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\centerline{{\bf Limit theorems and infinitely divisible
distributions.}  {\rm Part I.}}
\smallskip
\centerline{\it by P\'eter Major}
\centerline{The Mathematical Institute of the Hungarian Academy of
Sciences} 

\medskip\noindent
{\narrower {\narrower
{\it Summary:}\/ In this work we discuss when the appropriately
normalized partial sums of independent random variables or more
generally the sums of the random variables in a triangular array
have a limit distribution and also describe the limit.
The first part is of introductory character. Here we introduce the
most important notions, formulate the questions we are interested in
and recall some classical results. We show how the so-called
infinitely distributed random variables, the random variables whose
distributions are the natural (and right) candidates for the limit
distribution in these limit theorems, can be constructed as the
(regularized) sums of the elements of a Poisson process. We also show
that infinitely divisible distributions can be characterized by means
of the L\'evy--Hinchin formula. At the end, beside the construction of
infinitely divisible distribution we also construct infinitely
divisible processes with nice trajectories. The first part also has
an Appendix which contains some useful results like a simple
construction of a Poisson process and limit theorems with
Poissonian limit.
 
The second part of this work contains the proof of the necessary
and sufficient condition for the existence of a limit distribution
in the problem mentioned above together with some interesting
consequences of this result. The third part contains
the functional limit theorem version of this result. \par} \par}
 
\bigskip\noindent
{\bf 1. Introduction.}

\medskip\noindent
One of the basic questions of the probability theory is the
following problem:  

\medskip
Let  $\xi_1, \xi_2,\dots$, be a sequence of independent random
variables, and let $S_n=\summ_{k=1}^n\xi_k$, $n=1,2,\dots$, be the
sequence of partial sums made from these random variables. Let us
consider the normalized partial sums $\frac{S_n-A_n}{B_n}$ with an
appropriate normalization. When do the distributions of these
normalized partial sums behave  for large $n$ similarly to each other,
i.e.\ when do these normalized partial sums have a limit distribution
if $n\to\infty$? How should we choose the norming constants $A_n$ and
$B_n$? What kind of distributions appear as a limit distribution?
 
The same question arises in a natural way if
$\xi_1, \xi_2,\dots$, is a sequence of independent and
{\it identically distributed}\/ random variables. We are interested
in the question whether this additional restriction modifies the
possible normalizations and the set of limit distributions. A natural
modification of the problem is the following question formulated
about triangular array. Before its formulation let us first recall
the notion of triangular arrays.
 
\vfill\eject
 
\headline{\ifodd\pageno \hfill {\it Limit theorems and infinitely
divisible distributions.} {\rm Part I.} \hfill \else
\hfill {\it P\'eter Major} \hfill \fi}
 
 
\noindent {\bf The definition of triangular arrays.} {\it The set of
random variables
$$
\align
&\xi_{1,1},\dots,\xi_{1,n_1}\\
&\vdots\qquad \quad \;\vdots   \\
&\xi_{k,1},\dots,\xi_{k,n_k}\\
&\vdots\qquad \quad \;\vdots
\endalign
$$
$k\to\infty$, is a triangular array if the random variables
$\xi_{k,1},\dots,\xi_{k,n_k}$ in a row are independent.
(We assume nothing about the relation among random variables in
different rows.)} 

\medskip
Let $S_k=\summ_{j=1}^{n_k}\xi_{k,j}$, where $\xi_{k,j}$, $1\le j\le
n_k$, $k=1,2,\dots$ be a triangular array. We are interested in the
question what kind of limit theorems can the sums
$S_k$ or (normalized) sums $S_k-A_k$ satisfy. What kind of limit
theorems can appear if the random variables in a fixed row of the
triangular array are not only independent but also identically
distributed?
 
The following relation can be established between the investigation  of
the limit theorems for normalized partial sums of
independent random variables and limit theorems for the sums of the
random variables in a row of a triangular array.
 
Let $\xi_1,\xi_2,\dots$ be a sequence of independent (and possibly
identically distributed) random variables. Let us define the
triangular array $\xi_{k,j}=\frac{\xi_j}{B_k}$, $1\le j\le k$,
$k=1,2,\dots$. (Here the numbers $B_k$ agree with the norming constants
$B_k$ of the partial sums, and $n_k=k$.) With the help of this
construction the investigation of partial sums of independent random
variables can be considered as the investigation of the sums of the
random variables in a row of specially chosen triangular arrays.
 
In the investigation of limit theorems we want to exclude some
trivially non-interesting cases. Such a case appears for instance if
$\xi_1=\xi$, and $\xi_k\equiv0$ if $k\ge2$. In this case
$S_n=\xi$, and $\frac {S_n-0}1\to\xi$ if $n\to\infty$.
More generally, we want to exclude the possibility that one or a fixed
number of random variables played a dominant role in the limit behaviour
of the partial sums. To exclude such possibilities we introduce the
notion of uniform smallness of the random variables in partial sums of
sequences of independent random variables or in the sums of rows of
triangular arrays. 

\medskip \noindent
{\bf Definition of uniform smallness.}  {\it Let $\xi_1,\xi_2,\dots$ be
a sequence of independent random variables, and let us consider their
normalized partial sums with a norming (dividing) constant
$B_n$. We say that this sequence of random variables (with this norming
factor) satisfies the condition of uniform smallness if for all
$\e>0$
$$
\sup_{1\le j\le n}P(|\xi_j|>\e B_n)<\e, \quad\text{if }n>n_0(\e).
$$
 
A triangular array $\xi_{k,j}$, $1\le j\le n_k$, satisfies the
uniform smallness condition if for all $\e>0$ there exists a
threshold index $k_0=k_0(\e)$ such that
$$
\sup_{1\le j\le n_k}P(|\xi_{k,j}|>\e)<\e, \quad\text{if }k>k_0(\e)
$$
for all $\e>0$.} 

\medskip
In the sequel we shall investigate limit theorems for partial sums of
random variables or triangular arrays which satisfy the uniform
smallness condition. We formulate some well-known classical results.

\medskip\noindent
{\bf Central limit theorem.}
 
\noindent
{\it a.) For partial sums of independent random variables: Let
$\xi_1,\xi_2,\dots,$ be a sequence of independent random variables,
$E\xi_j=0$, $E\xi_j^2=\sigma_j^2$, $j=1,2,\dots$,
$D_n^2=\summ_{j=1}^n\sigma_j^2$, and let this sequence satisfy the
following Lindeberg condition:
$$
\lim_{n\to\infty}\frac1{D_n^2}\sum_{j=1}^n E\xi_j^2I(|\xi_j|>\e D_n)=0
$$
for all numbers $\e>0$. Let us consider the sequence of partial sums
$S_n=\summ_{j=1}^n\xi_j$. Their appropriate normalizations, the sequence
$\frac{S_n}{D_n}$ converges in distribution to the standard
normal distribution.
 
\noindent b.) For triangular arrays:  Let
$\xi_{k,j}$, $1\le j\le n_k$,, $k=1,2,\dots$, be a triangular array
such that $E\xi_{k,j}=0$, $E\xi_{k,j}^2=\sigma_{k,j}^2$, $1\le k\le
n_k$, $k=1,2,\dots$. Set $S_k=\summ_{j=1}^{n_k}\xi_{k,j}$. Let us
assume that $\limm_{k\to\infty}\summ_{j=1}^{n_k}E\xi_{k,j}^2=1$ and
the following Lindeberg condition is satisfied:
$$
\lim_{k\to\infty}\sum_{j=1}^{n_k} E\xi_{k,j}^2I(|\xi_{k,j}|>\e)=0\quad
\text{for all numbers }\e>0.
$$
Then the random variables $S_k$ converge in distribution to the
standard normal distribution as $k\to\infty$.} 

\medskip\noindent
{\it Remark:}  The Lindeberg condition also implies the validity of
the uniform smallness condition.
 
\medskip\noindent
{\bf Convergence to the Poisson distribution.} {\it Let
$$
\align
&\xi_{1,1}\dots,\xi_{1,n_1}\\
&\vdots\qquad \quad \;\vdots   \\
&\xi_{k,1}\dots,\xi_{k,n_k}\\
&\vdots\qquad \quad \;\vdots
\endalign
$$
be a triangular array which satisfies the following conditions:

\medskip
\item{1.)} The random variables $\xi_{k,j}$ take non-negative integer
values.
\item{2.)} $P(\xi_{k,j}=1)=\lambda_{k,j}$,
$\limm_{k\to\infty}\summ_{j=1}^{n_k}\lambda_{k,j}=\lambda>0$.
\item{3.)} $\supp_{1\le j\le n_k}\lambda_{k,j}\to0$ if $k\to\infty$,
and $\summ_{j=1}^{n_k}P\(\xi_{k,j}\ge2\)\to0$ if $k\to\infty$.

\medskip
Then the distributions of the random variables
$S_k=\summ_{j=1}^{n_k}\xi_{k,j}$ converge to the Poisson distribution
with parameter $\lambda$ as $k\to\infty$.}
\medskip
In the Appendix we shall prove this result.
 
\beginsection 2. The first question to be discussed.
 
Let us first consider the following question: Let
$\xi_{k,1},\dots,\xi_{k,n_k}$ be such a triangular array whose elements
in a row are not only independent but also identically distributed. Let
us assume that the normalized versions of the sums
$S_k=\summ_{j=1}^{n_k}\xi_{k,j}$, the random variables  $S_k-A_k$ with
some appropriate constants $A_k$ converge in distribution to a
distribution function~$F$. What kind of limit distributions $F$ may
appear in such a case? The goal of the following heuristic argument is
to justify the introduction of infinitely divisible distributions
as the natural candidates for the limit distributions.
 
