the above discussed
method by working with the Fourier series whose coefficients are
the values of this Poisson distribution. Let us show with the help of
this method that
$$
n!=\(\frac ne\)^n\frac{2\pi}{\int_{-\pi}^\pi e^{n(e^{it}-1-it)}\,dt}.
\tag3
$$
Let us show that
$$
\int_{-\pi}^\pi e^{n(e^{it}-1-it)}\,dt=\frac{\sqrt{2\pi}}{\sqrt
n}\(1+O\(\frac1n\)\) \tag4a
$$
and
$$
n!=\sqrt{2\pi n}\(\frac ne\)^n\(1+O\(\frac 1n\)\). \tag4b
$$
Let us prove the following improvement of the previous two statements:
$$
\int_{-\pi}^\pi e^{n(e^{it}-1-it)}\,dt=\frac{\sqrt{2\pi}}{\sqrt
n}\(1+\frac {c_1}{n^{1/2}}+\frac{c_2}n+\cdots
+\frac{c_k}{n^{k/2}}+O\(\frac1{n^{(k+1)/2}}\)\) \tag4c
$$
for arbitrary integer $k\ge1$ with explicitly calculable coefficients
$c_k$. In particular, $c_1=0$. Furthermore,
$$
n!=\sqrt{2\pi n}\(\frac ne\)^n \(1+\frac {\bar c_1}{n^{1/2}}
+\frac{\bar c_2}n+\cdots+\frac{\bar
c_k}{n^{k/2}}+O\(\frac1{n^{(k+1)/2}}\)\) \tag4d
$$
for arbitrary integer $k\ge1$ with explicitly calculable coefficients
$\bar c_k$. In particular, $\bar c_1=0$.

\medskip
Similarly to the solution of problem 2 a good asymptotic formula can be
given for the probability of the event that the sum of independent and
identically distributed random variables takes a definite value. This
investigation is based on formula~(2) where the integral of a Fourier
series with periodicity $2\pi$ has to be estimated. But such cases may
appear when this Fourier series also has a smaller periodicity. Such an
example appears for instance if the partial sums of such random
variables are considered which take only even or only odd values. The
Fourier series determined by the distribution of such random variables
has a smaller periodicity $\pi$. If we want to study the asymptotic
behaviour of the distribution of partial sums of independent and
identically distributed random variables in a systematic way, then we
have to clarify which is the smallest period of the Fourier series in
formula~(2). In particular, we want to tell when the smallest
periodicity of this Fourier series is precisely $2\pi$. This is the
reason why we introduce the following definition. Then we study
the case when the smallest periodicity of the Fourier series in
formula~(2) is $2\pi$. The investigation of partial sums of independent
and identically distributed random variables whose distribution takes
values on a lattice $kh+b$, $k=0,\pm1,\pm2,\dots$, with some numbers $h$
and $b$ can be relatively simply reduced to this special case.

\medskip\noindent
{\bf Definition A.} {\it The values of a random variable $\xi$
are concentrated on the lattice of integers (as on the rarest lattice)
if $\summ_{k=-\infty}^\infty P(\xi=k)=1$, and for arbitrary integers
$A>1$ and $B$  $\summ_{k=-\infty}^\infty P(\xi=Ak+B)<1$.
 
More generally, we call a random variable $\xi$ lattice valued if its
values are concentrated with probability one on a set of the form
$\{b+kh,\colon\; k=0,\pm1,\pm2,\dots\}$ with some real numbers $h>0$ and $b$.
We say that the values of a random variable $\xi$ are concentrated on
a lattice of width $h$, $h>0$, (as on the rarest lattice) if there
exists a real number $b$ such that $\summ_{k=-\infty}^\infty
P(\xi=kh+b)=1$, and $\summ_{k=-\infty}^\infty P(\xi=Akh+B)<1$ for all
integers $A>1$ and real numbers $B$.} 

\medskip
 To carry out the further investigation we need the following result.

\medskip
\item{3.)} If $\xi$ is a non-constant random variable concentrated
on a lattice, then there exist a smallest $h>0$ number such that
it is concentrated on a lattice of width $h$ as the rarest lattice. 

\item{} Let $\xi$  be a random variable concentrated on a lattice of
width $h$ (as on the rarest lattice). Let us choose a real number
$b>0$ such that $\summ_{n=-\infty}^\infty P(\xi=nh+b)=1$, and let us
consider the Fourier series $P(t)=\summ_{n=-\infty}^\infty
e^{inh}P(\xi-b=nh)$ determined by the distribution of the random
variable $\xi-b$. The periodicity of the Fourier series $P(t)$ equals
$\frac{2\pi}h$, $P(0)=1$, $|P(t)|\le1$ for all real numbers $t$, and
$|P(t)|<1$ if $|t|\le\frac\pi h$ and $t\neq0$. If the absolute value
of the random variable $\xi-b$ has $k$-th moment, that is
$E|\xi-b|^k<\infty$, then the function $P(t)$ is $k$ times continuously
differentiable, and
$\left.\frac{dP^{k}(t)}{dt^k}\right|_{t=0}=i^kE(\xi-b)^k$, (where
$i=\sqrt{-1}$).
\item{4.)} Let $\xi_1,\xi_2,\dots$, be a sequence of independent and
identically distributed random variables whose values are concentrated
with probability one on the lattice of the integers (as on the rarest
lattice). Let $E\xi_1=m$, $E\xi_1^2=m_2<\infty$, (that is we assume that
the second moment of the random variable $\xi_1$ is finite), and put
$\sigma^2=m_2-m_1^2$. (The number $\sigma^2$ denotes the variance of
the random variable $\xi_1$.) Let us consider the partial sums
$S_n=\xi_1+\cdots+\xi_n$, $n=1,2,\dots$. Then
$$
P(S_n=k)=\frac1{\sqrt{2\pi n}\sigma}\exp\left\{-\frac{(k-nm)^2}
{2n\sigma^2} \right\}+o\(\frac1{\sqrt n}\),\quad k=0,\pm1,\pm2,\dots,
$$
and $o(\cdot)$ is uniform in the variable $k$.
\item{5.)} Let us consider such a sequence $\xi_1,\xi_2,\dots$,
of independent and identically distributed random variables which
satisfies the conditions of the previous problem together with the
condition  $E|\xi_1|^3<\infty$. Under these conditions prove the
following sharper form of the previous problem.
$$
P(S_n=k)=\frac1{\sqrt{2\pi n}\sigma}\exp\left\{-\frac{(k-nm)^2}
{2n\sigma^2} \right\}+\e(n,k),\quad k=0,\pm1,\pm2,\dots,
$$
where $|\e(n,k)|\le\frac Kn$, and the constant $K$ depends only on the
distribution of the random variable $\xi_1$.

\medskip
Historically first that special case of the above limit theorems was
considered where the random variable $\xi_1$ is binomially distributed,
i.e.\ $P(\xi_1=1)=1-P(\xi_1=0)=p$, $0<p<1$. In this case the
distribution of the sum $S_n$ can be expressed in a simple explicit
form, and this expression can be well bounded with the help of the
Stirling formula. As this special case has a particular importance in
combinatorial applications, it deserves special discussion. This is
done in the following problem. 

\medskip
\item{5a.)} Let us give an elementary proof of the statement of problem
5 in the special case when $\xi_1$ is binomially distributed. (Here we
want to give a proof where the Stirling formula is applied instead of
the Fourier analysis.)
\item{6.)} Let a sequence $\xi_1,\xi_2,\dots$ of independent and
identically distributed random variables satisfy the conditions of
problem~4 with the difference that now we assume that the values of the
random variable $\xi_1$ are concentrated with probability~1 on a lattice
of width $h$, $h>0$, (which is the width of the rarest lattice). Let the
values of the random variable $\xi_1$ be concentrated on the numbers
$kh+b$, $k=\pm1,\pm2,\dots$, with some real number $b$. Then with the
notations of problem~4
$$
\aligned
P(S_n=kh+nb)&=\frac h{\sqrt{2\pi n}\sigma}\exp\left\{
-\frac{(kh+nb-nm)^2} {2n\sigma^2} \right\}+o\(\frac1{\sqrt n}\),\\
&\qquad\qquad k=0,\pm1,\pm2,\dots,
\endaligned \tag5
$$
where $o(\cdot)$ is uniform in the variable $k$. If also the condition
$E|\xi_1|^3<\infty$ is satisfied, then
$$
\align
P(S_n=kh+nb)&=\frac h{\sqrt{2\pi n}\sigma}\exp
\left\{-\frac{(kh+nb-nm)^2}{2n\sigma^2} \right\}+\e(k,n), \\
&\qquad\qquad k=0,\pm1,\pm2,\dots,
\endalign
$$
where $|\e(n,k)|\le\frac Kn$, and the constant $K$ depends only on the
distribution of the random variable $\xi_1$.
\item{7.)} If a sequence of random variables $S_n$, $n=1,2,\dots$,
satisfies relation (5), (it has no importance what additional properties
these random variables have), then
$$
\lim_{n\to\infty}P\(\frac{S_n-nm}{\sqrt n\sigma}<x\)
=\int_{-\infty}^x \frac1{\sqrt{2\pi}}e^{-u^2/2}\,du
$$
for all real numbers $x$. The convergence is uniform with respect to the
parameter~$x$.

\medskip
The above results supply good information on the probabilities the
partial sums of lattice valued independent and identically distributed
random variables take different values. Then by ``integrating out"
this relation in problem~7 we have obtained an estimate about the
limit behaviour of the distribution of the appropriately normalized
partial sums of independent and identically distributed random variables
if the summands are lattice distributed. These results can be sharpened,
but first we want to generalize the above result and to investigate the
limit behaviour of the distribution or density (if the density function
exists) function of appropriately normalized partial sums of
independent random variables which are not necessarily lattice
distributed.
 
Let $\xi_1,\xi_2,\dots$, be a sequence of independent random
variables, and let us consider the partial sums
$S_n=\xi_1+\cdots+\xi_n$, $n=1,2,\dots$. We want to show that if the
random variable $\xi_1$ has a nice density function, then a good
asymptotic formula can be given for the density function of the random
sum $S_n$. Besides, we want to show under general conditions that
the appropriate normalized version of the sums have a limit
distribution as $n\to \infty$, and also want to describe this limit
distribution function. In the case when the random variables $\xi_n$
are lattice valued then the corresponding problems could be solved by
means of the investigation of an appropriate Fourier series. The
question arises whether this method can be adapted to solve the
analogous problem in the general case. To carry out such a program we
should have a relatively simple inversion formula which enables to
calculate a density or distribution function by means of its Fourier
transform.
 
In case of nice density functions there exists such a simple inversion
formula. In case of general distribution function there are only
complicated inversion formulas which are not really well applicable in
the investigations we have in mind. To prove limit theorems for the
appropriately normalized partial sums of independent random variables
first we have to understand what means exactly convergence in
distribution. Then we can prove limit theorems for the normalized
partial sums of independent random variables by means of some basic
results of the Fourier analysis. Before doing this we formulate an
inversion formula for the Fourier transform which is useful in the
investigation of the density functions. 

\medskip\noindent
{\bf Inversion formula for Fourier transforms.} {\it Let $f(u)$  be an
integrable function on the real line, i.e. let us assume that
$\int_{-\infty}^\infty |f(u)|\,du<\infty$.  Let us consider the
Fourier transform
$$
\tilde f(t)=\int_{-\infty}^\infty e^{itu}f(u)\,du, \quad
-\infty<t<\infty \tag6a
$$
of this function $f(\cdot)$. If the function $\tilde f$ is also
integrable, i.e. $\int_{-\infty}^\infty|\tilde f(t)|\,dt<\infty$, 
then the identity
$$
f(u)=\frac1{2\pi}\int_{-\infty}^\infty e^{-itu}\tilde f(t)\,dt\quad
\tag6
$$
holds in almost all points of the real line with respect to the Lebesgue
measure. Moreover, even the following stronger statement holds. Let
$\mu$ be a finite measure, i.e. let $\mu(R^1)<\infty$. Let $\tilde
f(t)=\int e^{itu}\,d\mu(u)$ denote the Fourier transform of this measure
$\mu$. If the function $\tilde f(\cdot)$ is integrable, then the measure
$\mu$ has a density function, and it agrees with the function
$f(\cdot)$ defined in formula~(6). This result can be slightly
generalized. It also holds if $\mu$ is a signed measure with bounded
variation, i.e. it is the difference of two finite measures.}

\medskip
The above definition of the Fourier transform slightly differs from
the  definition usually given in the literature, where the integral in
formula~(6a) and its analog which defines the Fourier transform of
a measure~$\mu$ is divided by $\sqrt{2\pi}$. With such a normalization
the Fourier transform and the formula expressing its inverse are more
similar to each other. For us the formula given in~(6a) is more
convenient. We shall prove this inversion formula for Fourier transforms
in the Appendix. Let us remark that the restriction that formula~(6)
holds only for almost all points $u$ with respect to the Lebesgue
measure is natural. Indeed, by modifying a function on a set of
Lebesgue measure zero we get a new function which has the same
Fourier transform.
 
Let us finally remark that the formula about the calculation of the
Fourier coefficients by means of the Fourier transform suggest the
inversion formula~(6). Namely, let us approximate a function in the
points $k\e$, $k=0,\pm1,\pm2,\dots$, in a natural way and let us write
the Fourier series with these coefficients (in this case we consider
the Fourier series in the interval $\[-\frac\pi\e,\frac\pi\e\]$).
Then by expressing the Fourier coefficients of this Fourier series
with the help of the Fourier series and by applying the limit procedure
$\e\to0$ we get, at least on a formal level, formula~(6).

\medskip
\item{8.)} Let $\xi_1,\xi_2,\dots$, be a sequence of independent and
identically distributed random variables, $E\xi_1=0$,$E\xi_1^2=1$, 
(i.e.\ we consider random variables with expectation zero and 
variance one), and put $S_n=\xi_1+\cdots+\xi_n$, $n=1,2,\dots$. 
Let us also assume that the random variable $\xi_1$ has a density 
function $f(x)$, and the Fourier transform $\varphi(t)$ of the 
density function $f(x)$ is integrable or at least it satisfies the
following weakened condition: There exists an integer $k\ge1$ 
such that $\varphi^k(t)$ is integrable. Then the random variable
$\frac{S_n}{\sqrt n}$ has a density function $f_n(x)$ which 
satisfies the relation
$$
\lim_{n\to\infty}f_n(x)=\frac1{\sqrt{2\pi}}e^{-x^2/2} \quad
\text{for all real numbers }x, \tag7
$$
and the convergence is uniform in the parameter~$x$.
\item{9.)} If a sequence of distribution functions
$F_n(x)$, $n=1,2,\dots$ have  density functions $f_n(x)$,
and they satisfy formula~(7), then
$$
\lim_{n\to\infty}F_n(x)=\int_{-\infty}^x\frac1{\sqrt{2\pi}}
e^{-u^2/2}\,du \quad \text{for all real numbers } x.
$$
The convergence is uniform with respect to the variable~$x$.
\item{10.)} If the conditions of problem~8 are satisfied, and also the
relation $E|\xi_1|^3<\infty$ holds, then  relation~(7) holds with a
better error term. Namely,
$$
f_n(x)=\frac1{\sqrt{2\pi}}e^{-x^2/2}+O\(\frac1{\sqrt n}\),
$$
and $O(\cdot)$ is uniform with respect to the parameter~$x$.

\medskip\noindent
{\it Remark:}\/ We have proved in problems~6 and~10 that if the
absolute value of the random variables have finite third moment,
then the density function of the appropriately normalized sums can
be approximated with the normal density function with an accuracy of
order $O\(n^{-1/2}\)$, where $n$ is the number of summands in the
sum. In the second part of this series of problems we shall prove an
analog of this result about the approximation of the appropriately
normalized distribution function of partial sums of independent and
identically distributed random variables. This is an important result
of the probability theory, which is called the Berry--Esseen inequality.
It yields an estimate of the same order $O\(n^{-1/2}\)$ for the
approximation of the normalized distribution function by the standard
normal distribution function. Nevertheless, there is an essential
difference between the estimate supplied by the Berry--Esseen
inequality and the results of problems~6 and~10.
 
The Berry--Esseen inequality gives an upper bound of the form $\const
\mu_3 n^{-1/2}$, where $\mu_3$ is the third moment of the absolute
value of the summands, and the $\const$ is a universal number which
can be chosen independently of the distribution of the summands. One
cannot give a similar universal estimate on the normal approximation
of the density function. Indeed, if for instance the density function
of the summands take extremely large values in a small neighbourhood
of the origin and the probability of this neighbourhood is relatively
large, say larger than $\frac1{10}$, then for relatively small indices
$n$ the density function of the normalized sum of $n$ independent
random variables cannot be well approximated by a normal density
function, although the third moment of the absolute value of the
summands may be not large. A similar example can be given for the
density function of the partial sums of lattice valued random
variables if the width of the lattice may be very small.
 
The above sketched examples show that the analog of the Berry--Esseen
inequality for density functions does not hold. This difference is
related to the fact that in the first case we study distribution
functions while in the second case their derivatives. It is a quite
common experience in analysis that for the nice behaviour of a
sequence of functions we have to impose much weaker conditions than for
the nice behaviour of their derivatives.

\medskip
A deficiency of the results of problems~8 and~9 is that it imposes
conditions not for the density but for the characteristic function
of the summands, while in typical applications we have some direct
information about the density and not on the characteristic function.
Hence we have to understand the relation between the properties of
a density and the corresponding characteristic function. Later we
shall return to this question. We shall see that relation~(7)
holds for the density function of normalized sums of independent
and identically distributed random variables if the summands have
``nice" density function.

\medskip\noindent
{\script B.) The definition of the normal distribution function and
of the characteristic function. Some important results related to
these notions.} 

\medskip
Let us introduce the definition of the (standard) normal distribution.
The solutions of the previous problems suggest that in limit
distributions the normal limit distribution appears as the limit.

\medskip\noindent
{\bf The definition of the normal distribution.} {\it The $\Phi(x)$
standard normal distribution function is the distribution function
whose density function the standard density function $\varphi(x)$ is of
the form $\varphi(x)=\frac1{\sqrt{2\pi}}e^{-x^2/2}$, that is
$$
\Phi(x)=\int_{-\infty}^x\frac1{\sqrt{2\pi}}e^{-u^2/2}\,du.
$$
More generally, a linear transform of the standard normal distribution
function $\Phi(x)$ is called a normal distribution function. More
explicitly, a normal  distribution function can be characterized by two
parameters, by a real number $m$ and a positive real number $\sigma$.
The normal distribution function with parameters $m$ and $\sigma$ equals
$\Phi_{m,\sigma}(x)=\Phi\(\frac{x-m}\sigma\)$. Its density function is
$\varphi_{m,\sigma}(x)=\frac1\sigma\varphi\(\frac{x-m}\sigma\)
=\frac1{\sqrt{2\pi}\sigma}e^{-(x-m)^2/2\sigma^2}$.}

\medskip\noindent 
It follows from the solution of problem~1 that
$\Phi(x)$ is really a distribution function. The normal distribution is
also called the Gaussian distribution in the literature. \medskip
\item{11.)} Let us show  that the expected value of a standard
normal random variable equals zero, and its variance equals one. The
expected value of a random variable with distribution function
$\Phi_{m,\sigma}(x)$ equals $m$ and its variance equals
$\sigma^2$.

\medskip
Let us consider the appropriately normalized partial sums of
independent and identically distributed random variables with a nice
density function. We have seen that the density functions of these
normalized partial sums tend to the normal density and their
distribution functions to the normal distribution function. A similar
result holds for sums of lattice valued random variables. We say that
the appropriate normalizations $\bar S_n=\frac{S_n-a_n}{b_n}$,
$n=1,2,\dots$, of a sequence of random variables $S_n$, $n=1,2,\dots$,
satisfy the local central limit theorem if either the random variables
$\bar S_n$ have a density function for large $n$, and they converge
uniformly to the standard normal distribution function, or for all
large $n$ there exists a lattice $kh_n+b_n$ of width $h_n$,
$k=0,\pm1,\pm2,\dots$, such that $h_n\to0$ if $n\to\infty$, the random
variable $\bar S_n$ is concentrated on this lattice (as on the rarest
lattice), and $P(S_n=kh_n+b_n)=h_n\varphi(kh_n+b_n)+o(h_n)$,
$k=0,\pm1,\pm2,\dots$, $o(\cdot)$ is uniform in the variable~$k$, where
$\varphi(x)$ is the standard normal density function.
 
We have seen in problems 6 and 8 that the appropriately normalized sums
of independent and identically distributed random variables with
density function or lattice valued distribution satisfy the local
central limit theorem under fairly general conditions. The
normalization is made in a natural way. With the notation of the
previous paragraph we choose $a_n=nE\xi_1=ES_n$, $b_n=\sqrt {n\,
\text{Var}\,\xi_n}=\sqrt {\,\text{Var}\,S_n}$, i.e.\ such a
normalization is chosen with which the normalized partial sums have
expected value zero and variance~one. We also have seen that the local
central limit theorem implies the (global) central limit theorem
(problems~7 and~9). One would expect that the central limit theorem may
hold also in such cases when its local version fails to hold. We want
to show that this belief is right. Moreover, we want to prove the
central limit theorem in more general cases when appropriately
normalized partial sums of independent but not necessarily identically
distributed random variables are considered.
 
We have proved the local central limit theorem with the help of the
Fourier transform of the distribution functions we have investigated.
We want to show that this method can be adapted to the proof of
(global) central limit theorems. To do this we recall the definition of
the Fourier transform of general (probability) measures. We shall call
them, following the tradition of probability theory terminology,
characteristic functions. Since later we also want to study partial
sums of vector valued random variables, hence --- to avoid some
repetition --- we define the multi-dimensional version of the
characteristic functions.  

\medskip\noindent
{\bf The definition of characteristic functions.} {\it Let
$F(u)=F(u_1,\dots,u_k)=P(\xi_1<u_1,\dots,\xi_k<u_k)$ denote the
distribution function of a $k$-dimensional random vector
$(\xi_1,\dots,\xi_k)$. The $\varphi(t)=\varphi(t_1,\dots,t_k)$,
$t=(t_1,\dots,t_k)$, characteristic function of a $k$-dimensional
distribution function~$F$ or of an $F$ distributed $k$-dimensional
random vector $\xi=(\xi_1,\dots,\xi_k)$, $k\ge1$, is defined by the
formula
$$
\align
\varphi(t)=\varphi(t_1,\dots,t_k)&=Ee^{i(t,\xi)}=
Ee^{i(t_1\xi_1+\cdots+t_k\xi_k)}\\
&=\int e^{i(t,u)}F(\,du) =\int
e^{i(t_1u_1+\cdots+t_ku_k)}F(\,du_1,\dots,\,du_k),
\endalign
$$
where $t=(t_1,\dots,t_k)$ is an arbitrary point of the $k$-dimensional
Euclidean space.
(Here we applied the following notation: If
$u=(u_1,\dots,u_k)$ and $v=(v_1,\dots,v_k)$ are two $k$-dimensional
vector, then
$(u,v)=u_1v_1+\cdots+u_kv_k$ denotes the scalar product of the vectors
$u$ and $v$.)} 

\medskip \noindent
\item{12.)} A characteristic function $\varphi(t_1,\dots,t_k)$
is uniformly continuous in the $k$-dimensional Euclidean space
$R^k$, $\varphi(0,\dots,0)=1$ and
$$
|\varphi(t_1,\dots,t_k)|\le1\quad\text{in all points }
(t_1,\dots,t_k)\in R^k.
$$
Let $\varphi(t)=\varphi(t_1,\dots,t_k)$, $t=(t_1,\dots,t_k)$, be the
characteristic function of a random vector $\xi=(\xi_1,\dots,\xi_k)$.
Let $a$ be a real number and $m=(m_1,\dots,m_k)$ a
$k$-dimensional vector. Then the characteristic function of the random
vector $a\xi+m=(a\xi_1+m_1,\dots,a\xi_k+m_k)$ equals
$$
e^{i(m,t)}\varphi(at)
=e^{i(m_1t_1+\cdots+m_kt_k)}\varphi(at_1,\dots,at_k).
$$
\item{} If $\xi_1,\dots,\xi_n$, $1\le j\le n$,  is a sequence of
independent $k$-dimensional random vectors with the notation
$\xi_j=(\xi_j^{(1)},\dots,\xi_j^{(k)})$, and
$\varphi_j(t)=\varphi_j(t_1,\dots,t_k)$, $j=1,\dots,n$,
$t=(t_1,\dots,t_k)$, denotes their characteristic functions, then the
characteristic function of the sum $\xi_1+\cdots+\xi_n$ equals
$\prodd_{j=1}^n\varphi_j(t)=\prodd_{j=1}^n\varphi_j(t_1,\dots,t_k)$.

\medskip
Let us calculate the characteristic function of some important
distribution functions.
\item{13.)} Let us show that if the random variable $\xi$
\item{a.)} has standard normal distribution, i.e.\ it has a density
function of the form $f(u)=\frac1{\sqrt{2\pi}}e^{-u^2/2}$, then its
characteristic function equals $\varphi(t)=e^{-t^2/2}$.
\item{b.)} has uniform distribution in the interval $[0,1]$, i.e.\ it
has a density function of the form $f(u)=1$ if $0\le u\le1$ and
$f(u)=0$ is otherwise, then its characteristic function equals
$\varphi(t) =\frac{e^{it}-1}{it}$.
\item{c.)} has exponential distribution with parameter $\lambda>0$,
i.e.\ it has a density function of the form $f(u)=\lambda e^{-\lambda
u}$ if $u\ge0$ and $f(u)=0$ if $u<0$, then its characteristic function
equals $\varphi(t)=\frac\lambda{\lambda-it}$.
\item{d.)} has Cauchy distribution, i.e.\ it has a density function of
the form $f(u)=\frac1\pi\frac1{1+u^2}$, then its characteristic
function equals $\varphi(t)=e^{-|t|}$.
\item{e.)}  has Poisson distribution with parameter $\lambda>0$,
i.e.\ $P(\xi=k)=\frac{\lambda^k}{k!}e^{-\lambda}$, $k=0,1,2,\dots$,
then its characteristic function equals
$\varphi(t)=\exp\left\{\lambda(e^{it}-1)\right\}$.
\item{f.)}  has binomial distribution with parameters $n$ and $p$
where $n\ge1$ is an integer, $0<p<1$, i.e.\ $P(\xi=k)=\binom nk
p^k(1-p)^{n-k}$, $k=0,1,\dots,n$, then its characteristic function
equals $\varphi(t)=(1-p+pe^{it})^n$.
\item{g.)}  has negative binomial distribution with parameters $n$ and
$p$, where $n\ge1$ is an integer, $0<p<1$, i.e.\
$P(\xi=k)=\binom{n+k-1}{k}p^k(1-p)^n$, $k=0,1,2,\dots$, then its
characteristic function equals
$\varphi(t)=\(\frac{1-p}{1-pe^{it}}\)^n$.
\item{h.)} has $\gamma$ distribution with parameter $s$, $s>0$, i.e.,
its density function is $\gamma_s(u)=\frac1{\Gamma(s)}u^{s-1}e^{-u}$, 
if $u>0$, and $\gamma_s(u)=0$, if $u<0$, where 
$\Gamma(s)=\int_0^\infty u^{s-1} e^{-u}\,du$, then its characteristic
function equals $\varphi_s(t)=\frac1{(1-it)^s}$.

\medskip
It may be worth mentioning that the density function $\gamma_s(u)$,
considered in point~h.) shows some similarity with the Poissonian 
distribution, and this can be exploited to prove, similarly to 
the solution of Problem~2, a good asymptotic formula for the function
$\Gamma(s)$ if $s\to\infty$. This means a generalization of
Stirling's formula, since one prove with the help of some partial 
integration that $\Gamma(n)=(n-1)!$.

Let us observe that $\gamma_{s_1}(u)*\gamma_{s_2}(u)=\gamma_{s_1+s_2}(u)$,
where $*$ denotes  convolution. This can be seen e.g. with the
help of the form of the characteristic function of $\gamma_s$ and
some important properties of the characteristic functions discussed
later. Further, we can write the identity
$\gamma_s(s)=\frac{s^{s-1}e^{-s}}{\Gamma(s)}
=\frac1{2\pi}\int_{-\infty}^\infty \frac{e^{-ts}}{(1-it)^s}\,ds$
with the help of the inverse Fourier transform formula for density 
functions. Then we get by giving a good asymptotic formula for 
the expression at right-hand side of this identity (similarly 
to the solution of  Problem~2) that
$\Gamma(s)\sim\sqrt{2\pi(s-1)}(\frac{s-1}e)^{s-1}$ if $s\to\infty$. 

%\vfill\eject
 
\medskip\noindent
{\script C.) The definition of the convolution and some of its
important properties.}

\medskip
In this series of problems we investigate the asymptotic behaviour
of the distribution and density function of appropriately normalized
partial sums of independent random variables. These distribution or
density functions can be directly expressed by means of the
distribution of density functions of the random variables in these
partial sums. Hence the limit theorems discussed in this text also can
be expressed in the language of distribution (and density) functions
without speaking of sums of independent random variables. To do this we
have to introduce the notion of the convolution operator. We introduce
this notion in a slightly more general form and define the convolution
of integrable (not necessary density) functions and signed measures. We
shall not use the notion of the convolution in this series of problems.
Hence this section could be omitted. Nevertheless, the discussion of
limit theorems without the introduction of the convolution would not
be complete, hence we introduce it. In several investigations of the
analysis and probability theory this notion appears in a natural way,
and it appears in the more general form introduced in this text.

\medskip\noindent 
{\bf The definition of the convolution operator.}
{\it If $f(x_1,\dots,x_k)$ and $g(x_1,\dots,x_k)$ are two
$k$-dimensional measurable functions of $k$ variables which are
integrable, i.e.\ $\int |f(x_1,\dots,x_k)|\,dx_1\dots\,dx_k<\infty$ and
$\int |g(x_1,\dots,x_k)|\,dx_1\dots\,dx_k<\infty$, then the convolution
$f*g$ of the functions $f$ and $g$ is the function of $k$ variables
defined by the formula
$$
f*g(x_1,\dots,x_k)=\int
f(u_1,\dots,u_k)g(x_1-u_1,\dots,x_k-u_k)\,du_1\dots\,du_k \tag8
$$
in all such points $(x_1,\dots,x_k)$, where this integral is meaningful.
(In the remaining points we can define the function $f*g$ in an
arbitrary way.)
 
Let $\mu$ and $\nu$  be two signed measures of bounded variation on the
measurable sets of the $k$-dimensional Euclidean space  $R^k$.
This means that we assume that there exist two representations
$\mu=\mu_1-\mu_2$ and $\nu=\nu_1-\nu_2$ such that $\mu_i$ and $\nu_i$,
$i=1,2$, are finite measures on the measurable subsets of the space
$R^k$, i.e.\ $\mu_k(R^k)<\infty$ and $\nu_k(R^k)<\infty$, $k=1,2$.
Let $\mu\times\nu$ denote the direct product of these signed measures
$\mu$ and $\nu$ on the product space $R^k\times R^k=R^{2k}$. Then the
convolution $\mu*\nu$ is the signed measure on the measurable subsets
of the space $R^k$ defined by the formula
$$
\mu*\nu(A)=\mu\times\nu \{(u,v)\colon\; u+v\in A\}\quad\text{for all
measurable sets } A\subset R^k.
$$
In other words, $\mu*\nu$ is the pre-image of the product (signed)
measure $\mu\times\nu$ induced by the transformation $\bold T\colon\;
R^k\times R^k\to R^k$ defined by the formula $\bold T(u,v)=u+v$.
 
Let $f(x_1,\dots,x_k)$ be a measurable and integrable function of $k$
variables, $\nu$ a measure of bounded variation on the measurable subsets
of $R^k$. The convolution $f*\nu(x_1,\dots,x_k)$ is the following
function of $k$~variables:
$$
f*\nu(x_1,\dots,x_k)=\int
f(u_1,\dots,u_k)\nu(x_1-\,du_1,\dots,x_k-\,du_k)
$$
in all points $(x_1,\dots,x_k)\in R^j$, where this integral is
meaningful. This means that we integrate the function $f(\cdot)$
with respect to the measure $\bar\nu_{x_1,\dots,x_k}$ defined by the
formula
$$
\bar\nu_{x_1,\dots,x_k}(A)=\nu((x_1,\dots,x_k)-A).
$$
(In the remaining points where this integral is not meaningful we define
the function $f*\nu$ in an arbitrary way.)} 

\medskip
We have not defined the convolution $f*g(x_1,\dots,x_k)$  of two
functions $f$ and $g$ or the convolution $f*\nu(x_1,\dots,x_k)$ of a
 function $f$ and a measure $\nu$ in all points $(x_1,\dots,x_k)\in
R^k$. But this restriction is not so disturbing as one might think at
the first sight. As we shall see in problem~14, these convolutions
exist for almost all points $(x_1,\dots,x_k)\in R^k$ with respect to
the Lebesgue measure in the $k$-dimensional Euclidean space. On the
other hand, in typical applications these convolutions appear as the
density function of a signed measure which is absolutely continuous
with respect to the Lebesgue measure, and such density functions are
determined only almost everywhere.

\medskip
\item{14.)} If $f(x_1,\dots,x_k)$ and $g(x_1,\dots,x_k)$ are two
measurable and integrable functions on the space $R^k$,  then the
integral in formula (8) defining the convolution $f*g(x_1,\dots,x_k)$ is
meaningful for almost all points $(x_1,\dots,x_k)\in R^k$ with respect
to the Lebesgue measure. The convolution $f*g$ is a finite and
integrable function on~$R^k$.
\item{} If $\mu$ and $\nu$ are two signed measures of bounded variation
on the space $R^k$, then their convolution $\mu*\nu$  has the same
property.
\item{} If $\mu$ and $\nu$ are two signed measures of bounded
variation on the space $R^k$, and the measure $\mu$ has a density
function $f(u_1,\dots,u_k)$, that is $\mu(A)=\int_A
f(u_1,\dots,u_k)\,du$ for all measurable sets $A\subset R^k$, then the
convolution of these signed measures, the signed measure $\mu*\nu$ has
a density function, and it equals the function $f*\nu$. In particular,
the function $f*\nu(x)$ is integrable. If both signed measures $\mu$
and $\nu$ have a density function $f(u_1,\dots,u_k)$ and
$g(u_1,\dots,u_k)$, then their convolution, the signed measure $\mu*\nu$
also has a density function, and it equals the convolution~$f*g$.
\item{15.)} If $\xi$ and $\eta$ are two independent random vectors on
the space $R^k$, the distribution of $\xi$ is $\mu$, the distribution
of $\eta$ is $\nu$, then the distribution of the sum $\xi+\eta$ is the
convolution $\mu*\nu$. If the random vector $\xi$ has a density
function $f(u_1,\dots,u_k)$, then the sum $\xi+\eta$ also has a
density function,  and it equals $f*\nu$. If $\xi$ has a density
function $f$ and $\nu$ a density function $g$, then the sum $\xi+\eta$
also has a density function, and it equals the convolution $f*g$.
\item{} As a consequence, if $\xi_j$, $j=1,\dots,n$, are independent
random vectors with distribution functions $F_j(x)=F_j(x_1,\dots,x_k)$,
$j=1,\dots,n$, $\bar S_n=\frac{\summ_{j=1}^n\xi_j-A}B$ with some
norming factors $A=(A_1,\dots,A_k)$ and $B>0$, then the distribution of
the expression $\bar S_n$ equals $F_1*\cdots*F_n(Bx+A)$. If the random
vectors $\xi_j$ have density functions $f_j$, $1\le j\le n$, then
$\bar S_n$ has a density function of the form $Bf_1*\cdots*f_n(Bx+A)$.
\item{} $f*g=g*f$, $\mu*\nu=\nu*\mu$, $(f*g)*h=f*(g*h)$,
$(\mu_1*\mu_2)*\mu_3=\mu_1*(\mu_2*\mu_3)$, i.e. the convolution 
operator is commutative and associative.

