ON THE NUMBER OF LATTICE POINTS BETWEEN TWO

ENLARGED AND RANDOMLY SHIFTED COPIES OF AN OVAL

Zheming Cheng^{(1)}, Joel L. Lebowitz^{(2)} and Péter Major^{(3)}

(1)Rutgers University, Department of Mathematics, New Brunswick, NJ 08903, USA (2)Rutgers University, Department of Mathematics and Physics, New Brunswick, NJ 08903, USA (3)Mathematical Institute of the Hungarian Academy of Sciences, P.O.B. 127, H–1364, Budapest, Hungary

Summary: Let A be an oval with a nice boundary in ℝ^{2}, R a large positive number, c > 0 some fixed number and α a uniformly distributed random vector in the unit square [0, 1]^{2}. We are interested in the number of lattice points in the shifted annular region consisting of the difference of the sets \{(R + \frac{c}{R}) A − α\} and \{(R − \frac{c}{R}) A − α\}. We prove that when R tends to infinity, the expectation and the variance of this random variable tend to 4c times the area of the set A, i.e. to the area of the domain where we are counting the number of lattice points. This is consistent with computer studies in the case of a circle or an ellipse which indicate that the distribution of this random variable tends to the Poisson law. We also make some comments about possible generalizations.

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1. Introduction Using computer simulation Cheng and Lebowitz [4] studied the distribution of the number of lattice points in the domain between two concentric circles of radii R + \frac{c}{R} and

R − \frac{c}{R} whose center is uniformly distributed on the unit square [0, 1]^{2}. (By lattice

points we mean points from ℤ^{2}, i.e. from the set of points in ℝ^{2} with integer coordinates.) This computer study, motivated by works of Sinai [9], [10], Bleher [2] and Major [8], suggested that for large R this distribution is asymptotically Poissonian with parameter 4πc, i.e. with the area of the domain where we are counting the number of lattice points. A first step to check the correctness of this statement is to investigate whether the variance of this distribution is asymptotically 4πc, i.e. whether the variance behaves as the Poissonian limit suggests. We answer this question in the affirmative. A similar statement holds for a class of ovals defined as follows.

Definition of an oval. A closed bounded convex set A is an oval in ℝ^{2} if it contains the origin in its interior, and its boundary is a smooth four times differentiable Jordan curve whose curvature is positive at all points.

We also introduce the following notations. Let |A| denote the area of a measurable set A in ℝ^{2}. Given some set A ⊂ ℝ^{2}, α ∈ ℝ^{2} and number R ∈ ℝ^{+} define the set RA−α as

RA − α = {u ∈ ℝ^{2}; u = Rv − α, v ∈ A}

and for c > 0, R^{2} > c introduce the difference set

𝕆_{R}(c, α) = [(R + \frac{c}{R}) A − α] \ [(R − \frac{c}{R}) A − α] .  (1.1)

Clearly, |𝕆_{R}(c, α)| = 4c|A| for all α. The following Theorem is the main result of this paper.

Theorem. Let A be an oval, c > 0 some fixed positive number and α a uniformly distributed random variable on [0, 1]^{2}. For R > \sqrt{c} let ξ_{R} = ξ_{R}(α) denote the number of lattice points in the set 𝕆_{R}(c, α) defined in (1.1). Then the relations

Eξ_{R} = |𝕆_{R}| = 4c|A|  (1.2)
lim_{R→∞} Var ξ_{R} = |𝕆_{R}| = 4c|A|  (1.3)

hold. Here 𝕆_{R} denotes 𝕆_{R}(c, α) with α = 0.