Let us split the sequence $\xi_{k,1},\dots,\xi_{k,n_k}$ to $L$ blocks of
the same length with some integer~$L$. (That circumstance that the
numbers~$n_k$ may be not divisible by the number $L$ does not cause a
hard problem. We can exploit for instance that because of the uniform
smallness condition by leaving out finally many (less than $L$) terms
from the rows we get a new triangular array for which the sums of the
random variables in fixed rows satisfy the same limit theorem as the
sums of the rows in the original triangular array. With the help of
this fact we can restrict our attention to such triangular arrays
whose rows have a length divisible by~$L$.) Let $\eta_1^{(k)}$,
\dots, $\eta_L^{(k)}$ be the sum of the random variables in the
blocks of the $k$-th row minus the number $\frac {A_k}L$. Then
$\eta_1^{(k)}$, \dots, $\eta_L^{(k)}$ are independent and
identically distributed random variables, and
$$
\eta_1^{(k)}+\cdots+\eta_L^{(k)}\Rightarrow S \quad\text{in
distribution}.
$$
Carrying out  the limit procedure $k\to\infty$ we get that
$$
\eta_1+\cdots+\eta_L\overset\Delta\to= S,
$$
where $\overset\Delta\to=$  denotes identity in distribution, and
$\eta_1, \dots,\eta_L$ are independent and identically distributed
random variables, whose distribution agrees with the limit distribution
of the random variables $\eta_1^{(k)}$ as $k\to\infty$. (Actually, this
step would demand a more detailed explanation. The main problem is to
justify that the random variables $\eta_1^{(k)}$ converge in
distribution, or more precisely it is enough to show that this sequence
of random variables have a convergent subsequence in distribution.
This property could be proved, but we omit it from this heuristic
argument.) 

\medskip\noindent
{\bf Definition of infinitely divisible distributions.} {\it A
distribution function~$F$, (or an $F$ distributed random variable~$S$)
is infinitely divisible if for all positive integers~$L$ there exist
independent, and identically distributed random variables
$\eta_1,\dots,\eta_L$ such that the sum
$\eta_1+\cdots+\eta_L$ is $F$ distributed.} \medskip\noindent
{\bf An equivalent definition of infinitely divisible
distributions.} {\it A distribution function $F$ is an infinitely
divisible distribution if and only if its characteristic function
$\varphi(t)=\int e^{itx}F(\,dx)$ can be written for all positive
integers $L$ in the form $\omega(\cdot)^L=\varphi(\cdot)$, where the
function $\omega(\cdot)$ is also a} characteristic function.

\medskip\noindent
{\bf Questions to be investigated:}
 
\item{a.)} Characterization of infinitely divisible distributions.
\item{b.)} To prove that only infinitely divisible distributions can
appear as limit distribution in limit distributions for normalized sums
of independent random variables.

\medskip\noindent 
{\bf Questions to be investigated later:}
 
If we are interested in the limit distribution of the normalized partial
sums of independent and identically distributed random variables
$\xi_1,\xi_2,\dots$, then the characterization of an important subclass
of the infinitely divisible random variables appears, the
characterization of those distribution functions which appear as the
limit in this particular case. This leads to the introduction of the
so-called stable distributions. 

\medskip\noindent
{\bf Definition of stable distributions.} {\it A distribution function
$F$ is stable, if for all positive integers
$L$ there exist such norming constants
$A_L$ and $B_L$ for which the distribution functions
$F_L(x)=F(B_Lx+A_L)$ satisfy th identity
$$
\underbrace{F_L*\cdots* F_L}_{L\text{-times convolution}}=F. \tag*
$$
In an equivalent formulation: If $\eta_1,\eta_2,\dots$ are independent
and identically distributed random variables with distribution
function $F$, then $\eta_1\overset{\Delta}\to=
\frac{(\eta_1-A_L)+\cdots+(\eta_L-A_L)}{B_L}$, or in a different
formulation: $\varphi(t)=\(e^{-tA_L/B_L}\varphi\(\frac
t{B_L}\)\)^L$, where $\varphi(t)=\int e^{itx}F(dx)$ is the
characteristic function of the distribution function~$F$. (In this
definition $\overset{\Delta}\to=$ denotes again identity in
distribution.) Actually we may impose the following restriction in this
definition. We assume that the norming constant $B_L$ in the definition of
$F_L$ is of the form $B_L=L^\alpha$ with some $\alpha>0$. Some deeper results
show that only such norming constants can be chosen in the definition of the
stable distributions.}

\medskip\noindent
{\bf Further questions to be investigated:}

\medskip
\item{a.)} The characterization of stable distributions and the norming
constants $A_L$ and $B_L$ in their definition.
\item{b.)} Description of the domain of attraction of stable
distributions, and the calculation of the right norming constants
in limit theorems for normalized partial sums of independent and
identically distributed random variables.
 
\beginsection 3. Examples of infinitely divisible and stable
distributions.
 
\item{a.)} {\it Normal distribution.}\/ This is an infinitely divisible
and even stable distribution. Indeed, the distribution of a random
variable $\eta$ with expectation zero and variance $\sigma^2$
agrees with the distribution of the sum of~$L$ independent normally
distributed random variables with expectation zero and variance
$\frac{\sigma^2}L$. The distribution of the members in this sum agrees
with the distribution of the random variable. $\frac\eta{\sqrt L}$.
This means that the distribution of $\eta$ is stable with constants
$A_L=0$ and $B_L=\sqrt L$.
\item{b.)}  {\it Poisson distribution.}\/
It is an infinitely divisible but not stable law. The sum of two
independent Poisson distributed random variables with parameters
$\lambda$ and $\mu$ is a Poisson distributed random variable with
parameter $\lambda+\mu$. Hence the distribution of a Poisson
distributed random variable with parameter~$\lambda$ agrees with the
distribution of $L$ independent Poisson distributed random variables
with parameter $\frac \lambda L$. This means that the Poisson
distributed random variables are infinitely divisible. On the other
hand, a Poisson distributed random variable with parameter
$\frac \lambda L$ cannot be written as the linear transform of a
Poisson distributed random variable with parameter~$\lambda$, and the
Poisson distribution is not stable. Let us remark that the
characteristic function of a Poisson distributed random variable is
nowhere zero, hence its logarithm can be defined on the whole real
line. This implies that if the sum of independent and identically
distributed random variables is Poisson distributed, then the
parameter of this Poisson distribution determines the characteristic
function, hence also the distribution of the summands.
 
The above two examples are the most important infinitely divisible
distributions.
 
\medskip\noindent
{\it Further examples:}

\item{a.)} The Cauchy distribution. The density function of this
distribution is $f(x)=\frac 1{\pi(1+x^2)}$, and its characteristic
function is $\varphi(t)=e^{-|t|}$. As $e^{-|t|}=\(e^{-|t|/L}\)^L$,
i.e.\ $\varphi(t)=\varphi\(\frac tL\)^L$, the Cauchy distribution is
stable with the choice $A_L=0$ and $B_L=L$.
\item{b.)} $\Gamma$-distributions. The density functions of these
distribution functions are
$$
f_{\alpha,\nu}(x)=\cases
\frac1{\Gamma(\nu)}\alpha^\nu x^{\nu-1}e^{-\alpha x} &\text{ if }
x\ge0\\
0 &\text{if }x<0
\endcases,
$$
where $\alpha>0$ and $\nu>0$ are two parameters, and
$\Gamma(t)=\int_0^\infty
x^{t-1}e^{-x}\,dx$ is the $\Gamma$~function. Some calculation yields
that $f_{\alpha,\mu+\nu}=f_{\alpha,\mu}*f_{\alpha,\nu}$. Hence
$$
f_{\alpha,\nu}=\underbrace{f_{\alpha,\frac\nu L}*\cdots *
f_{\alpha,\frac\nu L}}_{L\text{-times convolution}},
$$
and $f_{\alpha,\nu}$ is the density function of an infinitely divisible
law. Let us also remark that the exponential distribution also belongs
to this class with the choice~$\nu=1$. It may be also worth remarking
that the characteristic function of a distribution with density
function $f_{\alpha,\nu}$ can be given explicitly, and it equals
$\varphi_{\alpha,\nu}(t)=(1-i\frac t\alpha)^{-\nu}$. The last identity
written up for convolution follows from this formula.

\medskip
The class of infinitely divisible distributions can be described
explicitly. The L\'evy--Hinchin formula supplies such a description.
Before its formulation we show its probabilistic content. Namely, we
show that with the help of the Gaussian and Poissonian distributions
new infinitely divisible distributions can be constructed in a natural
way. The main content of the L\'evy--Hinchin formula is that this
construction supplies all infinitely divisible laws.
 
\beginsection 4. Construction of random variables with infinitely
divisible distribution.
 
Let us observe that if $\xi_1,\cdots,\xi_k$ are independent random
variables with infinitely divisible distribution, then their linear
combination $\alpha_1\xi_1+\cdots+\alpha_k\xi_k+A$ is also a random
variable with infinitely divisible distribution. Further, if $F_n$,
$n=1,2,\dots$, is a sequence of infinitely distributed random variables,
$F_n\Rightarrow F$, where $\Rightarrow$ denotes convergence in
distribution, then the limit distribution $F$ is also an infinitely
distributed random variable. Indeed, by writing the representation
$F_n=\underbrace{G_{n,L}*\cdot*G_{n,L}}_{L\text{ fold convolution}}$
for arbitrary positive integer~$L$, then taking the limit
procedure $n\to\infty$ we get the desired identity
$F=\underbrace{G_L*\cdot*G_L}_{L\text{ fold convolution}}$. Actually,
we should have some special argument to justify the possibility to
carry out this limiting procedure. But we shall apply this procedure
only in such special cases, where the possibility of such a limiting
procedure can be simply justified.

We shall work with Poisson distributed random variables, and our
procedures can be carried out more simply by means of Poisson
fields. Hence we recall a result about the existence of Poisson
fields. In the Appendix we shall present a simple construction of
Poisson fields. 