\medskip
The next problem is about the relation between the convolution operator
and Fourier transform. 
\medskip
\item{16.)} If $f(u_1,\dots,u_k)$ and $g(u_1,\dots,u_k)$ are two
integrable functions on~$R^k$ with Fourier transforms
$$
\varphi(t_1,\dots,t_k)=\int
e^{i(t_1u_1+\cdots+t_ku_k)}f(u_1,\dots,u_k)\,du_1\dots\,du_k
$$
and
$$
\psi(t_1,\dots,t_k)=\int
e^{i(t_1u_1+\cdots+t_ku_k)}g(u_1,\dots,u_k)\,du_1\dots\,du_k,
$$
then the Fourier transform of the convolution $f*g(u_1,\dots,u_k)$
is the function $\varphi(t_1,\dots,t_k)\psi(t_1,\dots,t_k)$.
\item{} If $\mu$ and $\nu$ are two signed measures on $R^k$ of bounded
variation with Fourier transforms (or in other terminology with
characteristic functions
$$
\varphi(t_1,\dots,t_k)=\int
e^{i(t_1u_1+\cdots+t_ku_k)}\mu(\,du_1,\dots,\,du_k)
$$
and
$$
\psi(t_1,\dots,t_k)=\int
e^{i(t_1u_1+\cdots+t_ku_k)}\nu(\,du_1,\dots,\,du_k)
$$
then the Fourier transform of the convolution $\mu*\nu$ equals
$\varphi(t_1,\dots,t_k)\psi(t_1,\dots,t_k)$. 

\medskip
In an informal way the following problem can be formulated as
the statement  that the convolution is a smoothing operator. The
convolution of two smooth functions is an even smoother function.
For the sake of simplicity  we shall consider only functions of one
variable.
\item{17.)} Let $f(u)$ and $g(u)$ be two integrable functions. Let
us assume that the derivatives $\frac{d^j f(u)}{du^j}$ exist, they
are integrable functions, and $\limm_{u\to-\infty}\frac{d^j
f(u)}{du^j}=0$ for all integers $0\le j\le k$ with some integer
$k\ge1$. Let us also assume that the derivatives $\frac{d^j
g(u)}{du^j}$ also exits, they are also integrable functions, and
$\limm_{u\to-\infty}\frac{d^j g(u)}{du^j}=0$ for all integers $0\le
j\le l$ with all integers $l\ge1$. Then the derivative
$\frac{d^{k+l}f*g(u)}{du^{k+l}}$ also exists, it is an integrable
function, and $\limm_{u\to-\infty}\frac{d^{k+l}f*g(u)}{du^{k+l}}=0$.
\item{} Let $f(u)$ and $g(u)$ be two integrable functions such that the
function $f(u)$ has an analytic continuation to the domain 
$\{z\colon\; |\Im z|<A\}$ with some number $A>0$. Let us further 
assume that $\int |f(u+ix)|\,du<\infty$ for all numbers $|x|<A$, 
and for all numbers $\e>0$ and $B>0$ there exists a constant 
$K=K(A,B,\e)$ such that $\int_{|u|>K}|f(y-u+ix)g(u)|\,du<\e$ if 
$|y|<B$ and $|x|<A$ . Then the convolution $f*g$ is also an analytic 
function in the domain $\{z\colon\; |\Im z|<A\}$.
 
\medskip\noindent
{\script D.) Convergence in distribution.}

\medskip
Although there is no convenient inversion formula for the expression of
a distribution function by means of its characteristic function, the
proof of the local limit theorems suggests that the convergence of a
sequence of distribution functions can be proved by means of the
convergence of their characteristic functions. This statement really
holds, but to prove it we have to understand the situation better.
 
First we have to understand the meaning of the convergence of
distribution functions. Before giving the formal definition let us
consider a simple example which may explain some details of the
definition.
 
Let us consider a sequence of negative numbers $x_n$, $n=1,2,\dots$,
which satisfies the relation $\limm_{n\to\infty}x_n=0$, and put $x_0=0$.
Let us introduce the (degenerated) probability measures $\mu_n$,
$n=0,1,2,\dots$, on the real line which are concentrated in the points
$x_n$, $n=0,1,2,\dots$, that is $\mu_n(\{x_n\})=1$, $n=0,1,2,\dots$. It
is natural to expect that by a right definition of convergence of
probability distributions the sequence of these measures $\mu_n$
converges to the measure $\mu_0$ as $n\to\infty$. Let us remark that
the measures $\mu_n$, $n=0,1,2,\dots$, are determined by the
distribution functions defined by the formula $F_n(u)=0$ if $u\le
x_n$, and $F_n(u)=1$ if $u>x_n$, $n=0,1,2,\dots$, i.e.\
$\mu_n([a,b))=F_n(b)-F_n(a)$ for arbitrary pairs of numbers $a<b$.
Observe that the relation $\limm_{n\to\infty}F_n(x)=F_0(x)$ holds for
all numbers $x\neq0$, but this relation does not hold for $x=0$, since
$F_0(0)=0$, and $F_n(0)=1$ if $n\neq0$. This example shows that the
naive picture by which the convergence of a sequence of distribution
functions to a limit distribution function would mean that these
distribution functions converge to the limit distribution function in
all points of the real line would not supply a good definition. In the
previous example the convergence does not hold in the point $x=0$,
where the limit distribution function is not continuous. The right
definition of the convergence of distribution functions given below
does not demand convergence in the points of discontinuity of the
limit distribution function. We shall give this definition in the
multi-dimensional case. 

\medskip\noindent
{\bf Definition of convergence of distribution functions.} {\it Let
$F_n(x_1,\dots,x_k)$, $n=0,1,2,\dots$, be a sequence of
$k$-dimensional, $k\ge1$, distribution functions. We say that the
distribution functions  $F_n$ converge to a distribution function
$F_0$ in distribution, or in other words the probability measures
$\mu_n$, determined by the distribution functions $F_n$, $n=1,2,\dots$,
converge in distribution to the measure $\mu_0$ determined by the
distribution function $F_0$, or in a third terminology the random
vectors $\xi_n=(\xi_n^{(1)},\dots,\xi_n^{(k)})$, $n=1,2,\dots$, with
distribution function $F_n$ converge in distribution to a random
vector $\xi_0=(\xi_0^{(1)},\dots,\xi_0^{(k)})$, with distribution
function $F_0$ as $n\to\infty$, if
$$
\lim_{n\to\infty} F_n(x_1,\dots,x_k)=F_0(x_1,\dots,x_k)
$$
in all points of continuity $(x_1,\dots,x_k)\in R^k$ of the distribution
function $F_0(x_1,\dots,x_k)$. (Sometimes the convergence in
distribution is called weak convergence in the literature.)} 

\medskip
In later investigations the following Theorem~A formulated below plays
an important role. This result gives an equivalent condition for the
convergence in distribution. The next problem is the proof of Theorem~A.
We remark that the same problem also appears in problem~1 of the series
of problems {\it Weak convergence of probability  measures in metric
space.}\/ (For the time being it exists only in Hungarian.)

\medskip\noindent
{\bf Theorem A.} {\it The distribution functions $F_n(x_1,\dots,x_k)$
converge in distribution to the distribution function
$F_0(x_1,\dots,x_k)$ if and only if
$$
\int f(x_1,\dots,x_k)\,dF_n(x_1,\dots,x_k)\to
\int f(x_1,\dots,x_k)\,dF_0(x_1,\dots,x_k)
\quad\text{if }n\to\infty \tag9
$$
for all continuous and bounded functions $f(x_1,\dots,x_k)$ in the
$k$-dimensional Euclidean space~$R^k$.}

\medskip
\item{18.)} Let us prove Theorem A.

\medskip
Let us remark that the characterization of the convergence in
distribution given in Theorem~A also plays an important role in other
investigations. It helps to find the good definition of convergence in
distribution in general topological spaces. The original definition
is closely related to the notion of distribution functions which
exploits the simple geometrical structure of the Euclidean space.
Hence this definition has no natural generalization to more
sophisticated spaces. The result of Theorem~A helps to overcome this
difficulty, and this is the reason why Theorem~A appears in the series
of problems mentioned above.
 
It is also worth mentioning that condition~(9) in Theorem~A can also be
interpreted as a particular case of the weak convergence introduced in
functional analysis. The probability measures in the $k$-dimensional
Euclidean space can also be considered as continuous linear functionals
on the space of bounded and continuous functions on $R^k$ endowed with
the usual supremum norm. Then relation (9) means that the probability
measures $\mu_n$ determined by the distribution functions $F_n$ weakly
converge to the measure $\mu_0$ determined by the distribution function
$\mu_0$. Here we identify the measures $\mu_n$ with the linear
functionals on the space of continuous and bounded functions they
induce and apply the usual terminology of weak convergence in
 functional analysis. This fact may explain why convergence in
distribution is sometimes called weak convergence.
 
Theorem A suggests a natural approach to study limit theorems in
distribution with the help of Fourier analysis. For arbitrary
$k$-dimensional vector the trigonometrical function
$e_t(x)=e_{t_1,\dots,t_k}(x_1,\dots,x_k)
=e^{i(t,x)}=e^{i(t_1x_1+\cdots+t_kx_k)}$ is continuous and bounded.
Hence relation~(9) demands in particular that for all vectors
$(t_1,\dots,t_k)$ the relation
$$
\varphi_n(t_1,\dots,t_k)\to\varphi_0(t_1,\dots,t_k),\quad\text{if }
n\to\infty \tag$9'$
$$
should hold, where $\varphi_n(t_1,\dots,t_k)$ is the characteristic
function of the distribution function $F_n(t_1,\dots,t_k)$,
$n=0,1,2,\dots$.
 
Because of the nice properties of trigonometrical functions it is
simpler to check relation~($9'$) than formula~(9) dealing with
general bounded and continuous functions. The question arises whether
relation $(9')$ is sufficient to guarantee the validity of relation
of~(9) i.e.\ to prove convergence in distribution. We shall prove a
more refined positive result in this direction, and this result serves
as the basic tool in investigation of limit theorems in distribution.
To prove this result we need such a result which states that the
trigonometrical functions $e_{t_1,\dots,t_k}(x_1,\dots,x_k)
=e^{i(t_1x_1+\cdots+t_kx_k)}$, or more precisely their finite linear
combinations are a sufficiently rich sub-class of the space of
continuous and bounded functions. In our discussion we shall apply
Weierstrass second approximation theorem which is a result in this
spirit. In the Appendix we also supply a proof of Weierstrass second
approximation theorem. 

\medskip\noindent
{\bf Weierstrass second approximation theorem.} {\it For all continuous
and periodic by~$2\pi$  functions $f(t)$ and real numbers $\e>0$ there
exists a trigonometrical polynomial $P_n(t)=\summ_{k=-n}^n a_k e^{ikt}$
such that
$$
\sup_{-\infty<t<\infty} |f(t)-P_n(t)|<\e.
$$
(The degree of the polynomial $P_n$ and the coefficients $a_k$ in it
depends both on the function $f(\cdot)$ and the real number $\e>0$. If
the function $f(\cdot)$ is real valued, then the coefficients $a_k$
can be chosen in such a way that $a_{-k}=\bar a_k$ for all indices
$k=0,1,\dots,n$, where $\bar z$ is the conjugate of the number $z$.
Then also the polynomial $P_n(t)$ is real valued.)
 
Also the following multi-dimensional version of this result holds.
If $f(t_1,\dots,t_k)$ is a continuous function in the $k$-dimensional
Euclidean space which is periodic in all of its variables, i.e.
$f(t_1+2j_1\pi,\dots,t_k+2j_k\pi)=f(t_1,\dots,t_k)$ for all integers
$j_1,\dots j_k$, and a real number $\e>0$ is fixed, then there exists a
trigonometrical polynomial of $k$ variables
$$
P_n(t_1,\dots,t_k)=\sum_{(j_1,\dots,j_k)\colon\; |j_1|+\cdots+|j_k|\le n}
a_{j_1,\dots,j_k}e^{i(j_1t_1+\cdots+j_kt_k)},
$$
where $j_1,\dots,j_k$ are integers such that
$$
|f(t_1,\dots,t_k)-P_n(t_1,\dots,t_k)|<\e \quad \text {for all real
numbers } t_1,\dots,t_k.
$$
}

\medskip
An application of Weierstrass second approximation theorem together
with an appropriate scaling makes possible to approximate a
$k$-dimensional continuous function $f(x_1,\dots,x_k)$ with arbitrary
accuracy with a finite linear combination of trigonometrical sum
$\sum a(t_1,\dots,t_k)e^{i(t_1x_1+\cdots+t_kx_k)}$ in an arbitrary
finite domain. On the other hand, given a probability measure~$\mu$
the integral of the approximating trigonometrical can be expressed
by means of the characteristic function $\varphi(t_1,\dots,t_k)$ of
the measure $\mu$ as $\sum a(t_1,\dots,t_k)\varphi(t_1,\dots,t_k)$.
If the domain where the trigonometrical sum well approximates the
original function $f$ is chosen sufficiently large, then the above
expression gives a natural approximation of the integral $\int
f(x_1,\dots,x_k)\,d\mu(x_1,\dots,x_k)$.
 
In such a way Weierstrass second approximation gives a good
approximation of a continuous function in a finite domain by a
trigonometrical sum whose integral with respect to a probability
measure can be expressed by means of the characteristic function of
this measure. But it yields no information about the contribution of
a ``small neighbourhood of infinity" to the integrals in formula~(9).
This is not the deficiency of Weierstrass second approximation
theorem, but this difficulty belongs to the essence of the problem we
investigate. In the following problems we shall show that a sequence
of distribution functions may converge in distribution only if the
contribution of an appropriate neighbourhood of the infinity in
formula~(9) is uniformly small for all distribution functions we
consider. We shall formulate this statement in a more precise form,
and we shall also show that the problem about convergence of
distribution functions can be well investigated by means of their
characteristic functions. First we introduce the following definition.

\medskip \noindent 
{\bf Definition of the relative compactness and
tightness of distribution functions.} {\it Let a sequence of
distribution functions $F_n(t_1,\dots,t_k)$, $n=1,2,\dots$, be given
in the $k$-dimensional Euclidean space, and let $\mu_n$ denote the
probability measure on $R^k$ determined by the distribution function
$F_n$. We say that the sequence of distribution functions $F_n$ or
probability measures $\mu_n$ is relatively compact if all subsequences
$F_{n_k}$ (or $\mu_{n_k}$), $k=1,2,\dots$, of the original sequence
$F_n$ or $\mu_n$ have a (sub)subsequence $F_{n_{k_j}}$ (or
$\mu_{n_{k_j}}$), $j=1,2,\dots$, which is convergent in distribution.
 
We say that a sequence of distribution functions $F_n$ or of probability
measures $\mu_n$ determined by these distribution functions is tight if
for all numbers $\e>0$ there exists a number $K=K(\e)$ such that the
$k$-dimensional cube $\bold K(K)^k=\underbrace
{[-K,K]\times\cdots\times [-K,K]} _{k\text{\rm{-fold product}}}$
satisfies the inequality $\mu_n(\bold K(K)^k)\ge1-\e$ {\rm for all
indices} $n=1,2,\dots$.}

\medskip
\item{19.)} Let $\mu$ be a probability measure on the Borel measurable
subsets of the $k$-dimensional Euclidean space, and let us fix a real
number $\e>0$.  Then there exists a number $K=K(\mu,\e)>0$ in such a
way that the $k$-dimensional cube $\bold K^k(K)=\underbrace
{-[K,K]\times\cdots\times [-K,K]}_{k\text{-fold product}}$
satisfies the inequality $\mu(\bold K(K)^k)>1-\e$. Let us show with the
help of this statement and Weierstrass second approximation theorem that
a probability measure on $R^k$ is uniquely determined by its
characteristic function, that is if two probability measures
$\mu_1$ and $\mu_2$ on $R^k$ have the same characteristic function, then
$\mu_1=\mu_2$.
\item{} Furthermore, all signed measures $\mu$ of bounded variation on
the $k$-dimensional space $R^k$ is uniquely determined by their Fourier
transform
$$
\varphi(t_1,\dots,t_k)=\int e^{i(t_1x_1+\cdots+t_kx_k)}
\mu(\,dx_1,\dots,dx_k),\quad (t_1,\dots,t_k)\in R^k.
$$
\item{20.)} Let us give an example which shows that the characteristic
function of a distribution function in a finite interval does not
determine this distribution function. That is let us show that for all
numbers $T>0$ there exist two different characteristic functions
$F_1(\cdot)$ and $F_2(\cdot)$ such that their characteristic functions
$\varphi_i(t)=\int e^{itu}\,dF(u)$, $i=1,2$, satisfy the identity
$\varphi_1(t)=\varphi_2(t)$ for all real numbers $-T\le t\le T$.
\item{21.)} Let $\mu_n$, $n=1,2,\dots$, be a sequence of probability
measures on the $k$-dimensional Euclidean space $R^k$. This sequence of
probability measures is relatively compact if and only if it is tight.
In particular, all sequences of probability measures on $R^k$ which are
convergent in distribution are tight.

\medskip
The hard part in the proof of problem 21 is to show that tightness
implies relative compactness. To prove this the limit
of an appropriate subsequence of distribution functions has to be
constructed, and it has to be checked that the limit is really a
distribution function. The proof could be simplified with the
help of a classical result of functional analysis (Riesz theorem) which
represents finite measures on a compact topological space as the linear
functionals on the space of continuous functions on this topological
space. This result together with a one-point compactification of the
Euclidean space and the characterization of the convergence in
distribution given in Theorem~A may yield a simpler proof. The
tightness condition would guarantee that the limit measure we obtain
has no mass in the ``infinity". Nevertheless, I have described an
elementary but more complicated proof.

 
\medskip\noindent
I make a small d\'etour. I describe a natural generalization of the
results discussed in this section to sequences of probability
measures on complete separable metric spaces. We shall not use
these results in this note, but they are useful e.g. in the proof
of the so-called functional central limit theorem. I shall explain
these results without proof, although the results discussed in
these section may be useful in these proofs.

Theorem~A and formula~(9) in it may give a natural idea for the
definition of convergence in distribution of probability measures
on a separable metric space. We say that a sequence of probability 
measures defined on a separable metric space $(X,\rho)$ converges 
in distribution (or weakly, as this convergence is often called in 
the literature) to a probability measure $\mu$ on this metric 
space, if for all continuous and bounded functions on $(X,\rho)$ 
the relation
$$
\limm_{n\to\infty}\int f(x)\mu_n(\,dx)=\int f(x)\mu(\,dx).
$$
holds.

It may be worth remarking that a meausure $\mu$ on a separable metric
space is uniquely determined by the integrals $\int f(x)\mu(\,dx)$ 
of all continuous, bounded functions $f(x)$. This fact implies in
particular, that the limit of probability measures on a metric 
space is uniquely determined. We can also define the relative
compactness and tightness of sequences of probability measures on
general metric spaces, and we can formulate theorems similar to
the results of this section for them. At this point it is useful
to restrict our attention to complete separable metric spaces.

A sequence of probability measurs $\mu_n$ on a complete separable
metric space $(X,\rho)$ is called relatively compact if each
subsequence $\mu_{n_k}$ of $\mu_n$ has a sub-subsequence
$\mu_{n_{k_j}}$ convergent in distribution (or in other words 
weakly). It can be proved that in a complete separable
metric space of all probability measures $\mu$ and real
numbers $\e>0$ there exists a compact set $\bold K$ such that 
$\mu(\bold K)>1-\e$. Hence we can also define the tightness of
a sequence of probability measures on a complete separable metric
space. We say that a sequence $\mu_n$, $n=1,2,\dots$, of probability
meausures on a complete metric space $(X,\rho)$ is tight if for all
numbers $\e>0$ there exists a compact set $\bold K=\bold K(\e)$ 
such that $\mu_n(\bold K)>1-\e$ for all indices $n=1,2,\dots$.
The following theorem holds.

\medskip\noindent
{\bf Theorem about the relation of tightness and relative
compactness of probability measure sequences.} {\it Let $\mu_n$, 
$n=1,2,\dots$, be a sequence of probability measures on a complete
separable metric space $(X,\rho)$. This sequence of probability
meausures is relatively compact if and only if it is tight. In
particular, all sequences of probability measures on a compact, 
complete separable metric space are (relatively) compact.}


\medskip
This result is useful in the study of convergence of probabilty
measures on general metric spaces. (These problems are interesting,
the functional central limit theorem e.g. can be obtained with 
the help of such an investigation.) To apply the above results in 
a concrete metric space we need a description of the compact sets 
of this metric space. The description of the compact sets in the 
$C([0,1])$ space (in the Banach space of continuous functions on 
the interval $[0,1]$ with the supremum norm.) We shall describe 
them in the next theorem.

\medskip\noindent
{\bf Theorem about the charactrization of the compact sets in
 the space $C([0,1])$.} {\it The compact sets of $C([0,1])$ 
(and in general of all metric spaces are closed.) A closed set 
$\bold F$ in the space $C(0,1])$ is compact if and only if


\medskip
\item{(a)} The set $C=\{x(0)\colon\; x(\cdot)\in \bold F\}$ 
is bounded, and
\item{(b)} The functions  $x(\cdot)$ contained in the set
$\bold F$ are universally uniformly continuous, i.e. for all
numbers $\delta>0$ there is such a number $\eta=\eta(\delta)>0$ 
for which the relation
$$
\supp_{x(\cdot)\in\bold F}
\supp_{0\le s,t\le 1,\,|t-s|\le\eta}|x(t)-x(s)|\le\delta.
$$
holds.}

\medskip
In certain investigations, e.g. in the proof of the functional
central limit theorem the following consequence of the above
results is useful.

\medskip\noindent
{\bf Consequence.} {\it Let $X_n(t)$, $0\le t\le 1$, $n=1,2,\dots$, 
be a sequence of $C([0,1])$-valued random variables. The sequence 
$X_n(t)$ converges weakly to some $C([0,1])$ valued
random variable $X(t)$, $0\le t\le 1$, as $n\to\infty$ if and only if

\item{(a)} For all finite sequences of numbers 
$0\le t_1<\cdots< t_k\le 1$ the random vectors 
$(X_n(t_1),\dots,X_n(t_k))$ converge in distribution as $n\to\infty$.
\item{(b)} For all numbers $\e>0$ and $\delta>0$ there is a number 
$\eta=\eta(\delta,\e)>0$ such that
$$
P\(\oo\colon\;
\supp_{0\le s,t\le 1,\,|t-s|\le\eta}|X_n(t,\oo)-X_n(s,\oo)|)\le\delta\)>1-\e
$$
for all indices $n=1,2,\dots$.
}

\medskip\noindent
{\it Remark.} Point (a) of the above consequence guarantees that
the distributionos of all convergent subsequence of the $C([0,1])$ 
valued random variables $X_n(\cdot)$ have the same limit. 
Furthermore, it can be proved with the help of point~(b) and the 
convergence of the (real valued) random variables $X_n(0)$, $n=1,2,\dots$,
that the $\mu_n$ distributions of the random variables $X_n(\cdot)$
constitute a relatively compact set.

\medskip \noindent
{\script E.) The relation between the convergence of distribution
functions and the convergence of their characteristic functions.}
 
\medskip 
To express the conditions of convergence in distribution by
means of characteristic functions it is useful to give the necessary
and sufficient condition of tightness (or of relative compactness which
property is equivalent to tightness by the result of problem~20)
of probability measures with the help of distribution functions. This is
the content of the next problem.

\medskip
\item{22.)} Let $F_n(u)$, $n=1,2,\dots$, be a sequence of distribution
functions on the real line with characteristic functions $\varphi_n(t)$,
$n=1,2,\dots$. This sequence of distribution functions
$F_n(\cdot)$ is tight if and only if
$$
\lim_{\delta\to0}\limsup_{n\to\infty}\frac1{2\delta}\int_{-\delta}
^\delta\Re \(1-\varphi_n(t)\)\,dt=0, \tag10
$$
where $\Re z$ denotes the real part of the complex number~$z$.
\item{} Although we shall not need the following observation let us also
show that relation (10) is equivalent to the following relation $(10')$:
$$
\lim_{\delta\to0}\sup_n\frac1{2\delta}\int_{-\delta}^\delta
\Re\(1-\varphi_n(t)\)\,dt=0. \tag$10'$
$$

\medskip
With the help of the result of problem~22 and Weierstrass second
approximation theorem we can give the necessary and sufficient
condition of convergence of a sequence of distribution function in
distribution in the language of characteristic function. We formulate
and prove this result which we call because of its importance
fundamental theorem. 

\medskip\noindent
{\bf The Fundamental Theorem about the convergence of distribution
functions.} {\it Let $F_n(u_1,\dots,u_k)$ be a sequence of
distribution functions on the $k$-dimensional Euclidean space $R^k$
with characteristic functions $\varphi_n(t_1,\dots,t_k)$,
$n=1,2,\dots$. If the characteristic functions
$\varphi_n(t_1,\dots,t_k)$ have the limit $\varphi_0(t_1,\dots,t_k)
=\limm_{n\to\infty}\varphi_n(t_1,\dots,t_k)$ in all points
$(t_1,\dots,t_k)\in R^k$, and the limit function
$\varphi_0(t_1,\dots,t_k)$ is continuous in the origin, then there
exists a distribution function $F_0(u_1,\dots,u_k)$ in the
$k$-dimensional space whose characteristic function is this limit
function $\varphi_0(t_1,\dots,t_k)$. Besides, the distribution
functions $F_n(u_1,\dots,u_k)$ converge in distribution to this
distribution function $F_0(u_1,\dots,u_k)$. Moreover, the condition
about the continuity of the limit function $\varphi_0(t_1,\dots,t_k)$
in the origin can be slightly weakened. The above statements also hold
if we only demand that the restriction of the function
$\varphi_0(t_1,\dots,t_k)$ to each coordinate axis is
continuous in the origin.
 
In the converse direction we state that if a sequence of
$k$-dimensional  distribution functions $F_n(u_1,\dots,u_k)$,
$n=1,2,\dots$, converges in distribution to a $k$-dimensional
distribution function $F_0(u_1,\dots,u_k)$, and
$\varphi_n(t_1,\dots,t_k)$, $n=0,1,\dots$, denotes the characteristic
function of the distribution function $F_n(u_1,\dots,u_k)$, then
$\varphi_0(t_1,\dots,t_k)=\limm_{n\to\infty} \varphi_n(t_1,\dots,t_k)$
in all points $(t_1,\dots,t_k)\in R^k$. Furthermore, this convergence
is uniform in all compact subsets of the space~$R^k$.}

\medskip
We have formulated the above fundamental theorem in a slightly stronger
form than it is done in the literature. We remarked that if we want
to deduce the convergence of the distribution functions from the
convergence of the characteristic functions, then it is enough to know
that the restriction of this limit function to the coordinate axes is
continuous in the origin. We have made this remark because its proof
causes no problem, and it is useful in certain multi-dimensional
application of the result. In particular, it simplifies the solution of
problem~46 in this series of problems.
 
We want to prove the above formulated Fundamental Theorem. To do this
let us first prove the following problem interesting in itself.

\medskip
\item{23.)} Let $\bold\xi^{(n)}=(\xi_1^{(n)},\dots,\xi_k^{(n)})$ be a
sequence of $k$-dimensional random vectors and let
$\varphi_n(t_1,\dots,t_k)$ denote their characteristic functions. Prove
(with the help of problem~22) that if the characteristic functions
$\varphi_n(t_1,\dots,t_k)$ converge to a function
$\varphi(t_1,\dots,t_k)$ in a small neighbourhood of the origin which
is continuous in the origin, then the distribution functions of the
random vectors $\bold\xi^{(n)}=(\xi_1^{(n)},\dots,\xi_k^{(n)})$ are
tight. Moreover, this tightness property also holds under the weaker
condition that the restrictions of the limit function
$\varphi(t_1,\dots,t_k)$ to the coordinate axes are continuous in the
origin.
\item{24.)} Let us prove the Fundamental Theorem about the convergence
of distribution functions.
\item{25.)} Let us show an example for a sequence of characteristic
functions $\varphi_n(t)$, $n=0,1,2,\dots$, on the real line such that
$\limm_{n\to\infty}\varphi_n(t)=\varphi_0(t)$, this convergence is
uniform on all finite intervals, but it is not uniform on the whole real
line. 

\medskip\noindent
Let us make some comments about the above formulated Fundamental
Theorem. 

\medskip
\item{i.)} The fundamental theorem gives conditions which
automatically guarantee that the limit of characteristic functions is
also a characteristic function. This result is very useful in a more
systematic investigation of limit theorems. For instance it makes
possible to describe all possible limit distributions which appear as
the limit of appropriately normalized partial sums of independent
random variables.
\item{ii.)} In the Fundamental Theorem we stated the limit of
characteristic functions is again a characteristic function if it is
continuous in the origin. On the other hand, we know that the
characteristic functions are uniformly continuous in the whole
space~$R^k$ (see for instances the result of problem~12). Let us
remark that the class of the characteristic functions is a special
class of all continuous functions. There is a non-trivial
characterization of characteristic functions. A famous result of the
analysis, the Bochner theorem yields such a characterization. This
result plays an important role both in the analysis and in the
probability theory. Although we shall not use this result in this
series of problems, we shall prove Bochner's theorem in the second part
of this series of problems. The information obtained in the proof of
limit theorems are useful in the proof of Bochner's theorem.
\item{iii.)} The condition that the limit of the characteristic
functions is continuous in the origin guarantees that the distribution
functions corresponding to them are tight. Informally saying this means
that ``no mass flows out to the infinity". If  some mass ``may flow out
to the infinity", then the asymptotic behaviour of the characteristic
functions does not supply such a simple description of the asymptotic
behaviour of the distribution functions corresponding to them as the
Fundamental Theorem. The content of the next problem is to show this
with the help of an example. 

\medskip
\item{26.)} Let us show an example for a sequence of probability
measures $\mu_n$, $n=1,2,\dots$, on the real line which satisfy the
relation $\limm_{n\to\infty}\mu_n(\bold K)=0$ for all bounded sets
$\bold K$, and the characteristic functions $\varphi_n(t)$ of the
measures $\mu_n$ either 
\item{a.)} have a limit in all points of the real line \quad\quad or 
\item{b.)} do not have a limit function on the real line.
 
\medskip\noindent
{\script F.) The relation between the behaviour of a function or
measure and its Fourier transform.} 

\medskip
If we want to investigate the behaviour of distribution functions by
means of their characteristic functions, or more generally if we want
to study the properties of a function or a measure by means of its
Fourier transform, then we need a good ``dictionary" which describes
how the properties of a function or a measure are reflected in their
Fourier transform, and what kind of properties of the original
function or measure follow from certain properties of their Fourier
transform. The ``dictionary" described in this section will be by
no means complete, but it also contains some results which we do not
need in our further investigation. Actually, if we only wanted to get
a proof of the central limit theorem and were not interested
in the more intricate questions related to this problem, then the
proof of the first statement in problem~27 in this section would
suffice for our purposes. The results of this section will be
formulated only in the one-dimensional case, although the proof of
their multi-dimensional generalization would cause no essential
difficulty.
 
Informally the content of this ``dictionary" can be formulated in the
following way: The smoother a function or a measure (more precisely
its density function) on the real line is, the faster its Fourier
transform tends to zero at infinity. The less mass a probability
measure on the real line contains in a small neighbourhood of the
infinity (or in other words the more moments the absolute value of a
random variable with this distribution has), the smoother the Fourier
transform of this probability measure is. Besides, the derivatives
of the Fourier transform at the origin determine the moments of a
random variable whose distribution this probability measure is.
 
In the other direction: If the characteristic function of a
distribution function is smooth, and it is enough to assume this
smoothness property in a small neighbourhood of the origin,
then the smoother the characteristic function is the less mass
the probability measure with this characteristic function contains
in the vicinity of the infinity. On the other hand, the faster the
characteristic function tends to zero as its argument tends to infinity,
the smoother the original distribution function is. Furthermore, some
results also show that if a function is very smooth everywhere except
some points where the function has a singularity, then the behaviour
of its Fourier transform in the vicinity of the infinity very well
reflects the character of these singularities. We shall discuss this
last statement, --- which we shall not need later --- only
superficially, and omit the proof.
 
Actually, the statement of problem~22 can also be considered as
an element of this dictionary, and its content is in full
agreement with the above sketched heuristic picture. The statement
that a class of distribution functions is tight means that there is
an appropriate neighbourhood of the infinity which has a small
measure with respect to all measures determined by this class of
distribution functions. On the other hand, formula~(10) which is
equivalent to this property has the content that the characteristic
functions of this distribution functions satisfy some sort of uniform
continuity in a small neighbourhood of the origin. (Let us recall that
$1-\varphi(0)=0$.)
 
The results of problems~17 and 16 are also consistent with the above
heuristics. The result of problem 17 formulates a statement by which
the convolution of two smooth density functions is a function which
is even smoother than the functions taking part in the convolution.
On the other hand, the Fourier transform of the convolution of two
functions equals the product of their Fourier transforms. By the above
sketched heuristic argument the smoothness of a function depends on
how fast its Fourier transform tends to zero in the neighbourhood of
infinity. On the other hand, the product of two functions tending
to zero in the vicinity of the infinity tends to zero faster than the
functions in this product. This fact corresponds to the smoothing
property of the convolution operator in the language of Fourier
transforms. 

\medskip
\item{27.)} If the absolute value of the random variable $\xi$ has
finite $k$-th moment, i.e.\ $E|\xi^k|<\infty$, then the characteristic
function $\varphi(t)=Ee^{it\xi}$ of the random variable $\xi$ is
$k$ times continuously differentiable, and $\left.\frac{d^j \varphi(t)}
{dt^j}\right|_{t=0}=i^j E\xi^j$ for all numbers $0\le j\le k$.
\item{} A random variable $\xi$ has exponential moments in a small
neighbourhood of the origin if and only if the distribution function of
the random variable $|\xi|$ tends to~1 exponentially fast in the
infinity, i.e.\ $Ee^{u\xi}<\infty$ for all numbers $|u|\le t$ with some
sufficiently small  number $t>0$ if and only if $P(|\xi|>x)<
Ce^{-\alpha x}$ for all $x>0$ with some appropriate constants $C>0$ and
$\alpha>0$.  In this case the characteristic function
$\varphi(t)=Ee^{it\xi}$ has an analytic extension to the domain 
$\{z\colon\; |\Re z|<a\}$ with some appropriate number~$a>0$.
\item{28.)} If the absolute value of the function $f(u)$ is integrable
on the real line, then the Fourier transform $\varphi(t)=\int
e^{itu}f(u)\,du$ of the function $f(u)$
satisfies the relations $\limm_{t\to\infty}\varphi(t)=0$ and
$\limm_{t\to-\infty}\varphi(t)=0$. (Riemann lemma.)
\item{} If the absolute value of the function $f(u)$ is integrable,
the function $f(u)$ is $k$-times differentiable, and the absolute value
of the functions $\frac{d^j f(u)}{du^j}$ is integrable for all
numbers $1\le j\le k$, then $\int_{-\infty}^\infty e^{itu}
f(u)\,du=o\((1+|t|)^{-k}\)$ in case $|t|\to\pm\infty$.
\item{} If the function $f(u)$ has an analytic continuation to the
domain $\{z\colon\; |\Re z|<A\}$ with some $A>0$, $|\int |f(u+iv)|\,du
<\infty$, if $|v|<A$, then $\varphi(t)=\int e^{itu}f(u)\,du=
O\(e^{-\alpha|t|}\)$ as $t\to\pm\infty$ with some constant $\alpha>0$.
\item{29.)} Let $\xi$ be a random variable with non-degenerated
distribution, (i.e.\ assume that $\xi$ does not equal a constant with
probability~1), and let $\varphi(t)=Ee^{it\xi}$ denote its
characteristic function. There exists a constant $t\neq0$ such that
$|\varphi(t)|=1$ if and only if $\xi$ is a lattice valued random
variable. (The notion of lattice valued random variable is introduced
in Definition~A.) If the values of the random variable $\xi$
are concentrated on a lattice of width~$h$ (as on the rarest lattice),
then the relation $|\varphi(t)|=1$ holds for the characteristic
function of the random variable~$\xi$ if and only if $t=2\pi\frac kh$
with some $k=0,\pm1,\pm2,\dots$. If $\xi$ is a lattice valued, non 
constant random variable, then there exits a rarest lattice of
width $h$ where the distibution of $\xi$ is concentrated. If the 
random variable $\xi$ is not lattice valued, then  
$\supp_{A\le |t|\le B}|\varphi(t)|<1$ for all pairs of numbers 
$0<A<B<\infty$.
\item{30.)} Let the distribution of the random variable $\xi$ be given
by the formula $P(\xi=1)=P(\xi=\sqrt 2)=P(\xi=-1)=P(\xi=-\sqrt 2)
=\frac14$. Then the characteristic function $\varphi(t)=Ee^{it\xi}$
satisfies the inequality $|\varphi(t)|<1$ if $t\neq0$, and also the
relation $\limsupp_{t\to\infty}| \varphi(t)|=1$ holds. The relations
$\limsupp_{t\to\infty}|\varphi(t)|=1$ and $|\varphi(t)|<1$ if $t\neq0$
also hold in the more general case when the values of the random
variable $\xi$ are concentrated in finitely or countably many points
but $\xi$ is not a lattice valued random variable.

\medskip\noindent
{\it Remark 1:}\/ By a classical result of measure theory all (finite)
measures can be decomposed (in a unique way) as the sum of an absolute
continuous, a discrete (i.e. concentrated in countably many points)
and a  singular measure. (An absolute continuous measure has a density
function with respect to the Lebesgue measure, a singular measure is
concentrated on a set of Lebesgue measure zero, but all points have
measure zero with respect to it.) By the result of problem~28 the
Fourier transform $\varphi(t)$ of an absolute continuous measure tends
to zero as $t\to\pm\infty$, and the smoother density function
the measure has the faster this convergence is. The results of
problems~29 and~30 tell that the Fourier transform of a discrete
measure behaves in the opposite way in the neighbourhood of infinity.
The $\limsup$ of the absolute value of the Fourier transform of a
discrete measure equals the measure of the real line with respect
to this measure. The behaviour of a singular measure in the
neighbourhood of the infinity cannot be characterized in a similar
simple way.

\medskip\noindent
{\it Remark 2:}\/ Let us consider in the neighbourhood
of infinity the behaviour of the Fourier transform $\tilde f(u)$ of an
integrable function $f(s)$ which is sufficiently many times
differentiable everywhere except a point~$a\in R^1$. Let the
function $f$ have a singularity of the form $f(s)\sim C|s-a|^\alpha$,
$\alpha>-1$, $\alpha\neq2k$, $k=0,1,2,\dots$, in the neighbourhood of
the point~$a$. In this case the Fourier transform of the function~$f$
behaves asymptotically as $\tilde f(u)\sim \bar C e^{iua}u^{-\alpha-1}$
if $u\to\infty$, where the constant $\bar C\neq0$ can be given
explicitly. We give a short heuristic explanation of this result.
 
Because of the smoothness properties of the function $f(s)$ the
asymptotic behaviour of its Fourier transform in the neighbourhood of
the infinity is determined by the singularity of the function $f$ in
the point~$a$, and the Fourier transform equals asymptotically the
expression
$$
\align
\tilde g(u)&=C\int_{-\infty}^\infty
e^{ius}|s-a|^{\alpha}\,ds=
Ce^{iau}\int_{-\infty}^\infty e^{ius}|s|^{\alpha}\,ds\\
&=Ce^{iau}u^{-\alpha-1} \int_{-\infty}^\infty e^{is}|s|^{\alpha}\,ds=
\bar Ce^{iau}u^{-\alpha-1}
\endalign
$$
as $u\to\infty$.
 
The above calculation was rather incorrect. The main problem is that
the integrals considered here are meaningless because of the factor
$|s|^\alpha$ in the integrand, at least as usual Lebesgue integrals.
Nevertheless, this calculation supplies a correct result. Moreover,
by replacing the integral to the imaginary axis we can express the
constant $\bar C$ by means of the $\Gamma(\cdot)$ function. If the
function $f$ has several similar singularities, then their effects sum
up if the Fourier transform of the function in the neighbourhood of the
infinity is described.
 
Although we shall not apply the results sketched in this Remark~2,
such kind of results play an important role in certain investigations
of probability theory and analysis. The cases $\alpha=2k$,
$k=0,1,\dots$, had to be excluded, because in this case the function
$f(s)$ is smooth in a small neighbourhood of the point~$a$.
The exceptional behaviour of the function $|s-a|^\alpha$ with such
parameters $\alpha$ has deep consequences in certain problems of
statistical physics.

\medskip
The results of the subsequent problems 31--34 yield information about
the behaviour of a probability measure by means of the properties of
its Fourier transform.

\medskip
\item{31.)} If the characteristic function $\varphi(t)=Ee^{it\xi}$
of a random variable $\xi$ is twice differentiable in the origin, then
the random variable $\xi$ has finite second moment, i.e.\
$E\xi^2<\infty$. Let us prove by induction that if the characteristic
function of the random variable $\xi$ is $2k$-times differentiable in
the origin, then $E\xi^{2k}<\infty$.
\item{32.)} If the characteristic function $\varphi(t)=Ee^{it\xi}$
of the random variable $\xi$ is differentiable in a small neighbourhood
of the origin, and the derivative is a Lipschitz~$\alpha$ function in a
small neighbourhood of the origin with some parameter $\alpha>0$,
i.e.\
$|\varphi'(t)-\varphi'(s)|<C|t-s|^{\alpha}$ with an appropriate constant
$C>0$ if $|s|<\e$, $|t|<\e$ with some number  $\e>0$, then
$E|\xi|<\infty$. Under these conditions also the following somewhat
stronger result holds: $P(|\xi|>u)<\const u^{-1-\alpha}$ for all
numbers~$u>0$.
\item{} If the characteristic function $\varphi(t)$ is $2k+1$-times
differentiable in a small neighbourhood of the origin, and the
 $2k+1$-th derivative is a Lipschitz $\alpha$ function with some
constant $\alpha>0$, then $E|\xi|^{2k+1}<\infty$. 

\medskip\noindent
{\it Remark:}\/ It follows from the results of problems~31 and~27
that the second moment of a random variable is finite if and only if
the characteristic function of the random variable is twice
differentiable in the origin. If the absolute value of a random variable
has finite expected value, then the characteristic function of this
random variable is differentiable in the origin (and everywhere). But to
guarantee the existence of a finite expected value of the absolute value
of the random variable $\xi$ we have imposed a stronger condition than
the differentiability of the characteristic function in the origin.
(This would be the natural analog of the statement in problem~31.)
Although the conditions of~32 could be weakened, the differentiability
of the characteristic function of a random variable in the origin does
not suffice for the finiteness of the expectation of the absolute value
of this random variable.
 
Indeed, a necessary and sufficient condition of the differentiability
of the characteristic function of a random variable can be expressed
with the help of the distribution function of this random variable in a
relatively simple way. Since this problem arises in a natural way in
the investigation of the weak law of large numbers, this result is also
contained in problem~12 of the series of problems {\it Convergence in
probability and with probability~one}. (For the time being it exists
only in Hungarian). Let us formulate this result. The characteristic
function of a random variable with distribution function~$F$ has a
finite derivative $ia$, $-\infty<a<\infty$, in the origin if and only if
$$
\lim_{x\to\infty}x\[F(-x)+\(1-F(x)\)\]=0, \quad \text{and} \quad
\lim_{u\to\infty}\int_{-u}^u xF(\,dx)=a.
$$

\medskip
\item{33.)} If the characteristic function of a random variable $\xi$
has an analytic continuation to a small neighbourhood of the origin,
then there exists a number $\alpha>0$ such that
$P(|\xi|>x)\le\const e^{-\alpha x}$ for all numbers~$x>0$.
\item{34.)} If the characteristic function $\varphi(u)=Ee^{iu\xi}$ of
a random variable $\xi$ is integrable, i.e.\
$\int|\varphi(u)|\,du<\infty$, then the random variable $\xi$ has a
density function $f(x)$. If $|\varphi(u)|<\const |u|^{-(k+1+\e)}$ with
some number $\e>0$, then the density function $f(x)$ has continuous and
bounded $k$-th derivative. If $|\varphi(u)|<\const e^{-\alpha|u|}$ with
some number $\alpha>0$, then the density function $f(x)$ is analytic.

\medskip\noindent
{\it Remark 1:}\/ The statement of problem~34 can be slightly
generalized. To guarantee the existence of a density function of the
random variable $\xi$ it is enough to assume that the square of the
characteristic function $\varphi(u)=Ee^{iu\xi}$ is integrable. This
statement holds because the inverse Fourier formula which expresses
the density function of a random variable by means of its
characteristic function in the form written down in relation~(6)
also holds if the characteristic function is only square integrable.
Only in this case the integral in formula~(6) has to be considered not
as a usual Lebesgue integral. It has to be defined by the extension
of the $L_2$ isomorphism between a function and its Fourier transform
which follows from the Parseval formula. This is a classical result of
the Fourier analysis, but we do not discuss it here. It implies the
result of this Remark~1. 

\medskip\noindent
{\it Remark 2:}\/ We remarked in the formulation of problem~8 that
to guarantee the local central limit theorem the condition about the
integrability of the characteristic function of the random variables
we have considered can be weakened. It is enough to assume that a
sufficiently large power of the characteristic function is integrable.
The exponential distribution is an example for such a distribution
function whose characteristic function is not integrable, but it is
square integrable (see problem~13.c). Hence only the strengthened
form of the result of problem~8 yields the local central limit
theorem in this case. The results of this section may explain the
deeper background of this example.
 
Let us observe that the density function of the exponential
distribution function has a discontinuity in the origin, and the
non-integrability of its characteristic function is caused by this
discontinuity. The convolution of the exponential distribution
function with itself yields a distribution function with a smoother
density function whose characteristic function is the square of the
exponential distribution function. This fact may explain why the
square of the Fourier characteristic function of the exponential
distribution is integrable. A similar picture arises in the cases of
random variables with such a density function which is sufficiently
smooth except some exceptional points where the density function has
a not too strong singularity.
 
\medskip\noindent
{\script G.) The central limit theorem.}

\medskip
The previous results enable us to prove the central limit theorem by
means of characteristic functions. 

\medskip
\item{35.)} Let $\xi_1,\xi_2,\dots$, be a sequence of independent an
identically distributed random variables, $E\xi_1=0$, $E\xi^2_1=1$.
Put $S_n=\xi_1+\cdots+\xi_n$. Let us show that the random variables
$\frac{S_n}{\sqrt n}$, $n=1,2,\dots$, converge in distribution to the
standard normal distribution function. 

\medskip
We want to prove the central limit theorem in the more general case
when the normalized partial sums of independent but not necessarily
identically distributed random variables are considered. It is worth
formulating this question in an even more general setting when for all
positive integers~$k$ we consider the sum of independent random
variables indexed with this number~$k$, but we assume nothing about
the relation between random variables with different indices~$k$. If
the random variables indexed with the same number~$k$ are uniformly
small, then under some weak conditions the sums of the random
variables with the same index~$k$ have a Gaussian limit as
$k\to\infty$.
 
Essentially we can give the necessary and sufficient condition for
the convergence of the distributions of the above sums to a Gaussian
law. To formulate this statement in an explicit form we introduce the
notion of triangular arrays. The central limit theorem for normalized
partial sums of independent but not necessarily identically
distributed random variables can be deduced as the special case of
the central limit theorem for triangular arrays.

\medskip\noindent
{\bf The definition of triangular arrays.} {\it A system of random
variables
$$
\align
&\xi_{1,1},\dots,\xi_{1,n_1}\\
&\vdots\qquad \quad \;\vdots   \\
&\xi_{k,1},\dots,\xi_{k,n_k}\\
&\vdots\qquad \quad \;\vdots
\endalign
$$
with  $k=1,2,\dots$ is called a triangular array, if the random
variables $\xi_{k,1}$,\dots, $\xi_{k,n_k}$ with the same first
index~$k$ are independent. (We assume nothing about the relation of
the random variables with different first index $k$.)} 

\medskip
To prove the central limit theorem by means of characteristic
function it is worth proving the following technical lemma. It gives an
estimate of the difference of the function $e^{it}$ and the sum of the
first $k$ terms of its Taylor series.
\item{36.)} For all non-negative integers $k$ and real numbers~$t$
$$
\left|e^{it}-\(1+\frac{it}{1!}+\cdots+\frac{(it)^k}{k!}\)\right|\le
\frac{|t|^{k+1}}{(k+1)!}. \tag11
$$

\medskip
If we want to prove the central limit theorem for triangular arrays
with the help of characteristic functions, then the following approach
is natural. Let us consider the characteristic functions
$\varphi_{k,j}(\cdot)$  of the random variables $\xi_{k,j}$, $1\le j\le
n_k$. We have to show that the product of these characteristic functions
converge to the characteristic function of a Gaussian random variable.
Let us consider the logarithm of these products. Then we have to
investigate the sum of the functions $\log \varphi_{k,j}(t)$ with a
fixed index $k$ and fixed argument~$t$. If we have a condition which
says that the random variables $\xi_{k,j}$ are small in an appropriate
sense, then it is natural to expect that
$\varphi_{j,k}(t)\sim1$ for a fixed number~$t$, and the error of the
approximation $\log\varphi_{k,j}(t)\sim(1-\varphi_{k,j}(t))$ gives a
negligible error. This enables to reduce the proof to the investigation
of a simpler sum. The goal of the next problem is to give a precise
formulation and justification of the above heuristic argument. It makes
possible to study both the necessary and sufficient conditions of the
central limit theorem. 

\medskip
\item{37.)} Let us consider the characteristic function $\varphi(t)$ of
a random variable~$\xi$ with a fixed argument~$t$.
\itemitem {a.)} If $E\xi=0$ and $E\xi^2\le \e$ with a sufficiently small
positive number $\e=\e(t)>0$, then $|1-\varphi(t)|\le
\frac{t^2}2E\xi^2$, and $|\log \varphi(t)+(1-\varphi(t))| \le
t^4\(E\xi^2\)^2$.
\itemitem{b.)} Let $\xi_{k,j}$, $k=1,2,\dots$, $1\le j\le n_k$, be
a triangular array such that $E\xi_{k,j}=0$, $k=1,2,\dots$, $1\le j\le
n_k$, $\limm_{k\to\infty}\summ_{j=1}^{n_k}E\xi_{k,j}^2=1$,  and let the
elements of triangular array satisfy the uniform smallness condition
$\limm_{k\to\infty}\(\supp_{1\le j\le n_k}E\xi_{k,j}^2\)=0$. Let
$\varphi_{k,j}(t)=Ee^{it\xi_{k,j}}$, $-\infty<t<\infty$, denote the
characteristic function of the random variable $\xi_{k,j}$,
$k=1,2,\dots$, $1\le j\le n_k$. The random sums
$S_k=\summ_{j=1}^{n_k}\xi_{k,j}$, $1\le k<\infty$, converge in
distribution to the Gaussian distribution with expectation
$m$ and variance $\sigma^2$ as $k\to\infty$ if and only if
$$
\limm_{k\to\infty}\summ_{j=1}^{n_k}(\varphi_{k,j}(t)-1)=-\frac{\sigma^2
t^2}2+imt  \tag12
$$
for all numbers $-\infty<t<\infty$. 

\medskip
Although we are first of all interested in the question when the
partial sums $S_k$ of a triangular array satisfying the conditions of
problem~37 converge in distribution to the standard normal distribution,
it was also useful to formulate the condition for the convergence of
these random sums to a general Gaussian distribution with expectation
$m$ and variance $\sigma^2$. This knowledge helps us to find also the
necessary conditions of the central limit theorem. We shall also show
such examples where the appropriately normalized partial sums of
independent random variables with expectation zero converge to a
Gaussian distribution with expectation $m\neq0$.
 
In the investigation of the central limit theorem for triangular arrays
the so-called Lindeberg condition formulated below plays a fundamental
role. We shall see that the partial sums
$S_k=\summ_{j=1}^{n_k}\xi_{k,j}$ of a triangular array satisfying
part~b) of problem~37 converge to the standard normal distribution
function if and only if this triangular array satisfies the
Lindeberg conditions.

\medskip\noindent
{\bf The definition of the Lindeberg condition:} {\it Let $\xi_{k,j}$,
$k=1,2,\dots$, $1\le j\le n_k$, be such a triangular array for which
$E\xi_{k,j}=0$, $k=1,2,\dots$, $1\le j\le n_k$, and
$\limm_{k\to\infty}\summ_{j=1}^{n_k}E\xi_{k,j}^2=1$. This triangular
array satisfies the Lindeberg condition if and only if for all positive
real numbers $\e>0$
$$
\lim_{k\to\infty}\sum_{j=1}^{n_k}E\xi_{k,j}^2I\(\{|\xi_{k,j}|>\e\}\)=0,
$$
where $I(A)$ denotes the indicator function of a set $A$.} 

\medskip
\item{38.)} Let $\xi_{k,j}$, $k=1,2,\dots$, $1\le j\le n_k$, be a
triangular array which satisfies the relations
$E\xi_{k,j}=0$, $k=1,2,\dots$, $1\le j\le n_k$,
$\limm_{k\to\infty}\summ_{j=1}^{n_k}E\xi_{k,j}^2=1$ and the Lindeberg
condition. Then
\itemitem{a.)} the triangular array $\xi_{k,j}$, $k=1,2,\dots$, $1\le
j\le n_k$,  also satisfies the uniform smallness condition
$\limm_{k\to\infty}\(\supp_{1\le j\le n_k}E\xi_{k,j}^2\)=0$.
\itemitem{b.)} The random sums $S_k=\summ_{j=1}^{n_k}\xi_{k,j}$, $1\le
k<\infty$, converge in distribution to the standard normal distribution
(i.e. to the normal distribution  with expectation zero and
variance~1) if $k\to\infty$.

\medskip 
The above result can be reversed in the following way.

\medskip
\item {39.)} Let $\xi_{k,j}$, $k=1,2,\dots$, $1\le j\le n_k$, be a
triangular array such that $E\xi_{k,j}=0$, $k=1,2,\dots$, $1\le j\le
n_k$, $\limm_{k\to\infty}\summ_{j=1}^{n_k}E\xi_{k,j}^2=1$, and it
satisfies the uniform smallness condition
$\limm_{k\to\infty}\(\supp_{1\le j\le
n_k}E\xi_{k,j}^2\)=0$. Let us also assume that the random sums
$S_k=\summ_{j=1}^{n_k}\xi_{k,j}$, $1\le k<\infty$, converge in
distribution to a normal distribution function with variance~1 and
arbitrary expected value if $k\to\infty$. Then the triangular array
$\xi_{k,j}$, $1\le k<\infty$, $1\le j\le n_k$, also satisfies the
Lindeberg condition.

\medskip
The content of the Lindeberg condition is that it guarantees that the
too large values (which have the same order as the square-root of
the variance of the sum) have a negligible influence in the central
limit theorem. The contribution of such extremely large values have a
small influence both on the variance and the distribution of the sum we
investigate. The next problem expresses such a fact. This result
together with the result of problem~42 formulated later have the
following consequence: Let us consider a triangular array which
satisfies the Lindeberg condition. If we truncate the too large values
(those which are greater than a fixed positive number $\e>0$) and then
normalize the truncated random variables in such a way that their
expected value be zero, then the sum of these modified random variables
have the same limit in distribution as the sums of the original random
variables.
 
It is worth mentioning that some modification of the above argument
may help to find a different proof of the central limit theorem
if the Lindeberg condition holds. It can be seen that we may also
truncate and normalize the elements of the $k$-th row of the
triangular array at a level $\e_k>0$ with such a sequence $\e_k\to0$
which converges to zero sufficiently slowly instead of a fixed
number $\e>0$ and we can guarantee that the new triangular
array is equiconvergent with the original one. Then some not too hard
calculation shows that all moments of the sums made from this truncated
and normalized random variables converge to the corresponding moments
of a random variable with standard normal distribution. On the other
hand, it is known from the general theory that this fact implies the
central limit theorem. Now we omit the details. We return to this
remark in Part~II.\ of this series of problems.

\medskip
\item{40.)} Let a triangular array $\xi_{k,j}$, $1\le j\le n_k$,
$\limm_{k\to\infty}\summ_{j=1}^{n_k}E\xi_{k,j}^2=1$, $E\xi_{k,j}=0$,
satisfy the Lindeberg condition. Let us fix a real positive number
$\e>0$, and define the random variable $\bar\xi_{k,j}
=\bar\xi_{k,j}(\e)=\xi_{k,j} I(|\xi_{k,j}|<\e)-E\xi_{k,j}
I(|\xi_{k,j}|<\e)$. Then $\limm_{k\to\infty}\summ_{j=1}^{n_k}
E\bar\xi_{k,j}^2=1$. Let us also define the partial sums $\bar
S_{k}=\summ_{j=1}^{n_k}\bar \xi_{k,j}$ and $S_{k}=\summ_{j=1}^{n_k}
\xi_{k,j}$. Then the differences $S_k-\bar S_k$ converge stochastically
to zero if $k\to\infty$. 

\medskip
Let us emphasize that to guarantee the validity of the Lindeberg
condition in problem~39 we have imposed not only the condition that
the random sums $S_k$ converge in distribution to a normal law. We
also demanded that the limit distribution have the ``right" variance~1.
We also shall show examples where the triangular array $\xi_{k,j}$,
$1\le k<\infty$, $1\le j\le n_k$, satisfies the uniform smallness
condition $\limm_{k\to\infty}\(\supp_{1\le j\le n_k}E\xi_{k,j}^2\)=0$,
it does not satisfy the Lindeberg condition, and the random sums $S_k$
converge in distribution to a normal distribution. But in this example
the variance of the limit distribution is less than~1. Before the
discussion of such examples we present some results which give useful
sufficient conditions for the validity of the Lindeberg conditions.
We shall consider only normalized partial sums of independent random
variables instead of triangular arrays. First we formulate the
appropriate version of the Lindeberg condition for sequences of
independent random variables. 

\medskip\noindent
{\bf The definition of the Lindeberg condition for sequences of
independent random variables.} {\it Let $\xi_n$, $n=1,2,\dots$, be a
sequence of independent random variables, for which $E\xi_n=0$,
$\sigma_n^2=E\xi_n^2<\infty$, $n=1,2,\dots$, and the sequence of numbers
$s^2_n=\summ_{k=1}^n\sigma_k^2$, $n=1,2,\dots$, satisfies the relation
$\limm_{n\to\infty}s_n^2=\infty$. The sequence $\xi_n$, $n=1,2,\dots$,
satisfies the Lindeberg condition if
$$
\lim_{n\to\infty}\frac1{s_n^2}\sum_{k=1}^nE\xi_k^2I(\{|\xi_k|>\e
s_n\})=0.
$$
for all numbers $\e>0$.}

\medskip
Given a sequence $\xi_n$, $n=1,2,\dots$, of independent random variables
for which $E\xi_n=0$, $\sigma_n^2=E\xi_n^2<\infty$, and the sequence
$s_n^2=\summ_{k=1}^n \sigma_k^2$ satisfies the relation
$\limm_{n\to\infty}s_n^2=\infty$, define the triangular array
$\xi_{k,j}$, $k=1,2,\dots$, $j=1,\dots,n_k$, by the formulas
$n_k=k$ and $\xi_{k,j}=\frac{\xi_j}{s_k}$, if $1\le j\le k$ with the
help of this sequence. The original sequence of random variables
$\xi_n$, $n=1,2,\dots$ satisfies the Lindeberg condition if and only if
the triangular array $\xi_{k,j}$  defined by its help satisfies it.
Hence the central limit theorem and its converse formulated for
triangular arrays can be reformulated for normalized partial sums of
independent random variables. (In this reformulation the uniform
smallness property of the summands have the form $\limm_{n\to\infty}
\frac{\max\limits_{1\le k\le n}\sigma_k^2}{s_n^2}=0$.) In the next
problem we formulate some properties which imply the Lindeberg
condition.  

\medskip
\item{41.)} Let $\xi_n$, $n=1,2,\dots$, be a sequence of independent
random variables for which $E\xi_n=0$, $\sigma_n^2=E\xi_n^2<\infty$,
$n=1,2,\dots$, $\limm_{n\to\infty}s_n^2=\infty$, where
$s_n^2=\summ_{k=1}^n\sigma_k^2$. This sequence of random variables
satisfies the Lindeberg condition if one of the following properties
holds.
\item{a.)} $E|\xi_k|^{2+\alpha}<\infty$, for all numbers $k=1,2,\dots$
with some constant $\alpha>0$, and $\limm_{n\to\infty}\frac
{\(\summ_{k=1}^n E|\xi_k|^{2+\alpha}\)^{2/(2+\alpha)}}{s_n^2}=0$.
In particular, this condition holds if $E\xi_k^2\ge K$ with some
constant $K>0$ for all indices $k=1,2,\dots$, and besides
the relation $E\limm_{k\to\infty}k^{-\alpha/2}
E|\xi_k|^{2+\alpha}=0$ holds.
\item{b.)} The independent random variables $\xi_n$, $n=1,2,\dots$,
are uniformly distributed. (This means that the result of problem~35
follows from the central limit theorem formulated under general
conditions.) 

\medskip
We want to show an example for a sequence of independent random
variables with expectation zero and finite variance which satisfies
the uniform smallness condition, the normalized partial sums made
with the help of this sequence converge in distribution to the
standard normal distribution, but the normalization is not the natural
one, that is we divide the partial sums not with the square-root of
their variances. Such an example shows that the normalized partial sums
of independent random variables may satisfy the central limit theorem
(with a non-usual normalization) also if the Lindeberg condition does
not hold. Before the construction we prove a simple result which is
useful also in other investigations. This result states that a
sequence of random variables convergent in distribution has the same
limit as its small perturbations. Let us remark that a similar result
also holds for random variables taking values in more general spaces.

\medskip
\item{42.)} Let two sequences of random variables $S_n$ and $T_n$,
$n=1,2,\dots$, be given such that the sequence of random variables
$S_n$, $n=1,2,\dots$, converges in distribution to a distribution
function $F$ and the sequence $T_n$, $n=1,2,\dots$, converges
stochastically to zero, i.e.\ $P(|T_n|>\e)\to0$ for all numbers
$\e>0$ if $n\to\infty$. Then the sequence of random variables
$S_n+T_n$, $n=1,2,\dots$, converges to the distribution function~$F$.
\item{43.)} Let us construct a sequence of independent random
variables $\xi_n$, $n=1,2,\dots$, for which $E\xi_n=0$, $E\xi_n^2=1$,
$n=1,2,\dots$, and the sequence of partial sums
$S_n=\summ_{k=1}^n\xi_k$, $n=1,2,\dots$, satisfy one of the
following statement:
\item{a.)} The sequence of normalized partial sums $\sqrt{\frac
2n}S_n$, $n=1,2,\dots$, converge in distribution to the central normal
distribution.
\item{b.)} The sequence of normalized partial sums $\sqrt{\frac
2n}S_n$, $n=1,2,\dots$, converges in distribution to a normal random
variable with expected value $m\neq0$ and variance~1.

\medskip
The result of problem 38 implies that an example satisfying problem~43
cannot satisfy the Lindeberg condition. Indeed, if the Lindeberg
condition held, then the distribution of the normalized partial sums
$\frac{S_n}{\sqrt n}$ would converge to the standard normal
distribution function. Let us also remark that in the counter examples
of problem~43 the limit distribution had a {\it strictly smaller}
variance than the variance of the normalized partial sums. The following
problem shows that no counter-example exists where the limit
distribution has too large variance. 

\medskip
\item{44.)} Let a sequence of distribution functions $F_n$,
$n=1,2,\dots$, converge in distribution to a distribution function
$F_0(x)$. Then $\liminff_{n\to\infty}\int u^2F_n(\,du)\ge \int
u^2F_0(\,du)$.
 
\medskip\noindent
{\script H.) The multi-dimensional central limit theorem.}

\medskip
The multi-dimensional limit theorems can be investigated by means of
the characteristic functions similarly to their one-dimensional
analog. Moreover, the characteristic function method enables us to
reduce the investigation of the multi-dimensional limit theorems to
the one-dimensional case. This is the content of the subsequent two
problems. 

\medskip
\item{45.)} Let $\bold Z=(Z_1,\dots,Z_m)$, $n=1,2,\dots$, be an
$m$-dimensional random vector. Let us consider for all $m$-dimensional
vectors $(a_1,\dots,a_m)\in R^m$ the random variable
$Z=Z(a_1,\dots,a_m)=\summ_{j=1}^m a_jZ_j$. The distribution functions
of all one-dimensional random variables $Z=Z(a_1,\dots,a_m)$
also detbermine the distribution of the random vector
$\bold Z=(Z_1,\dots,Z_m)$.
\item{46.)} Let $\bold Z_n=(Z_{1,n},\dots,Z_{m,n})$, $n=1,2,\dots$,
be a sequence of $m$-dimensional random vectors. The random vectors
$\bold Z_n$ converge in distribution to an $m$-dimensional
distribution as $n\to\infty$ if and only if for all vectors
$(a_1,\dots, a_m)\in R^m$ the one-dimensional random variables
$Z_n=Z_n(a_1,\dots,a_m)=\summ_{j=1}^m a_jZ_{j,n}$, $n=1,2,\dots$,
converge in distribution as $n\to\infty$. If the random vectors
$\bold Z_n$ converge in distribution to a probability measure~$\mu$
in the $m$-dimensional space, then this limit measure $\mu$ can be
characterized in the following way. If $\bold Z=(Z_1,\dots,Z_m)$
is a random vector with distribution $\mu$, and $(a_1,\dots,a_m)\in
R^m$ is an arbitrary $m$-dimensional vector, then the distribution
function of the random variable $Z=Z(a_1,\dots,a_m)=\summ_{j=1}^m
a_jZ_j$ equals the limit of the distribution functions of the random
$Z_n=Z_n(a_1,\dots,a_m)=\summ_{j=1}^m a_jZ_{j,n}$. In particular,
we claim that these relations uniquely determine the measure $\mu$.

\medskip
Let us define the notion of multi-dimensional normal (Gaussian)
distributions. It will be the natural analog of the one-dimensional
normal distribution. The multi-dimensional normal distributions
with expectation zero are exactly those distributions which appear as
the limit in the multi-dimensional version of the central
limit theorem. Before their formal definition let us recall some
notions.
 
The expected value of an $m$-dimensional random vector $\bold
Z=(Z_1,\dots,Z_m)$ is the  $m$-dimensional vector $\bold
M=(M_1,\dots,M_m)$ whose coordinates are the numbers $M_j=E\xi_j$,
$1\le j\le m$. The covariance matrix of the random vector $\bold
Z=(Z_1,\dots,Z_m)$ is the $m\times m$ matrix $(D_{j,k})$, $1\le j,k\le
m$ whose elements are the numbers
$D_{j,k}=EZ_jZ_k-EZ_jEZ_k=E(Z_j-EZ_j)(Z_k-EZ_k)$. In our notations
a vector $\bold b=(b_1,\dots,b_m)$ will mean a row-vector, and we
shall denote its transpose, which is a column vector, by
$\bold b^*$. If $\bold x=(x_1,\dots,x_m)\in R^m$ and $\bold
y=(y_1,\dots,y_m)\in R^n$ are two $m$-dimensional vectors, then
$(\bold x,\bold y)$ denotes their scalar product, i.e.\
$(\bold x,\bold y)=\summ_{j=1}^mx_jy_j$. 

\medskip\noindent
{\bf The definition of the multi-dimensional normal distribution.}
{\it Let $\xi_j$, $1\le j\le m$, be independent random variables
with standard normal distribution. Then the random vector
$\bold \xi=(\xi_1,\dots,\xi_m)$ will be called an
$m$-dimensional standard normal random vector and its distribution
the $m$-dimensional standard normal distribution. If
$B=(b_{j,k})$, $1\le j,k\le m$, is an $m\times m$ matrix,
$\bold M=(M_1,\dots,M_m)$ is an $m$-dimensional vector and
$\bold \xi=(\xi_1,\dots,\xi_m)$ is an $m$-dimensional standard
normal random vector, then $\bold\xi B+\bold M$ is called an
$m$-dimensional vector with normal distribution. A probability
measure $\mu$ on the measurable sets of the $m$-dimensional Euclidean
space $R^m$ is called an $m$-dimensional normal distribution if and
only if it equals the distribution of an $m$-dimensional random
vector with normal distribution defined in the above way with an
appropriate $m\times m$ matrix $B$ and a vector $\bold M\in R^m$.}

\medskip\noindent 
Let us first characterize the multi-dimensional
normal distributions. 

\medskip
\item{47.)} The covariance matrix $\Sigma$ of all $m$-dimensional
random vectors is positive semi-definite, i.e.\ for all $m$-dimensional
vectors $\bold x=(x_1,\dots,x_m)$ $\bold x\Sigma\bold x^*=(\bold
x\Sigma,\bold x)\ge0$. In the converse direction, for all
$m$-dimensional vectors $\bold M=(M_1,\dots,M_m)\in R^m$ and $m\times
m$ symmetric, positive semi-definite matrices $\Sigma$ there exists an
$m$-dimensional random vector with normal distribution whose expected
value is this vector $\bold M$ and whose covariance matrix is this
matrix $\Sigma$. The characteristic function of an $m$-dimensional
normal distribution with expected value $\bold M$ and covariance
matrix $\Sigma$ equals
$$
\varphi(t_1,\dots,t_m)=Ee^{i(\bold t,\bold \xi)}=
Ee^{i(t_1\xi_1+\cdots+t_m\xi_m)}=
\exp\left\{-\frac{(\bold t\Sigma,\bold t)}2+i(\bold t,\bold M)\right\},
\tag13
$$
where $\bold t=(t_1,\dots,t_m)$, and $\bold \xi=(\xi_1,\dots,\xi_m)$
denotes an $m$-dimensional random vector with expected value $\bold M$
and covariance matrix $\Sigma$. In particular, an $m$-dimensional
normal distribution is uniquely determined by its expected value $\bold
M$ and covariance matrix $\Sigma$. 

\medskip
The special form of the characteristic function of an
$m$-dimensional characteristic function given in formula~(13) has
several simple but important consequences. It implies the most important
properties of Gaussian distributions. Although these results do not 
belong to the subject of this series of problems, I discuss two
problems that may be useful in such investigations.