\medskip\noindent
{\bf Theorem about the existence of Poisson fields.} {\it Let
$(X,\Cal A,\mu)$ be a measurable space with a $\sigma$-finite measure
$\mu$. Then there exits a Poisson field with counting measure~$\mu$.
More explicitly, a probability space $(\Omega,\Cal B,P)$  can be
constructed together with a finite or countably infinite set of random
variables $\{x_1(\oo),x_2(\oo),\dots\}$, $\oo\in \Omega$, which take
their values in the space $X$ and satisfy the following properties:
\item{1.)} With probability~1 the Poisson field has only
finitely many points in all measurable sets $A$ with finite
$\mu$-measure, i.e.\ $P(\{\oo\: \#\{n\:x_n(\oo)\in A\}<\infty\})=1$, if
$\mu(A)<\infty$.
\item{2.)} If $A_1\in\Cal A$, $A_2\in \Cal A,\dots, A_k\in \Cal A$ are
disjoint sets, and $\mu(A_j)<\infty$, $j=1,\dots,k$, then the number of
the points $x_n(\oo)$ which fall into the sets $A_1$, $A_2$\dots,
and $A_k$ are independent Poisson distributed random variables with
parameters $\mu(A_j)$, $j=1,\dots,k$, i.e.\ if we fix some non-negative
integers $l_1,\dots,l_k$ and define the sets $B_j=\{\oo\:
\#\{n\:x_n(\oo)\in A_j\}=l_j\}$, $1\le j\le k$, then
$P\(\bigcapp_{j=1}^k B_j\)
=\prodd_{j=1}^k\frac{\mu(A_j)^{l_j}}{l_j!}e^{-\mu(A_j)}$. 

\medskip}
In the sequel we shall sometimes call a Poisson field a Poisson process
if the measurable space $X$ is the real line or a subset of it. 

\medskip
Let $\mu$ be a $\sigma$-finite measure on the set $\bold R$, ($\bold R$
denotes the real line in the sequel), and let
$(x_1(\oo),x_2(\oo),\dots)$ be a Poisson field on the space
$(\bold R, \Cal A)$ with counting measure~$\mu$.
(Here $\Cal A$ denotes the Borel $\sigma$-algebra.) We should like to
show that the sum $\xi(\oo)=\summ_{n=1}^\infty x_n(\oo)$, i.e.\ the sum
of the coordinates of the  Poisson field is a random variable with
infinitely divisible distribution. It is natural to expect this. Indeed,
for all positive integers~$L$ let us consider $L$ independent Poisson
fields $(y_1^j(\oo),y_2^j(\oo),\dots)$, $j=1,2,\dots,L$ on the space
$(\bold R,\Cal A)$ with counting measure $\frac \mu L$,
and define the random variables $\eta_j=\summ_{n=1}^\infty y_n^j(\oo)$.
We expect that the random variable $\xi(\oo)$ and the sum
$\eta_1(\oo)+\cdots+\eta_L(\oo)$ have the same distribution, since the
random variable $\eta_1(\oo)+\cdots+\eta_L(\oo)$ is the sum of all
coordinates $y_n^j(\oo)$, $n=1,2,\dots$, $j=1,\dots,L$, of the Poisson
fields $(y_1^j(\oo),y_2^j(\oo),\dots)$, $j=1,\dots,k$. But the union of
these Poisson fields is a Poisson field with counting measure $\mu$,
because the number of points falling into disjoint sets $A_s$ are
independent random variables with parameters $\mu(A_s)$. Indeed the
number of these points are the sum of~$L$ independent Poisson
distributed random variables with parameters $\frac\mu L$.
 
There is one weak point in the above heuristic argument. Namely, the
random sums defining the random variables $\xi(\oo)$ and $\eta_j(\oo)$
may be meaningless. On the other hand, we show that under appropriate
not too restrictive conditions for the measure $\mu$ the above sums
can be defined in a meaningful way with the help of a good
regularization. With the help of this regularization we really get
random sums with infinitely divisible distribution. Furthermore, as the
later described L\'evy--Hinchin formula states in such a way we get a
sufficiently rich class of infinitely divisible distributions.
 
Let us assume that the measure $\mu$ satisfies the following condition:
$$
\aligned
&\mu([a,\infty))<\infty \qquad \mu((-\infty,-a])<\infty \\
&\int_0^a x^2\mu(\,dx)<\infty, \qquad
\int_{-a}^{0} x^2\mu(\,dx)<\infty
\endaligned
\qquad\text{for all numbers } a>0.
\tag$**$
$$
Let us define for all $N=0,1,\dots$ the sums $\xi_N(\oo)=\summ_{n\:
|x_n(\oo)|>2^{-N}}x_n(\oo)$ containing randomly many terms
These random variables are meaningful, since by the property
$$
\mu((-\infty,-2^{-N})\cup(2^{-N},\infty))<\infty
$$
the set
$(-\infty,-2^{-N})\cup(2^{-N},\infty)$ contains only finitely many
points of the Poisson field with probability~1, and the sum defining
the random variable $\xi_N(\oo)$ contains only finitely many terms. We
also claim that the limit $\xi(\oo)=\limm_{N\to\infty}\xi_N(\oo)
-(E\xi_N(\oo)-E\xi_1(\oo))$ exists with probability~1. As $\xi_N(\oo)
-(E\xi_N(\oo)-E\xi_1(\oo))=\xi_0(\oo)+\summ_{k=0}^N\zeta_k(\oo)$,
where
$$
\zeta_k(\oo)=\zeta'_k(\oo)-E\zeta'_k(\oo),\quad \text{and}\quad
\zeta_k'(\oo)=\summ_{n\:2^{-k-1}<|\xi_n(\oo)|\le 2^{-k}} \xi_n(\oo),
$$
it is enough to show that the sum $\summ_{k=0}^\infty \zeta_k(\oo)$ is
convergent with probability~1. On the other hand, the random variables
$\zeta_k$ are independent. (Here we sum up the coordinates of the
Poisson process lying in disjoint sets, and this implies the
independence.) The convergence of the sum
$\summ_{k=0}^N\zeta_k(\oo)$ can be proved by means of a
classical result of the probability theory, by means of the so-called
three series theorem (actually we only need the simpler sufficiency part
of this result.) Because of this result it is enough to show that
$\summ_{k=0}^\infty \text{Var}\,\zeta_k<\infty$. (By the definition
of the random variables $\zeta_k$ $E\zeta_k=0$.) This inequality
follows from the following Lemma~1 and relation~$(**)$.

\medskip\noindent
{\bf Lemma~1.} {\it Let $\mu$ be a finite measure on the
$\sigma$-algebra of the Borel  measurable sets of a finite
interval $(a,b]$. Let $x_1(\oo), \dots, x_{k(\oo)}(\oo)$
(with a random index $k=k(\oo)$) be a Poisson field on the space
$((a,b],\Cal A)$ with counting measure $\mu$, and put
$S(\oo)=\summ_{j=1}^{k(\oo)}x_j(\oo)$.
Then
$$
ES=\int_a^b x\mu(\,dx),\qquad \text{\rm Var}\,S=\int _a^b x^2\mu(\,dx).
$$
Further, the logarithm of the characteristic function of the random
variable $S$ (which always exists) satisfies the identity
$$
\log Ee^{itS}=\int_a^b\(e^{itx}-1\)\mu(\,dx) \quad\text{for all } t\in
\bold R.
$$
} 

\medskip\noindent
{\it Remark:} In Lemma~1 we spoke about the logarithm of the
characteristic function of a random variable. Let us explain its precise
definition. The reason why this problem deserves some attention is that
the logarithm of a complex number is a multi-valued function. If
$\log z=z_1$, then we can take $\log z=z_1+2ki\pi$,
$k=0,\pm1,\pm2,\dots$, with the same right. On the other hand, if a
characteristic function $\varphi(\cdot)$ does not take the value zero
in an interval $[a,t]$ such that $0\in[a,b]$, then  the function
$\log \varphi(t)$, $t\in [a,b]$ can be defined in a simple unique way.
Namely, put $\log\varphi(0)=0$, and let us define the logarithm of the
(continuous) function $\varphi(t)$ as a continuous function on the
interval $[a,b]$. In such a way we can tell which branch of the
logarithm of the function $\varphi(t)$ we choose.
 
By Lemma~1 and relation $(**)$
$$
\summ_{k=0}^\infty \text{Var}\,\zeta_k=
\summ_{k=0}^\infty \int_{2^{-k-1}<|x|\le 2^{-k}}x^2\mu(\,dx)
= \int_{0<|x|\le1 }x^2\mu(\,dx)<\infty,
$$
and this implies the desired convergence.

\medskip\noindent
{\it Proof of Lemma~1:}\/ If the measure $\mu$ is concentrated to
finitely many points $u_1,\dots,u_n$, $\mu(u_j)=\mu_j$, $j=1,\dots,n$,
then $S=u_1Z_1+\dots+u_nZ_n$, where $Z_1,\dots,Z_n$ are independent
Poisson distributed random variables with parameters
$\mu_1,\dots,\mu_n$. Hence in this case
$$
\align
ES&=\sum u_jEZ_j=\sum u_j \mu_j=\int x\mu(\,dx)\\
\text{Var}\,S&=\sum u_j^2\text{Var}\,Z_j=\sum u_j^2\mu_j=\int
x^2\mu(\,dx)\\
\log Ee^{itS}&=\sum\log Ee^{itu_jZ_j}=\sum
\mu_j\(e^{itu_j}-1\)=\int\(e^{itx}-1\)\mu(\,dx).
\endalign
$$
If $\mu$ is an arbitrary finite measure on the interval $(a,b]$, then
let us fix an integer $T>0$, and define the measures $\mu_T$ so that
they are concentrated in the points $a+\frac {b-a}Tt$,
$t=1,\dots,T$, and
$$
\mu_T\left\{a+\frac{b-a}Tt\right\}
=\mu\left\{\(a+\frac{b-a}T(t-1), a+\frac{b-a}Tt\]\right\}.
$$
If $x_1(\oo),\dots,x_{k(\oo)}(\oo)$ is a Poisson field on the space
$((a,b],\Cal A)$ with counting measure $\mu$,  then
let us define the point process $x_{j,T}(\oo)=a+\frac{b-a}T
t_j$ if the point $x_j(\oo)$ satisfies the inequality
$a+\frac{b-a}T(t_j-1)<x_j(\oo)\le a+\frac{b-a}Tt_j$, $1\le
j\le k(\oo)$. Then $x_{1,T}(\oo),\dots,x_{k(\oo),T}$ is a Poisson field
on the space $((a,b],\Cal A)$ with counting measure $\mu_T$.
 