\medskip
\item{48.)} Let $\xi_j$, $1\le j\le m$, be $m$ independent 
random variables with standard normal distribution, 
$\bold M=(M_1,\dots,M_l)$ an $l$-dimensional 
(deterministic) vector, and $B$ a rectangular matrix 
of size $l\times m$. Then
$(\eta_1,\dots,\eta_l)=(\xi_1,\dots,\xi_m)B+\bold M$ is an
$l$-dimensional random vector with normal distribution.
In particular, if we preserve only $l$ coordinate of a
a normal random vector $\eta$ of dimension~$m$, then we
get a normal random vector of dimension~$l$.
\item{49.)} Let $\eta=(\eta_1,\dots,\eta_m)$ be such a random 
vector of dimension~$m$ with normal distribution whose 
covariance matrix $\Sigma=(\sigma_{p,q})$, $1\le p,q\le m$,
has the following property: The set $\{1,\dots,m\}$ has
a partition  $\{1,\dots,m\}=\bigcupp_{j=1}^k L_j$, 
$1\le j\le k$, such that the non zero elements of the matrix
$\Sigma$ are concentrated in the union of the squares 
$L_1\times L_1$,\dots, $L_k\times l_k$, i.e., $\sigma_{p,q}=0$, 
if  $p\in L_j$, $q\in L_{j'}$, and $j\neq j'$.  In this case
the random vectors $\bar\eta_j=(\eta_p,\;p\in L_j)$, $1\le j\le k$, 
obtained by an appropriate grouping of the coordinates of the
vector $\eta$ are Gaussian random vectors, independent of
each other.

\medskip
Finally we formulate the multi-dimensional central limit theorem for
appropriately normalized partial sums of independent vectors. We do
not formulate this result in its most general form, and do not
discuss its version for triangular arrays, although this would be
also possible.

\medskip
\item{50.)} Let $\bold\xi_k=(\xi_{1,k},\dots,\xi_{m,k})$,
$k=1,2,\dots$, be a sequence of independent $m$-dimensional random
vectors with expectation zero, i.e.\ we assume that the relation
$\bold M_k=(E\xi_{1,k},\dots,E\xi_{m,k})=(0,\dots,0)$ holds for all
numbers $k=1,2,\dots$. Let us also assume that the random vectors
$\bold\xi_k=(\xi_{1,k},\dots,\xi_{m,k})$ have a finite covariance
matrix $\Sigma_k$ for all indices $k=1,2,\dots$, and the relation
$\limm_{n\to\infty}\frac1{A_n^2}\summ_{k=1}^n \Sigma_k=\Sigma$ holds
with an $m\times m$ matrix $\Sigma$ and with norming constants
$A_n$, $n=1,2,\dots$, such that $A_n\to\infty$ if $n\to\infty$. If
beside this all coordinates of the random vectors $\bold\xi_k$,
$k=1,\dots$, satisfy the Lindeberg condition, i.e.\
$$
\lim_{n\to\infty}\frac1{A_n^2}\sum_{k=1}^nE\xi_{p,k}^2I(|\xi_{p,k}|>\e
A_n)=0, \quad \text{ for all numbers } p=1,\dots m, \tag14
$$
for all $\e>0$, then  the normalized partial sums
$\frac1{A_n}\bold S_n=\frac1{A_n}(S_{1,n},\dots,S_{m,n})=
\frac1{A_n}\summ_{k=1}^n(\xi_{1,k},\dots, \xi_{m,k})$ converge in
distribution to the normal distribution function with expected value
$M=(0,\dots,0)$ and covariance matrix $\Sigma$.
\item{} In particular, if $\bold\xi_k=(\xi_{1,k},\dots,\xi_{m,k})$,
$k=1,2,\dots$, is a sequence of independent and identically distributed
$m$-dimensional random vectors with expectation zero and finite
covariance matrix $\Sigma$, then the normalized partial sums
$\frac1{\sqrt n} (S_1,\dots,S_m)=\frac1{\sqrt n}
\summ_{k=1}^n(\xi_{1,k},\dots, \xi_{m,k})$ converge in distribution to
the normal distribution function with expectation zero and covariance
$\Sigma$.
\item{51.)} Let $\bold\xi_n=(\xi_{1,n},\dots,\xi_{m,n})$,
$n=1,2,\dots$, be a sequence of $m$-dimensional random vectors with
expectation zero and such that all random vectors
$\bold\xi_n=(\xi_{1,n},\dots,\xi_{m,n})$
have a finite covariance matrix $\Sigma_n$ for all
indices $n=1,2,\dots$. Let us further assume that the relation
$\limm_{n\to\infty}\frac1{A_n^2} \summ_{k=1}^n\Sigma_n=\Sigma$
with an appropriate matrix $\Sigma$ and norming factors $A_n$,
$n=1,2,\dots$, such that $A_n\to\infty$, ha $n\to\infty$. Besides,
we also assume that the uniform smallness condition
$\limm_{n\to\infty}\frac{\max\limits_{1\le j\le
m} \max\limits_{1\le k\le n}E\xi_{j,k}^2}{A_n^2}=0$ holds. If the
sequence of random vectors $\frac{S_n}{A_n}$ converges in
distribution to an $m$-dimensional normal distribution whose
covariance matrix is $\Sigma$, then also the Lindeberg condition
formulated in relation~(14) holds.
 
\parskip=2.5pt plus 1.2pt
 
\beginsection Some additional remarks
 
We formulate some problems whose investigation is a natural continuation
of the study of the problems in this paper.

\item{1.)} The central limit theorem for appropriately normalized
partial sums of independent random variables was deduced from the
convergence of their characteristic functions. Actually not only the
convergence of the characteristic functions can be proved, but also
the speed of convergence can be given. Besides, the
characteristic functions of the normalized partial sums can be
better approximated if we add some correction terms to the
characteristic function of the approximating normal distribution. These
correction terms can be found by means of a natural Taylor expansion. It
is natural to expect that these estimates also supply a good estimate
about the accuracy of the normal approximation of the distribution
function of normalized partial sums of independent random variables.
Moreover, a better approximation of their characteristic functions
also yields a better approximation of the distribution function of
normalized partial sums of independent random variables if some
appropriate correction terms are added to the normal approximation.
 
\item{} A natural analog of the above question is to give a good
estimate on the Gaussian approximation of the density function of
appropriately normalized partial sums of independent random variables
and to find a better approximation for this density function by means
of an appropriate asymptotic expansion. Naturally, in this
investigation we have to impose some additional assumptions on the
distribution of the random variables whose partial sums we investigate
to guarantee the existence of a nice density function. This problem
about the behaviour of density functions is simpler, because there
exists a relatively simple inverse Fourier transformation formula to
express a density function by means of its Fourier transform. On the
other hand, there is no really useful formula for the calculation of
distribution functions with the help of their characteristic function
in a simple, well applicable form. But by convolving the distribution
function of the normalized partial sums we want to investigate with an
appropriate density function (essentially concentrated in a small
neighbourhood of the origin) we get a new distribution function close
to the original one which has a density function, and as a consequence
this modified (smoothed) distribution function can be well estimated as
the integral of its density function. This approach together with some
additional ideas enables us to study the problems mentioned in the
previous paragraph. These problems will be the main part of the
(essentially shorter) second part of this series of problems. 

\medskip
\item{2.)} We have seen that the appropriately normalized partial sums
of independent random variables converge in distribution to a normal
distribution function under some mild conditions. On the other hand, we
would like to get a complete picture about all possible limit theorems
for the distribution of appropriately normalized partial sums or in a
more general way, for the sums of the elements in the same rows of a
triangular array of random variables. It is natural to impose some
kind of uniform smallness condition which should guarantee that there
are no such dominating terms in the random sums we investigate
whose magnitude is comparable with the magnitude of the whole sum.
 
\item{} A fairly complete picture can be given about all possible
limit theorems for the normalized partial sums of independent random
variables. In the proof the {\it Fundamental Theorem about the
convergence of distribution functions}\/ proved also in this series of
problems plays a key role. It makes possible to reformulate the problem
to the language of characteristic functions. The first question to be
understood in this problem is the description of all possible limit
theorems. This leads to the study of the fixed points of certain
operators in the space of distribution functions defined by means of
convolution and rescaling. This study leads to the description of the
so-called infinitely divisible distributions which appear as the limits
in limit theorems. Further investigations make possible also to
describe in which limit theorems a certain infinitely divisible
distribution appears.
 
\item{} These problems are solved by means of certain methods of the
analysis about the study of the behaviour of the characteristic
functions, but the proof also contains several probabilistic
ideas. We also remark that although there are several different kind
of limit theorems, the central limit theorem is the only ``universal
law" for the limit distribution of normalized partial sums of
independent random variables, where the limit distribution function
``forgets" the distribution of the single terms in the partial sums.
Here we do not explain the more precise meaning of this rather loose
statement. A fairly complete discussion with complete proofs of the
problems mentioned in this Section~2 can be found together with the
explanation of the ideas behind the proofs in the text {\it Limit
theorems and infinitely divisible distribution} \/ on my homepage.
(For the time being it exists only in Hungarian.) It also contains the
explanation of the statement that the central limit theorem is the
only ``universal law" among the limit theorems for the limit
distribution of partial sums of independent variables. 

\medskip
\item{3.)} Let us consider a sequence $\xi_1,\xi_2,\dots$,
of independent, identically distributed random variables with
expectation zero and finite variance together with the partial sums
$S_n=\summ_{k=1}^n\xi_n$, $n=1,2,\dots$, defined with their help.
The central limit theorem gives a good estimate on the probability
$P\(\frac{S_n}{\sqrt n}>x\)$ for large indices~$n$ and fixed
numbers~$x$. We may ask whether a similar good estimate can be given for
the probabilities $P\(\frac{S_n}{\sqrt n}>x_n\)$, that is we are
interested in a good estimate on the probability of a similar event
when the number $x$ is replaced by a number $x_n$ which may depend
on~$n$. The special case $x_n=x\sqrt n$, that is the investigation of
the probability of the event that the average of independent,
identically distributed random variables is larger than a fixed
number~$x$ is a particularly important question. This problem belongs
to an important part of the probability theory, called the theory of
large deviations. The above mentioned problem is discussed together
with some additional questions also in a series of problems {\it The
theory of large deviation; the case of partial sums of real valued
random variables}\/ on my homepage. (For the time being it exists only
in Hungarian). Let us remark that the estimates in the large
deviations problems do not agree with the estimates suggested by a
formal extension of the central limit theorem.
 
\item{} It would be natural to try to estimate the probability
$P(S_n>nx)$ similarly to the estimate of the event $P(S_n>\sqrt nx)$
in the proof of the central limit theorem. Nevertheless, this method
in itself does not suffice to estimate this probability. It is worth
understanding why the method of the proof of the central limit theorem
is not sufficient to estimate this probability. Let us first consider
the case when the distribution function $P(S_n>nx)$ also has a nice
density function $f_n(x)=\frac d{dx}P(S_n> nx)$, and let us first
estimate this density function. Then we have to investigate the
integral $f_n(x)=\frac n{2\pi}\int e^{-intx}\varphi^n(t)\,dt$,
where $\varphi(t)$ is the characteristic function of the random
variable~$\xi_1$. This integral is similar to the integral appearing
in the proof of the local central limit theorems. The only difference
is that the factor $e^{it\sqrt nx}$ in the integral investigated in the
proof of the local central limit theorem is replaced by the term
$e^{it\sqrt nx}$ in this case.
 
\item{} In the proof of the local central limit theorem a singular
integral had to be investigated which was strongly concentrated in a
small neighbourhood of the origin. In the analogous large deviation
problem a similar singular integral has to be investigated. But there
is a small difference which makes the latter problem more difficult.
The cause of this difficulty is that although the absolute value of
the complex number valued integrand whose integral has to be estimated
in the large deviation problem has a strongly localized maximum in
the origin, but its imaginary part has a strong fluctuation in a
small neighbourhood of the origin. This fluctuation causes a strong
cancellation, hence we cannot claim that the contribution of a small
neighbourhood of the origin yields the main contribution to the
integral we are estimating. In the proof of the local central limit
theorem this difficulty did not appear, since in this case the
fluctuation of the imaginary part of the integrand is not so strong in
a small neighbourhood of the origin. The reason for this difference is
that the term $e^{-i\sqrt ntx}$ appears instead of the term $e^{-intx}$
in the integral investigated in the proof of the local central limit
theorem and $\left.\frac{d\varphi(t)}{dt}\right|_{t=0}=E\xi_1=0$ in
this case.
 
\item{} Such kind of problems appear often in the analysis, and an
important technique, the saddle point method was worked out to
investigate such problems. To apply the saddle point method let us
rewrite the integral we are interested in in the form $f_n(x)=\frac
n{2\pi}\int e^{n(-itx+\log\varphi(t))}\,dt$. (In this heuristic
explanation let us disregard such technical difficulties as the problem
that we cannot always take logarithm of a function.) If the function
$n(-itx+\log \varphi(t))$ in the exponent of the integrand we study is
an analytic function of the variable~$t$, then the saddle-point method
suggests to replace the domain of integration to an appropriate new
curve which goes through the saddle point, i.e.\ through a
point~$z$ which satisfies the identity
$$
\frac{d(-izx+\log\varphi(z))}{dz}=0. \tag$*$
$$
The integral we are investigating does not change by this replacement.
On the other hand this new integral can be better estimated since the
absolute value of the integrand on this new curve has a strongly
localized (local) maximum in the saddle point, and its fluctuation
around the saddle point is small. The proof of the large deviation
estimates for averages of independent and identically distributed
random variables are based on this idea. Naturally, it is useful to
combine it with some additional observations. For instance, it is
enough to look for the saddle point on the imaginary axis where the
imaginary part of the left-hand side of the saddle point equation~$(*)$
is automatically zero. If the saddle point equation has no solution,
then an appropriate approximation and some additional ideas are needed
to solve the large deviation problems.
 
\item{} Let us remark that in most probability text-books on the
theory of large deviations the saddle point method is not discussed.
In these text-books the notion of the so-called conjugated
distributions are introduced, and the problem is solved with their
help. On the other hand, a better understanding of the problem may
help us to understand that the introduction of conjugated
distributions can be interpreted as the application of the saddle
point method to the investigation of large deviation problems
formulated in the language of probability theory.
 
\item{} Finally we make the following remark. The above sketched
saddle point method can be applied for the investigation of large
deviation problems only if the characteristic function of the
random variables whose averages we investigate has an analytic
continuation. One may ask whether this condition does not mean an
unnecessary restriction. A detailed investigation shows that the
answer to this question is negative. It turns out that this condition
about the analiticity of the characteristic function has a
probabilistic content, and the behaviour of the probabilities we
investigate in large deviation problems heavily depends on whether
the characteristic function is analytic. If this analiticity
property does not hold, then the probabilities investigated in the
large deviation problems have an essentially different behaviour.
 
\vfill\eject
 
\parskip=3pt plus 1pt
 
\beginsection Solutions.
 
\item{1.)} Put $I=\frac1{\sqrt {2a\pi}}\int_{-\infty}^\infty
e^{-u^2/2a}\,du$. By expressing the number $I^2$ as a double integral
and rewriting it in polar coordinate system we get that
$$
\align
I^2&=\frac1{{2a\pi}}\int_{-\infty}^\infty \int_{-\infty}^\infty
e^{-(u^2+v^2)/2a}\,du \\
&=\int_{-\pi}^\pi\int_0^\infty
\frac1{2\pi}\cdot\frac ra e^{-r^2/2a}\,dr\,d\varphi
=\int_{-\pi}^\pi\frac1{2\pi}\,d\varphi=1.
\endalign
$$
This relation implies the first statement of the problem.
\item{} Let us fix a number $a>0$. Put
$F(z)=\frac1{\sqrt {2a\pi}}\int_{-\infty}^\infty
e^{-(u-z)^2/2a}\,du$. With the substitution $\bar u=u-z$
we get that
$$
F(z)=\frac1{\sqrt {2a\pi}}\int_{-\infty}^\infty e^{-\bar
u^2/2a}\,d\bar u=1
$$
for all {\it real numbers}\/ $z$. We can generalize this identity for
all complex numbers~$z$ with the help of one of the following argument
of the theory of analytic functions.
\item{} {\it First argument:}\/ Both functions
$F(z)$ and $G(z)\equiv1$ are analytic function on the plane of complex
numbers. As they agree for all real numbers they also agree for all
complex numbers. To see that the function $F(z)$ is really analytic,
observe that it can be obtained as the limit of analytic functions in
such a way that the convergence is uniform on all compact subsets
of the plane of complex numbers. Then the limit of these analytic
functions, i.e.\ the function $F(z)$ is also analytic. Such analytic
functions converging to $F(z)$ can be obtained by approximating the
integrals $F(z)$ by the natural approximating sums of these integrals.
\item{} {\it Second argument:}\/ We get with the change of variable
$\bar u=u-z$ that
$$
F(z)=\frac1{\sqrt {2a\pi}}\int_{-\infty-\Im z}^{\infty-\Im z}
e^{-\bar u^2/2a}\,d\bar u=1.
$$
To prove the above identity observe that
$\limm_{|u|\to\infty}e^{-(u+iv)^2/2a}=0$, and the
convergence is uniform with respect to the variable~$v$ if $|v|$ is in
a bounded interval. This fact together with the result of analytic
functions by which the contour integral of an analytic function
(containing no singular point) on a closed curve is zero imply that the
value of the above integral does not  change if we replace the route of
integration $[-\infty-\Im z,\infty-\Im z]$ to the line
$[-\infty,\infty]$. This implies the identity we wanted to prove.
\item{2.)} Let $\xi$ be a random variable with Poissonian distribution
with parameter $\lambda=n$, i.e.\ $P(\xi=k)=\frac{n^k}{k!}e^{-n}$,
$k=0,1,2,\dots$. The Fourier series corresponding to the distribution of
this random variable~$\xi$ equals
$$
P_n(t)=\sum_{k=0}^\infty P(\xi=k)e^{itk}=\sum_{k=0}^\infty
\frac{n^k}{k!}e^{-n+ikt}=e^{-n+ne^{it}}.
$$
This identity together with relation~(2) imply relation~(3) 
with the choice $k=n$.
\item{} Let us observe that relation~(4b) is a simple 
consequence of relations~(3) and~(4a) and the asymptotic formula
$\frac1{1+x}=1-x+O(x^2)=1+O(x)$, if $|x|\le \frac12$.
\item{} To prove formula (4a) let us first give an upper bound 
about the contribution of the domain $\{t\colon\; |t|\ge n^{-1/3}\}$ 
to the integral at the left-hand side of formula~(4a). Then let us 
consider the restriction of this integral to the domain 
$\{t\colon\; |t|<n^{-1/3}\}$ and let us give a good asymptotical 
estimate on it.
\item{} To carry out the first estimate let us observe that
$$
\left|e^{n(e^{it}-1-it)}\right|= e^{n\Re (e^{it}-1-it)}=e^{n(\cos
t-1)}<e^{-nt^2/4}<e^{-n^{1/3}/4}
\text{ if }n^{-1/3}\le t\le \pi,
$$
and this implies that
$$
\left|\int_{\{n^{-1/3}\le |t|\le \pi\}}
e^{n(e^{it}-1-it)}\,dt\right|=O\(e^{-\const n^{1/3}}\). \tag2.1
$$
To carry out the second estimation let us give a good asymptotical
formula on the integrand of relation~(4a) in a small neighbourhood of
the origin by means of a Taylor expansion. We get that
$n(e^{it}-1-it)=n\(-\frac{t^2}2-i\frac{t^3}6+O(t^4)\)$ and
$$
\align
e^{n(e^{it}-1-it)}&=e^{-nt^2/2}e^{-in{t^3}/6+O(nt^4)}\\
&=e^{-nt^2/2}\(1-\frac{i(\sqrt nt)^3}{6\sqrt n}+O\(\frac{(\sqrt
nt)^4}n\)+O\(\frac{\sqrt nt)^6}n\)\),
\endalign
$$
if $|t|\le n^{-1/3}$. By exploiting this estimate and making the
substitution of variables $\bar t=\sqrt nt$ we get that
$$
\align
\int_{-n^{-1/3}}^{n^{-1/3}} e^{n(e^{it}-1-it)}\,dt&=
\frac1{\sqrt n}\int_{-n^{1/6}}^{n^{1/6}}
e^{-{\bar{t}}^2/2}\(1-i\frac{{\bar {t}}^3}{6\sqrt
n}+\frac{O\({\bar t}^4+{\bar t}^6\)}n\)\,d\bar t\\
&=\frac1{\sqrt n}\(\int_{-\infty}^\infty-\int_{|\bar t|>n^{1/6}}\).
\endalign
$$
On the other hand,
$$
\int_{|t|\ge n^{1/6}} e^{-t^2/2}\(1-i\frac{t^3}{6\sqrt
n}+\frac{O(t^4+t^6)}n\)\,dt=O\(e^{-n^{1/3}/4}\),
$$
and
$$
\align
&\int_{-\infty}^\infty e^{-t^2/2}\(1-i\frac{t^3}{6\sqrt
n}+\frac{O(t^4+t^6)}n\)\,dt\\
&\qquad=\int_{-\infty}^\infty e^{-t^2/2}\,dt-
\int_{-\infty}^\infty \frac i{6\sqrt n} t^3e^{-t^2/2}\,dt+O\(\frac1n\)
=\sqrt{2\pi}+O\(\frac1n\)
\endalign
$$
because of the result of the first problem, and since
the function $t^3e^{-t^2}$ is odd. These relations imply that
$$
\int_{-n^{-1/3}}^{n^{1/3}} e^{n(e^{it}-1-it)}\,dt=\frac{\sqrt{2\pi}}{
\sqrt n} \(1+O\(\frac1n\)\). \tag2.2
$$
Then relation (4a) is a consequence of formulas (2.1) and (2.2).
\item{} Relation (4d) follows similarly from formulas (4c)  as
formula~(4a) implies relation~(4b). The only difference is that now we
apply a better approximation
$\frac1{1+x}=1-x+x^2-\cdots+(-1)^{k}x^k+O\(|x|^{k+1}\)$
for $|x|\le\frac12$. Also formula~(4c) can be proved similarly to
formula~(4a). The difference is that now we consider the Taylor
expansion of the functions $n(e^{it}-1-it)$ and $P_n(t)$ defined by
the identity $e^{n(e^{it}-1-it)}=e^{-nt^2/2}(1+P_n(\sqrt nt))$
up to the $k$-th and not only to the first term in the interval
$|t|\le n^{-1/3}$. Since $P_n(t)=\exp\left\{\frac{t^3}{\sqrt
n}R\(\frac t{\sqrt n}\)\right\}$ with an analytic function $R(t)$
bounded on the real line, these calculations enable us to give an
estimate for the integrand of the integral in~(4.1) with an accuracy of
$e^{-nt^2/2}O\(\frac{\sum_{j=1}^k(\sqrt n|t|)^{j(l(j)}}
{n^{(k+1)/2}}\)=O\(\frac{e^{-nt^2/4}}{n^{(k+1)/2}}\)$, where
$l(j)=\min\{ l\colon\;lj\ge k+1\}$ with a function of the form
$e^{-nt^2/2}\(1+Q_n(t)\)$. The function $Q_n(t)$ can be written
as $Q_n(t)=\summ_{j=1}^k \frac {\bar Q_{j}(\sqrt nt)}{n^{j/2}}$,
with some polynomials $\bar Q_j$ which do not depend on the
parameter $n$, and they can be calculated explicitly. In particular,
$\bar Q_1(t)=\frac{-i}6 t^3$. Then by carrying out the substitution
$\bar t=\sqrt n t$ we get the improvement~(4c) of formula~(4a).
\item{3.)} Let us first show that the distribution of a random variable
$\xi$ has the periodicity $h$ if and only if $|Ee^{2\pi i\xi/h}|=1$.
Indeed, if $|Ee^{2\pi tei\xi/h}|=1$, then 
$Ee^{2\pi i\xi/h}=e^{2\pi i/b}$ with some
real number $b$. But this is possible if and only if the distribution
of the random variable $2\pi\frac{\xi-b} h$ is concentrated in the points
$n=0,\pm1,\pm2,\dots$, i.e.\ the random variable $\xi$ is
concentrated on the lattice $nh+b$, $n=0,\pm1,\pm2,\dots$.
To see the statement in the converse direction observe that
if the distribution of a random variable $\xi$ is concentrated on a
lattice $nh+b$, $n=0,\pm1,\dots$, of width~$h$, then $|Ee^{2\pi i\xi}|=1$.
Furthermore, if the random variable $\xi$ concentrated on some lattice
is not concentrated in a single point, then there are two numbers $a$
and $b$, $a\neq b$, such that $P(\xi=a)>0$ and $P(\xi=b)>0$. Then
we have $|Ee^{itx}|<1$ for all sufficiently small $t>0$, and we can
even state that there exists a smallest number $t>0$ such that
$|Ee^{it\xi}|=1$.