Let $S_T(\oo)=\summ_{j=1}^{k(\oo)}x_{j,T}(\oo)$. Then $S_T(\oo)\to
S(\oo)$ if $T\to\infty$. Hence
$$
\limm_{T\to\infty}ES_T=ES,\quad
\limm_{T\to\infty}\text{Var}\,S_T=\text{Var}\,S\quad\text{and} \quad
\limm_{T\to\infty}\log Ee^{itES_T}=\log Ee^{itES},
$$
and by taking the limit $T\to\infty$ we get the statements of Lemma~1
for arbitrary finite measure $\mu$.

\medskip\noindent
{\it Remark:}\/ The formula expressing the logarithm of the
characteristic function remains valid also for
$a=-\infty$ and $b=\infty$ if
$\mu([a,b])<\infty$. Indeed, by applying this formula for such intervals
$(a_n,b_n]$ for which $-\infty<a_n<b_n<\infty$, $a_n\to a$
and  $b_n\to b$, then we obtain this identity with the help of the limit
procedure $n\to\infty$ in the general case.

\medskip
By means of Lemma~1 we can describe the logarithm of the
characteristic function of the above defined random variable~$\xi$.
Namely,
$$
\log \varphi(t)=\log E e^{it\xi}=\int_{\bold
R\setminus\{0\}}\(e^{itx}-1-itA(x)\)\mu(\,dx),
$$
where $\mu$  is a measure satisfying condition~$(**)$, and
$A(x)=x$ if $|x|\le 1$, and $A(x)=0$ if $|x|>1$.
 
This formula follows from the following observation: As
$\xi(\oo)=\xi_0(\oo)+\summ_{k=0}^\infty\zeta_k(\oo)$, and the members
of the sum are independent random variables, hence we get the logarithm
of the characteristic function of the random variable $\xi(\oo)$ by
summing up the logarithm of the characteristic functions of these terms.
Furthermore, by means of Lemma~1
$$
\log Ee^{it\zeta_k}=\log E^{it\zeta'_k}-itE\zeta'_k
=\int_{2^{-k-1}<|x|\le 2^{-k}}\(e^{itx}-1-itx\)\mu(\,dx).
$$
 
The definition of the above random variables $\xi(\oo)$ can be written
in a slightly more general form. Let $B(N)\to0$, $C(N)\to\infty$ if
$N\to\infty$, $0<B(N)<C(N)\le \infty$ be arbitrary monotone
deterministic sequences, and introduce the random variables
$\xi_N(\oo)=\summ_{n\: B(N)<|x_n(\oo)|<C(N)}x_n(\oo)$,where
$x_n(\oo)$, $n=1.2.\dots$, is a Poisson field on the space $(\bold
R,\Cal A)$ with counting measure $\mu$ which satisfies
condition~$(**)$. Then the following regularized sum exists with
probability~1:
$$
\align
\xi(\oo)&=\lim_{N\to\infty}\text{Reg}\,\xi_N(\oo)\\
&=\lim_{N\to\infty}
\!\!\!\( \sum_{n\: B(N)<|x_n(\oo)|<C(N)} \!\!\! x_n(\oo)-
E\sum_{n\: B(N)<|x_n(\oo)|<1} \!\!\!  x_n(\oo)\).
\endalign
$$
Then for arbitrary positive integer~$L$ we can define similarly
$L$ independent copies of a random variable $\eta_{j,L}(\oo)$, $1\le
j\le L$, with the help of a Poisson field on the space
$(R,\Cal A)$ with counting measure $\frac\mu L$ as the regularized
sum of the coordinates of this random field. Then
$\eta_{1.L}+\cdots+\eta_{L,L}\overset\Delta\to=\xi$, where
$\overset\Delta\to=$ denotes identity in distribution. Hence $\xi(\oo)$
is a random variable with infinitely divisible distribution.
 
We  get a new random variable with infinitely divisible distribution if
we consider instead of the above constructed random variable $\xi(\oo)$
a random variable of the form $\xi(\oo)+\eta(\oo)+D$, where $\eta$ is
independent of $\xi$, and it is a Gaussian random variable with
expectation zero with some variance $\sigma^2\ge0$. The logarithm of
the characteristic function of this new random variable has the form
$$
\log \bar\varphi(t)=\log\varphi(t)-\frac{\sigma^2t^2}2+itD
=\int_{\bold R\setminus\{0\}}\(e^{itx}-1-itA(x)\)\mu(\,dx)
-\frac{\sigma^2t^2}2+itD,    \tag1
$$
where $\varphi(t)$ is the characteristic function of the random variable
$\xi$.
 
Now we formulate the L\'evy-Hinchin formula. Roughly speaking, it states
that the distribution of the above constructed infinitely divisible
random variables give all possible infinitely divisible distributions.
It is expressed in different equivalent form in different works,
and there exists no ``best version of the L\'evy--Hinchin formula".
Here we formulate it in the form as it is done in the 17-th chapter of
the book {\it An Introduction to the Probability Theory and Its
Application II.}\/ of William Feller. Then we shall show that this
representation of infinitely divisible distribution is equivalent to
that found by our construction. To formulate this result first
we introduce the following definition. 

\medskip\noindent
{\bf Definition of canonical measures on the real line.} {\it A
measure $M$ on the Borel $\sigma$-algebra of the real line is a
canonical measure if for all finite intervals
$[a,b]\subset\bold R$ the relation $M\{[a,b]\}<\infty$ holds, and for
all numbers $a>0$}
$$
\int_{a}^\infty\frac1{x^2}M(\,dx)<\infty,\quad \text{and}\quad
\int_{-\infty}^{-a}\frac1{x^2}M(\,dx)<\infty.
$$

\medskip\noindent
{\bf Theorem. L\'evy--Hinchin formula.} {\it  A probability distribution
$F$ is infinitely divisible if and only if its characteristic function
$\varphi(t)=\int e^{itx}F(\,dx)$ has a logarithm, (\i.e.
$\varphi(t)\neq0$ for all $t\in \bold R$, and this logarithm can be
written in the form
$$
\log \varphi(t)=\int_{-\infty}^\infty \frac {e^{itx}-1-it\sin
x}{x^2}M(\,dx)+itB, \tag2
$$
where $M$ is a canonical measure. The infinitely divisible
distribution function $F$ determines the canonical measure $M$
and number $B$ in the formula describing the logarithm of its
characteristic function.}

\medskip\noindent
{\it Remark:}\/ To explain completely the content of the above formula
we have to explain the value of the integrand in the origin.
We define the value of the integrand in the origin by extending this
function to a continuous function on the
whole real line. Hence we define
$$
\left. \frac {e^{itx}-1-it\sin
x}{x^2}\right|_{t=0}=-\frac{t^2}2.
$$
After this remark we can compare the L\'evy-Hinchin formula with
formula~(1) found by means of our construction. 

\medskip\noindent
Let us choose $M(0)=\sigma^2$ and introduce the measure
$\mu(dx)=\frac{M(dx)}{x^2}$ if $x\neq0$ with the help of the
quantities in formula~(2). Then some consideration shows the
equivalence of formulas~(1) and (2). Indeed, the measure
$\mu(dx)=\frac{M(dx)}{x^2}$, $x\in \bold R\setminus\{0\}$, satisfies
the condition $(**)$ if and only if $M(\cdot)$ is a canonical measure,
and the term $-\frac{\sigma^2t^2}2$ in formula (1) equals to the
contribution of the origin to the integral~(2). By rewriting the
integral in formula~(1) as an integral with respect to the $M$ instead
of the measure $\mu$ we get that the difference of the expressions in
formulas~(1) and~(2) equals $\int_{-\infty}^{\infty}it
\frac{A(x)-\sin x}{x^2}M(\,dx)+i(B-D)t$. The integral in this formula
is finite, since
$$
\sup_{x\neq0}|A(x)-\sin(x)|<\infty,\quad\text{and}\quad \limm_{x\to0}
\frac{A(x)-\sin x}{x^2}=0.
$$
Hence we can make the difference of these to expressions zero by an
appropriate choice or the constant $B$ or~$D$. The above argument also
showed that the contribution of the origin to the integral in
formula~(2) gave the Gaussian part of the infinitely divisible
distributions.
 
The difference between different representations of infinitely
distributed random variables, beside the application of different
measures $\mu$ and~$M$, is caused by the fact that in the definition
of the characteristic functions we have to guarantee that the integrals
appearing in these formulas are finite. To achieve this we
have to make some kind of regularization. This can be done in different
ways, and different regularizations were applied in formulas~(1)
and~(2). Let us remark that there exists no ``most natural
regularization".
 