\item{} Take the smallest number $t>0$ such that $|Ee^{it\xi}|=1$. Then
$h=\frac{2\pi}t$ is the greatest number $h$ such that the distribution
of the random variable $\xi$ is concentrated on a lattice $nh+b$,
$n=0,\pm1,\pm2,\dots$, of width $h$. Then the periodicity of the Fourier
series $P(t)=Ee^{it(\xi-b)}=\summ_{n=-\infty}^\infty
P(\xi-b=nh)e^{itnh}$ of the random variable $\xi-b$ is $\frac{2\pi}h$.
Further, from the definition of the number~$h$ follows that $|P(t)|<1$,
if $0<t<\frac{2\pi}h$. Since $P(-t)=P\(\frac{2\pi}h-t\)$ the above
relation can be rewritten as $|P(t)|<1$, if $0<|t|<\frac\pi h$. Clearly,
$P(0)=1$.
\item{} Formal differentiation by terms of the infinite sum $P(t)$
yields that $\frac{P^{(k)}(t)}{dt^k}=\summ_{n=-\infty}^\infty
i^k(nh)^ke^{itnh} P(\xi-b=nh)$, and $\left.\frac{P^{(k)}(t)}
{dt^k}\right|_{t=0}=\summ_{n=-\infty}^\infty i^k(nh)^kP(\xi-b=nh)=i^k
E(\xi-b)^k$. Under the condition $E|\xi-b|^k<\infty$ this formal
calculation is allowed. Indeed, the approximating partial sums of
the $k$-th derivative of the function $P(t)$ satisfies the inequality
$\summ_{n=-N}^N |i^k(nh)^ke^{itnh}P(\xi-b=nh)|\le E|\xi-b|^k$, and
this property allows ($k$-fold) differentiation by terms.
\item{4.)} By relation (2) and the formula written after it
$$
P(S_n=k)=\int_{-\pi}^\pi \frac1{2\pi} e^{-ikt}
P^n(t)\,dt=\int_{|t|<\frac1{\e\sqrt n}}
+\int_{\frac1{\e\sqrt n}<|t|<\e} +\int_{\e<|t|<\pi}=I_1+I_1+I_3,
$$
where $P(t)=\summ_{k=-\infty}^\infty e^{ikt}P(\xi_1=k)$,
and $\e>0$ is an arbitrary (small) positive real number. We solve the
problem if we give a good estimate for the integrals $I_1$, $I_2$ and
$I_3$ for small~$\e>0$.
\item{} It is easy to bound the integral $I_3$. It follows from the
result of problem~3 and the continuity of the function $P(t)$ that
$\supp_{\e\le |t|<\pi}|P(t)|<1$ with some number $0<q=q(\e)<1$. Hence
$$
|I_3|=\left|\int_{\e<|t|<\pi} \frac1{2\pi} e^{-ikt}
P^n(t)\,dt\right|<q^n
$$
with some number $0<q<1$. To estimate the integrals $I_1$ and $I_2$
we need a good bound on the function $P^n(t)$ if $|t|<\e$. It is more
convenient to work with the function $\log P(t)$ instead. (For
small numbers $\e>0$ this is allowed, since in this case the value of
the function $P(t)$ in the interval $[-\e,\e]$ is in a small
neighbourhood of the number~1.) Simple calculation yields that
$$
\align
\frac{d\log P(t)}{dt}&=\frac{P'(t)}{P(t)},\qquad
\left.\frac{d\log P(t)}{dt}\right|_{t=0}=im,\\
\frac{d^2\log P(t)}{dt^2}&=\frac{P''(t)P(t)-P'(t)^2}{P^2(t)},\qquad
\left.\frac{d^2\log P(t)}{dt^2}\right|_{t=0}=-m_2+m^2=-\sigma^2,
\endalign
$$
hence a Taylor expansion around the origin yields that
$$
\log P(t)=imt-\frac{\sigma^2}2t^2+o(t^2), \quad \text{if }|t|<\e.
$$
This estimate together with the relation $|P^n(t)|=e^{n\Re \log
P(t)}=e^{-n\sigma^2t^2/2+no(t^2)}\le e^{-n\sigma^2t^2/3}$, if $|t|<\e$
and $n\ge n(\e)$ imply that
$$
\align
|I_2|\le \frac1{2\pi}\int_{\frac1{\e\sqrt n}<|t|<\e} |P^n(t)|\,dt &\le
\frac1{2\pi}\int_{\frac1{\e\sqrt n}<|t|<\e}e^{-n\sigma^2t^2/3}\,dt \\
&\le\frac1{\pi\sqrt n}\int_{-\frac1\e}^\infty e^{-\sigma^2t^2/3}\,dt
\le \frac{e^{-\sigma^2/4\e^2}}{\sqrt n}.
\endalign
$$
Furthermore,
$$
\align
I_1&=\int_{-\frac1{\e\sqrt n}}^{\frac1{\e\sqrt n}} \frac1{2\pi}
e^{-ikt+inmt-n\sigma^2t^2/2+o(nt^2)}\,dt\\
&=\int_{-\frac1\e}^{\frac1\e} \frac1{2\pi\sqrt n}
e^{i(mn-k)t/\sqrt n-\sigma^2t^2/2+o(t^2)}\,dt \\
&=\int_{-\frac1\e}^{\frac1\e} \frac1{2\pi\sqrt n}
e^{i(mn-k)t/\sqrt n-\sigma^2t^2/2}(1+o(t^2))\,dt \\
&=\int_{-\infty}^{\infty} \frac1{2\pi\sqrt n}
e^{i(nm-k)t/\sqrt n-\sigma^2t^2/2}\,dt
-\int_{|t|>\frac1\e} \frac1{2\pi\sqrt n}
e^{i(nm-k)t/\sqrt n-\sigma^2t^2/2}\,dt \\
&\qquad\qquad +o\(\frac1{\sqrt n}\).
\endalign
$$
On the other hand
$$
\left|\int_{|t|>\frac1\e} \frac1{2\pi\sqrt n}
e^{i(nm-k)t/\sqrt n-\sigma^2t^2/2}\,dt \right|\le\frac
{e^{-\sigma^2/4\e^2}}{\sqrt n},
$$
and by completing the quadratic term in the exponent of the last formula
we get that
$$
\align
&\int_{-\infty}^{\infty} \frac1{2\pi\sqrt n}
e^{i(nm-k)t/\sqrt n-\sigma^2t^2/2}\,dt \\
&\qquad =\frac{e^{-(nm-k)^2/2n\sigma^2}} {2\pi\sqrt n}
\int_{-\infty}^{\infty}
\exp\left\{-\frac{\sigma^2}2\(t-i\frac{nm-k}{\sqrt
n}\)^2\right\}\,dt =\frac{e^{-(nm-k)^2/2n\sigma^2}}
{\sqrt{2\pi n}\sigma}
\endalign
$$
by the result of problem~1. These estimates imply that
$$
\left| I_1-\frac1{\sqrt{2\pi n}\sigma}\exp\left\{-\frac{(k-nm)^2}
{2n\sigma^2} \right\}\right|\le \const\frac{e^{-\sigma^2/4\e^2}}
{\sqrt n}
$$
if $n>n(\e)$. As the estimates given for the expressions $I_1$, $I_2$
and $I_3$ hold for all $\e>0$ if $n$ is sufficiently large, hence they
imply the result of problem~4.
\item{5.)} The solution of this problem is similar to that of problem~4.
Since the random variable $\xi_1$ has three finite moments in the
present case, hence we can make the following better approximation of
the function $\log P(t)$: $\log P(t)=imt-\frac{\sigma^2}2t^2+O(t^3)$.
Hence $P^n(t)=e^{imnt-n\sigma^2t^2/2+O(nt^3)}$. Then we can solve
problem~5 by making estimations similar to those in the proof of
problem~4. The only essential difference is that now we define the
domain of integration in the definition of the expressions in $I_1$
and~$I_2$ in a different way. Now put $I_1=\int_{|t|<n^{-1/3}}$ and
$I_2=\int_{n^{-1/3}\le |t|<\e}$. The reason the domain of integration
in the definition of expression~$I_1$ was chosen in the above way is
that in the domain is that $e^{O(nt^3)}=1+O(nt^3)$ in the domain
$\{|t|<n^{-1/3}\}$, hence the approximation
$e^{-ikt}P^n(t)=e^{i(mn-k)t-n\sigma^2t^2/2}\(1+O(nt^3)\)$
holds in this domain. Then by applying a natural adaptation of the
calculation in problem~4 that by omitting the error term $O(nt^3)$
from the integral approximating the expression $I_1$ we get an error
of order  $O\(\frac1n\)$. The error of all other remaining
estimations is essentially smaller. They are of order
$e^{O(-\const n^{1/3})}$. Hence these estimates yield the solution of
problem~5.
\item{5a.)} We have to estimate the expression $P(S_n=k)=\binom
nkp^k(1-p)^{n-k}$. Let us first consider the case $\alpha n<k<\beta n$
with some numbers $0<\alpha<\beta<1$. By the Stirling formula
$$
\aligned
\binom nk&=\frac{n!}{k!(n-k)!}=\frac{\(\frac ne\)^n}
{\(\frac {n-k}e\)^{n-k}\(\frac ke\)^k}
\sqrt{\frac n{2\pi k(n-k)}}\(1+O\(\frac 1n\)\)\\
&=\(1-\frac kn\)^{-(n-k)}\(\frac kn\)^{-k}\frac 1{\sqrt{2\pi n}
\sqrt{\frac kn\(1-\frac kn\)}} \(1+O\(\frac 1n\)\).
\endaligned
$$
A Taylor expansion of the function $\log x$ around the point $p$ yields
that
$$
\align
p^k\(\frac kn\)^{-k}&=\exp\left\{k\(\log p-\log \frac kn\)\right\}\\
&=\exp\left\{-\frac kp\(\frac kn-p\)+\frac k{2p^2}\(\frac
kn-p\)^2+O\(n\(\frac kn-p\)^3\)\right\}.
\endalign
$$
Similarly,
$$
\align
&(1-p)^{n-k}\(\frac{n-k}n\)^{-(n-k)} \\
&\qquad=\exp\left\{\frac{n-k}{1-p}\(\frac kn-p\)+\frac
{n-k}{2(1-p)^2}\(\frac kn-p\)^2+O\(n\(\frac kn-p\)^3\)\right\}.
\endalign
$$
By multiplying the last two expressions and exploiting that in the
formula got in such a way the coefficient of the term $\(\frac
kn-p\)^2$ is
$$
\frac k{2p^2}+\frac{n-k}{2(1-p)^2}=\frac
{(k-np)^2}{2p(1-p)}+(1-2p)\frac{(k-np)^3}{n^2}
$$
we get that
$$
p^k\(\frac kn\)^{-k}(1-p)^{n-k}\(\frac{n-k}n\)^{-(n-k)}
=\exp\left\{-\frac{(kn-p)^2}{2np(1-p)}+
O\(n\(\frac kn-p\)^3\)\right\}.
$$
Since $\frac kn\(1-\frac kn\)=p(1-p)\(1+O\(\frac{k-np}n\)\)$ the
above calculations yield that
$$
P(S_n=k)=\frac{\exp\left\{-\frac{(kn-p)^2}{2np(1-p)}+ O\(n\(\frac
kn-p\)^3\)+O\(\frac kn-p\)+O\(\frac 1n\)\right\}} {\sqrt{2\pi
p(1-p)n}}
$$
Since $E\xi_1=p$, $\text{Var}\,\xi_1=p(1-p)$ the last formula yields an
estimate stronger than we want in the case $|k-np|<\gamma n$ with a
sufficiently small number~$\gamma>0$. Indeed, by introducing the
quantity $z=\frac{k-np}{\sqrt n}$ we get the demanded estimate with an
error term $\e(n)$ of the following form:
$$
\e(n)=\e(n,z)\le\frac C{\sqrt n}e^{-K_1z^2}\[\exp\left\{K_2
\frac{|z|^3}{\sqrt n}+K_3\frac{|z|}{\sqrt n}+\frac{K_4}n\right\}-1\]
$$
with appropriate constants $C>0$, and $K_j>0$, $j=1,\dots,4$.
Then we have to show that $\e(z,n)\le\frac{\const}n$, if $|z|\le\gamma
\sqrt n$. This estimate holds for $n^{1/6}>|z|<\gamma \sqrt n$, since in
this case $\e(n,z)\le e^{-K_1z^2/2}$. If $|z|\le n^{1/6}$ then
$e(n,z)\le\frac C{\sqrt n}e^{-K_1z^2}\(\frac {|z|^3+|z|+1}{\sqrt
n}\)\le\frac{\const}n$, that is the demanded estimate holds also in
this case.
\item{} If $|k-np|\ge \gamma n$, then the result of problem~5a follows
from the relations $P(S_n=k)\le \frac{\const}n$ and
$e^{-(k-np)^2/2np(1-p)}<\frac{\const}{n^2}$. Actually, even stronger
estimates could be proved. The first estimate is a consequence of the
Chebishev inequality, since $P(S_n=k)\le P(|S_n-ES_n|\ge \gamma n)\le
\frac{\text{Var\,}S_n}{\gamma^2n^2}\le\frac{\const}n$. The second
inequality is obvious.
\item{6.)} Let us introduce the random variables
$\bar\xi_j=\frac{\xi_j-b}h$, $j=1,\dots,n$, and $\bar
S_n=\summ_{j=1}^n\bar \xi_j$. Then
$E\bar\xi_j=\frac{m-b}h$ and $\text{Var}\,\bar\xi_j
=\frac{\sigma^2}{h^2}$. Since $P(S_n=kh+nb)=P(\bar S_n=k)$, and the
random variable $\bar S_n$ is concentrated on the lattice of the
integers as on the rarest lattice, statement formulated in this problem
follows from the results of problems~4  and~5.
\item{7.)} It follows from formula (5) that the relation
$$
\lim_{n\to\infty}P\(A<\frac{S_n-nm}{\sqrt n\sigma}<B\)
=\int_{A}^B \frac1{\sqrt{2\pi}}e^{-u^2/2}\,du, \tag2.3
$$
holds if $-\infty<A<B<\infty$, and the limit in this relation is
uniform for all pairs of numbers $(A,B)$ such that $C_1\le A<B\le C_2$
with some fixed numbers $-\infty<C_1<C_2<\infty$. Indeed,
$$
\align
&P\(A<\frac{S_n-nm}{\sqrt n\sigma}<B\)=P(A\sqrt n\sigma+nm-nb<S_n-nb
<B\sqrt n\sigma+nm-nb)\\
&\qquad =\summ_{k\colon\;k\in \Cal K(A,B)}P(S_n=kh+nb) \\
&\qquad =\frac h{\sqrt{2\pi n}\sigma}\summ_{k\colon\; k\in \Cal K(A,B)}
\exp\left\{-\frac{(kh+nb-nm)^2} {2n\sigma^2} \right\}
+\sqrt no\(\frac1{\sqrt n}\)\\
&\qquad =\frac h{\sqrt n\sigma}\summ_{l(k,n)\in\Cal L(A,B)}
\frac1{\sqrt2\pi} e^{-l(k,n)^2/2}+o(1)
\endalign
$$
where $\Cal K(A,B)=\{k\colon\; A\sqrt n\sigma<kh+nb-nm>B\sqrt n\sigma\}$,
$l(k,n)=\frac{kh-nm+nb}{\sqrt n\sigma}$, and
$$
\Cal L(A,B)=\left\{l(k,n)=\frac{kh-nm+nb}{\sqrt
n\sigma},\;k=0,\pm1,\pm2,\dots\right\}\cap (A,B),
$$
i.e.\ the points of the set $\Cal L(A,B)$ are the points of the
lattice of width $\frac h{\sqrt n\sigma}$  and containing the point
$\frac{nb-nm}{\sqrt n\sigma}$ which fall into the interval $(A,B)$.
This implies formula (2.3), since for a fixed number~$n$ the probability
at the left-hand side is an approximating sum of the integral at the
right-hand side plus an error which tends to zero as $n\to\infty$.
\item{} We prove that relation (2.3) also holds for $A=-\infty$.
Indeed, for all $\e>0$ we can choose a number $K=K(\e)$ such that
$\int_{-K}^K\frac1{\sqrt{2\pi}}e^{-u^2/2}\,du>1-\e$. Then
$\limm_{n\to\infty}P\(\left|\frac{S_n-nm}{\sqrt n\sigma}\right|<K\)
>1-\e$, and
$\limm_{n\to\infty}P\(\frac{S_n-nm}{\sqrt n\sigma}<-K\)<\e$. Then
$$
\align
&\limsupp_{n\to\infty}\left|
P\(\frac{S_n-nm}{\sqrt n\sigma}<x\)
-\int_{-\infty}^x\frac1{\sqrt{2\pi}}e^{-u^2/2}\,du\right| \\
&\qquad \le\limsupp_{n\to\infty}\left|
P\(-K\le \frac{S_n-nm}{\sqrt n\sigma}<x\)
-\int_{-K}^x\frac1{\sqrt{2\pi}}e^{-u^2/2}\,du\right|+\e\le\e.
\endalign
$$
Since the last relation holds for all $\e>0$, it implies the statement
of problem~7.
\item{8.)} Let $\varphi(t)=\int_{-\infty}^{\infty}e^{itx}f(x)\,dx
=Ee^{it\xi_1}$ denote the Fourier transform of the density function of
the random variable $\xi_1$. Then the Fourier transform of the density
function of the random sum $S_n=\xi_1+\cdots+\xi_n$ equals
$Ee^{it(\xi_1+\cdots+\xi_n)}=\(Ee^{it\xi_1}\)^n=\varphi^n(t)$. Since
$|\varphi(t)|\le1$, hence under the conditions of problem (8) the
function $\varphi^n(t)$ is integrable for $k\ge n$, and the density
function $f_n(t)$ of the random sum can be expressed by formula~(6) if
we replace the function $\varphi(t)$ by $\varphi^n(t)$. Moreover, this
relation also holds if we only assume that the function $\varphi^k(t)$
is integrable, and $n\ge k$. The above calculation makes possible to
prove problem~8 similarly to problem~4 with some natural modification.
Now we have to estimate the integral (6) instead of the integral~ (2)
(with the modification that we write $\varphi^n(t)$ instead of the
function $\varphi(t)$ in~(6). Further, because of the condition
a $E\xi^2<\infty$ the Fourier transform $\varphi(t)$ is twice
differentiable, $\varphi'(0)=iE\xi_1$,  $\varphi(0)''=-E\xi_1^2$. This
means that the analogs of the relations applied in the solution of
problem~4 holds in this case. (Later we shall discuss the properties of
the Fourier transform $\varphi(t)$ in the general case.)
\item{} The only essential difference in the estimation of the
integral we have to investigate is that the integral
$I_1=\int_{\e<|t|<\pi}e^{-ikt}P^n(t)\,dt$ introduced in the solution
of problem~4 now we write $I_1'=I_1'(x)=\int_{\e<|t|<\infty}e^{-itx}
\varphi^n(t)\,dt$. Observe that
$$
\align
I_1'\le\int_{\e<|t|<\infty}|\varphi(t)|^n\,dt&\le
\sup_{\e\le |t|<\infty}|\varphi(t)|^{n-k}
\int_{\e<|t|<\infty}|\varphi(t)|^k\,dt \\
&\le \const \sup_{\e\le |t|<\infty}\varphi(t)|^{n-k},
\endalign
$$
since $\varphi^k(\cdot)$  is an integrable function. For a fixed
number $t$, $t\neq0$, $|\varphi(t)|<1$. Further, by the Riemann lemma
$\limm_{|t|\to\infty}|\varphi(t)|=0$, and $\varphi(t)$ is a continuous
function. (This series of problem also contains the proof of the
Riemann lemma.) These facts imply that $\supp_{\e<|t|<\infty}
|\varphi(t)|<q<1$. This relation together with the previous estimates
imply that $|I_1'|\le\const q^n$. The only further difference in the
solution of problem~8 when compared to the solution of problem~4 is
that now in the integrals defining the expressions $I_1$ and $I_2$
the functions $e^{-ikt}P^n(t)$ are replaced by the function
$e^{-itx}\varphi^n(t)$. These expressions can be estimated in the
same way as the analogous integrals in problem~4.
\item{9.)} The solution of this problem is similar to that of
problem~6, only the notations are simpler. The condition directly
implies that
$$
\lim_{n\to\infty}(F_n(B)-F_n(A))=\int_{A}^B\frac1{\sqrt{2\pi}}
e^{-u^2/2}\,du,
$$
and the convergence is uniform for pairs of numbers $(A,B)$ in a
bounded set. Then we can show similarly to the argument in problem~6
that the number $A$ in the above formula can be replaced by $-\infty$.
\item{10.)} The proof is a slight modification of the proof sketched in
the solution of problem~ 8. This modification is similar to the
modification made in the solution of problem~5 compared to problem~4.
Since $E|\xi_1|^3<\infty$, hence the approximation
$$
\log \varphi(t)=it E\xi_1-\frac {t^2}x E\xi_1^2+O(t^3)
$$
holds in a small neighbourhood of the origin. This makes possible to
get the solution of the problem by modifying the domain of integration
in the expressions $I_2$ and $I_3$ as it was done in the solution of
problem~5.
\item{11.)} Let $\xi$ be a random variable with standard normal
distribution. Then
$$
E\xi=\int_{-\infty}^\infty\frac1{\sqrt{2\pi}}
ue^{-u^2/2}\,du=0,
$$
since the integrand is an odd function. On the other hand, integration by
parts yields that
$$
\align
E\xi^2&=\int_{-\infty}^\infty\frac1{\sqrt{2\pi}}u^2e^{-u^2/2}\,du
=-\int_{-\infty}^\infty\frac1{\sqrt{2\pi}} u\(\frac
d{du}e^{-u^2/2}\)\,du\\
&=\int_{-\infty}^\infty\frac1{\sqrt{2\pi}} e^{-u^2/2}\,du=1.
\endalign
$$
If $\xi$ is a $\Phi_{m,\sigma}$ distributed random variable, then its
transform $\frac{\xi-m}\sigma$ has standard normal distribution, \i.e.
it has expectation zero and variance~1. Hence $\xi$ has expectation $m$
and variance $\sigma^2$.
\item{12.)} Let $\varphi(t)=\varphi(t_1,\dots,t_k)
=Ee^{i(t,\xi)}=Ee^{i(t_1\xi_1+\cdots+t_k\xi_k)}$ denote the
characteristic function of a $k$-dimensional random vector
$\xi=(\xi_1,\dots,\xi_k)$, where $t=(t_1,\dots,t_k)$, and
$(t,\xi)=\summ_{j=1}^k t_j\xi_j$. Then $|\varphi(t)|\le
E|e^{i(t,\xi)}|=1$. For arbitrary number $\e>0$ there exists a constant
$R=R(\e)$ such that $P(|\xi|>R)=P\(\summ_{j=1}^k\xi_j^2>R^2\)
<\frac\e2$. Put $\delta=\frac{\e}{2R(\e)}$ and consider such numbers
$t=(t_1,\dots,t_k)$ for which $|t|^2=\summ_{j=1}^kt_j^2<\delta$. Then
$|e^{i(t,\xi)}-1|\le|(t,\xi)|\le\frac\e2$ for $|x|<R(\e)$, and
$|\varphi(t)-\varphi(\bar t)|=| Ee^{i(t,\xi)}-
Ee^{i(\bar t,\xi)}|\le E|e^{i(t-\bar t,\xi)}-1|\le E|e^{i(t-\bar
t,\xi)}-1|I(|\xi|\le R)+P(|\xi|>R)\le E|(t-\bar t,\xi)|I(|\xi|\le
R)+\frac \e2\le\e$ if $|t-\bar t|\le\delta$, where $I(A)$ denotes the
indicator function of a set~$A$. Hence the function $\varphi(t)$ is
uniformly continuous.
\item{} The characteristic function of a random vector $a\xi+m$
in a point $t\in R^k$, where $a\in R$, $m\in R^k$ is the function
$Ee^{i(t,a\xi+m)}=e^{(it,m)}Ee^{i(at,\xi)}=e^{(it,m)}\varphi(at)$,
where $\varphi$ denotes the characteristic function of the random
vector~$\xi$.
\item{} If $\xi_j$, $j=1,\dots,n$, are independent random vectors with
characteristic functions $\varphi_j(t)$, then the characteristic
function of the random sum $\xi_1+\cdots+\xi_n$ in a point $t\in R^k$
equals $Ee^{i(t,\xi_1+\cdots+\xi_n)}=Ee^{i(t,\xi_1)}\cdots
e^{i(t,\xi_n)}=Ee^{i(t,\xi_1)}\cdots
Ee^{i(t,\xi_n)}=\prodd_{j=1}^k\varphi_j(t)$.
\itemitem{13.) $\,$a.)} If the random variable $\xi$ has
standard normal distribution, then
$$
Ee^{it\xi}=\int_{-\infty}^\infty \frac1{\sqrt{2\pi}}e^{itu-u^2/2}\,du
=e^{-t^2/2}\int_{-\infty}^\infty \frac1{\sqrt{2\pi}}e^{-(it-u)^2/2}\,du
=e^{-t^2/2}
$$
by the result of problem 1.
\itemitem{b.)} If the random variable $\xi$ has uniform distribution
in the interval $[0,1]$, then
$$
Ee^{it\xi}=\int_0^1e^{itu}\,du=\frac{e^{it}-1}{it}.
$$
\itemitem{c.)} If the random variable $\xi$ has exponential distribution
with parameter $\lambda>0$, then
$$
Ee^{it\xi}=\int_0^\infty \lambda e^{itu-\lambda u}\,du
=\frac{\lambda}{\lambda-it}.
$$
\itemitem{d.)} If $\xi$ is a random variable with Cauchy distribution,
then
$$
Ee^{it\xi}=\int_{-\infty}^\infty \frac1\pi \frac{e^{itu}}{1+u^2}\,du.
$$
This integral can be calculated by means of the residium theorem in the
theory of analytic functions.
\itemitem{} The function $g(z)=g(z,t)=\frac{e^{itz}}{\pi(1+z^2)}$
is analytic in the plane of complex numbers with two poles $z=\pm i$.
The residium of the function $g(z)$ in the point $i$ equals $e^{-t}$,
and in the point $-i$ it equals $e^t$. Let us consider the following
contour integral. Let us first integrate the function $g(z)=g(z,t)$ on
the interval $[-R,R]$  and then on the half-circle $|z|=R$, $\Im z\ge0$
if $t\ge0$ and on the half-circle $|z|=R$, $\Im z\le 0$ if $t\le0$. The
above contour integral equals the residium of the function $g(z)$ in
the point $i$ if $t>0$ and the residium of this function in the point
$-i$ if $t<0$. On the other hand the restriction of the integral to the
half-circle of radius $R$ tends to zero if $R\to0$. This implies that
$Ee^{it\xi}=\int_{-\infty}^\infty g(t,u)\,du=e^{-|t|}$.
\itemitem{} The following argument gives another different a little
bit artificial but correct proof of this statement. The characteristic
function of the density function $f(x)=\frac12e^{-|x|}$ equals
$$
\frac12\int_{-\infty}^\infty e^{-|x|+itx}\,dx=\frac12
\(\frac1{1+it}+\frac1{1-it}\)=\frac1{1+t^2}.
$$
Since the function $\frac1{1+t^2}$ is integrable the inverse Fourier
transformation formula~(6) implies the desired statement.
\itemitem{e.)} If the random variable $\xi$ has Poissonian distribution
with parameter~$\lambda>0$, then
$$
Ee^{it\xi}=e^{-\lambda} \sum_{k=0}^\infty
\frac{\lambda^k}{k!}e^{ikt}=\exp\left\{\lambda(e^{it}-1)\right\}.
$$
\itemitem{f.)} If the random variable $\xi$ has binomial distribution
with parameters~$n$ and~$p$, then
$$
Ee^{it\xi}=e^{-\lambda} \sum_{k=0}^n \binom nkp^ke^{ikt}(1-p)^{n-k}
=\(1-p+pe^{it}\).
$$
\itemitem{g.)} If $\xi$ is a random variable with negative binomial
distribution with parameters $n$ and $p$, then its distribution agrees
with the distribution of the random sum  $\xi_1+\cdots+\xi_n$, where
$\xi_j$, $1\le j\le n$, are independent random variables with negative
binomial distribution with parameters~$1$ and~$p$. (To see this
property observe that a possible the random variable~$\xi$ has the
following probabilistic interpretation: If we make independent experiment
after each other which are successful with probability~$p$, then $\xi$
denotes the number of unsuccessful experiments up to the~$n$-th
successful experiment. If $\xi_j$ denotes the number of the
unsuccessful experiments between the $j-1$-th and $j$-th successful
experiments, then we get the above representation.) Hence
$Ee^{it\xi}=\(Ee^{it\xi_1}\)^n$. On the other hand,
$$
Ee^{it\xi_1}=\sum_{k=0}^\infty
(1-p)p^ke^{itk}=\frac{1-p}{1-pe^{it}}.
$$
\itemitem{h.)} We get with the change of variables $\bar u=(1-it)u$ 
that
$$
\varphi_s(t)=\frac1{\Gamma(s)}\int_0^\infty e^{-u+itu}u^{s-1}\,du
=\frac1{\Gamma(s)}\frac1{1-it)^s}\int_0^\infty e^{-\bar u}\bar u^{s-1}\,d\bar u
=\frac1{(1-it)^s}.
$$
In this calculation we applied some complex analysis argument. At
the change of variables step $\bar u=(1-it)u$ of the calulation
the domain of integration became the line $(1-it)u$, $u>0$, instead
of the positive abscissa axis. But we can turn back the domain of
integration to the positive abscissa axis by means of a usual complex
analysis argument, by which the integral of an analytic function on
a closed curve equals zero. At this step we have to exploit that
the function  $e^{-z}$ tends to zero fast as $\Re z\to\infty$.
\item{14.)} If $f(x_1,\dots,x_k)$ and
$g(x_1,\dots,x_k)$ are two integrable functions, then
$$
\align
\infty&>\int\int |f(x_1,\dots,x_k)||g(u_1,\dots,u_k)|\,dx_1\cdots
dx_kdu_1\dots du_k \\
&\hskip6truecm (\text{with substitution }\bar u_j=x_j+u_j
\;j=1,\dots,k ,) \\
&=\int\int |f(x_1,\dots,x_k)| |g(\bar u_1-x_1,\dots,\bar u_k-x_k)|
\,dx_1\cdots \,dx_k\,d\bar u_1\dots \,d\bar u_k \\
&=\int\(\int |f(x_1,\dots,x_k)|
|g(\bar u_1-x_1,\dots,\bar u_k-x_k)|\,dx_1\cdots dx_k\)d\bar u_1\dots
du_k\\
&=\int |f|*|g|(x_1,\dots,x_k)\,dx_1\dots dx_k.
\endalign
$$
This relation implies that the function $|f*g(x_1,\dots,x_k)|\le |f|*|g|
(x_1,\dots,x_k)$ is bounded in almost all points $(x_1,\dots,x_k)\in
R^k$. It also implies that $f*g$ is an integrable function. In the
sequel we write $x$ instead of $(x_1,\dots,x_k)$ and $u$ instead of
$(u_1,\dots,u_k)$.
\item{} If $\mu$ and $\nu$ are two measures of bounded variation, then
there exists a representation $\mu=\mu_1-\mu_2$, $\nu=\nu_1-\nu_2$
such that $\mu_i$ and $\nu_i$, $i=1,2$, are finite measures. The
identity $\mu*\nu=(\mu_1*\nu_1+\mu_2*\nu_2)-(\mu_1*\nu_2+\mu_2*\nu_1)$
holds. Since $\mu_i*\nu_j(R^k)<\infty$ for all indices $i,j=1,2$, this
implies that $\mu*\nu$ is a measure of bounded variation.
\item{} If the measure $\mu$ has a density function $f$, then we have
for all measurable sets $A\subset R^k$
$$ \allowdisplaybreaks
\align
&\int_A f*\nu(x)\,dx=\int_A\(\int f(u)\nu(x-\,du)\)\,dx
=\int_A\(\int f(x-u)\nu(\,du)\)\,dx\\
&\qquad=\int \(\int_A f(x-u)\,dx\)\nu(\,du)=\int\(\int I({x\colon\;x\in
A})f(x-u)\,dx\)\nu(\,du)\\
&\qquad=\int\(\int I({v\colon\;u+v\in A})f(v)\,dv\)\nu(\,du)\\
&\qquad=\int\int I({v\colon\;u+v\in A})\mu(\,dv)\nu(\,du)\\
&\qquad=\mu\times\nu\(\{(u,v)\colon\; u+v\in A\}\)=\mu*\nu(A)
\endalign
$$
and this means that the function $f*\nu$ is the density function of the
convolution $\mu*\nu$ of the measures $\mu$ and~$\nu$.
\item{} Let us observe that the above calculations also imply that the
function $f*\nu(x)$ is finite in almost all points $x\in R^k$, moreover
it is integrable. Indeed, the above calculation implies
this result with the choice $A=R^k$ if $\mu$ and $\nu$ are (bounded)
positive measures. The general case can be reduced to this case if
we decompose the measures $\mu$ and $\nu$ as the difference of two
positive finite measures. (We may assume that the measures in the
decomposition $\mu=\mu_1-\mu_2$ have density function.)
\item{} If the measure $\mu$ has a density function $f$ and the
measure $\nu$ has a density function $g$ then let us define the
measures $\bar \nu_x(A)=\nu(x-A)$ for all $x\in R^k$. The density
function of the measure $\bar \nu_x(A)$ equals $g(x-u)$ in the point
$u\in R^k$, and the density function of the measure $\mu*\nu$ in the
point~$x$ equals
$$
\int f(u)\bar \nu_x(du)=\int f(u)g(x-u)\,du=f*g(x)
$$
by the already proved results.
\item{15.)} It follows from the definition of the convolution that if
$\xi$ an $\eta$ are independent random variables with distributions
$\mu$ and $\nu$, then the distribution of the random sum $\xi+\eta$
equals $\mu*\nu$. It follows from the result of the previous problem
that if the distribution $\mu$ of the random variable $\xi$ has a
density function $f$, then the distribution $\mu*\nu$ of the random
variable $\xi+\eta$ has a density function $f*\nu$. If also the measure
$\nu$ has a density function $g$, then this density function equals
$f*g$.
\item{} If the distribution of the random variable~$Z$ is $F(x)$, then
the distribution of the random variable $\bar Z=\frac{Z-A}B$ with
$B>0$ equals $F(Bx+A)$. If the random variable $Z$ has a density
function $f(x)$, then the random variable $\bar Z$ has a density
function $Bf(Bx+A)$. The previous results imply the statements
formulated for the distribution and density function of the random sum
$\bar S_n$.
\item{} The relation $\mu*\nu=\nu*\mu$ follows from the definition of
the convolution. The statement $(\mu_1*\mu_2)*\mu_3=\mu_1*(\mu_2*\mu_3)$
follows from the fact that the identity
$$
(\mu_1*\mu_2)*\mu_3(A)=\mu_1*(\mu_2*\mu_3)(A)
=\mu_1\times\mu_2\times\mu_3(\{(u,v,w)\colon\;u+v+w\in A\})
$$
holds for all measurable sets~$A$. The analog statements about the
convolution of functions can be reduced to these statement if we
represent the convolution of functions as the convolution of the density
function of the corresponding signed measures. Otherwise these statement
also can be simply proved by simple calculation.
\item{16.)} If $\mu$ and $\nu$ are two signed measures of bounded
variation with Fourier transforms $\tilde f(t_1,\dots,t_k)$ and
$\tilde g(t_1,\dots,t_k)$, then
$$
\align
&\tilde f(t_1,\dots,t_k)\tilde g(t_1,\dots,t_k) \\
&\qquad=\int e^{i(t_1(u_1+v_1)+\cdots+t_k(u_k+v_k))}
\mu(\,du_1,\dots,\,du_k) \nu(\,dv_1,\dots,\,dv_k).
\endalign
$$
The transformation $\bold T(u_1,\dots,u_k,v_1,\dots,v_k)
=(u_1+v_1,\dots,u_k+v_k)$, $(u_1,\dots,u_k)\in R^k$, $(v_1,\dots,v_k)\in
R^k$, is a measurable transformation from the space \hfill\break
$(R^k\times R^k,
\Cal B_{2k},\mu\times\nu)$ to the space $(R^k,B_k,\mu*\nu)$, 
where $\Cal B_{2k}$ and $\Cal B_k$ denote the
$\sigma$-algebras in the spaces $R^{2k}$ and~$R^k$. By applying the
measure theoretical result which describe how measurable
transformations transform integrals for the functions
$$
h(x_1,\dots,x_k)=e^{it(x_1+\cdots+x_k)})
$$
and
$$
%\align
g(u_1,\dots,u_k,v_1,\dots,v_k)=
h(\bold T(u_1,\dots,u_k,v_1,\dots,v_k))=
e^{i(t_1(u_1+v_1)+\cdots+t_k(u_k+v_k))}
% \\
%&=e^{i\((t_1,\dots,t_k), \bold T(u_1,\dots,u_k,v_1,\dots,v_k)\)},
%\endalign
$$
%where $\((x_1,\dots,x_k),(y_1,\dots,y_k)\)=\summ_{j=1}^k x_jy_j$
%denotes scalar product, and 
with the above defined transformation $\bold
T(u_1,\dots,u_k,v_1,\dots,v_k)$ we get from the relation written at the
beginning of the solution that
$$
\tilde f(t_1,\dots,t_k)\tilde g(t_1,\dots,t_k)
=\int e^{i(t_1x_1+\cdots+t_kx_k)}
\mu*\nu(\,dx_1,\dots,\,dx_k).
$$
This implies the statement about the Fourier transform of the
convolution of measures. The analogous statement about the Fourier
transform of the convolution of density functions follows from this
statement and the relation between the convolution of measures and
their density functions.
\item{17.)} By differentiating the identity $f*g(x)=\int f(x-u)g(u)\,du$
$k$ times we get that
$$
\frac {df*g^k(x)}{dx^k}=\int \left.\frac
{df^k(v)}{dv^k}\right|_{v=x-u}g(u)\,du
=\int \left.\frac {df^k(v)}{dv^k}\right|_{v=u}g(x-u)\,du.
$$
The conditions of the problem allow the above successive
differentiations. Further, working with the right-hand side of the last
formula we can carry out~$l$ additional differentiations and get that
$$
\frac {df*g^{k+l}(x)}{dx^{k+l}} =\int\left.\frac {df^k(v)}{dv^k}
\right|_{v=u}\left.\frac{dg^l(v)}{dv^l}\right|_{v=x-u}\,du.
$$
\item{} If $f(u)$ is analytic function which also satisfies the other
conditions of the problem, then the function
$$
F(z)=\int f(z-u)g(u)\,du
$$
is an analytic continuation of the convolution $f*g(x)$ to the domain
$\{z\colon\; \Im z<A\}$.
\item{18.)} a.) Convergence in distribution implies the
convergence of the integrals:
\item{} Since $F(x_1,\dots,x_k)\to 1$ if $x_j\to\infty$ for all
$j=1,\dots,k$, and $F(x_1,\dots,x_k)\to 0$ if $x_j\to-\infty$ for one
of the indices $1\le j\le k$ hence for all $\e>0$ there exists
a $k$-dimensional rectangle $\bold K=\bold K(\e)$ such that
$\mu_F(\bold K)>1-\e$. (Given a distribution function $F$ in the sequel
we shall denote by $\mu_F$ the probability measure on $R^k$ induced by
the distribution function $F$.) We also may assume, by enlarging the
rectangle $\bold K$ if it is necessary, that the boundary of the
rectangles $\bold K$ has zero $\mu_F$ measure. Indeed, the projection
of the distribution function $F$ to the $j$-th coordinate is a
one-dimensional distribution function, and as a consequence it has at
least countably many atoms (points with positive measure with respect
the measure induced by this distribution) for all indices $j=1,\dots,k$.
This implies that we can choose a larger rectangle $\bold K$ if this
is necessary whose boundary has $\mu_F$ measure zero.
\item{} Because of the boundedness of the function $f$ the relation
$$
\left|\int_{R^k\setminus \bold
K}f(x_1,\dots,x_k)\,dF(x_1,\dots,x_k)\right| <\const\e
$$
holds, and also $\limsupp_{n\to\infty}\left|\int_{R^k\setminus \bold
K}f(x_1,\dots,x_k) \,dF_n(x_1,\dots,x_k)\right|<\const\e$ for all
$n=1,2,\dots$ because of the zero $\mu_F$ boundary of the rectangle
$\bold K$ and the convergence of the distribution functions $F_n$ to
the distribution function~$F$. Furthermore, the function $f$ is
uniformly continuous on the rectangle $\bold K$ hence there exists
some constant $\delta>0$ such that $|f(x)-f(y)|<\e$ if
$|x-y|\le\delta$, and $x,y\in \bold K$. The rectangle $\bold K$ can be
decomposed to finitely many rectangles $\Delta_j$, $j=1,\dots, p(\bold
K)$, of diameter less than $\delta$ without joint interior points, and
such that all these rectangles $\Delta_j$ have boundaries with $\mu_F$
measure zero. These properties imply that $\lim\limits_{n\to\infty}
\mu_{F_n}(\Delta_j)=\mu_F(\Delta_j)$ for all indices $j=1,\dots,
p(\bold K)$, and because of the uniform continuity of the function~$f$
on the rectangle~$\bold K$
$$
\limsup\left|\int_{\bold K}f\,dF_n-\int_{\bold K}f\,dF\right|<\e.
$$
The above inequalities imply that
$\limsup\left|\int f\,dF_n-\int f\,dF\right|<\const\e$ with a $\const$
independent of $\e$. Since this inequality holds for all $\e>0$,
it implies the statement we wanted to prove. 