Let us finally remark that there exists a third equivalent and
frequently used description of the characteristic function of
infinitely divisible distributions. We shall also give this
representation. Actually, in the second and third part of this work
we shall use this representation of infinitely divisible distributions,
because we can better work with it. To describe this representation we
fix a positive number $a>0$ and define the function
$$
\tau(x)=\tau_a(x)=\cases
x&\text{if }|x|\le a\\
a&\text{if }x\ge a  \\
-a&\text{if }x\le -a.
\endcases \tag3
$$
Then the logarithm of the characteristic functions of the infinitely
divisible distributions can be written with the help of a canonical
measure~$M$ and real number$B$ in the following form:
$$
\log \varphi(t)=\int_{-\infty}^\infty \frac {e^{itx}-1-it\tau(x)}
{x^2}M(\,dx)+itB. \tag4
$$
The representation (4) is very similar to the representation~(1).
The main difference is that here we work with the function
$\tau(\cdot)=\tau_a(\cdot)$ instead of the function~$A(\cdot)$.
Let us emphasize that the function $\tau(\cdot)$, unlike the
function $A(\cdot)$, is a continuous (and bounded) function. This
property will simplify the limiting procedures in later proofs.
 
\beginsection 5. Construction of infinitely divisible process.
 
Let us consider an $F$ infinitely divisible distribution which has no
Gaussian component. Then the logarithm of its characteristic function
$\varphi(t)=\int e^{itx}F(\,dx)$ can be written by the L\'evy--Hinchin
formula as
$$
\log \varphi(t)=\int_{x\neq0}(e^{itx}-1-it\tau(x))\mu(\,dx)+iDt \tag5
$$
with the function $\tau(x)$ defined in formula (3) and a measure $\mu$
which satisfies the condition $(**)$. (Here we applied the
L\'evy--Hinchin formula in the form~(4), and replaced the canonical
measure $M$ by the measure $\mu(\,dx)=\frac{M(\,dx)}{x^2}$.)
 
Now we show that the ideas of Section~4  enables one to construct
not only  random variables with infinitely divisible distribution,
but also so-called infinitely divisible stochastic processes with
nice, so-called c\`adl\`ag trajectories. Such processes are closely
related to infinitely divisible distributions. Infinitely divisible
processes play a role similar to the role of infinitely distributed
random variables if we want to prove not only limit theorems for sums
of independent random variables but also limit theorems for random
broken lines made from partial sums of independent random variables
in a natural way. Such results are called functional limit theorems,
and they will be the subject of the third part of this text.
Before formulating the result about the existence of infinitely
divisible processes we introduce the following definition.

\medskip\noindent
{\bf Definition of c\`adl\`ag functions.} {\it We call a real valued
function $x(\cdot)$ on the interval $[0,1]$ a c\`ad\`ag (continue \`a
droite, limite \`a gauche) function if it is continuous from the right
and has a left-side limit in all points $t\in[0,1]$.}

\medskip
The reason for introducing the notion of c\`adl\`ag function is that
we want to take such a version of the stochastic processes we shall
work with which has nice trajectories. In very nice cases we can have
a version with continuous trajectories. General infinitely divisible
do not have such a nice version, but as we shall show they have a
version whose trajectories are c\`adl\`ag functions.
 
We shall prove that there exists a stochastic process on a probability
space $(\Omega,\Cal A,P)$ $\xi(t)=\xi(t,\oo)$, $\oo\in\Omega$, $0\le
t<\infty$, which satisfies the following properties:

\medskip
\item{1.} The random variable $\xi(1,\oo)$ has a prescribed infinitely
divisible distribution $F$ (without Gaussian component) whose
characteristic function is defined in formula~(5).
\item{2.} The stochastic process $\xi(t,\oo)$  has independent and
stationary increments, i.e.\ for all numbers $0\le t_1<t_2<\cdots
<t_k\le1$, $\xi(0)\equiv0$, the random variables $\xi(t_1)$,
$\xi(t_2)-\xi(t_1)$,\dots, $\xi(t_k)-\xi(t_{k-1})$ are independent,
and for all pairs of non-negative numbers $s$ and $t$ such that
$s+t\le1$ the distribution of the random variable $\xi(t+s)-\xi(t)$
does not depend on the parameter~$t$.
\item{3.} For  almost all $\oo\in\Omega$ the trajectory $\xi(\cdot,\oo)$
is a c\`adl\`ag function on the interval $[0,1]$.
 
\medskip\noindent
{\it Remark:}\/ There exists a Gaussian process $W(t,\oo)$,
$0\le t\le1$, with independent and stationary increments whose
trajectories are continuous functions $W(0,\oo)\equiv0$, and
$W(1,\oo)$ is a random variable with standard normal distribution.
Such stochastic processes are called Wiener processes in the
literature. The existence of Wiener processes enables us to embed
a general infinitely distribution function $F$ to an infinitely
divisible process $X(t)$, $0\le t\le1$, with properties~2.) and~3.)
such that $X(1)$ has distribution $F$. Indeed, by the previously
formulated statement a process of the form $X(t)=\xi(t)+\sigma W(t)$
with an appropriate infinitely divisible process $\xi(t)$ and
number $\sigma$ satisfies the desired properties. 

\medskip
We can construct the infinitely divisible stochastic process with
the desired properties in the following way. Introduce the measure
$\bar\mu=\mu\times \lambda$ on the set $\bold R\times [0,1]$, the
product of the measure $\mu$ appearing in formula~(5) and the Lebesgue
measure on the unit interval. Let us then consider a Poisson field
$$
(x_1(\oo),x_2(\oo),\dots)
=((x_1^{(1)}(\oo),x_1^{(2)}(\oo)),(x_2^{(1)}(\oo),x_2^{(2)}(\oo),\dots)
$$
on the space $(\bold R\times [0,1], \Cal A)$ with counting measure
$\bar\mu$, where $\Cal A$ denotes the Borel $\sigma$-algebra on
$\bold R\times [0,1]$.
 
We apply a construction similar to the regularized sum of a Poisson
process carried out in the previous Section by which we constructed a
random variable with infinitely divisible distribution. We also make
some small changes in the construction. Thus, now we only consider an
appropriate monotone decreasing $B(N)$, $\limm_{N\to\infty}B(N)=0$, and
put $C(N)=\infty$ for all $N$. Besides, we work with the function
$\tau(\cdot)$ instead of the function $A(\cdot)$ in the regularization
in order to get a representation of the characteristic function in the
form~(5). We shall define the random variables $\xi(t)$ simultaneously
for all $0\le t\le 1$ by taking the (regularized) sum of the first
coordinate of the Poisson field with counting measure $\bar\mu$. But
for a fixed $0\le t\le1$  we take only those points of the Poisson
field in the definition of the random variable $\xi(t)$ whose second
coordinates are less than or equal to~$t$. More explicitly,
$$
\align
\xi(t,\oo)&=\lim_{N\to\infty}\text{Reg}\,\xi_N(t,\oo)\\
&=\lim_{N\to\infty}
\!\!\! \(\sum\Sb n\: B(N)<|x_n^{(1)}(\oo)|\\ x_n^{(2)}(\oo)\le t\endSb
\!\!\!
x^{(1)}_n(\oo)- E\sum\Sb n\: B(N)<|x_n^{(1)}(\oo)|\\ x_n^{(2)}(\oo)\le
t\endSb \!\!\!  \tau(x_n^{(1)}(\oo))\).
\endalign
$$
 
The argument of the previous Section yields that the limit
$\limm_{N\to\infty}\text{Reg}\,\xi_N(t,\oo)$ exists with probability~1
for all fixed $0\le t\le1$. Moreover, this argument also shows that
$\xi_1(\oo)$  satisfies formula~(5) with $D=0$. To satisfy formula~(5)
in the case of a general constant~$D$ we have to consider the process
$\xi(t)+Dt$, and if the process $\xi(\cdot)$ satisfies Properties~2.)
and~3.), then the new process satisfies all Properties 1.)---3.).
 
Hence we have to check the validity of Properties~2.) and~3.). Let
us first consider the approximating stochastic processes
$\text{Reg}\,\xi_N(t,\oo)$, $0\le t\le1$. They satisfy
Property~2.) because of the independence property of the Poisson
field and the invariance property of its distribution with respect
to the shift of the second coordinate. They also satisfy Property~3.),
and besides the definition of the sum $\text{Reg}\,\xi_N(t,\oo)$
is meaningful. To see this observe that the Poisson process
has only finitely many points in the domain $((-\infty,-B(N))\cup
(B(N),\infty))\times[0,1]$. Hence the regularized sums are meaningful,
because they consist of finitely many terms, and the trajectories
$\text{Reg}\,\xi_N(t,\oo)$, $0\le t\le1$, have only finitely many
jumps, where they are continuous from the right.
 
A simple limiting procedure $N\to\infty$ shows that the stochastic
process $\xi(t,\oo)$ also satisfies  Property~2.). To prove
Property~3.), that is to show that the trajectories $\xi(\cdot,\oo)$
are c\`adl\`ag functions, we have to make a more careful limiting
procedure. Let us exploit our freedom in the choice of the
sequence~$B(N)$, and choose this sequence in such a way that the
inequality $\int_{\{x\:0<|x|<B(N)\}}x^2\mu(\,dx)\le 4^{-N}$
holds.
 
Let us observe that the processes $\text{Reg}\,\xi_N(t,\oo)-
\text{Reg}\,\xi_{N+1}(t,\oo)$, $0\le t\le1$, are stochastic processes
with independent increments and zero expectation which satisfies the
inequality
$$
E\(\text{Reg}\,\xi_{N+1}(1,\oo)-\text{Reg}\,\xi_N(1,\oo)\)^2=
\int_{\{x\:B(N+1)<|x|<B(N)\}}x^2\mu(\,dx)\le 4^{-N}.
$$
for all $N\ge1$ because of Lemma~1. Hence, by the Kolmogorov inequality
$$
\align
P&\(\sup_{0\le t\le
1}\left|\text{Reg}\,\xi_{N+1}(t,\oo)-\text{Reg}\,\xi_N(t,\oo)\right|
\ge 2^{-N/2}\)\\
&\qquad\le 2^N \int_{\{x\:0<|x|<B(N)\}}x^2\mu(\,dx)\le 2^{-N}.
\endalign
$$
Since the sum we get by summing up the expressions at the right-hand
side of the last formula for all~$N$ is convergent, and
$$
\xi(t,\oo)=\text{Reg}\,\xi_1(t,\oo)+\sum_{N=1}^\infty
\[\text{Reg}\,\xi_{N+1}(\oo)-\text{Reg}\,\xi_N(t,\oo)\],
$$
hence the Borel--Cantelli lemma implies that
$$
\sup_{0\le t\le1}\left|\xi(t,\oo)-\text{Reg}\,\xi_N(t,\oo)\right|
\le\frac{2^{-N/2}}{\sqrt 2-1}\quad\text{if }N\ge N_0(\oo)
$$
for almost all $\oo\in\Omega$.
 
The last relation means that for almost all $\oo\in\Omega$ the
trajectories $\text{Reg}\,\xi_N(t,\oo)$ converge to the trajectory
$\xi(\cdot,\oo)$ in the interval $[0,1]$ supremum norm. This implies
that not only the functions $\text{Reg}\,\xi_N(t,\oo)$ but also the
functions $\xi(\cdot,\oo)$ are c\`adl\`ag functions, i.e. Property~3.)
holds.
 
Actually we need a more general result in Part~III of this work. In
that part more general processes appear as the limit process in the
functional limit theorem. Those processes $X(t)$, $0\le t\le1$, are
also stochastic processes with independent increments and c\'adl\`ag
trajectories and such that $X(t)$ is an infinitely divisible random
variable for all $0\le t\le1$. But these processes may have not
stationary increments. Such processes appear if we  consider the limit
of such random broken lines which are determined by partial sums of
independent random variables with not necessarily identical
distributions. We formulate below the statement about the existence of
the stochastic processes we shall need in Part~III of this work.
\
medskip\noindent
{\bf Lemma 2.} {\it Let $\mu$ be a measure on the strip $\bold
R\times[0,1]$ such that its projection $\mu^{(1)}$ to the first
coordinate, defined by the formula $\mu^{(1)}(B)=\mu(B\times[0,1])$
for all measurable sets $B\in\bold R$ satisfies the relation~$(**)$, and
$\mu(\bold R^1\times\{0\})=0$. Then there exists a stochastic process
$\xi(t,\oo)$ which satisfies the following properties:

\medskip
\item{1.} All differences $\xi(t_2,\oo)-\xi(t_1,\oo)$ are random
variables with infinitely divisible distribution functions, and their
characteristic functions
$\varphi_{t_1,t_2}(u)=Ee^{iu(\xi(t_2,\oo)-\xi(t_1,\oo))}$
satisfy the identity
$$
\log\varphi_{t_1,t_2}(u)=\int_{\{(y,s)\: -\infty<y<\infty,\;t_1< s<\le
t_2\}}\(e^{iuy}-1-iu\tau(y)\)\mu(\,dy,\,ds),  \quad u\in\bold R.
$$
for all $0\le t_1<t_2\le 1$. (This integral depends on the argument~$s$
through the determination of the domain of integration.)
\item{2.} The stochastic process $\xi(t,\oo)$ has independent
increments, i.e.\ for all numbers $0\le t_1<t_2<\cdots
<t_k\le1$, $\xi(0)\equiv0$, the random variables $\xi(t_1)$,
$\xi(t_2)-\xi(t_1)$,\dots, $\xi(t_k)-\xi(t_{k-1})$ are independent.
\item{3.} For  almost all $\oo\in\Omega$ the trajectory $\xi(\cdot,\oo)$
is a c\`adl\`ag function on the interval $[0,1]$.
}

\medskip
The proof of Lemma~2 together with the construction of an appropriate
stochastic process $\xi(t,\oo)$ is a natural adaptation of the
argument of this section. Let us consider a Poisson field
$(x_1(\oo),x_2(\oo),\dots)
=((x_1^{(1)}(\oo),x_1^{(2)}(\oo)),(x_2^{(1)}(\oo),x_2^{(2)}(\oo),\dots)$
on the space $(\bold R\times [0,1], \Cal A)$ with counting measure
$\mu$, where $\Cal A$ denotes the Borel $\sigma$-algebra on the
strip $\bold R\times [0,1]$. Then we define the process $\xi(t,\oo)$
with  the help of this Poisson field and the formula
$$
\align
\xi(t,\oo)&=\lim_{N\to\infty}\text{Reg}\,\xi_N(t,\oo)\\
&=\lim_{N\to\infty}
\!\!\! \(\sum\Sb n\: B(N)<|x_n^{(1)}(\oo)|\\ x_n^{(2)}(\oo)\le t\endSb
\!\!\!
x^{(1)}_n(\oo)- E\sum\Sb n\: B(N)<|x_n^{(1)}(\oo)|\\ x_n^{(2)}(\oo)\le
t\endSb \!\!\!  \tau(x_n^{(1)}(\oo))\),
\endalign
$$
with the above Poisson field and a monotone decreasing function
$B(N)$ such that $\limm_{N\to\infty}B(N)=0$, and
$\int_{\{(x,t)\:0<|x|<B(N),\,\,0\le t\le1\}}\!\!\!\!\!
x^2\mu(\,dx) \le 4^{-N}$. Then the argument described in this Section
supplies with slight modifications the proof of Lemma~2.
 
\beginsection Appendix
 
{\script A.) Stable distributions:} 

\medskip
The stable distributions and their embedding to such infinitely
divisible processes whose one-dimensional distributions are stable
distributions can be simply constructed by means of the results
described in this text. They are infinitely divisible  distributions
which are determined in the L\'evy--Hinchin formula by such measures
$\mu$  which have appropriate homogeneity properties. Namely, define the
characteristic function of the stable distributions by formula~(5) with
a measure $\mu$ whose density function is
$$
\frac {d\mu}{dx}(x)=\cases
C_1x^{-\alpha} &\text{if } x>0\\
C_2|x|^{-\alpha} &\text{if } x<0
\endcases
$$
where $C_1\ge0$, $C_2\ge0$, $C_1+C_2>0$, $-3<\alpha<-1$. The  last
condition  satisfies that the measure~$\mu$ satisfies property~$(**)$.

A detailed investigation shows that  the above formula describes all
stable distributions. Moreover, the domain of attraction of stable
distributions, together with the appropriate normalization can be given
explicitly in a relatively simple way. Let us also mention that the
integral (5) defining the characteristic functions can be calculated
explicitly, and it is a homogeneous function. Nevertheless, in most
investigations it is simpler to work with the original representation 
of the characteristic function and not with its shorter, integrated 
form. Here we shall not discuss the details. Let us remark that a 
complete description of stable distributions  and their domain of 
attraction is based on the results described in  the second part of 
this work. Besides, the investigation has still another important 
ingredient. It is the investigation of the so-called slowly varying 
functions,  that is of such  functions $L(\cdot)$ in the interval 
$[1,\infty]$ which satisfy  the relation 
$\limm_{t\to\infty}\frac{L(st)}{L(t)}=s^{\alpha}$ for all 
$0<s<\infty$ with some $-\infty<\alpha<\infty$.
 
\medskip\noindent
{\script B.) A simple construction of Poisson processes.} 

\medskip
 One of the basic properties of the Poisson distribution is that the
sum of two independent Poisson distributed random variables with
parameters $\lambda$ and $\mu$ is Poisson distributed with parameter
$\lambda+\mu$. The following Lemma~B is a reverse statement to this
result. It helps to construct Poisson fields. 

\medskip\noindent
{\bf Lemma~B.} {\it Let $k$ urns be given, and let us throw
a random number of balls into them. Let us denote the number of balls
thrown into these urns by $\xi$, and let us assume that $\xi$ is a
Poisson distributed random variable with parameter~$\lambda>0$. Let us
throw all balls independently of each other and the random variable
$\xi$,  and let each ball fall into the $j$-th urn  with probability
$p_j\ge0$, $j=1,\dots,k$, $\sum\limits_{j=1}^k p_j=1$.  Let $\eta_j$
denote the number of balls thrown into the $j$-th urn, $1\le j\le k$.
Then the random variables $\eta_j$, $j=1,\dots,k$, are independent, and
$\eta_j$ is Poisson distributed with parameter $p_j\lambda$,
$j=1,\dots,k$.}

\medskip\noindent
{\it Proof  of Lemma~B:}
$$
\align
P(\eta_1=l_1,\dots,\eta_k=l_k)
&=P(\xi=l_1+\cdots+l_k)\frac{(l_1+\cdots+l_k)!}{l_1!\cdots l_k!}
p_1^{l_1}\cdots p_k^{l_k}\\
&=\frac{\lambda^{(l_1+\cdots+l_k)}}{l_1!\cdots l_k!}
p_1^{l_1}\cdots p_k^{l_k}e^{-\lambda}
=\prod_{j=1}^k\frac{ (\lambda p_j)^{l_j}}{l_j!}e^{-\lambda p_j}
\endalign
$$
for arbitrary integers $l_1\ge0$, \dots, $l_k\ge0$. This identity
implies Lemma~B.

\medskip
We formulate the following Corollary of Lemma~B.

\medskip\noindent 
{\bf Corollary of Lemma~B.} {\it Let a measurable space 
$(X,\Cal A)$  be given together with a probability measure 
$\mu$ on it. Let $\xi$ be a Poisson distributed random variable 
with parameter $\lambda>0$. Let us choose $\xi$ points 
$x_1,\dots,x_\xi$ on the space $X$ independently of each other,
and the random variable $\xi$ in such a way that the distribution of
the random points $x_l$ satisfies the identity $P(x_l\in\A)=\mu(\A)$ for
all sets $\A\in\Cal A$ and $l=1,\dots,\xi$. Then for all disjoint sets
$\A_1\in \Cal A$,\dots, $\A_k\in \Cal A$ the number of the points
$x_l$, $1\le l\le \xi$ contained in the sets $\A_j$, $j=1,\dots,k$, are
independent Poisson distributed random variables with parameters
$\lambda\mu(\A_j)$.
 