\medskip
\item{b.)} The convergence of the integrals implies convergence in
distribution:
\item{} Let $x=(x_1,\dots,x_k)$ be a point of continuity of the
distribution function~$F$. Then for all numbers $\e>0$ there exists a
number $\delta=\delta(\e)>0$ such that the points $y=(y_1,\dots,y_k)
=(x_1-\delta,\dots,x_k-\delta)$ and $z=(z_1,\dots,z_k)
=(x_1+\delta,\dots,x_k+\delta)$ satisfy the inequalities
$F(y)>F(x)-\e$ and $F(z)<F(x)+\e$. There exist continuous functions
$f_1(u)$ and $f_2(u)$ an the $k$-dimensional Euclidean space $R^k$
which satisfy the following properties: $0\le f_i(u)\le1$ for all $u\in
R^k$, $i=1,2$. Further $f_1(u)=1$ for $u=(u_1,\dots,u_k)$ if
$u_j\le y_j$ for all indices $j=1,\dots,k$, and $f_1(u)=0$ if $u_j\ge
x_j$ for some of the indices $1\le j\le k$. The function $f_2(\cdot)$
satisfies the following relations: $f_2(u)=1$ if $u_j\le x_j$ for all
$j=1,\dots,k$, and $f_2(u)=0$ if $u_j\ge z_j$ for some $1\le j\le k$.
Then
$$
\align
\limsup_{n\to\infty} F_n(x)&\ge \lim_{n\to\infty}\int
f_1(u)\,dF_n(u)=\int f_1(u)\,dF(u)\ge F(x)-\e\\
\liminf_{n\to\infty}F_n(x)&\le \lim_{n\to\infty} \int
f_2(u)\,dF_n(u)=\int f_2(u)\,dF(u)\le F(x)+\e.
\endalign
$$
Since these relations hold for all $\e>0$, they imply the statement of
part~b.).
\item{19.)}  Since $\bigcupp_{K=1}^\infty \bold K(K)^k= R^k$, and the
rectangles $\bold K(K)^k$, $K=1,2,\dots$, constitute a monotone
increasing series of sets, hence $\limm_{K\to\infty}\mu(\bold
K(K)^k)=\mu(R^k)=1$, i.e.\ $\mu(\bold K(K)^k)\ge 1-\e$ if $K\ge K(\e)$.
\item{} To show that the characteristic function of the probability
measure $\mu$ is determined by its characteristic function let us first
observe that the integrals $\int f(u)\,d\mu(u)$ determine the measure
$\mu$ if we take all continuous functions $f(\cdot)$ with a bounded
support. Indeed, the measure $\mu$ of those rectangles
$\bold P=[K_1,L_1)\times\cdots\times [K_k,L_k)$ whose boundary has
$\mu$ measure zero determine the measure $\mu$. Besides, we claim
that for all numbers $\e>0$ and rectangles $\bold P$ there exists a
function $f_{\e,\bold P}(\cdot)$ such that $0\le f_{\e,\bold P}(u)\le1$
for all points $u\in R^k$, $f_{\e,\bold P}(u)=1$ if $u\in \bold P$, and
$f_{\e,\bold P}(u)=0$ if $\rho(u,\bold P)>\e$. (In the sequel
$\rho(\cdot,\cdot)$ denotes the usual Euclidean distance in the
space~$R^k$.) Then the relation $\mu(\bold P)=\limm_{\e\to0}\int
f_{\bold P,\e}\,d\mu(u)$ implies the above property.
\item{} A possible construction of a function $f_{\bold P,\e}$
with the above properties is the following: Put $f_{\bold P,\e}(u)
=1-g_{\bold P,\e}(u)$ and $g_{\bold P,\e}(u)=\min\(1,\frac1\e
\rho(u,\bold P)\)$.
\item{} Given a continuous function $f(\cdot)$ of compact support
together with a sufficiently large number $K>0$ for which the cube
$[-K,K]\times\cdots\times [-K,K]$ contains the support of the function
$f(\cdot)$ let us define the periodic extension of the function
$f(\cdot)$ with period $2K$ by the formula
$f_K(u_1+2Kj_1,\cdots,u_k+2Kj_k)=f(u_1,\cdots,u_k)$, $-K\le u_j<K$,
$l_j=0,\pm1,\pm2,\dots$, $j=1,\dots,k$. The integrals $\int
f_K(u)\,d\mu(u)$ of these functions obtained by periodic extension
determine the measure $\mu$, since $\int f(u)\,d\mu(u)
=\limm_{K\to\infty}\int f_K(u)\,d\mu(u)$ because of the
tightness property of the measure $\mu$ formulated in the already proven
part of problem~19.
\item{} Finally, by Weierstrass second approximation theorem for all
continuous functions $f_K(\cdots)$ $K$ with period~$K$ and real number
$\e>0$ there exists a trigonometrical polynomial
$$
g_{\e}=g_{\e,f_K}(u_1,\cdots,u_k)=\summ
c^{\e}_{j_1,\dots,j_k}e^{i\pi(j_1u_1+\cdots+j_ku_k)/K}
$$
such that $\supp_{u\in R^k}|f_K(u)-g_\e(u)| \le\e$. Hence
$$
\left|\int f_K(u)\,d\mu(u)-\int g_\e(u)\,d\mu(u)\right|\le\e.
$$
On the other hand, $\int g_\e(u)\,d\mu(u)=\summ c^{\e}_{j_1,\dots,j_k}
\varphi\(\frac{\pi j_1}{K},\dots,\frac{\pi j_k}{K}\)$, that is the
above integral can be calculated by means of the characteristic function
of the measure~$\mu$. This implies that this characteristic function
determines the integrals of the form $\int f_K(u)\,d\mu(u)$. Hence it
also determines the measure~$\mu$.
\item{} The proof can be generalized without any essential modification
to arbitrary signed measures~$\mu$ with bounded variation.
\item{20.)} First we show that for all numbers $a>0$ there exists an
even density function $f(u)$ whose Fourier transform $\varphi(t)$ is
sufficiently smooth, e.g.\ it is twice differentiable, and it equals
zero outside the interval $[-a,a]$.
\item{} Indeed, let us consider a continuously differentiable function
$g(u)$ which is concentrated in the interval $\[-\frac a2,\frac a2\]$,
$g^-(u)=g(-u)$. Then put $h(u)=g*g^-(u)$, $f(u)=\frac{2\pi}M\int
e^{itu}h(u)\,du$, where $*$ denotes convolution, and $M=h(0)=\int
|f(u)|^2\,du$. We claim that this function $f$ is a density
function, and its characteristic function is the function
$\frac{h^-(u)}M$, $h^-(u)=h(-u)$, which vanishes outside the interval
$[-a,a]$. Indeed, the function $h(\cdot)$ is twice differentiable,
(see problem~17), hence its Fourier transform tends to zero in
plus--minus infinity with order $|t|^{-2}$ (see e.g.\ problem~28 of
this series of problems discussed later),  hence the above defined
Fourier transform $f(\cdot)$ of the function $\frac {h(u)}M$ is
integrable, and we can apply the inverse Fourier transform for it.
Since $f(\cdot)$ is en even function, this means that the function
$\frac{h^-(u)}M=\int e^{itx}f(u)\,du$ is the Fourier transform
of the function $f(u)$. In particular, $\frac{h(0)}M=1=\int
f(u)\,du$. Finally, $f(t)\ge0$ for all numbers $t\in R^1$, since the
Fourier transform of the function $g*g^-(\cdot)$ equals $\int
e^{itu}g*g^-(u)\,du=\int e^{itu}g(u)\,du\int e^{itu}g^-(u)\,du
=\left|\int e^{itu}g(u)\,du\right|^2\ge0$. These properties mean that
the function $f(\cdot)$ is a density function. (We shall return to the
above problem in the second part of this series of problems where such
a construction will be useful in a different context.)
\item{} Let us consider an even density function $f(u)$ whose
characteristic function $\varphi(t)$ is twice differentiable, and
vanishes outside of a finite interval $[-a,a]$. Consider a number
$T>a$, and let us define the numbers $a_k=\frac1{4\pi T}\int
e^{-i\pi tk/T} \varphi(t)\,dt$, $k=0,\pm1,\pm2,\dots$. Let us put
weights $a_k$ in the points $\frac{\pi k}T$, $k=0,\pm1,\pm2,\dots$.
We claim that in such a way we constructed a probability distribution
on the lattice $\frac{\pi k}T$, $k=0,\pm1,\pm2,\cdots$, whose
characteristic function to the interval $[-T,T]$ agrees with the
restriction of the characteristic function $\varphi(t)$ to this
interval. The characteristic function of this discrete distribution
is the periodic extension of the restriction of $\varphi(t)$ to the
interval~$[-T,T]$ with period $2T$. This statement means in
particular, that the above defined discrete distribution together with
the distribution with density function $f(\cdot)$ yield an example
satisfying the statement of problem~20.
\item{} The statement formulated for the discrete distribution with
weights~$a_k$ holds, because for one hand a comparison of the
definition of the number~$a_k$ with the inverse Fourier transformation
formula expressing the function $f(\cdot)$ yields that
$a_k=\frac1{2T}f\(\frac{\pi k}T\)\ge0$. On the other hand, the
trigonometrical sum $\summ_{k=-\infty}^\infty a_ke^{\pi ik/T}$ is the
Fourier series of the function $\varphi(t)$ restricted to the
interval~$[-T,T]$. In particular, $\varphi(0)=1=\summ_{k=-\infty}
^\infty a_k$. (As $\varphi(\cdot)$ is a twice differentiable function,
hence it equals his Fourier series in all points.)
\item{21.)} Let us first show that if the sequence of the probability
measures $\mu_n$, $n=1,2,\dots$, is relatively compact, then it is
also tight.
\item{} Let us assume indirectly this sequence of measures $\mu_n$,
$n=1,2,\dots$, is not tight. Then there exists a constant $\e>0$, a
subsequence $\mu_{n_k}$ of the sequence of probability measures
$\mu_n$ and a sequence of positive numbers $K_n$, $n=1,2,\dots$,
such that $K_n\to\infty$, and $\mu_{n_k}([-K_n,K_n]\times\cdots
\times[-K_n,K_n])<1-\e$. We shall show that this subsequence
$\mu_{n_k}$ of the sequence of measures $\mu_n$ has no sub-subsequence
convergent in distribution. This means that the above formulated
indirect assumption leads to contradiction.
\item{} Indeed, let us assume indirectly that the sequence of measures
$\mu_{n_k}$ has a subsequence $\mu_{n_{k_j}}$ which converges in
distribution to a probability measure $\mu$. Then, there exists a
constant $K>0$ such that $\mu([-K,K]\times\cdots\times[-K,K])>
1-\frac\e2$, and the hyperplanes $u_j=\pm K$, $j=1,2,\dots,k$, have
$\mu$ measure zero. Then also the relation
$\limm_{j\to\infty}\mu_{n_{k_j}} ([-K,K]\times\cdots\times[-K,K])
=\mu([-K,K]\times\cdots\times[-K,K])$ should hold. But this is not
possible, since the $\limsup$ of the probabilities at the left-hand
side is smaller than $1-\e$, while the right-hand side is greater than
$1-\frac\e2$.
\item{} Let us show that if the sequence of measures $\mu_n$ is tight
then it is also relatively compact.
\item{} We have to show that an arbitrary subsequence of the sequence
$\mu_n$ has a sub-subsequence convergent in distribution. For the sake
of simpler notations let us denote by reindexing the elements of the
subsequence again by $\mu_n$. We have to show that this (also tight)
sequence of probability measures $\mu_n$ has a subsequence convergent
in distribution.
\item{} Let $F_n(u)=F_n(u_1,\dots,u_k)$ denote the distribution function
$F_n(u_1,\dots,u_k)=\mu_n(\{(v_1,\dots,v_k)\colon\; v_j<u_j,\;j=1,\dots,k\})$
determined by the measure $\mu_n$. Let
$$
u^{(p)}=\(u_1^{(p)},\dots,u_k^{(p)}\),\quad p=1,2,\dots,
$$
denote  the (countable) set of points $u^{(p)}\in R^k$ with rational
coordinates with some indexing. First we show with the help of the
so-called diagonal procedure that there exists an appropriate sequence
of positive integers $n_j$, $j=1,2,\dots$, such that the limit
$$
\limm_{j\to\infty} F_{n_j}\(u_1^{(p)},\dots,u_k^{(p)}\)= \tilde
F\(u_1^{(p)},\dots,u_k^{(p)}\) \tag2.4
$$
exists for all numbers $p=1,2,\dots$.
\item{} Indeed, as $0\le F_n(u)\le1$, there is a subsequence
$\bar n_j=(n_{j,1})$ of the integers such that the limit
$\limm_{j\to\infty}F_{n_{j,1}}(u^{(1)})=\tilde F(u^{(1)})$
exists. This sequence has a subsequence $n_{j,2}$ such that the
limit $\limm_{j\to\infty}F_{n_{j,2}}(u^{(2)})=\tilde F(u^{(2)})$ also
exists. Following this procedure we can construct sequences $n_{j,p}$,
$j=1,2,\dots$, $p=1,2,\dots$ in such a way that the $p+1$-th sequence
is a subsequence of the $p$-th sequence, i.e.\ $\{n_{j,p+1},
\;j=1,2,\dots\}\subset \{n_{j,p},\;j=1,2,\dots\}$, $p=1,2,\dots$, and
the limit $\limm_{j\to\infty}F_{n_{j,p}}(u^{(p)})=\tilde F(u^{(p)})$
exists for all numbers $p=1,2,\dots$. Then the sequence $n_j=n_{j,j}$
satisfies relation~(2.4).
\item{} Let us introduce the function
$$
F(u_1,\dots,u_k)=\sup_{\left\{u^{(p)}=(u_1^{(p)},\dots,u_k^{(p)})
\colon\;u_s^{(p)}<u_s, \;s=1,\dots,k\right\}}
\tilde F\(u_1^{(p)},\dots,u_k^{(p)}\), \tag2.5
$$
where the function $\tilde F$ satisfies relation~(2.4) with an
appropriate (fixed) sequence of positive integers~$n_j$, and the points
$u^{(p)}$ are the points of the $k$-dimensional space with rational
coordinates. We claim that the above defined function $F(u_1,\dots,u_k)$
is a distribution function, and the distribution functions
$F_{n_j}(u_1,\dots,u_k)$ converge in distribution to it. If we prove
this statement then we complete the solution of problem~21.
\item{} To show that the function $F(u_1,\dots,u_k)$ is a distribution
function we recall an ``internal" characterization result of
distribution functions which describes the distribution functions only
with the help of their property. The result we recall states that a
function $F(u_1,\dots,u_k)$ is a distribution function if and only if it
satisfies the following four properties:

\medskip
\item\item {(i)} The function $F(u_1,\dots,u_k)$ is a function
continuous from the left in all of its arguments.
\itemitem {(ii)} $\lim\limits\Sb u_j\to\infty \\ \text{for all indices }
j=1,\dots,k\endSb F(u_1,\dots,u_k)=1$.
\itemitem{(iii)} $\lim\limits\Sb u_j\to-\infty \\ \text{for some index }
1\le j\le k\endSb F(u_1,\dots,u_k)=0$.
\itemitem{} Finally, to formulate the last condition let us define for a
function $F$ on the space $R^k$ and a rectangle
$\bold K=[a_1,b_1)\times\cdots\times[a_k,b_k)$ the number
$$
\mu(\bold K)=\mu_F(\bold K)=\sum\Sb u_j=\; a_j\text{ vagy
}b_j\\j=1,\dots,k\endSb (-1)^{\chi(u_1,\dots,u_k)}F(u_1,\dots,u_k),
$$
where $\chi(u_1,\dots, u_k)$ denotes the quantity of the numbers~$a_j$
in the sequence $u_1,\dots,u_k$. Then
\itemitem{(iv)} $\mu_F(\bold K)\ge 0$ for all rectangles $\bold K$.

\medskip
\item{} Since the function $\tilde F(u_1,\dots,u_k)$ defined in points
of rational coordinates is a monotone increasing function in all of its
coordinates, the function $F(\cdot)$ defined in formula~(2.5) also
satisfies property~(i). Furthermore, this monotonicity also
implies that we can replace the $\sup$ by $\lim$ in formula~(2.5) if
we consider such sequences of indices $\(u_1^{(p)},\dots,u_k^{(p)}\)$,
$p=1,2,\dots$, in the limit for which $u_j^{(p)}<u_j$ for all numbers
$1\le j\le k$ and $p=1,2,\dots$, and $\limm_{p\to\infty}
u_j^{(p)}=u_j$ for all indices $j=1,\dots,k$.
\item{} Let us consider such rectangles $\bold K(p)$ whose edges
have rational coordinates, and the edges of these coordinates converge
in a monotone increasing way to the coordinates of the corresponding
edges of the rectangle $\bold K$. Then $\mu_{\tilde F}(\bold
K(p))\ge0$, since $\tilde F$ is the limit of distribution functions.
Hence $\mu_F(\bold K)=\limm_{p\to\infty} \mu_{\tilde F}(\bold
K(p))\ge0$, i.e.\ the function $F$ satisfies property~(iv). Let us
observe that properties (ii) and~(iii) hold if the functions $F_n$ are
replaced by the function $\tilde F$, and the limit is taken only in
rational points. (We applied  at this point of the proof the tightness
of the measures~$\mu_n$.) This property together with formula~(2.5)
imply that the function~$F$ satisfies properties (ii) and~(iii).
Property~(i) also holds. This is a simple consequence of it definition
in formula~(2.5).
\item{} To show that the distribution functions $F_{n_j}$ converge in
distribution to the distribution function~$F$ let us consider a point
of continuity $u=(u_1,\dots,u_k)$ of the function~$F$, and let us then
choose for all numbers $\e>0$ such a constant number
$\delta=\delta(\e)>0$ for which $F(u)-\e\le F(u-\delta)\le F(u)\le
F(u+\delta)\le F(u)+\e$, where $u\pm\delta=(u_1\pm \delta,\dots,u_k
\pm\delta)$. Let us then choose two points $r=(r_1,\dots,r_k)\in R^k$
and $\bar r=(\bar r_1,\dots,\bar r_k)\in R^k$ with rational coordinates
such that $u_j-\delta <r_j<u_j<\bar r_j<u_j+\delta$ for all indices
$j=1,\dots,k$. Then by the monotonicity properties of the function
$\tilde F(\cdot)$ and the definition of the function~$F$
$$
F(u)-\e\le F(u-\delta)<\tilde F(r)\le F(u)\le \tilde F(\bar r)\le
F(u+\delta)\le F(u)+\e.
$$
This relation together with the definition of the function $\tilde F$
and the monotonicity property of the functions~$F_{n_j}$ imply that
$$
\align
F(u)-\e&\le \lim_{j\to\infty}F_{n_j}(r) \le
\liminf_{j\to\infty}F_{n_j}(u)\\
&\le\limsup_{j\to\infty}F_{n_j}(u)\le
\lim_{j\to\infty}F_{n_j}(\bar r)\le F(u)+\e,
\endalign
$$
hence
$$
-\e\le \liminf_{j\to\infty}F_{n_j}(u)-F(u) \le
\limsup_{j\to\infty}F_{n_j(u)}-F(u)\le\e.
$$
Since this relation holds for all $\e>0$, it implies that
$\limm_{j\to\infty}F_{n_j}(u)=F(u)$.
\item{22.)} Fix some $\delta>0$ and write the following identity:
$$
\aligned
&\frac1{2\delta}\int_{-\delta}^\delta
\Re[1-\varphi_n(t)]\,dt=\int_{-\delta}^\delta\frac1{2\delta}
\int_{-\infty}^\infty \[1-\cos tx\]\,dF_n(x)\,dt\\
&\qquad=\int_{-\infty}^\infty\frac1{2\delta}\int_{-\delta}^\delta
[1-\cos tx]\,dt\,dF_n(x)
=\int_{-\infty}^\infty\[\frac t{2\delta}-\frac{\sin tx}{2\delta
x}\]_{t=-\delta}^{t=\delta}\,dF_n(x)\\
&\qquad=\int_{-\infty}^\infty\(1-\frac{\sin\delta x}{\delta
x}\)\,dF_n(x) =\int_{-K}^K\(1-\frac{\sin\delta x}{\delta x}\)\,dF_n(x)
\\ &\qquad\qquad+\int_{|x|>K}\(1-\frac{\sin\delta x}{\delta x}\)
\,dF_n(x)=I^{\delta}_{1,n}(K)+I^{\delta}_{2,n}(K).
\endaligned \tag2.6
$$
\item{} First we show with the help of relation~(2.6) that the
validity of formula~(10) implies that the sequence of distribution
functions~$F_n$ is tight. Since $\(1-\frac{\sin\delta x}{\delta
x}\)\ge0$ for all $x$ and $\delta$, hence the left-hand side of
formula~(2.6) yields an upper bound on the expression
$I^{\delta}_{2,n}(K)$ for all numbers $\delta>0$, $n\ge1$ and $K>0$.
If formula~(10) holds, then for all $\e>0$ there exists a number
$\delta=\delta(\e)>0$ and threshold index $n_0=n_0(\delta,\e)$ such that
$\frac\e2\ge\int_{|x|>K} \(1-\frac{\sin\delta x}{\delta
x}\)\,dF_n(x)$ for $n\ge n_0$. Put $K=\frac2\delta$. Then
$1-\frac{\sin\delta x}{\delta x}\ge\frac12$ for all $|x|\ge K$. Hence
the previous estimate implies that $\frac\e2\ge
\int_{|x|>K}\(1-\frac{\sin\delta x}{\delta x}\)\,dF_n(x)\ge
\frac12[(1-F_n(K))+F_n(-K)]$, i.e.\ $\e\ge [(1-F_n(K))+F_n(-K)]$ with
this number $K$ if $n\ge n_0$. By increasing the number $K>0$ if it is
necessary we can achieve that the above inequality holds for all
indices $n\ge1$. This means that the distribution functions
$F_n$,~$n=1,2,\dots$, are tight.
\item{} Let us prove with the help of formula~(2.6) that the tightness
of the distribution functions $F_n$ implies formula~(10) and even its
slightly stronger version, formula~$(10')$ where $\limsupp_{n\to\infty}$
is replaced by $\supp_{n\ge1}$. Since $\left|1-\frac{\sin\delta
x}{\delta x}\right| \le2$, the tightness of the distribution functions
$F_n$ makes possible to choose a number $K=K(\e)>0$ for all $\e>0$ such
that $|I^{\delta}_{2,n}(K)|=\left|\int_{|x|>K}
\(1-\frac{\sin\delta x}{\delta x}\)\,dF_n(x)\right|\le\frac\e2$ for
all numbers $\delta>0$ and $n=1,2,\dots$. After fixing the number
$K=K(\e)>0$ we can choose a number $\bar\delta=\bar\delta(\e,K)>0$
such that the inequality $\e\ge1-\frac{\sin\delta x}{\delta x}\ge0$
holds for all numbers $|x|<K$ and $0<\delta<\bar\delta$. Hence
$|I^{\delta}_{1,n}(K)|=\left|\int_{-K}^K\(1-\frac{\sin\delta
x}{\delta x}\)\,dF_n(x)\right|\le\frac\e2$. These estimates together
with relations~(2.6) imply that $\left|\frac1{2\delta}
\int_{-\delta}^\delta\Re[1-\varphi_n(t)]\,dt\right|\le\e$ for
all $n\ge1$ if $\delta\le\bar\delta(\e)$. The statements of
problem~22 are proved.
\item{23.)} Let us consider the $j$-th coordinate, $1\le j\le k$,
of the random vectors $\bold\xi^{(n)}$, i.e.\  the random variables
$\bold\xi^{(n)}_j$ for all indices $n=1,2,\dots$. The characteristic
function of the random variable $\xi^{(n)}_j$ is the function
$$
\varphi^{(j)}_{n}(t)=\varphi_n(\underbrace{0,\cdots,0,}\Sb j-1 \\
\text{0 coordinates}\endSb
t,\underbrace{0,\cdots,0}\Sb n-j-1\\ \text{0 coordinates}\endSb).
$$
By the conditions of the problem the functions $\varphi^{(j)}_n(t)$
converge to a function $\varphi^{(j)}(t)$ continuous in zero in a
small neighbourhood of the origin. Let us remark that
$\varphi^{(j)}(0)=\limm_{n\to\infty}\varphi^{(j)}_n(0)=1$.
Hence the continuity of the function $\varphi^{(j)}(t)$ in the origin
implies that for all numbers $\e>0$ there exists a threshold index
$\bar\delta=\bar\delta(\e)>0$ such that for all numbers
$0<\delta<\bar\delta$
$$
0\le\frac1{2\delta}\int_{-\delta}^\delta
\Re[1-\varphi_n^{(j)}(t)]\,dt<\e.
$$
Furthermore, as $\limm_{n\to\infty}\Re[1-\varphi^{(j)}_n(t)]=
\Re[1-\varphi^{(j)}(t)]$ if $|t|<\delta<\bar\delta$ (we choose a smaller
threshold $\bar\delta>0$ if it is necessary), and
$0\le\Re[1-\varphi^{(j)}_n(t)]\le2$, it follows from  Lebesgue's
dominated convergence theorem that $0\le\limsupp_{n\to\infty}
\frac1{2\delta}\int_{-\delta}^\delta\Re[1-\varphi^{(j)}(t)]\,dt<\e$.
Hence the result of problem~22 shows that the distribution functions of
the random variables $\xi^{(j)}_n$, $n=1,2,\dots$, are tight, i.e.\
for all $\e>0$ there exists a constant $K=K(\e)>0$ such that the
inequality $P\(\left|\xi^{(j)}_n\right|>K\) <\frac\e k$ holds.
Since this statement holds for all numbers $j=1,\dots,k$, it implies
that the distributions of the random vectors
$\bar\xi_n= (\xi_1^{(n)},\dots,\dots,\xi_k^{(n)})$ are tight.
\item{24.)} It follows from the results of problems~21 and~23 that
the sequence of distribution functions $F_n(u_1,\dots,u_k)$,
$n=1,2,\dots$, is relatively compact, i.e.\ an arbitrary subsequence
of the sequence of the distribution functions $F_n$ has a
sub-subsequence convergent in distribution if the characteristic
functions of the distribution functions $F_n$ converge to a function
continuous in the origin. (Moreover, it is enough to assume that this
property holds for the restriction of the characteristic functions to
the coordinate axes.) To see that under the conditions of the first
part of problem~24 the distribution functions $F_n$ converge in
distribution it is enough to show that in this case all convergent
subsequences of this sequence of distribution functions have the same
limit. To justify this reduction of the problem let us choose a
convergent subsequence~$F_{n_l}$ which converges to some distribution
function $F(u_1,\dots,u_k)$. If this distribution function $F(\cdot)$
were not the limit of the distribution functions $F_n$, then the
distribution function $F(u_1,\dots,u_k)$ would have a point of
continuity $u=(u_1,\dots,u_k)$ together with a constant $\e>0$ and a
sequence of indices $n_j$, $j=1,2,\dots$, such that
$|F_{n_j}(u_1,\dots,u_k)-F(u_1,\dots,u_k)|>\e$ for all $j=1,2,\dots$.
But this would mean that a convergent subsequence of the sequence of
distribution functions $F_{n_j}$, $j=1,2,\dots$, would have a limit
different of the distribution function $F(u_1,\dots,u_k)$.
\item{} The statement that all convergent subsequences of the
distribution functions $F_n$ have the same limit follows from the
results of Theorem~A and problem~19. Indeed, it follows from Theorem~A
that the characteristic function of the limit distribution function of a
convergent subsequence of the distribution functions $F_n$ is the limit
of the characteristic functions of the distribution functions in this
subsequence. The limit of these characteristic functions does not depend
on which convergent subsequence we have considered. But by the result of
problem~19 a distribution function is determined by its distribution
function. Hence the condition that the characteristic functions of a
sequence of distribution functions converge to a function continuous in
the origin implies that these distribution functions converge in
distribution to a distribution function, and besides the
characteristic function of the limit distribution function equals the
limit of the characteristic functions of the distribution functions we
have considered.
\item{} If a sequence of distribution functions $F_n(u_1,\dots,u_k)$
converges in distribution to a distribution function
$F_0(u_1,\dots,u_k)$, then it follows from Theorem~A that the
characteristic functions $\varphi_n(t_1,\dots,t_k)$ of these
distribution functions converge to the characteristic function
$\varphi_0(t_1,\dots,t_k)$ of the distribution function $F_0$ in all
points $(t_1,\dots,t_k)\in R^k$. To complete the proof of the
Fundamental Theorem we have still to show that this convergence is
uniform in all compact subset of the Euclidean space~$R^k$.
\item{} To prove this statement let us observe that since the
distribution functions $F_n$ converge in distribution they are tight.
Hence for all $\e>0$ there exists a constant $K=K(\e)$ such that a
sequence of random vectors $\bold\xi_n=(\xi^{(1)}_n,\dots,\xi^{(k)}_n)$,
$n=1,2,\dots$, with distribution functions $F_n$ satisfy the inequality
$P(|\xi_n|>K)<\frac\e3$ for all indices $n=0,1,2,\dots$. (In the
further part of the proof $\xi(\oo)$, $t\in R^k$, $u\in R^k$ denote
points of the $k$-dimensional space, and $(u,t)$, $u\in R^k$,
$t\in R^k$, denotes the scalar product of the vectors $u$ and $t$.)
Let us choose  a finite set of points $\bold
T=\left\{t^{(1)},\dots,t^{(s)}\right\}\subset \bold K$, $s=s(\bold
K,\delta)$, in a compact set $\bold K\subset R^k$ which is
$\delta$-dense in the set~$\bold K$, i.e.\ for all $t\in \bold K$
there is a point $t^{(j)}\in \bold T$ such that
$\rho(t,t^{(j)})<\delta$. Then
$$
\align
\left|\varphi_n(t)-\varphi_n(t^{(j)})\right|&=
\left|Ee^{i(t,\bold\xi_n)}-e^{i(t^{(j)},\bold\xi_n)}\right| \\
&\le E\left|e^{i(t-t^{(j)},\bold\xi_n)}-1\right|
I\(|\bold\xi_n|\le K\)+P(|\bold\xi_n|>K) \le \frac{2\e}3
\endalign
$$
for all numbers $n=1,2,\dots$. Further we can choose a threshold index
$n_0=n_0(\e)$, such the inequality
$\supp_{n\ge n_0}\supp_{t^{(j)}\in\bold T}\left|\varphi_n(t^{(j)})-
\varphi_0(t^{(j)})\right|<\frac\e3$ holds. It follows from the last
inequalities that $\supp_{t\in\bold K}|\varphi_n(t)-\varphi_0(t)|<\e$
if $n\ge n_0$. This means that the convergence
$\varphi_n(t)\to\varphi_0(t)$ is uniform on all compact sets~$\bold K$.
\item{25.)} Let $\varphi_0(t)$ denote the characteristic function of
the uniform distribution in the interval $[-1,1]$, i.e.\
$\varphi_0(t)=\int_{-1}^1\frac12e^{itu}\,du
=\frac{e^{it}-e^{-it}}{2it}$. Let us define the characteristic
functions $\varphi_n(t)$ as the characteristic functions of the
following discretizations $\mu_n$, $n=1,2,\dots$, of the uniform
distribution on the interval $[-1,1]$: $\mu_n\(\frac
kn\)=\frac1{2n+1}$, $-n\le k\le n$. Then
$$
\varphi_n(t)=\frac1{2n+1}\summ_{k=-n}^n
e^{ikt/n}=\frac{e^{i(n+1)t/n} -e^{i(-n+1)t/n}}{(2n+1)(e^{it/n}-1)}.
$$
Simple calculation shows that $\varphi_n(t)\to\varphi_0(t)$ for all
points~$t\in R^1$, and the convergence is uniform in all finite
intervals. On the other hand, the convergence is not uniform on the
whole real line, since $\limm_{t\to\infty}\varphi_0(t)=0$, while
$\varphi_n(t)=1$ in the points of the form $t=2\pi kn$,
$k=0,\pm1,\pm2,\dots$. (The background of this construction: We have
approximated the characteristic function of a distribution function
having a density function with the characteristic function of more and
more closer discretization of this distribution function. These
discretized approximations of the original distribution functions had
lattice distributions. The characteristic functions of such
approximating distributions converge to the characteristic function of
the limit distribution by the Fundamental Theorem. Besides,
the characteristic function of a distribution function with a density
function tends to zero in the infinity by the Riemann lemma. On the
other hand the characteristic function of a lattice valued distribution
is a periodic function which has absolute value~1 in certain points.)
\item{26.)} An example for case a): Let the measures $\mu_n$ have
uniform distribution in the interval $[-n,n]$.
Then $\varphi_n(t)=\frac1{2n}\int_{-n}^{n}e^{itu}\,du
=\frac{e^{itn}-e^{-itn}}{2int}$. Hence
$\limm_{n\to\infty}\varphi_n(t)=0$ if $t\neq0$, and
$\limm_{n\to\infty}\varphi_n(0)=1$.
\item{} An example for case b): Let $\mu_{2n}(\{n\})
=\mu_{2n}(\{-n\})=\frac12$, and $\mu_{2n+1}$ be the probability measure
$\mu_n$ defined in case~a). Then $\varphi_{2n}(t)=
\frac12(e^{itn}+e^{-itn})$, and it equals~1 in the points of the form
$\frac{2k\pi}n$. This means that in the points of the form
$t=\varphi\(\frac{2k\pi}l\)$ $\varphi_{n_k}(t)=0$ for a certain
subsequence $n_k$, and $\limm_{k\to\infty}\varphi_{\bar n_k}(t)=1$ for
a certain subsequence $\bar n_k$.
\item{27.)} Let $F(x)$ denote the distribution function of the random
variable $\xi$. Then $\varphi(t)=\int e^{itu}dF(u)$. By successive
differentiation we get that $\frac{d^k\varphi(t)}{dt^k}=i^k\int
u^ke^{itu}\,dF(u)$, in particular $\left.\frac{d^k\varphi(t)}
{dt^k}\right|_{t=0}=i^k\int u^k\,dF(u)=i^kE\xi^k$ if the order of
derivation and integration can be changed in the above calculation.
The above calculation is legitim if the distribution $F$ is
concentrated in the interval $[-K,K]$, because the integrand in the
integral expressing the fraction $\frac{\varphi(t+h)-\varphi(t)}h$
satisfies the relation $\frac{e^{i(t+h)u}-e^{itu}}h=iue^{itu}+O(h)$, 
and for a fixed number $t$ the order $O(h)$ is uniform if $u\in [-K,K]$.
\item{} If $E|\xi|=\int|u|\,dF(u)<\infty$, then to prove the statement 
of problem~27 for the first derivative we introduce the functions
$$
G_n(t)=\int_{-n}^n e^{iut}\,dF(u), \quad
H_n(t)=i\int_{-n}^n ue^{iut}\,dF(u), \quad n=1,2,\dots,
$$ 
and 
$G(t)=\int_{-\infty}^\infty e^{iut}\,dF(u)$ \'es
$H(t)=i\int_{-\infty}^\infty ue^{iut}\,dF(u)$. Then
$\limm_{n\to\infty}G_n(t)=G(t)$, and
$\limm_{n\to\infty}\frac{dG_n(t)}{dt}=H(t)$ with
$H(t)=i\int_{-\infty}^\infty ue^{iut}\,dF(u)$. We show with the 
help of the above statements that the function $G(t)$ is
differentiable, and $\frac{dG(t)}{dt}=H(t)$. Indeed, 
$G(t)=\limm_{n\to\infty}G_n(t)=\limm_{n\to\infty}\left[G_n(0)+
\int_0^t H_n(s)\,ds\right]$, hence the relations 
$\limm_{n\to\infty}G_n(0)=G(0)$, $\limm_{n\to\infty}H_n(s)=H(s)$, 
and the validity of the inequality $|H_n(s)|\le E|\xi|$ for all 
numbers $n$ and $s$ together with Lebesgue dominated convergence
theorem imply that $G(t)=G(0)+\int_0^t H(s)\,ds$. Hence 
$\frac{dG(t)}{dt}=H(t)$, and this is what we had to prove.
\item{} The statement about the $k$-th derivative of the Fourier 
transform under the condition $E|\xi|^k<\infty$ can be proved 
similarly by induction with respect to the parameter~$k$ with
the help of the identity $\frac{dG^k(t)}{dt^k}
=\frac d{dt}(\frac{dG^{k-1}(t)}{dt^{k-1}})$. Only in this case
we have to work with the functions
$G_n(t)=i^{k-1}\int_{-\infty}^\infty u^{k-1}e^{iut}\,dF(u)$, 
$H_n(t)=i^k\int_{-n}^n u^ke^{iut}\,dF(u)$, $n=1,2,\dots$, and
$G(t)=i^{k-1}\int_{-\infty}^\infty u^{k-1}e^{iut}\,dF(u)$,
$H(t)=i^k\int_{-\infty}^\infty u^ke^{iut}\,dF(u)$. We prove 
the identity $G(t)=G(0)+\int_0^t H(s)\,ds$ also in this case.