Let a measurable space $(X,\Cal B)$ be given together with a
$\sigma$-finite measure $\mu$ on it. With the help of the above
construction such a set of points $x_1,x_2,\dots$ can be chosen in
the space $X$ which satisfies the following properties: For all sets
$A$ with finite $\mu$ measure the number of points from the set
$\{x_1,x_2,\dots\}$ which are contained in the set $A$ is
a Poisson distributed random variable with parameter $\mu(A)$. Beside
this, the number of the points from the set $\{x_1,x_2,\dots\}$
contained in disjoint sets of finite $\mu$ measure are independent
random variables.} 

\medskip\noindent
{\it Proof of the Corollary.}\/ Le us adjust to the sets $A_1,\dots,A_k$
the set $\A_{k+1}=X\setminus\bigcup\limits_{j=1}^k\A_j$, and put
$p_j=\mu_j(\A_j)$, $j=1,\dots,k+1$. Then by Lemma~B the number of points
falling into the sets $\A_j$ are independent, Poisson distributed random
variables with parameter $\lambda\mu(\A_j)$, and this is the statement
of the first paragraph in the Corollary.
 
To prove the second statement of the Corollary let us consider a
partition of the space $X$ such that
$X=\bigcup\limits_{j=1}^\infty X_j$, the sets $X_j$,
$j=1,2,\dots$, are disjoint, and $\mu(X_j)=\lambda_j<\infty$.
Let us construct with the help of the already proven part of the
Corollary (with the choice $\lambda=\lambda_j=\mu(X_j)$ and
probability measure $\bar\mu$, $\bar\mu(A)=\frac1{\lambda_j}\mu(A)$ on
the measurable sets $A\subset X_j$) a set of points
$\{x_{j,1},x_{j,2},\dots\}\subset X_j$ on each  set $X_j$ in such a way
that the number of the points from the set $\{x_{j,1},x_{j,2},\dots\}$
contained in a set $\A_j\subset X_j$ is Poisson distributed with
parameter $\mu(\A_j)$, and the number of the points contained in
disjoint subsets of the set~$X_j$ are independent random variables. Let
us choose these sets $\{x_{j,1},x_{j,2},\dots\}$ independently from
each other for different indices~$j$. Let us take the union
$\bigcupp_{j=1}^\infty \{x_{j,1},x_{j,2},\dots\}$ of these sets. We
claim that this set of points satisfies the Conditions of the
Corollary. Really, the number of the points of this set falling
into a set $A$ equals the sum of the numbers of points falling into
the sets $A\cap X_j$, $j=1,2,\dots$, which are independent Poisson
distributed random variables with parameters $\mu(A\cap X_j)$. Hence the
number of the points of the above constructed set falling into a set $A$
is a Poisson distributed random variable with parameter
$\summ_{j=1}^\infty \mu(A\cap X_j)=\mu(A)$. The needed independence
property can be checked similarly.

\medskip
The above Corollary contains actually a construction of a Poisson
field. Its first statements describes this construction in the case when
$\mu(X)<\infty$, i.e.\ if the measure of the space is finite. The second
statement of the Corollary reduces the general case where we only know
that the space is $\sigma$-finite to this former case by splitting the
space to countable many disjoint sets with finite measure.

\medskip\noindent
{\script C.) Proof of the Poissonian limit theorem for sums of
independent and integer valued random variables.}

\medskip\noindent
Here we prove the Poissonian limit theorem formulated in pages~3 and~4.
We shall give two different proofs. It may be instructive to consider
both of them, because they show a simple example of the two different
methods applied in this paper. The first method applied mainly in
Part~II is based on the characteristic function technique to prove limit
theorems. The second method applied in Part~III exploits the fact that
a small perturbation of a sequence of random variables does not change
the limit behaviour of this sequence. This fact combined with a good
coupling makes possible to reduce the problem we are interested in to a
much simpler problem. 

\medskip\noindent 
{\it First proof:} \/ We show that the characteristic
functions of the random variables $S_k$ converge to the characteristic
function of a Poisson distributed random variable with parameter
$\lambda$. The characteristic function of a Poisson distributed random
variable $\eta$ with parameter $\lambda$ equals
$Ee^{it\eta}=\sum\limits_{k=0}^\infty
\frac{\lambda^k}{k!}e^{-\lambda+ikt}=\exp\{-\lambda+\lambda e^{it}\}$.
Let $\varphi_{k,j}(t)$ denote the characteristic function of the random
variable $\xi_{k,j}$. Then by the Condition~1. of the Theorem
$$
\varphi_{k,j}(t)=P(\xi_{k,j}=0)+P(\xi_{k,j}=1)e^{it}+\e(k,j,t)
=1+\lambda_{k,j}(e^{it}-1)+\bar\e(k,j,t),
$$
where $|\e(k,j,t)|\le P(\xi_{k,j}\ge2)$, and
$|\bar\e(k,j,t)|\le 2P(\xi_{k,j}\ge2)$. Hence
$$ \allowdisplaybreaks
\align
Ee^{itS_k}&=\prod_{j=1}^{n_k}\varphi_{k,j}(t)
=\prod_{j=1}^{n_k}\(1+\lambda_{k,j}(e^{it}-1)+\bar\e(k,j,t)\)\\
&=\prod_{j=1}^{n_k}\exp\left\{\lambda_{k,j}(e^{it}-1)
+O(\lambda_{k,j}^2+\bar\e(k,j,t))\right\}\\
&=\exp\left\{(e^{it}-1)\(\sum_{j=1}^{n_k}\lambda_{k,j}\)+
O\(\sum_{j=1}^{n_k}\(\lambda_{k,j}^2+\bar\e(k,j,t)\)\)\right\}\to
\exp\{\lambda(e^{it}-1)\},
\endalign
$$
if $k\to\infty$, because
$\limm_{k\to\infty}\summ_{j=1}^{n_k}\lambda_{k,j}
=\lambda$, $\summ_{j=1}^{n_k}\lambda_{k,j}^2\le\supp_{1\le j\le
n_k} \lambda_{k,j}\cdot \summ_{j=1}^{n_k}\lambda_{k,j}\to0$ if
$k\to\infty$ by conditions~2 and~3 of the Theorem. Furthermore,
$\summ_{j=1}^{n_k}\bar\e(k,j,t)\le 2\summ_{j=1}^{n_k}P(\xi_{k,j}\ge
2)\to0$ if $k\to\infty$ by  condition~3. Since
$\exp\{\lambda(e^{it}-1)\}$ is the characteristic function of a
Poisson distributed random variable with parameter $\lambda$, these
relations imply the Theorem.  

\medskip
The second proof is based on a Lemma~C formulated bellow. We shall
prove Lemma~C in Part~III of this work.  

\medskip\noindent
{\bf Lemma C.} {\it Let $S_k$ and $\bar S_k$, $k=1,2,\dots$, be two
sequences of random variables such that the differences $S_k-\bar S_k$
converge stochastically to zero as $k\to\infty$. If the sequence of
random variables $\bar S_k$ converges in distribution to a
distribution $F$, then the sequence of random variables $S_k$ converges
to the same distribution function~$F$.}

\medskip\noindent
{\it Second proof:}\/ We shall prove that for all indices $k$ a 
sequence of independent and Poison distributed random variables 
$\bar\xi_{k,j}$, $1\le j\le n_k$, can be constructed with an 
appropriate parameter $\bar\lambda_{k,j}$ for which the 
differences of the random sums $\bar S_k=
\summ_{j=1}^{n_k}\bar\xi_{k,j}$ and $S_k=\summ_{j=1}^{n_k}\xi_{k,j}$,
the expression $\bar S_k-S_k$ converges stochastically to zero if
$k\to\infty$, and $\limm_{k\to\infty}\summ_{j=1}^{n_k}
\bar\lambda_{k,j}=\lambda$. This implies the Statement of the Theorem.
Indeed, $\bar S_k$ is a Poisson distributed random variable with
parameter $\summ_{j=1}^{n_k}\bar\lambda_{k,j}$. Hence the distributions
of the random variables $\bar S_k$ converge to the Poisson distribution
with parameter $\lambda$, and by Lemma~C the same statement holds for
the distributions of the random variables $S_k$.
 
Let $\bar\lambda_{k,j}$, the parameter of the Poisson distributed random
variable $\bar\xi_{k,j}$, be the solution of the equation
$\lambda_{k,j}=xe^{-x}$ in the interval $[0,1]$. If $\lambda_{k,j}\le
e^{-1}$, then such a solution exists, and $|\lambda_{k,j}-\bar
\lambda_{k,j}|=\bar \lambda_{k,j}|1-e^{-\bar \lambda_{k,j}}|\le \const
\bar \lambda_{k,j}^2\le \const\lambda_{k,j}^2$. Hence the relations
$\limm_{k\to\infty}\summ_{j=1}^{n_k} \lambda_{k,j}=\lambda$
and $\limm_{k\to\infty}\supp_{1\le j\le n_k} \lambda_{k,j}=0$
imply that $\limm_{k\to\infty}\summ_{j=1}^{n_k}\bar\lambda_{k,j}
=\lambda$, and $\limm_{k\to\infty}\supp_{1\le j\le n_k}
\bar\lambda_{k,j}=0$. Let us define the random variable $\bar\xi_{k,j}$
in such a way that the events $\bar\xi_{k,j}=1$ and $\xi_{k,j}=1$
agree. Let us remark that this is possible, since we defined the number
$\bar\lambda_{k,j}$ in such a way that a Poisson distributed random
variable $\bar\xi_{k,j}$ with parameter $\bar\lambda_{k,j}$ satisfies
the identity $P(\bar\xi_{k,j}=1)=\bar\lambda_{k,j}e^{-\bar\lambda_{k,j}}
=P(\xi_{k,j}=1)$. We shall define the events $\bar\xi_{k,j}=l$,
$l\neq1$ in such a way that $P(\bar\xi_{k,j}=l)
=\frac{\bar\lambda_{k,j}^l}{l!}e^{-\bar\lambda_{k,j}}$, and the
random variables $\bar\xi_{k,j}$, $1\le j\le n_k$, are independent for
a fixed index $k$.
 