\item{} If $Ee^{t\xi}<\infty$ with some number $t>0$, then
$P(\xi>x)=P(e^{t\xi}>e^{tx})\le e^{-tx}Ee^{t\xi}\le \const e^{-tx}$
for all numbers~$x\ge 0$. Similarly, $P(\xi<-x)\le \const e^{-tx}$,
if $Ee^{-t\xi}<\infty$. Hence $P(|\xi|>x)\le\const e^{-tx}$ if
$Ee^{ux}<\infty$ for $|u|\le t$. Conversely, if $G(u)=P(|\xi|>x)
\le\const e^{-\alpha x}$, then we get by partial differentiation that
$Ee^{t|\xi|}=\int_0^\infty e^{tu}\,dG(u)=[e^{tu}G(u)]_{0}^\infty
-\int_0^\infty te^{tu}G(u)\,du<\infty$ in the case  $0<t<\alpha$.
Hence $Ee^{\pm t\xi}\le1+Ee^{t|\xi|}<\infty$.
\item{} Finally, if $P(|\xi|>x)\le \const e^{-\alpha x}$, then the
function $G(z)=\int e^{izx}\,dF(x)$ is analytic in the domain 
$\{z\colon\;|\Im z|<\alpha\}$, because in an arbitrary compact 
set in the interior of this domain the function $G(z)$ can be 
represented as the uniform limit of analytic functions (finite 
sums approximating this integral). This function $G(z)$ is the 
analytic continuation of the function $\varphi(t)$ to the above domain.
\item{28.)} Let us first prove Riemann's lemma. If $g(u)=I([a,b])$ is
the indicator function of an interval $[a,b]$, then
$\int e^{itu}g(u)\,du=\frac{e^{ibt}-e^{iat}}{it}\to0$ if
$t\to\infty$ or $t\to-\infty$. This relation also holds if
$g(u)=\summ_{j=1}^k c_j I([a_j,b_j])$, \i.e.\ $g(\cdot)$ is a finite
linear combination of indicator functions of intervals. Such functions
constitute an everywhere dense subset of the integrable functions in
$L_1$ norm, that is for all numbers $\e>0$ and integrable function
$f(\cdot)$ there exists a function $g(\cdot)$ or the above form such
that $\int|f(u)-g(u)|\,du<\e$. This relation implies that $\left|\int
e^{itu} f(u)\,du-\int e^{itu}g(u)\,du\right|<\e$ for all numbers $t\in
R^1$. The Riemann lemma is a consequence of the above relations.
\item{} If the function $f(t)$ is $k$ times differentiable, and the
first $k$ derivatives are integrable functions on the real line, then
we get by successive partial differentiation that
$\varphi(t)=i^kt^{-k}\int e^{itu}\left.\frac
{df^k(s)}{ds^k}\right|_{s=u}\,du$. This relation together with
Riemann's lemma imply that in this case $\varphi(t)=o(t^{-k})$ if
$t\to\pm\infty$.
\item{} If the function $f(\cdot)$ is analytic in a strip
$\{z\colon\; \Re z\in [-A,A]\}$ around the real line and the function
$f(-ia+\cdot)$ is integrable for $a<A$, then we can write, because of
the inequality $\left.|e^{i(u-ia)t}\right|<e^{-at}$ that
$$
|\varphi(t)|=\left|\int_{-ia-\infty}^{-ia+\infty}e^{itu}f(u)\,du\right|
\le e^{-at}\int_{-\infty}^{\infty}|f(u-ia)|\,du\le\const e^{-at}
$$
for $t>0$. The case $t<-0$ can be handled similarly, only the integral
has to be replaced to the line $[-\infty+ia,\infty+ia]$ on the positive
half-space.
\item{29.)} The relation $|\varphi(t)|=1$ holds if and only if
$\varphi(t)=e^{ita}$, that is $Ee^{it(\xi-a)}=1$
with some real number $a$. This identity holds if and only if
$P(t(\xi-a)\in2\pi \{0,\pm1,\pm2,\dots,\})=1$. This means that
$|\varphi(t)|=1$ with some $t\neq0$  if the values of the random
variable $\xi$ are concentrated to a lattice $\left\{\frac {2\pi k}
t+a,\; k=0,\pm1,\pm2,\dots\right\}$ of width $\frac{2\pi}t$. In
particular, the relation $|\varphi(t)|=1$ for all $t\in R^1$ can 
hold if and only if the random variable takes a constant value with
probability~1. If $\xi$ is not a deterministic constant, and it is
lattice distributed, then there is a largest $h>0$ such that the
distribution of $\xi$ has period~$h$. Indeed, in this case there
are two numbers $a$ and $b$, $a\neq b$, such that $P(\xi=a)>0$, and
$P(\xi=b)>0$. Then the distribution of $\xi$ may have a period only
with $\frac{b-a}k$, where $k$ is a positive integer. Since the 
distribution of $\xi$ has a period $h>0$, the above statement holds.
\item{} To finish the solution of the problem it is enough to
observe that the characteristic function is continuous on the real line,
and the characteristic function $\varphi(\cdot)$ of a non-lattice
valued random variable satisfies the inequality $|\varphi(t)|<1$ if
$t\neq0$. Hence $\supp_{A\le |t|\le B}|\varphi(t)|<1$ in this case.
\item{30.)} The characteristic function of the random variable $\xi$
considered in this problem equals $\varphi(t)=\frac12\(\cos
t+\cos (\sqrt2t)\)$. There exist pairs of integers $(p_n,q_n)$,
$n=1,2,\dots$, such that $q_n\to\infty$, and $|\sqrt2q_n-p_n|\le
\frac1{q_n}$. (Such pairs of integers $(p_n,q_n)$ can be found as the
nominator and denominator of  the continued fraction of the number
$\sqrt 2$.) Choose $t_n=2\pi q_n$. Then $\cos t_n=1$, and
$\limm_{n\to\infty}\cos (\sqrt2t_n)=1$. Hence $t_n\to\infty$, and
$|\varphi(t_n)|\to1$, if $n\to\infty$. On the other hand,
$\varphi(t)\neq1$ if $t\neq0$.
\item{} If $\xi$ is a random variable whose values are concentrated in
a subset of the real line consisting of finitely or countably many
points, then for all numbers $\e>0$ there exists an integer
$s=s(\e)<\infty$ and points $u_1,\dots, u_s$ on the real line such that
$P(\xi\in\{u_1,\dots,u_s\})\ge1-\frac\e3$. Further, by a classical (and
simple) result of the number theory, by a result of Dirichlet, for all
numbers $N\ge1$ there exists an integer $1\le q_N\le N$ and a set of
$s$ integers $p_1,\dots,p_s$ such that $|q_N u_k-p_k|\le N^{-1/s}$, for
all indices $k=1,\dots,s$. Hence, by choosing the number $N$
sufficiently large we can achieve with the choice $t=2\pi q_N$ that
$\Re e^{itu_k}\ge1- \frac\e3$ for all indices $1\le k\le s$. Then $\Re
Ee^{it\xi}\ge \summ_{k=1}^s P(\xi=u_k)(1-\frac\e3)-\frac\e3 \ge1-\e$.
Since we can make this construction for all $\e>0$, there exists
a sequence of positive integers $q_N$, $N=1,2,\dots$, such that the
sequence $t_N=2\pi q_N$ satisfies the relation $\limm_{N\to\infty}
\varphi(t_N)=1$. Since the random variable $\xi$ is not lattice
distributed, hence $\supp_{2\pi\le t\le B}|\varphi_n(t)|\le q<1$ for
any $B>2\pi$ with an appropriate constant $q=q(B)<1$. This implies
that the above constructed sequence of positive real numbers $t_N$
satisfies the relation $t_N\to\infty$ if $N\to\infty$.
\item{31.)} Let us denote the distribution function of the random
variable $\xi$ by $F(u)$, its characteristic function by $\varphi(t)$
and put $\Re \varphi(t)=u(t)$. Then for all $h\ge0$
$$
\frac{1-u(h)}{h^2}=\int_{-\infty}^\infty \frac{1-\cos
hu}{h^2u^2}u^2\,dF(u),
$$
and since $\limm_{h\to0}\frac{1-\cos hu}{h^2u^2}=\frac12$,
$\frac{1-\cos hu}{h^2u^2}\ge0$ for all numbers $u\in R^1$ and $h\in
R^1$, hence $\liminff_{h\to0} \int_{-\infty}^\infty \frac{1-\cos
hu}{h^2u^2}u^2\,dF(u)\ge\frac12 \int_{-\infty}^{\infty}u^2\,dF(u)
=\frac12E\xi^2$ by Fatou's lemma. Hence, to solve the problem it is
enough to show that $\limsupp_{h\to0}\frac{1-u(h)}{h^2}<\infty$ if
the function $\varphi(t)$ is twice differentiable in the origin.
If the function $\varphi(t)$ is twice differentiable in the origin,
then the same relation holds for the function~$u(t)$. Then the
derivative $u'(t)=\frac{du(t)}{dt}$ exists in a small neighbourhood
of the origin, and $u'(0)=0$, as $u(\cdot)$ is an even function.
Further, $u(0)=1$, $u(t)\le 1$ for all numbers $t\in R^1$, hence
$0\le\frac{1-u(h)}{h^2}=\frac{u(0)-u(h)}{h^2}=-\frac
{u'(\vartheta h)}h\le \supp_{0\le s\le h} \frac{u'(0)-u'(s)}s<\infty$
for small number $h>0$, if $u(\cdot)$ is twice differentiable in the
origin, where $0\le\vartheta\le1$ is an appropriate number, and
$u'(\cdot)$ denotes the derivative of the function $u(\cdot)$.
Hence $\limsupp_{h\to0}\frac{1-u(h)}{h^2}<\infty$ in this case.
\item{} By induction with respect to the parameter $k$ we can see that
if $\xi$ is a random variable with distribution function $F$, and the
$2k$-th derivative of the characteristic function in the origin is
finite, then the random variable $\xi$ has finite $2k$-th moment.
Indeed, by the induction hypothesis there exists a distribution
function $F^{(k-1)}(du)=\frac{u^{(2k-2)}F(du)}{m_{2k-2}}$
where $m_{2k-2}=\int u^{2k-2}\,dF(u)$. Further, the characteristic
function of the distribution function $F^{(k-1)}$ has finite second
moment in the origin, since the characteristic function of this
distribution function equals the $2k-2$-th derivative of the
characteristic function of the distribution function $F$ multiplied by
$(-1)^{k-1}m_{2k-2}^{-1}$, and the characteristic function of a random
variable with distribution function $F^{(k-1)}(u)$ has finite second
derivative in the origin. Hence by the already proven part
of this problem a random variable with distribution function $F^{(k-1)}$
has finite second  moment. This is equivalent to the statement that a
random variable with distribution function $F$ has finite $2k$-th
moment.
\item{32.)} The solution of this problem applies similar idea as the
proof of problem~22. In problem~22 we deduced from some kind of
continuity of the characteristic function of a distribution function
in the origin some sort of estimate about the tail behaviour of the
distribution function. In this problem we exploit that if we know more
continuity about the distribution function in the neighbourhood of the
origin, then we get sharper estimates about the tail behaviour of the
distribution function.
\item{} Let $F(x)$ denote the distribution function of the random
variable $\xi$, and put $u(t)=\Re \varphi(t)$. The estimate
$\left|\frac{1-u(h)}h\right|=|u'(\vartheta h)|\le \const h^\alpha$
holds with an appropriate constant $0<\vartheta<1$ under the conditions
of the problem. Then the following analog of formula~(2.6) holds.
$$   \allowdisplaybreaks
\align
\const h^{1+\alpha}&>\int_{-h}^h \frac{1-u(t)}h\,dt=
\int_{-1/2h}^{1/2h}\(1-\frac{\sin h x}{hx}\)\,dF(x)\\
&\qquad +\int_{|x|>\frac1{2h}}\(1-\frac{\sin h x}{h x}\) \,dF(x)
\ge\int_{|x|>\frac1{2h}}\frac12 \,dF(x) \\
&=\frac12 P\(|\xi|>\frac1{2h}\).
\endalign
$$
This implies that $P(|\xi|>u|)\le \const u^{-1-\alpha}$ for all numbers
$u>0$. Let us introduce the function $G(u)=P(|\xi|>u|)$. Integration by
parts yields that
$$
E|\xi|=\int_0^\infty |u|\,dG(u) =\[uG(u)\]_0^\infty-\int
G(u)\,du<\infty.
$$
\item{} If the characteristic function $\varphi(t)$ is $2k+1$-times
differentiable in a small neighbourhood of the origin, and the
$2k+1$-th differential is a Lipschitz $\alpha$ function, $\alpha>0$,
in this small neighbourhood, then let us introduce the distribution
function $F^{(k)}(du) =\frac{u^{2k}F(du)}{m_k}$, where $F(\cdot)$
denotes the distribution function of the random variable $\xi$, and
$m_k=\int u^{2k}\,dF(u)$. The argument applied in the solution of the
previous problem can be adapted to the present case. The already
proven part of this problem can be applied to a random variable with
distribution function  $F^{(k)}$, and it yields that
$E|\xi|^{2k+1}<\infty$.
\item{33.)} The relation $(-1)^k E\xi^{2k}=\left.\frac{d^{2k}
\varphi(t)}{dt^{2k}}\right|_{t=0} =\frac{(2k)!}{2\pi i}
\oint_{z=R}\frac{\varphi(z)}{z^{2k+1}}\,dz$ holds for all
integers~$k$ by the result of problem~31 and the Cauchy integral formula
if the circle with center in the origin and radius $R$ is in the domain
of analiticity of the function $\varphi(z)$. Since
$\supp_{z=R}|\varphi(z)|<\infty$ the above relation implies that
$E\xi^{2k}\le (ak)^{2k}$ with some constant $a>0$, and
$P(|\xi|>x)\le\(\frac{ak}x\)^{2k}$ for all positive integers $k\ge1$.
If $x\ge C_0$ with some number $C_0>0$, then let us fix the constant
$k=\[\frac x{2a}\]$, where $[u]$ is the greatest integer smaller
than~$u$. This implies that the inequality $P(|\xi|>x)<\const
e^{-\alpha x}$ holds for all numbers $x>0$ with an appropriate
constant $\alpha>0$.
\item{} Let us remark that we had to apply the above relatively
complicated argument, because at the beginning of the proof we did
not know that the analytic continuation of the characteristic
function of a random variable $\xi$ is always the function
$\varphi(z)=Ee^{iz\xi}$.
\item{34.)} If the characteristic function of a random variable
$\xi$ is integrable, then the result about the inverse Fourier
transform can be applied. It implies that the random variable $\xi$
also has a density function $f(x)$, and the identity
$f(x)=\frac{1}{2\pi} \int e^{-itx}\varphi(t)\,dt$ holds.
Also the formula $\frac{d^kf(x)}{dx^k}=\frac{(-i)^k}{2\pi}
\int t^k e^{-itx}\varphi(t)\,dt$ holds if the order of integration
and differentiation can be changed in the inverse Fourier transform
formula. In the solution of problem~27 we have proved that this change
of order can be carried out if the functions $|t|^j|\varphi(t)|$ are
integrable for all indices $0\le j\le k$. Hence the statement of this
problem holds if $|\varphi(u)|<\const |u|^{-(k+1+\e)}$ with some number
$\e>0$. If $|\varphi(u)|<\const e^{-\alpha|u|}$ with some constant
$\alpha>0$, then the function $f(z)=\frac1{2\pi} \int
e^{-itz}\varphi(t)\,dt$ is an analytic continuation of the density
function $f(x)$ of the random variable~$\xi$.
\item{35.)} Let $\varphi(t)=E^{it\xi}$ denote the characteristic
function of the random variable $\xi_1$. The normalized partial sum
$\frac{S_n}{\sqrt n}$ has the characteristic function
$\varphi^n\(\frac t{\sqrt n}\)$. Hence by the Fundamental Theorem about
convergence in distribution and the result of problem~13a) it is enough
to show that $\varphi^n\(\frac t{\sqrt n}\)\to e^{-t^2/2}$ for all
numbers $t\in R^1$ if $n\to\infty$. On the other hand, a Taylor
expansion of the  function $\varphi(t)$ around the point $t=0$
and the result of problem~27 yield that
$\varphi(t)=1-\frac{t^2}2+o(t^2)$ if $t=o(1)$,  and 
$\varphi\(\frac t{\sqrt n}\)=1-\frac {t^2}n+o\(\frac1n\)
=e^{-t^2/2n+o(n^{-1})}$ if $t=O(1)$. The
last relation implies that $\varphi^n\(\frac t{\sqrt
n}\)=e^{-t^2/2+o(1)}\to e^{-t^2/2}$ for all fixed number~$t$, 
if $n\to\infty$.
\item{36.)} Let us introduce the function
$F(t)=e^{it}-\(1+\frac{it}{1!}+\cdots+\frac{(it)^k}{k!}\)$, and let us
consider its derivatives $F^{(j)}(t)$, $j=1,2,\dots,k$. Observe that
$F^{(j)}(0)=0$ of all $0\le j\le k$, and $|F^{(k+1)}(t)|=|e^{ikt}|=1$
for all $t\in R^1$. We get by induction with respect to the
parameter~$j$ that $|F^{(j)}(t)|\le \int_0^t
|F^{(j+1)}(s)|\,ds\le\int_0^t \frac
{|s|^{k-j}\,ds}{(k-j)!}=\frac{|t|^{k+1-j}}{(k+1-j)!}$ for all indices
$j=k+1,k,\dots,0$. In particular, $|F(t)|\le
\frac{|t|^{k+1}}{(k+1)!}$, and this is the statement of the problem.
\parindent=30pt
\item{37.a)} By applying formula (11) with $k=1$ we get that
$|e^{it\xi}-1-it\xi|\le \frac{t^2\xi^2}2$. Taking the expected value of
the expression at the left-hand side between the absolute value sign we
get that $|\varphi(t)-1|\le \frac{t^2}2E\xi^2$. If $E\xi^2<\e$ with a
sufficiently small number $\e=\e(t)>0$, then
$|1-\varphi(t)|\le \frac14$, and $|\log\varphi(t)+(1-\varphi(t))|=
|\log\(1-(1-\varphi(t))\)-(1-\varphi(t))|
\le |1-\varphi(t)|^2\le t^4\(E\xi^2\)^2$.
\item{b.)} The sequence of random variables $S_k$ converge in
distribution to a normal random variable with expected value $m$ and
variance $\sigma^2$ if and only if
$$
\limm_{k\to\infty}\prodd_{j=1}^{n_k}\varphi_{k,j}(t)
=e^{-\sigma^2t^2/2+imt} \quad\text{for all } t\in R^1.   \tag2.7
$$
Let us take the logarithm in the relation~(2.7). We claim that
formula~(2.7) is equivalent to formula $(2.7')$ formulated below.
$$
\limm_{k\to\infty} \summ_{j=1}^{n_k}\log\varphi_{k,j}(t)=-\frac
{\sigma^2t^2}2+imt \quad\text{for all } t\in R^1.
\tag$2.7'$
$$
The equivalence of relations (2.7) and $(2.7')$ is less obvious than it
may seem at first sight. Some difficulty arises because in the space
of complex numbers where we have to work the equation $e^{z_1}=e^{z_2}$
only implies that $z_1=z_2+i2k\pi$ with some integer $k$, but the
numbers $z_1$ and $z_2$ may be different. Hence although the implication
$(2.7')\Rightarrow (2.7)$ needed in the proof of the central limit
theorem is straightforward, the proof of the implication
$(2.7)\Rightarrow (2.7')$ needed in the proof of the converse of the
central limit theorem requires a more careful argument.
\item{} Before the proof of the implication $(2.7)\Rightarrow (2.7')$
let us remark that because of the uniform smallness condition for
a fixed value $t\in R^1$ the characteristic function $\varphi_{k,j}(t)$
we consider is in a small neighbourhood of the number~1 for all $1\le
j\le n_k$ if the index $k$ is sufficiently large. Hence for a
sufficiently large $k\ge k_0(t)>0$ there is a version
$\log\varphi_{k,j}(t)$ of the functions $\varphi_{k,j}(t)$ which is in
a small neighbourhood of the origin, say $|\log\varphi_{k,j}(t)|
<\frac12$. We take this version of the logarithm in formula~$(2.7')$.
\item{} To prove the implication $(2.7)\Rightarrow (2.7')$ let us
first make the following observation. We have seen in the proof of
problem~24 (in the proof of the Fundamental Theorem about convergence
of distribution functions) that the convergence in formula (2.7) is
uniform in all finite intervals. Besides, the right-hand side of
formula (2.7) is separated both from zero and infinity in a finite
interval. Hence we get by taking logarithm in formula (2.7) that for
all $\e>0$ and $T>0$ there exists a $k_0=k_0(\e,T)$ and
$\delta=\delta(\e,k,T)$ such that
$$
\left|\summ_{j=1}^{n_k}\log\varphi_{k,j}(t)-\(-\frac
{\sigma^2t^2}2+imt\)+i2\pi l_k(t) \right|<\e \quad\text{if } |t|\le T
\text { and } k\ge k_0 \tag$2.7''$
$$
with some integer $l_k(t)$ which may depend both on $k$ and $t$. We
have to show that $l_k(t)\equiv0$ in formula~$(2.7'')$. First we prove
the weaker statement that $l_k(t)=l_k$, i.e.\ this constant in
formula~$(2.7'')$ does not depend on $t$. Indeed, otherwise for all
$\delta=\delta_k>0$ a pair of constant $-T\le s,t\le T$, $|t-s|<\delta$
could be found such that $l_k(t)\neq l_k(s)$, i.e.\ $|2\pi i(l_k(t)
-l_k(s)|\ge 2\pi$. But this is not possible, because the function
$g_k(t)=\summ_{j=1}^{n_k}\log\varphi_{k,j}(t)$ is uniformly continuous
in the interval $[-T,T]$. Hence fixing a small $\e>0$ we can write
$|g_k(t)-g_k(s)|<\e$ for sufficiently small $\delta=\delta(k,\e,T)>0$.
Besides, also the inequality $\left|\(-\frac{\sigma^2t^2}2+imt\)-
\(-\frac{\sigma^2s^2}2+ims\)\right|<\e$ holds if $\delta>0$ is
sufficiently small. Let $\e<\frac13$. The above relations
together with formula $(2.7'')$ would contradict to the assumption
$|2\pi i(l_k(t)-l_k(s)|\ge 2\pi$. Hence $l_k(t)=l_k$. Finally, it is
easy to see that $l_k=l_k(0)=1$. Hence relation $(2.7'')$ holds for all
$\e>0$ with $l_k(t)\equiv0$, and this implies relation $(2.7')$.
\item{} Because of the uniform smallness condition of problem~37 and
the already proved part~a) of this problem we can write for $k\ge
k_0(t)$
$$
\left|\summ_{j=1}^{n_k}\log\varphi_{k,j}(t)-
\summ_{j=1}^{n_k}(\varphi_{k,j}(t)-1)\right|\le t^4\summ_{j=1}^{n_k}
\(E\xi^2_{k,j}\)^2\le\const t^4 \max_{1\le j\le n_k}E\xi_{k,j}^2,
$$
since $\limm_{k\to\infty}\summ_{j=1}^{n_k}E\xi_{j,k}^2=1$. Because of
this relation and the uniform smallness condition
$$
\limm_{k\to\infty}\left|\summ_{j=1}^{n_k}\log\varphi_{k,j}(t)-
\summ_{j=1}^{n_k}(\varphi_{k,j}(t)-1)\right|=0.
$$
These relations also imply part b) of problem 37.
\parindent=20pt
\item{38.)} Let us fix a number $\e>0$. Then
$$
E\xi_{k,j}^2=E\xi_{k,j}^2I(\{(|\xi_{k,j}|<\e\})+
E\xi_{k,j}^2I(\{|\xi_{k,j}|\ge\e\}) \le \e^2+\sum_{j=1}^{n_k}
E\xi_{k,j}^2I(\{|\xi_{k,j}|\ge\e\}),
$$
hence by the Lindeberg condition
$\limsupp_{k\to\infty}\supp_{1\le j\le n_k}E\xi_{k,j}^2\le\e^2$. Since
this formula holds for all numbers $\e>0$, it implies the uniform
smallness property.
\item{} By formula (12) (with the choice $m=0$ and $\sigma=1$)
to prove the central limit theorem it is enough to show that
under the conditions of this problem
$$
\lim_{k\to\infty}\sum_{j=1}^{n_k}\(\varphi_{k,j}(t)-1\)=
\lim_{k\to\infty}\summ_{k=1}^{n_k}
E\(e^{it\xi_{k,j}}-1-it\xi_{k,j}\)\to-\frac{t^2}2,
$$
or since $\limm_{k\to\infty}\summ_{j=1}^{n_k}E\xi^2_{k,j}=1$ it is
enough to show that
$$
\lim_{k\to\infty}\summ_{j=1}^{n_k}
E\(e^{it\xi_{k,j}}-1-it\xi_{k,j}+\frac{t^2}2\xi_{k,j}^2\)\to0.
$$
By applying formula~(11) for $k=2$ if $|tx|\le\e$ and for $k=1$ if
$|tx|\ge\e$ together with the Lindeberg condition we get that
$$
\align
\left|\summ_{j=1}^{n_k}
E\(e^{it\xi_{k,j}}-1-it\xi_{k,j}+\frac{t^2}2\xi_{k,j}^2\)
I(\{|\xi_{k,j}|\le \e\})\right| &\le
\summ_{j=1}^{n_k} E\frac{|t\xi_{k,j}|^3}6I(\{|\xi_{k,j}|\le \e\}) \\
&\le\e |t|^3 \summ_{j=1}^{n_k}E\frac{\xi_{k,j}^2}6\le \const\e,
\endalign
$$
and
$$
\align
&\lim_{k\to\infty}\left|\summ_{j=1}^{n_k}
E\(e^{it\xi_{k,j}}-1-it\xi_{k,j}+\frac{t^2}2\xi_{k,j}^2\))
I(\{|\xi_{k,j}|>\e\})\right| \\
&\qquad\qquad\le \lim_{k\to\infty}\summ_{j=1}^{n_k}
Et^2\xi_{k,j}^2 I(\{|\xi_{k,j}|>\e\})=0.
\endalign
$$
Since these relations hold for all $\e>0$, it implies formula
(12) with the choice $m=0$ and $\sigma^2=1$. In such a way we have
solved problem~38.
\item{39.)} If the conditions of the problem are satisfied, then
$\limm_{k\to\infty}\summ_{j=1}^{n_k}\Re(\varphi_{k,j}(t)-1)=
-\frac{t^2}2$. Further, since the sum of the random variables in the
$k$-th row is almost~1 for large index~$k$, hence
$$
\limm_{k\to\infty}\summ_{j=1}^{n_k} E\(\cos
(t\xi_{k,j})-1+\frac{t^2\xi_{k,j}^2}2\)=0\quad \text{for all numbers }
t\in R^1.
$$
Let us observe that $\cos u-1+\frac{u^2}2\ge 0$ for all numbers 
$u\in R^1$, since we have for the function 
$F(u)=\cos u-1+\frac{u^2}2\ge 0$, $F(0)=0$, $F'(0)=0$ and
$F''(u)=1-\cos u\ge0$ for all numbers $u\in\in R^1$. Besides,
$\cos u-1+\frac{u^2}2\ge \frac {u^2}4$ if $|u|>3$. The above 
inequalities imply that
$$
\limm_{k\to\infty}\summ_{j=1}^{n_k}
\frac{t^2}4E\xi_{k,j}^2 I\(\left\{|\xi_{k,j}|\ge \frac3t\right\}\)=0.
$$
We get the solution of problem~39 from this relation with the choice
$t=\frac3\e$.
\item{40.)} By the Schwarz inequality
$$
\(E\xi_{k,j}I(|\xi_{k,j}|\le\e\)^2=
\(E\xi_{k,j}I(|\xi_{k,j}|>\e\)^2\le E\xi_{k,j}^2I(|\xi_{k,j}|>\e).
$$
Hence the Lindeberg condition implies that
$$
\limm_{k\to\infty} \summ_{j=1}^{n_k}\(E\xi_{k,j}
I(|\xi_{k,j}|\le\e)\)^2\le \limm_{k\to\infty}
\summ_{j=1}^{n_k}E\xi_{k,j}^2I(|\xi_{k,j}|>\e)=0,
$$
and
$$
\limm_{k\to\infty}\summ_{j=1}^{n_k}E\xi_{k,j}^2
I(|\xi_{k,j}|\le\e)=1.
$$
These two relations imply the first statement of the problem.
\item{} To prove the second statement let us first observe that
$E(\bar \xi_{k,j}-\xi_{k,j})=E\bar \xi_{k,j}-E\xi_{k,j}=0$, $1\le j\le
n_k$. Hence the Chebishev inequality and Lindeberg condition imply
that for all $\e>0$
$$
\align
P(|S_k-\bar S_k|>\e)&\le\frac{\text{Var}\,(S_k-\bar S_k)}{\e^2}=
\frac1{\e^2}\sum_{j=1}^{n_k} \text{Var}\,(\xi_{k,j}-\bar\xi_{k,j})\\
&\le \frac1{\e^2}\sum_{j=1}^{n_k} E\xi_{k,j}^2I(|\xi_{k,j}|>0)\to0.
\endalign
$$
Since this relation holds for all $\e>0$ it implies the second
statement of the problem.
\item{41.)} By the H\"older inequality
$$
\align
\summ_{k=1}^n E\xi_k^2I(|\xi_k|>\e s_n)
&\le \sum_{k=1}^n \( E|\xi_k|^{(2+\alpha)}\)^{(2/\alpha+2)}
P(|\xi_k|>\e s_n)^{\alpha/(2+\alpha)}\\
&\le \( \summ_{k=1}^n
E|\xi_k|^{(2+\alpha)}\)^{(2/\alpha+2)}\(\summ_{k=1}^n P(|\xi_k|>\e
s_n)\)^{\alpha/(2+\alpha)}.
\endalign
$$
On the other hand, by the Chebishev inequality
$\summ_{k=1}^n
P(|\xi_k|>\e s_n)\le\summ_{k=1}^n\frac{E\xi_k^2}{\e^2s_n^2}
=\frac1{\e^2}$. Hence it the conditions of part~a) of problem~41 hold,
then
$$
\limm_{n\to\infty}\frac1{s_n^2}
\summ_{k=1}^n E\xi_k^2I(|\xi_k|>\e s_n)=0,
$$
i.e. these conditions imply the Lindeberg condition.
\item{} In the case considered at the end of part~a.) $s_n^2\ge \const
n$, and $\summ_{k=1}^nE|\xi_k|^{2+\alpha}=o\(n^{(\alpha+2)/2)}\)$
if $n\to\infty$, hence in this case the condition formulated in part~a)
of problem~41 is satisfied.
\item{} If $\xi_1,\xi_2,\dots$, is a sequence of independent and
identically distributed random variables, $E\xi_1=0$,
$0<E\xi_1^2<\infty$, then
$$
\frac1{s_n^2}\summ_{k=1}^n
E\xi_k^2I(|\xi_k|>\e s_n)=\frac1{E\xi_1^2} E\xi_1^2I\(|\xi_1|>\e
\sqrt{nE\xi_1^2}\)\to0
$$
if $n\to\infty$. This means that the Lindeberg condition holds also in
this case.
\item{42.)} If the point $x$ is a point of continuity of the limit
distribution function $F(\cdot)$, then for all $\e>0$ there exists a
$\delta>0$ such that $F(x)-\frac\e2<F(x-\delta)<F(x)<F(x+\delta)
<F(x+\delta)+\frac\e2$. Since the monotone increasing function
$F(\cdot)$ has at most countably infinite points of discontinuity,
hence  we may assume without violating the generality that we choose
the point $\delta>0$ in such a way that the points $x\pm\delta$ are
also points of continuity of the function
$F(\cdot)$. Then there exists an index $n_0=n_0(\delta,\e)$ such that
$P(S_n<x+\delta)<F(x+\delta)+\frac\e4$, $P(S_n>x-\delta)
<1-F(x-\delta)+\frac\e4$, and $P(|T_n|\ge \delta)<\frac\e4$ if $n\ge
n_0$. Then $P(S_n+T_n<x)\le P(S_n<x+\delta)+P(|T_n|>\delta)
<F(x+\delta)+\frac\e2<F(x)+\e$ if $n\ge n_0(\e,\delta)$. We get in a
similar way that $P(S_n+T_n>x)<1-F(x)+\e$ if $n\ge n_0(\e,\delta)$.
Since the above statements hold for all $\e>0$, they imply the
statement of the problem.
\item{43.)} Let the  independent random variables $\xi_n$,
$n=1,2,\dots$, have the following distribution:
$P(\xi_n=n)=P(\xi_n=-n)=\frac1{4n^2}$,
$P(\xi_n=2)=P(\xi_n=-2)=\frac14$, and
$P(\xi_n=0)=\frac12-\frac1{2n^2}$, $n=1,2,\dots$. Then $E\xi_n=0$,
$E\xi_n^2=1$. Put $X_k=\xi_k I(|\xi_k|\le 2)$, $Y_n=\xi_k
I(|\xi_k|>2)$ for all $k=1,2,\dots$. Consider the partial sums
$S_n=\summ_{k=n}^n X_k$ and $T_n=\summ_{k=1}^n Y_k$, $n=1,2,\dots$.
Then the normalized partial sums $\sqrt{\frac 2n} S_n$ converge in
distribution to the standard normal distribution. Indeed, the partial
sums of the random variables $X_k$, $k=1,2,\dots$, satisfy the
conditions of the central limit theorem, and $EX_k^2=\frac12$,
$k=1,2,\dots$. On the other hand, the expressions $\sqrt{\frac 2n}
T_n$ converge stochastically to zero if $n\to\infty$. Indeed,
$\summ_{k=1}^\infty P(Y_k\neq0)<\infty$, hence with probability~1
only finitely many terms $Y_k(\oo)$ do not equal zero, and
$\summ_{k=1}^\infty |Y_k(\oo)|\le K(\oo)$. Since $\sqrt{\frac2n}
\summ_{k=1}^n\xi_k=\sqrt{\frac2n}S_n+\sqrt{\frac2n} T_n$, the above
calculation and the result of problem~42 imply that the above
construction yields an example for the statement of part~a) of
problem~43.
\item{} Let us make some slight modifications in the construction of
the above random variables~$\xi_n$. Let us put similarly to the
previous construction $P(\xi_n=2)=P(\xi_n=-2)=\frac14$ and
$P(\xi_n=n)=\frac1{4n^2}$. Let us define further
$$
P\(\xi_n=\frac1{\sqrt n}\)=\frac12-\frac1{2n^2},\quad\text{and}
\quad P\(\xi_n=-n-2n^{3/2}\(1-\frac1{n^2}\)\)=\frac1{4n^2},
$$
$n=1,2,\dots$. Then $E\xi_n=0$, $n=1,2,\dots$. By applying the
truncation technique of part~a) and carrying out a natural modification
of the calculation following it we get that these random variable
$\xi_n$, $n=1,2,\dots$, yield an example for part~b) of problem~43.
\item{44.)} Let us choose an arbitrary number $L$ such that $\int u^2
F_0(\,du)>L$. It is enough to show that $\liminff_{n\to\infty}\int u^2
F_n(\,du)\ge L$. There exists such a bounded and continuous function
$g(u)=g_L(u)$ for which $g(u)\le u^2$, and $\int g(u)F_0(\,du)\ge L$.
Indeed, the function $g(u)=g_L(u)=\min (u^2,K)$ satisfies this property
if we choose a sufficiently large constant $K=K(L)>0$. Then the
characterization of the convergence in distribution given in Theorem~A
implies that $\liminff_{n\to\infty}\int u^2 F_n(\,du)\ge
\limm_{n\to\infty}\int g(u) F_n(\,du)=\int g(u)F_0(\,du)\ge L$.
\item{45.)} The distribution of the random vector $\bold
Z=(Z_1,\dots,Z_m)$ is determined by its characteristic function.
(See the result of problem~19.) On the other hand, the characteristic
function $E^{i(t_1Z_1+\cdots+t_mZ_m)}$ of the random vector
$(Z_1,\dots,Z_m)$ in the point $(t_1,\dots,t_m)$ agrees with the
characteristic function of the random variable $Z(t_1,\dots,t_m)$
in the point~1. Hence the characteristic function and distribution
function of the random vector $\bold Z$ is determined by the
distribution of the above considered one dimensional distributions.
\item{46.)} First we show with the help of the Fundamental Theorem about
the convergence of distribution functions that the random vectors
$\bold Z_n=(Z_{1,n},\dots,Z_{m,n})$, $n=1,2,\dots$, converge in
distribution to some $m$-dimensional distribution as $n\to\infty$ if
the one-dimensional random variables $Z_n=Z_n(a_1,\dots,a_m)$,
$n=1,2,\dots$, converge in distribution for all real numbers
$a_1,\dots, a_m$ as $n\to\infty$. Indeed, if the one-dimensional random
variables interested in this problem convergence in distribution,
then the characteristic functions of the random vectors $\bold Z_n$,
$n=1,2,\dots$, converge to a function $\varphi(t_1,\dots,t_m)$ in all
points $(t_1,\dots,t_m)\in R^m$, and the restriction of this limit
function to the coordinates axes is continuous. Hence by the Fundamental
Theorem the random vectors $\bold Z_n$, $n=1,2,\dots$, also converge in
distribution, and the characteristic function of the limit distribution
is the above limit function. The Fundamental Theorem also implies that
the convergence of the random vectors $\bold Z_n$ implies the
convergence of the random variables of the random variables
$Z_n(a_1,\dots,a_m)$ in distribution.
\item{} If the random vectors $\bold Z_n$ converge in distribution, then
the limit distribution is determined by its characteristic function
which is the limit of the characteristic functions of these random
variables. Similarly, the characteristic function of the limit of the
one-dimensional random variables we have considered equals the limit of
the characteristic function of these random variables. These facts
imply the characterization of the limit distribution $\mu$ given in
this problem together with the statement that the above
characterization determines the limit distribution in a unique way.
\item{47.)} Let $\Sigma=\(D_{j,k}\)$, $1\le j,k\le m$, denote the
covariance matrix of an $m$-dimensional random vector $(Z_1,\dots,Z_m)$,
i.e. let $D_{j,k}=E(Z_j-EZ_j)(Z_k-EZ_k)$, $1\le j,k\le m$. Then the
matrix $\Sigma$ is symmetrical. Besides, $\bold x\Sigma
\bold x^*=\summ_{j=1}^m\summ_{k=1}^m x_jE(Z_j-EZ_j)(Z_k-EZ_k)x_k=
E\(\summ_{j=1}^k x_j(Z_j-EZ_j)\)^2\ge0$ for all vectors
$x=(x_1,\dots,x_m)\in R^m$, and this means that the matrix
$\Sigma$ is positive semi-definite.
\item{} On the other hand, if $\Sigma$ is an arbitrary $m\times m$
positive semi-definite matrix, then the results of linear algebra imply
that there exists a matrix $B$ such that $\Sigma=B^*B$. (The
matrix $B$ satisfying this relation is not determined in a unique way.
A possible construction of a matrix~$B$ satisfying the above relation
can be given in the following way: It is known from linear algebra
that a symmetric matrix $\Sigma$ can be represented in the form
$\Sigma=U\Lambda U^*$ where the matrix $U$ is unitary and the matrix
$\Lambda$ is diagonal with some elements $\lambda_1,\dots,\lambda_m$ in
the diagonal. The matrix $\Sigma$ is positive semi-definite if and only
if all elements $\lambda_j$, $1\le j\le m$, in the above representation
are non-negative. If $\Sigma=U\Lambda U^*$ is a positive semi-definite
matrix, then let us define the symmetric matrix $B=U\sqrt \Lambda U^*$,
where $\sqrt \Lambda$ is the diagonal matrix with elements $\sqrt
{\lambda_j}$, $j=1,\dots,m$, in the diagonal. Then $\Sigma=B^2=B^*B$.)
\item{} Let $\Sigma=(D_{j,k})$, $1\le j,k\le m$, be an arbitrary
$m\times m$ positive semi-definite matrix, and
$\bold M=(M_1,\dots,M_m)\in R^m$ a vector in the Euclidean
space~$R^m$. Let $\bold\xi=(\xi_1,\dots,\xi_m)$ be an $m$-dimensional
random variable with standard normal distribution, $B=(b_{j,k})$, $1\le
j,k\le m$, an $m\times m$ matrix such that $B^*B=\Sigma$. Let us define
the $m$-dimensional random vector $\bold \eta=(\eta_1,\dots,\eta_m)
=\bold \xi B+\bold M$. Then $\bold \eta$ has normal distribution, and
we claim that it has expected value $\bold M$ and covariance matrix
$B^*B=\Sigma$. This implies that for all vectors $\bold M\in R^m$
and $m\times m$ positive semi-definite matrices $\Sigma$ there exists
an $m$-dimensional normally distributed random vector with expectation
zero and covariance matrix $\Sigma$.
\item{} Indeed, $E\bold \eta=(E\eta_1,\dots,E\eta_m)=(M_1,\dots,
M_m)=\bold M$, and the elements of the covariance matrix of $\eta$
can be calculated in the following way.
$$
E(\eta_j-E\eta_j)(\eta_k-E\eta_k)=E\(\summ_{l=1}^m
b_{j,l}\xi_l\)\(\summ_{p=1}^m b_{k,p}\xi_k\)=\summ_{l=1}^p
b_{j,l}b_{k,l}=D_{j,k}
$$
for all indices $1\le j,k\le m$, because $E\xi_l\xi_p=0$, if $l\neq p$,
$E\xi_l^2=1$. The last identity means the identity $B^*B=\Sigma$ in
coordinate form.
\item{} Let us consider an $m$-dimensional random vector
$\bold\eta=(\eta_1,\dots,\eta_m)=\bold\xi
B+\bold M$ with normal distribution function, where the random vector
$\bold\xi=(\xi_1,\dots,\xi_m)$ has standard normal distribution,
$B$ is an $m\times m$ matrix, $\bold M=(M_1,\dots,M_m)\in R^m$.
Put $B^*B=\Sigma=(D_{j,k})$, $1\le j,k\le m$. Let us calculate the
characteristic function $\varphi(t_1,\dots,t_m)
=Ee^{i(t_1\eta_1+\cdots+t_m\eta_m)}$ $\bold\eta$ of the random vector
$\bold \eta$. To calculate it let us introduce the random variable
$\zeta=t_1\eta_1+\cdots+t_m\eta_m$. Then $\zeta$ is a normally
distributed random variable with expected value $\bar M=\bar
M(t_1,\dots,t_m)=\summ_{k=1}^m t_kM_k=(\bold M,\bold t)$ and variance
$\sigma^2=\summ_{j=1}^m\summ_{k=1}^m t_jt_kE\eta_j\eta_k
=\summ_{j=1}^m\summ_{k=1}^m t_j t_kD_{j,k}=\bold t \Sigma \bold t^*$,
where $\bold t=(t_1,\dots,t_m)$. Hence the characteristic function of
the random variable $\zeta$ with normal distribution equals
$\psi(u)=Ee^{iu\zeta}=e^{-u^2 \bold t\Sigma t/2+i(\bold M,t) u}$. This
implies that $\varphi(t_1,\dots,t_m)=Ee^{i(t_1\eta_1+\cdots+t_m\eta_m)}
=\psi(1) =e^{-\bold t\Sigma \bold t^*/2+i(\bold M,\bold t)}$, that is
relation~(13) holds.
\item{} It follows from formula~(13) that if $\bold \eta$ is an
$m$-dimensional random variable with normal distribution, then the
characteristic function and as a consequence, the distribution function
of this random vector is determined by its expectation and covariance
matrix. Let us remark that there can be given two different
$m\times m$ matrices $B_1$ and $B_2$ such that $B_1^*B_1=B_2^*B_2$.
Let $\bold \xi$ be an $m$-dimensional random vector with standard
normal distribution, $B_1$ and $B_2$ two $m\times m$ matrices such that
$B_1^*B_1=B_2^*B_2$, an $\bold M\in R^m$ an arbitrary vector.
Let us define the random vectors $\eta_1=\bold \xi B_1+\bold M$ and
$\eta_2=\bold \xi B_2+\bold M$. Then the expectation and covariance
matrix, hence the distribution function of the random vectors $\eta_1$
and $\eta_2$ agree, although this statement is not self-evident because
of the relation $B_1\neq B_2$.
\item{48.)} We get similarly to the proof of formula~(13) in problem~47.,
that the characteristic function of the random vector
$\eta=(\eta_1,\dots,\eta_l)$ equals 
$Ee^{i(\bold t,\eta)}=e^{-\bold t\Sigma\bold t^*/2+i(\bold M,\bold t)}$,
where $\bold t=(t_1,\dots,t_l)$, and $\Sigma=B^*B$. It follows
from this formula that $\eta$ is a normally distributed random vector
with covariance matrix $\Sigma$ and expectation vector $\bold M$.
\item{} Given a normally distributed random vector $\eta$ 
of dimension~$m$, let us write it in the form $\eta=\xi B+\bold M$,
where $\xi$ is a random vector of dimension $m$ with standard
normal distribution. (It can be proved that such a representation
is always possible. Actually it would be enough for us such a
representation of a random vector with the same distribution as
$\eta$. The possibility of such a representation follows from
the definition of the normally distributed random vectors.)
If we omit some of the coordinates of the random vector $\eta$
and we preserve only $l$ coordinates, then the random vector
$\eta'$ we obtain after this delition of coordinates can be 
presented in the following way. Let us omit those rows of the
vector $B$ which have the same indices as the elements of $\eta$
we omitted. Similarly let us omit the coordinates of the vector
$\bold M$ with the same indices as the coordines omitted from
$\eta$. Let us denote the matrix and vector obtained in such a
way by $\bold M'$ and $B'$. Then we have $\eta'=\xi B'+\bold M'$, 
hence $\eta'$ is normally distributed by the already proved
part of the problem.
\item{49.)} I shall give two different solutions of the problem,
because, this may be instructive. 
\item{} First soulution: We can write up the characteristic
function  of the random vector $\eta$, as 
$Ee^{i(\bold t,\eta)}=e^{-\bold t\Sigma\bold t^*/2+i(\bold M,\bold t)}$ 
where $\bold t=(t_1,\dots,t_m)$,
and $\bold M=(M_1,\dots, M_m)$ is the expected value of $\eta$, 
with the introduction of some notations in the following way. 
Let $\bold t_j$ denote the restriction of the vector $\bold t$, 
and let $\bold M_j$ denote the restriction of vector $\bold M$ 
to the coordinates $p\in L_j$, $1\le j\le k$. Let us define 
similarly the matrix $\Sigma_j$ as the restriction 
of the matrix $\Sigma$ to the coordinates  $\sigma_{p,q}$, 
$p\in L_j$ and $q\in L_j$, $1\le j\le k$. With such a notation
$Ee^{i(\bold t,\eta)}
=\prodd_{j=1}^k e^{-\bold t_j\Sigma_j\bold t_j^*/2+i(\bold M_j,\bold t_j)}$.
(We exploited the properties of the matix $\Sigma$ at this point.)
Let $\eta_1',\dots,\eta_k'$ independent, normally distributed
random vectors with covariance matrix $\Sigma_j$ and expected
value $\bold M_j$, $1\le j\le k$. The characteristic function of
E$\eta'_j$ equals $Ee^{i(\bold t_j,\eta'_j)}
=e^{-\bold t_j\Sigma_j\bold t_j^*/2+i(\bold M_j,t_j)}$
for all indices $1\le j\le k$. Hence the characteristic function
of the random vectors $\eta'=(\eta'_1,\dots,\eta'_k)$ and $\eta$ 
agree. Hence the distribution of $\eta$ and $\eta'$ also agree,
and as a consequence the random vectors $\bar\eta_j$, 
$1\le j\le k$, are independent of each other, similarly to the
random vectors $\eta'_j$.
\item{} Second solution: Let us apply the notations introduced
in the previous solution. The random vector $\eta'$ defined there
is normally distributed with covariance matrix $\Sigma$ and
expected value $\bold M$. As the distribution of a normal random
vector is determined by its covariance matrix and expected value,
hence the distribution of $\eta$ \'es $\eta'$ agree. Therefore
the random vectors $\eta_1'$,\dots, $\eta_k'$ are independent, 
similarly to the random vectors $\bar\eta_1$,\dots, $\bar\eta_k$.
\item{50.)} Because of the result of problem~46 it is enough to show
that for all vectors $\bold a=(a_1,\dots,a_m)\in R^m$ the normalized
partial sums $\frac1{A_n}S_n=\frac1{A_n}S_n(a_1,\dots,a_m)
=\frac1{A_n}\summ_{p=1}^m a_pS_{p,n}$, $n=1,2,\dots$ converge in
distribution to the normal distribution with expectation zero and
variance $\sigma^2=\bold a\Sigma\bold a^*$ if $n\to\infty$. Let us
observe that $\frac1{A_n}S_n=\frac1{A_n}\summ_{k=1}^n \eta_k$,
$n=1,2,\dots$, where $\eta_k=\summ_{p=1}^m a_p\xi_{p,k}$, $k=1,2,\dots$.
Besides, the random variables $\eta_k$, $k=1,2,\dots$ are
independent, $E\eta_k=0$, $E\eta_k^2=\bold a\Sigma_k \bold a^*$.
\item{} Let us consider separately the cases $\bold a\Sigma\bold a^*=0$
and $\bold a\Sigma\bold a^*>0$. If $\bold a\Sigma\bold a^*=0$, then
$\frac1{A_n^2}ES_n^2=\frac1{A_k^2}\summ_{j=1}^k \bold a\Sigma_j\bold
a^*\to0$ if $k\to0$, hence $\frac{S_n}{A_n}$ converges to zero
stochastically if $n\to\infty$. That is, in this case the random
variables $\frac1{A_n}S_n$ converge in distribution to the
(degenerated) normal variable with expectation zero and variance
zero. In the case $\bold a\Sigma\bold a^*>0$ let us define the
triangular array $\eta_{k,n}=\frac{\eta_k}{A_n\sqrt{\bold a\Sigma\bold
a^*}}$, $1\le k\le n$, $n=1,2,\dots$. Then $\summ_{k=1}^n
E\eta_{k,n}^2=\frac1{A_n^2}\summ_{k=1}^n \frac{\bold a\Sigma_k\bold
a^* }{\bold a\Sigma\bold a^*}\to1$ if $n\to\infty$. Hence in the case
$\bold a\Sigma\bold a^*>0$ the convergence of the normalized partial
sums $\frac1{A_n}S_n$, $n=1,2,\dots$, to the normal distribution with
expectation zero and variance $\bold a\Sigma \bold a^*$ follows from
the central limit theorem for triangular arrays formulated in
problem~38 if we show that the above defined triangular array
$\eta_{k,n}$, $1\le k\le n$, $n=1,2,\dots$, satisfies the Lindeberg
condition. This implies the statement of the problem. To prove 
the Lindeberg condition first we show the following inequality~(2.8).
Here we shall apply the notation $K=\max\limits_{1\le p\le m}|a_p|$. 
$$
E(\eta_{k,n}^2I(|\eta_{k,n}|>\e)\le\frac{K^2m^2}
{A_n^2\bold a\Sigma \bold a^*}
\sum_{p=1}^m E \xi_{p,k}^2I(|\xi_{p,k}|>\bar\e A_n) \tag2.8
$$
for all indices $1\le k\le n$ and $n=1,2,\dots$, where
$\bar\e=\bar \e(k)=\frac{\e}{m\supp_{1\le p\le m}|a_p|}\cdot
\frac1{\sqrt{\bold a\Sigma\bold a^*}}$. By summing up these
inequalities for all indices $1\le k\le n$, and applying 
formula~(14) for all $1\le p\le m$, we get that the triangular 
array $\eta_{k,n}$, $1\le k\le n$, $n=1,2,\dots$ satisfies the
Lindeberg condition. 
\item{} To prove formula~(2.8) let us introduce the random
index $\bar p(k)=\bar p(k,\oo)$ which is the (smallest) number
$p$ such that
$|\xi_{\bar p(k),k}(\oo)|=\max\limits_{1\le p\le m}|\xi_{p,k}(\oo)|$.
Let us observe that
$$
\{\oo\colon\; |\eta_{k,n}(\oo)|>\e\}\subset\bigcupp_{p=1}^M\{\oo\colon\;
|\xi_{p,k}(\oo)| >\bar\e A_n\},
$$
hence
$I(|\eta_k(\oo)|>\e)\le I(|\xi_{\bar p(k,\oo),k}(\oo)|>\bar\e A_n)$.
Besides, $|\eta_k(\oo)|^2\le 
\frac{m^2K^2}{A_n^2\bold a\Sigma \bold a^*}\xi_{\bar p(k),k}^2$.
This implies that
$$
\align
\eta_k^2I(|\eta_k|>\e)
&\le \frac{m^2K^2}{A_n^2\bold a\Sigma\bold a^*}\xi_{\bar p(k),k}^2
I(|\xi_{\bar p(k,\oo),k}(\oo)|>\bar\e A_n) \\ 
&\le\frac{ m^2K^2}{A_n^2 \bold a\Sigma \bold a^*}
 \sum_{p=1}^m\xi_{p,k}^2 I(|\xi_{p,k}|>\bar\e A_n).
\endalign
$$
By taking expection in this inequality we get formula~(2.8).
\item{} Let us finally observe that if
$\bold\xi_k=(\xi_{1,k},\dots,\xi_{m,k})$, $k=1,2,\dots$, is a sequence
of independent and identically distributed $m$-dimensional random
vectors with expectation zero and finite covariance matrix
$\Sigma$, then this sequence of random vectors satisfies relation
(14) with the choice $A_n^2=n$. This statement was proved in part~b) of
problem~41 for those coordinates $p$ for which $E\xi_{p,1}^2>0$. For
those coordinates~$p$ for which $E\xi_{p,1}^2=0$, $\xi_{p,1}\equiv0$.
Hence these coordinates can be omitted.
\item{51.)} Let us consider first those numbers $p$, $1\le p\le m$ for
which the $p$-th element of the diagonal of the (semi-definite) matrix
$\Sigma$ $D_{p,p}$ satisfies the inequality $D_{p,p}>0$. Let us define
the triangular array $\eta_{k,j}=\eta_{k,j}(p)=\frac{\xi_{p,j}}{A_k
D_{p,p}}$, $1\le j\le k$, $k=1,2,\dots$. Then
$\limm_{k\to\infty}E\eta_{k,j}^2=0$, and the triangular array
$\eta_{k,j}$, $1\le j\le k$, $k=1,2,\dots$ satisfies the condition of
uniform smallness and the central limit theorem. Hence by the result
of problem~39 the Lindeberg condition formulated in formula~(14) holds
for all such indices~$p$.
\item{} Since the matrix $\Sigma$ is positive semi-definite, hence
$D_{p,p}\ge0$ for all $1\le p\le m$. Therefore we have still consider
those indices $p$ for which $D_{p,p}=0$. In this case
$\limm_{n\to\infty}\frac1{A_n^2}\summ_{k=1}^nE\xi_{p,k}^2=0$. Since
$E\xi_{p,k}^2\ge E\xi_{p,k}^2I(|\xi_{p,k}|>\e A_n)$, relation~(14) also
holds for such indices~$p$.
 
%\vfill\eject
 
\beginsection Appendix
 
{\bf The proof of the inversion formula for Fourier transforms.}

\medskip\noindent
Let us introduce the function $\hat f(u)=\int e^{-itu}\tilde f(u)\,du$.
To prove formula~(6) we have to show that $\hat f(u)=f(u)$ for almost
all numbers $u\in R^1$. This statement is equivalent to the identity
$\int_0^t f(u)\,du=\int_0^t \hat f(u)\,du$ for all numbers~$t\in R^1$.
Since
$$
\int_0^t \hat f(u)\,du=\frac1{2\pi}\int_{-\infty}^\infty \int_0^t
e^{-ius}\tilde f(u)\,ds\,du= \frac1{2\pi}\int_{-\infty}^\infty
\frac{e^{-itu}-1}{-itu} \tilde f(u)\,du, \tag A1
$$
we have to show the identity
$$
\int_{-\infty}^\infty  I_{[0,t]}(u)f(u)\,du=\frac1{2\pi}
\int_{-\infty}^\infty \frac{e^{-itu}-1}{-itu} \tilde f(u)\,du,  \tag A2
$$
where $I_{[0,t]}(\cdot)$ is the indicator function of the interval
$[0,t]$. The identity (A2) is a special case of an important identity of
the Fourier analysis, of the Parseval formula. Let us formulate it.

\medskip\noindent
{\bf Parseval formula.} {\it
$$
\int f(u)\bar g(u)\,du=\frac1{2\pi}\int \tilde f(u)\bar{\tilde
g}(u)\,du, \tag A3
$$
where $\tilde f(\cdot)$ denotes the Fourier transform and $\bar
f(\cdot)$ the conjugate of a function $f(\cdot)$. Formula~(A3) holds
if one of the following conditions is satisfied: 

\medskip
\item{a.)} Both functions $f$ and $g$ are square integrable.
\item{b.)} Both functions $\tilde f$ and $\tilde g$ are square
integrable.

\medskip
If one of the conditions a.) and b.) is satisfied, then also the other
condition holds. In this case the identity $\int
|f(u)|^2\,du=\frac1{2\pi}\int|\tilde f(u)|^2,du$ holds
(because of the Parseval formula). The transformation
$\bold Tf=f\to\frac1{\sqrt{2\pi}}\tilde f$ is an automorphism in the
space of square integrable function. This statement means not only the
validity of the identity $\int\bold Tf(u)\overline {\bold T
g}(u)\,du=\int f(u)\bar g(u)\,du$. It also states that all square
integrable functions $f$ can be represented in the form $f=\bold T h$
with a square integrable function~$h$.)
 
Let us finally remark that in a complete formulation of the Parseval
formula the notion of the Fourier transform has to be defined for all
square integrable but not necessarily integrable functions $f(\cdot)$.
This definition can be given by means of the above mentioned $L_2$
isomorphism. For all square integrable functions $f(\cdot)$ there exists
a sequence of integrable and square integrable functions $f_n(\cdot)$
which converges to the function $f(\cdot)$ in the $L_2$ norm of square
integrable functions, i.e.\ $\int |f_n(u)-f(u)|^2\,du\to0$ if
$n\to\infty$. Then the Fourier transform $\tilde f(\cdot)$ of the
function $f(\cdot)$ is the limit of the functions $\tilde f_n(\cdot)$
in the $L_2$ norm. This limit always exists, and it does not depend
on the choice of the sequence of functions $f_n(\cdot)$ converging to
the function $f(\cdot)$.} 

\medskip
In the Parseval formula formulated in this text a norming factor
$\frac1{2\pi}$ is present which does not appear in its formulation
in text books. The reason of this difference is that we have chosen
a different normalization in the definition of the Fourier transform.
(We have omitted the factor $\frac1{\sqrt{2\pi}}$ from the
definition.)
 
If the Fourier transform of an integrable function $f$ is integrable,
then it is also square integrable, since it is a bounded function.
Further, the Fourier transform of the function $g(u)=I_{[0,t]}(u)$ is
the square integrable function $\tilde g(v)=\int_0^t e^{iuv}\,du=
\frac{e^{iv}-1}{iv}$. Hence the formula (A2) (and therefore also
formula~(6)) is a consequence of the Parseval formula with the choice
of the above functions $f(\cdot)$ and $g(\cdot)$.
 
We can prove that a finite measure $\mu$ with an integrable Fourier
transform $\tilde f(u)=\int e^{itu}\mu(\,dt)$ has a density function
$f(\cdot)$ defined by formula~(6) with the help of the following
smoothing argument. Let us consider for all numbers $\e>0$ the
Gaussian measure $\nu_\e$ with expectation zero and variance
$\e$. This measure has density function $\varphi_\e(u)
=\frac1{\sqrt{2\pi\e}}e^{-u^2/2\e}$ and Fourier transform $e^{-\e
u^2/2}$. Let us introduce the convolution $\mu_\e=\mu*\nu_\e$,
i.e.\ $\mu_\e(A)=\mu*\nu_\e(A)=\int \mu(A-u)\varphi_\e(u)\,du$.
 
The measure $\mu_\e$ has a density function $f_\e(u)=\int
\varphi_\e(u-v)\mu(\,dv)$, and the Fourier transform of this measure
is the integrable function $\tilde f_\e(u)=e^{-\e u^2/2}\tilde f(u)$.
Hence the function $f_\e(u)$ can be expressed as the
inverse Fourier transform of the function $\tilde f_\e(u)$ defined in
formula~(6). If $\e\to0$, then $f_\e(u)\to f(u)$, where $f(u)$ is the
function defined in formula~(6), and this convergence is uniform in
the variable~$u$. On the other hand, the measure $\mu$ is the weak
limit of the measures $\mu_\e$ if $\e\to0$, that is the probability
measures $\frac {\mu_\e}{\mu(R^1)}$ converge weakly to the
probability measure $\frac {\mu}{\mu(R^1)}$. (Let us remark that
$\mu(R^1)=\mu_\e(R^1)$.) Hence we get by taking limit $\e\to0$ that
$\mu((a,b])=\int_a^b f(u)\,du$ if $\mu(\{a\})=\mu(\{b\})=0$, i.e.\ if
the points $a$ and $b$ are points of continuity of the measure~$\mu$.
This implies that the function $f$ is the density function of the
measure~$\mu$.
 
By some slight modification of the above argument we can prove
the above statement also in the case if $\mu$ is a signed
measure with bounded variation and integrable Fourier transform.
Let us remark that by refining the argument of the above limit
procedure and exploiting the~$L_2$ isomorphism property of the Fourier
transform, the above result about the density function of a measure
$\mu$ and its Fourier transform can be strengthened. It is enough to
assume that the Fourier transform of the (signed) measure
$\mu$ is square integrable. But in this case the inverse Fourier
transform defined in formula~(6) has to be defined by means of
the $L_2$ extension of the original integral, and we cannot claim
that the density function of the measure~$\mu$ is continuous.
 
\medskip\noindent
{\it The proof of the Parseval formula.}\/ Let us first prove the
Parseval formula for such simple pairs of functions $(f,g)$
which disappear outside of a finite interval $[-A,A]$, and which are
sufficiently smooth, say they have two continuous derivatives. Then
by considering the restriction of these functions to some interval
$[-\pi T,\pi T]\supset[-A,A]$, $T\pi\ge A$, and the discrete 
version of the Parseval formula we can write that
$$
\int f(u)g(u)\,du=2\pi T \sum_{k=-\infty}^\infty a_k(T)\bar b_k(T),
$$
where $a_k(T)=\frac1{2\pi T} \int e^{iku/T}f(u)\,du=
\frac1{2\pi T} \tilde f\(\frac kT\)$, and $b_k(T)=
\frac1{2\pi T} \tilde g\(\frac kT\)$. But the above expression
 $2\pi T \summ_{k=-\infty}^\infty a_k(T)\bar b_k(T)$ is an
approximating sum of the integral $\int\tilde f(u)\bar {\tilde
g}(u)\,du$, and the Fourier transforms $\tilde f(u)$ and $\tilde g(u)$
tend to zero fast as $|u|\to\infty$ because of the smoothness of the
functions $f$ and $g$. (See for instance the result of problem~28.)
Hence the limit procedure $T\to\infty$ yields formula~(A3) in this
special case.
 
The Parseval formula yields with the choice $f=g$ the identity
$\int |f(u)|^2\,du=\frac1{2\pi}\int |\tilde f(u)|^2\,du$, further
the functions $f$ for which we have proved these identity are
everywhere dense in the space of square integrable functions. Hence
we get proof of the Parseval formula by extending the isometry
$\bold T\colon\; f\to \bold T f=\frac 1{\sqrt {2\pi}}\tilde f$ in the
$L_2$ norm to the space of all square integrable functions. To
complete the proof we still have to show that this  extension of the
transformation $\bold T$ maps to the whole space of square integrable
functions.
 
To prove this missing part let us consider those functions $f$ which
are sufficiently smooth (say they are 10-times differentiable) and
to zero sufficiently fast in plus minus infinity (say $|f(u)|\le
\const \(1+|u|^{100}\)$). Since such functions constitute an
everywhere dense set in the space of square integrable functions it
is enough to show that they are in the image space of the operator
$\bold T$. We will show that the identity $\bold T \sqrt{2\pi}
\tilde{f^-}=f$, where $f^-(u)=f(-u)$, follows from the already proved
statements.
 
Also the function $\tilde f$ is smooth, and it tends to zero
fast. (This also follows from the statements of problems~27 and~28.
Actually the statement of problem~27 deals only with the Fourier
transform of probability measures, but it is not difficult to see
that this statement also holds for the Fourier transform of all signed
measures with bounded variance. We want to exploit that the measures
$\mu^{\pm}$, $\mu^\pm(A)=\int_A f^\pm(u)\,du$, $f^+(u)=\max(f(u),0)$,
$f^-(u)=-\min(f(u),0)$ have at least 8~moments.) Since both the
functions $f(u)$ and the indicator function $I_{[0,t]}(u)$ of the
interval $[0,t]$ are square integrable, formula~(A2) holds by the
already proved part of the Parseval formula. Since the faction $\hat
f$ defined with the help of the function $\tilde f(u)$ at the start of
this Appendix is integrable, also formula (A1) holds. Formulas~(A1)
and~(A2) together imply that the pair of functions $(f,\tilde f)$
satisfy relation~(6). But this relation is equivalent to the statement
that the function $f$ is the Fourier transform of the function $2\pi
\tilde {f^-}$, where $f^-(u)=f(-u)$, and this is what we wanted to
prove.
 
%\vfill\eject
 
\medskip\noindent
{\bf The proof of Weierstrass second approximation theorem.}

\medskip
The functions $\frac1{(2\pi)^{k/2}} e^{i(j_1t_1+\cdots+j_kt_k)}$
constitute a complete orthonormal system in the space of square
integrable functions which are periodic by $2\pi$ in all their
arguments. This important result of the theory of Fourier series
implies that all sufficiently smooth and in all their arguments
periodic functions are the uniform limits of their Fourier series.
Indeed, in this case the Fourier coefficients tend to zero fast,
and this implies the uniform convergence. Since such functions are
everywhere dense in the supremum norm in the space of continuous
functions, this statement implies Weierstrass second approximation
theorem. Nevertheless, instead of this argument we present a
direct proof of Weierstrass second approximation theorem which
does not apply the completeness of the trigonometrical functions in
the $L_2$ space. We shall prove Fej\'er's theorem, more precisely its
multi-dimensional version. Weierstrass second approximation theorem
is a direct consequence of this result. 

\medskip\noindent
{\bf Fej\'er's theorem.} {\it Let $f(x_1,\dots,x_k)$ be a continuous
function of $k$~arguments which is periodic by $2\pi$ in all of its
arguments. Let us define for all $k$-dimensional vectors
$(n_1,\dots,n_k)$ with non-negative integers the trigonometrical sum
$$
\align
s_{n_1,\dots,n_k}(f)(t_1,\dots,t_k)&=\sum_{j_1=-n_1}^{n_1}\cdots
\sum_{j_k=-n_k}^{n_k} A_{j_1,\dots, j_k}e^{i(j_1t_1+\cdots+j_kt_k)},\\
\intertext{where }
A_{j_1,\dots,j_k}&=\frac1{(2\pi)^k}\int_{-\pi}^\pi \cdots\int_{-\pi}^\pi
e^{-i(j_1u_1+\cdots+j_ku_k)} f(u_1,\dots,u_k)\,du_1\dots\,du_k.
\endalign
$$
Let us also consider the following Cesaro means $A_n(f)$,
$n=1,2,\dots$, of the above trigonometrical sums:
$$
A_n(f)(t_1,\dots,t_k)=\frac1{(n+1)^k}\sum \Sb 0\le n_j\le n\\
\text{for all indices }1\le j\le k \endSb
s_{n_1,\dots,n_k}(f)(t_1,\dots,t_k).
$$
Then $\limm_{n\to\infty}A_n(f)(t_1,\dots,t_k)=f(t_1,\dots,t_k)$, and the
above convergence is uniform in all of its arguments
$t_1,\dots,t_k$.} 

\medskip\noindent
{\it The proof of  Fej\'er's theorem.}\/ The proof of Fej\'er's theorem
is based on the following formula:
$$
A_n(f)(t_1,\dots,t_k)=\int_{-\pi}^\pi\cdots\int_{-\pi}^\pi
f(u_1,\dots,u_k)\bar K_n(t_1-u_1,\dots,t_k-u_k)\,du_1\dots\,du_k,
\tag A4
$$
where
$$
\bar K_n(u_1,\dots,u_k)=K_n(u_1)\cdots K_n(u_k), \tag A5
$$
and
$$
K_n(u)=\frac1{2\pi(n+1)}\sum_{k=-n}^n(n+1-|k|)e^{iuk}
=\frac{\sin^2\(\frac{n+1}2u\)} {2\pi(n+1)\sin^2\(\frac u2\)}. \tag A$5'$
$$
These relations hold, since by writing into the definitions of the
expressions $A_n(f)$ and $s_{n_1,\dots,n_k}(f)$ the definition of the
Fourier coefficient $A_{j_1,\dots,j_k}$ we get relation (A4) together
with formula (A5), where
$$
\align
\bar K_n(u_1,\dots,u_k)&=
\frac1{(2\pi(n+1))^k}\sum \Sb 0\le n_j\le n\\
\text{for all indices }1\le j\le k \endSb \sum \Sb |m_j|\le
n_j\\ \text{for all indices }1\le j\le k\endSb
e^{i(m_1u_1+\dots+m_ku_k)} \\
&=\frac1{(2\pi(n+1))^k}\prod_{j=1}^k \(
\sum_{n_j=0}^n \sum_{m_j=-n_j}^{n_j} e^{im_j u_j}\)=K_n(u_1)\dots
K_n(u_k),
\endalign
$$
and the function $K_n(u)$ is defined in the middle term of
formula~(A$5'$). This sum can be written in a closed form with the
help of the following calculation.
$$
\align
&\frac1{2\pi(n+1)}\sum_{k=-n}^n(n+1-|k|)e^{iuk}=
\frac1{2\pi(n+1)}\(\sum_{k=0}^n e^{iuk}\) \(\sum_{k=0}^n e^{-iuk}\)\\
&\qquad=\frac{1}{2\pi(n+1)}\left|\frac{e^{i(n+1)u}-1}
{e^{iu}-1}\right|^2 =\frac1{2\pi(n+1)}
\frac{\left|e^{i(n+1)u/2}-e^{-i(n+1)u/2}\right|^2}
{\left|e^{iu/2}-e^{-iu/2}\right|^2}\\
&\qquad=\frac{\sin^2\(\frac{n+1}2u\)} {2\pi(n+1)\sin^2\(\frac u2\)}.
\endalign
$$
The function $K_n(u)$ defined in formula (A$5'$) have the following
properties important for us:

\medskip
\item{(i)} $\int_{-\pi}^\pi K_n(u)\,du=1$. This statement follows from
the representation of the functions $K_n(\cdot)$ in the form of a sum.
\item{(ii)} $K_n(u)\ge0$ for all numbers $u\in R^1$.
\item{(iii)} $\limm_{n\to\infty}\supp_{\e\le |u|\le \pi}K_n(u)=0$ for
all numbers $\e>0$.
 
Statements (ii) and (iii) follow from the representation of the
functions $K_n(\cdot)$ given in a closed form.
 
Since a function continuous on a compact set is uniformly continuous,
there exists a constant $\delta=\delta(\e,f)$ for all $\e>0$ such that
the continuous and in its coordinates periodic function $f$ satisfies
the inequality $|f(x_1,\dots,x_k)-f(y_1,\dots,y_k)|<\e$ if
$|x_j-y_j|<\delta$ for all indices $j=1,\dots,k$. (In this
relation we identify the points $x_j+2\pi l$, $l=0\pm1,\pm2,\dots$,
and the inequality $|x_j-y_j|<\delta$ means that $|x_j-y_j+2\pi
l_l|<\delta$ with an appropriate integer $l_j$.) Let us introduce the
notation $\bold B(\delta,(t_1,\dots,t_k))=\{(u_1,\dots,u_k)\colon\;
|u_j-t_j|<\delta,\;-\pi\le  u_j<\pi,\; j=1,\dots,k\}$. Because of
property~(i)
$$
\align
&A_n(f)(t_1,\dots,t_k)-f(t_1,\dots,t_k)\\
&\qquad =\int_{[-\pi,\pi)^k}
\(f(u_1,\dots,u_k)-f(t_1,\dots,t_k)\)
K_n(t_1-u_1)\cdots K_n(t_k-u_k)\,du_1\dots\,du_k \\
&\qquad=\int_ {\bold B(\delta,(t_1,\dots,t_k))}
[\cdots]\,du_1\dots\,du_k + \int_ {[-\pi,\pi)^k\setminus\bold
B(\delta,(t_1,\dots,t_k))} [\cdots]\,du_1\dots\,du_k \\
&\qquad =I_{1,n}(t_1,\dots,t_k)+I_{2,n}(t_1,\dots,t_k).
\endalign
$$
It follows from the definition of the set $B(\delta,(t_1,\dots,u_t))$,
the number $\delta$ and properties~ (i) and~(ii) that
$$
\left|I_{1,n}(t_1,\dots,t_k)\right| \le \e\int_{[-\pi,\pi)^k}
K_n(t_1-u_1)\cdots K_n(t_k-u_k)\,du_1\dots\,du_k \le\e
$$
for all indices $n=1,2,\dots$ and points $(t_1,\dots,t_k)$.
On the other hand, by applying the notation
$\supp_{(u_1,\dots,u_k)}|f(u_1,\dots,u_k)|=
L$ and carrying out the substitutions $t_j-u_j=\bar u_j$
we can show with the help of relations (i), (ii) and~(iii) that
$$
\left|I_{2,n}(t_1,\dots,t_k)\right|\le 2L
\int_ {[-\pi,\pi)^k\setminus\bold
B(\delta,(0,\dots,0))} K_n(\bar u_1)\cdots K_n(\bar
u_k)\,d\bar u_1\dots\,d\bar u_k\to0,
$$
if $n\to\infty$, since
$$
\align
&\int\limits\Sb\delta<|\bar u_j|<\pi\\ -\pi\le\bar u_l<\pi,\,l\neq
j,\,1\le l\le k\endSb
K_n(\bar u_1)\cdots K_n(\bar u_k)\,d\bar u_1\dots\,d\bar
u_k=\int_{\delta<|u|<\pi} K_n(u)\,du\\
&\qquad\qquad \le 2\pi\sup_{\delta<|u|<\pi} K_n(u)\to0,
\quad \text {ha } n\to\infty.
\endalign
$$
for all numbers $1\le j\le k$. Since the above estimates hold for all
constants $\e>0$ (together with an appropriate number
$\delta=\delta(\e,f)$), they imply Fej\'er's theorem.
 
 \bye