In a sufficiently rich probability space such a construction is possible.
A possible construction is the following one: Let
$\eta_1,\dots,\eta_{n_k}$ be a sequence of independent random variables
with uniform distribution on the interval $[0,1]$ which random
variables are also independent of the random variables $\xi_{k,j}$,
$1\le j\le n_k$. Let us consider for all numbers $1\le j\le n_k$ a
partition $A_{0,j}=[0,a_{1,j}]$, $A_{l,j}=[a_{l-1,j},a_{l,j}]$,
$l=2,3,\dots$, of the interval $[0,1]$ (depending on the parameters
$k$ and $j$) in such a way that the length of the interval $A_{0,j}$ is
$a_{0,j}= \frac{e^{-\bar\lambda}}{1-\bar\lambda
e^{-\bar\lambda}}$ and the length of the interval $A_{l,j}$ is
$a_{l,j}-a_{l,j-1}=\frac{\bar \lambda^le^{-\bar\lambda}}
{l!(1-\bar\lambda e^{-\bar\lambda})}$, $l=2,3\dots$. In the case
$l\neq1$ let the set $\{\oo\:\bar\xi_{j,k}(\oo)=l\}$
agree with the set $\{\oo\:\eta_j(\oo)\in A_{l,j}\}$. The random
variables $\xi_{k,j}$  constructed in such a way are independent for
a fixed~$k$, and they have the prescribed distributions, since the
conditional probabilities $P(\bar\xi_{k,j}=l|\bar\xi_{k,j}\neq1)$
have the right values.
 
The assumption that the probability space where we made this
construction is sufficiently rich does not mean an unpleasant
restriction, because the validity of the statement to be proved
does not depend on the properties of the probability space
where we are working. We show that with the above constructed
random variables $\bar\xi_{k,j}$ the differences of the sums
$\bar S_{k}$ and $S_{k}$, the expressions $\bar S_{k}-S_{k}$, converge
stochastically to zero.
 
To prove this statement let us observe that for an arbitrary number
$\e>0$
$$
P(|S_k-\bar S_k|>\e)\le \summ_{j=1}^{n_k}P(\xi_{k,j}\ge2)
+ \summ_{j=1}^{n_k}P(\bar\xi_{k,j}\ge2),
$$
since the relation $S_k-\bar S_k\neq0$ can only hold if either
$\xi_{k,j}\ge2$ or $\bar\xi_{k,j}\ge2$ for some index $1\le j\le n_k$.
On the other hand, $\limm_{k\to\infty}\summ_{j=1}^{n_k}P(\xi_{k,j}\ge2)
=0$ because of the conditions of the Theorem. Furthermore,
$\limm_{k\to\infty}\summ_{j=1}^{n_k}P(\bar\xi_{k,j}\ge2)=0$, since
$\summ_{j=1}^{n_k}P(\bar\xi_{k,j}\ge2)\le\const\summ_{j=2}^\infty
\bar\lambda_{k,j}^2$, and also the relations
$\limm_{k\to\infty}\summ_{j=1}^{n_k}\bar\lambda_{k,j}=\lambda$ and
$\limm_{k\to\infty}\supp_{1\le j\le n_k}\bar\lambda_{k,j}=0$ hold.
Hence the desired inequality and also the statement of the Theorem
holds.
 
%\vfill\eject

\medskip\noindent
{\bf Example for a non-stable distribution that satisfies a
weakened form of the stability condition.} 

\medskip\noindent
We construct such a random variable $S$ whose distribution $F$  
does not satisfy the condition $(*)$ of stability, but it
satisfies a weakened version of it.

Let us fix a number $\alpha>\frac12$. We construct such a non-stable
distribution $F$ that satisfies the identity
$F(2^\alpha (x-A))*F(2^\alpha (x-A))=F(x)$ with an appropriate number
$A=A(\alpha)$, or in a different formulation, we show that if $S$ 
is an $F$ distributed random variable, and  $S'$ and $S''$ are
two independent random variables with the same distribution as $S$,
then $\frac {(S'+A)+(S''+A)}{2^\alpha}\overset\Delta\to=S$. Here 
$\overset\Delta\to=$ means that on the two sides of the formula
we have two random variables with the same distribution.


We construct the random variable $S$ with the desired distribution
$F$ in the following way. Let $\eta(2^n)$, $n=0,\pm1,\pm2,\dots$, 
be independent Poisson distributed random variables with parameter
$\mu=2^n$, and put 
$$
S=\sum_{n=-\infty}^0 2^{-n\alpha}\eta(2^n) 
+\sum_{n=1}^\infty 2^{-n\alpha} [\eta(2^n)-E(\eta(2^n)]=S_1+S_2.
$$ 
I claim that in the case $\alpha>1/2$ the above definition of
the random variable $S$ is meaningful, i.e. the random sum defining
it is convegent (with probability 1), and if  $S'$ and $S''$ are 
two independent random variables with the same distribution as $S$, 
then $\frac {(S'+A)+(S''+A)}{2^\alpha}\overset\Delta\to=S$.

To prove the above convergence let us observe that
$E2^{-n\alpha}[\eta(2^n)-E\eta(2^n)]=0$, and
$\text{Var}\,(2^{-n\alpha}[\eta(2^n)-E\eta(2^n)])=2^{-2n\alpha+n}$,
which implies that
$\summ_{n=1}^\infty\text{Var}\,( 2^{-n\alpha_n}[\eta(2^n)-E\eta(2^n)])
<\infty$, if $\alpha>\frac12$. Hence the partial sum
$\summ_{n=1}^\infty 2^{-n\alpha} [\eta(2^n)-E(\eta(2^n)]$ defining the
expression $S_2$ is convergent. On the other hand, in the case
$n\le0$ the estimate $P(\eta(2^n)\neq0)=1-e^{-2^n}\le\const 2^n$
holds. Hence $\summ_{n=-\infty}^0 P(\eta(2^n)\neq0)\le\const
\summ_{n=0}^\infty 2^{-n}<\infty$, and the Borel--Cantelli
lemma implies that the sum defining the expression $S_1$ contains
only finitely many term, and as a consequence it is also convergent.

Let $\eta'(2^n)$ and $\eta''(2^n)$, $n=0,\pm1,\pm2,\dots$, be
similarly to $\eta(2^n)$ independent Poisson distributed random
variables with parameter $\mu=2^n$, and define the random variables
$S'$ and $S''$ similarly to $S$, only let us reply the random
variables $\eta(2^n)$ in the sum defining them by the random 
variables $\eta'(2^n)$ and $\eta''(2^n)$ respectively. I claim
that $\frac {(S'+A)+(S''+A)}{2^\alpha}\overset\Delta\to=S$. To
prove this let us observe that 
$$
\frac{2^{-\alpha(n-1)}\eta'(2^{n-1})+2^{-\alpha(n-1)}\eta''(2^{n-1})}
{2^{\alpha}}
\overset\Delta\to=2^{-\alpha n}\eta(2^n),
$$ 
and
$$ 
\align
&\frac{2^{-\alpha(n-1)}(\eta'(2^{n-1})-E(\eta'(2^{n-1}))
+2^{-\alpha(n-1)}(\eta''(2^{n-1})-E\eta''(2^n))}
{2^{\alpha}} \\
&\qquad\qquad \overset\Delta\to=2^{-\alpha n}(\eta(2^n)-E\eta(2^n))
\endalign
$$ 
for all integers $n$, and the expression appearing in these
expressions are independent random variables for different
indices $n$. By summing up these identities for all indices~$n$
(by summing up the first identity for indices
$n\le0$ and the second identity for $n\ge 1$) we get
the identity we wanted to prove. The constant $A$ with
an appropriate choice appears in this identity, because
in the sum defining the random variable $S$ we took the 
term $\eta(2^0)=\eta(1)$ for the index $n=0$
and the term $\eta(2)-E\eta(2)$ for the index $n=1$.

We prove with the help of a result proved in Part~II. that
the above defined random variable $S$ is not stable distributed.
To do this let us observe that the distribution function of $S$ 
is infinitely distributed, and the logarithm of its characteristic
function is defined by such a measure $\mu$ which is concentrated
in the pointe $y_n=2^{n\alpha}$, $n=0,\pm1,\pm2,\dots$, and 
$\mu(2^{n\alpha})=2^{-n}$. In Part~II. of this work we shall prove
that an infinitely divisible distribution uniquely determines
the measure $\mu$ which appears in the L\'evy--Hinchin formula
representing its characteristic function. On the other hand,
observe that if $S_1$, $S_2$ are $S_3$ independent random 
variables with the same distribution as $S$, then the measure
$\bar\mu$ determing the characteristic function of the random
variable $\frac{S_1+S_2+S_3-B}{3^\alpha}$ with the help of the
L\'evy--Hinchin formula is different from the measure~$\mu$.
This measure $\bar\mu$ is defined by the identity
$\bar\mu(A)=3\mu(\frac A{3^\alpha})$ for all measurable sets~$A$. 
It is even concentrated in a different set as the measure~$\mu$.

There is the following idea behind the above construction. An
infiniely divisible distribution is stable if and only if the
measure $\mu$ (or the canonical measure $M$) defining its
characteristic function satisfies certain homogeneity
properties. If only a weakened, discretized version of this
homogeneity property holds, if the desired identitities hold
only for special parameters $t=k^n$ with a fixed integer $k$
and $n=0,\pm1,\pm2,\dots$, then a weakened version of the
stability property expressed in~$(*)$ may hold, but the
infinitely divisible distribution under consideration is not 
stable.

\bye