The investigation of the number of lattice points in a domain is a popular subject in number theory. See e.g. [6] for a recent treatment or [7]. This problem also has physical motivations, relating to the investigation of the statistics of eigenvalues in a quantum system with an integrable classical Hamiltonian. For example, if A is a circle, the lattice points n label energy levels E(n; α) = |n − α|^{2} of the Laplacian −(∇ − α)^{2} on the unit torus. These energies can be thought of as points on the real line and their statistics can be studied. This problem seems to be very hard. An easier problem is to consider not a fixed α but a random one distributed uniformly on the unit square [0, 1]^{2}, and this is what we have done. A widely accepted conjecture in the physics community is that the distribution of levels is, for typical systems in this class, locally Poissonian [1], i.e. the statistics of the energy levels in the interval [E, E + L], L is fixed and E is uniformly distributed in an interval [0, T] with T → ∞, behave like Poisson distributed points with density π. In our context the conjecture is the following:

Let P (n; R) be the probability that there are exactly n lattice points in 𝕆_{R}(c, α). Then

lim_{R→∞} P (n, R) = p(n) with p(n) = \frac{(4c |A|)^{n}}{n!} e^{−4c|A|n} .  (1.4)

Such a result was proved by Sinai [8] and Major [9] for the number of lattice points in scaled annuli domains bounded by very random curves. Here all the randomness comes from α, so the proof of (1.4), if indeed it is true, is far from trivial. Our theorem proves that the limit of the first and second moment of P (n; R) has the right behavior when R → ∞. In fact, our result also shows that the covariance of pairs of distinct random variables η_{R}^{(j)} = ξ_{R+jc} − |𝕆_{R}| vanishes as R → ∞. This suggests that

lim_{c→∞} lim_{R→∞} \frac{ξ_{R} − |𝕆_{R}|}{\sqrt{|𝕆_{R}|}} should be a Gaussian random variable. This is consistent with

taking the large parameter limit of the Poissonian distribution, but may be valid more generally.

2. Proof of the Theorem The proof of relation (1.2) is simple. We can write Eξ_{R}(α) as the the sum of the probabilities

Eξ_{R}(α) = ∑_{m∈ℤ^{2}} P (m ∈ 𝕆_{R}(c, α)) = ∑_{m∈ℤ^{2}} |𝕆_{R} ∩ ([0, 1]^{2} − m)| = |𝕆_{R}| .

Because of (1.2) formula (1.3) is equivalent to the relation

lim_{R→∞} Eξ_{R}(α)(ξ_{R}(α) − 1) = |𝕆_{R}|^{2} .  (2.1)

We claim that

Eξ_{R}(α)(ξ_{R}(α) − 1) = ∑_{m∈ℤ^{2}\{0}} |𝕆_{R} ∩ (𝕆_{R} − m)| .  (2.2)

Indeed,

Eξ_{R}(α)(ξ_{R}(α) − 1) = ∑_{m∈ℤ^{2}} ∑_{m_{1}∈ℤ^{2}\{m}} P (m ∈ 𝕆_{R}(c, α), m_{1} ∈ 𝕆_{R}(c, α))

= ∑_{m∈ℤ^{2}} ∑_{m_{1}∈ℤ^{2}\{m}} |[0, 1]^{2} ∩ (𝕆_{R} − m) ∩ (𝕆_{R} − m_{1})|

= ∑_{m∈ℤ^{2}} ∑_{m_{1}∈ℤ^{2}\{0}} |([0, 1]^{2} + m) ∩ 𝕆_{R} ∩ (𝕆_{R} − m_{1})|

= ∑_{m∈ℤ^{2}\{0}} |𝕆_{R} ∩ (𝕆_{R} − m)| .

Hence to prove the Theorem it is enough to prove relation (2.1) with the help of relation (2.2). This requires a good estimate on the area of 𝕆_{R} ∩ (𝕆_{R} − m). First we introduce some notations.

Let us denote the boundary of the set tA for t > 0, by γ_{t}, and let γ = γ_{1}. For some x ∈ γ_{t} let φ(x) denote the angle of the vector x and ψ(x) the angle of the normal of the curve γ_{t} at x (pointing outside of the domain tA) with the vector e_{1} = (1, 0). Given some z ∈ ℝ^{2} \ {0} let d(z, t) denote the diameter of the set tA in the direction z, i.e.

d(z, t) = max {|z_{1} − z_{2}|; z_{1} ∈ γ_{t}, z_{2} ∈ γ_{t}, z_{1} − z_{2} = λz, with some λ > 0} .

Let z_{d,r}(t) and z_{d,l}(t) be the end points of this maximal vector, i.e. z_{d,r}(t) ∈ γ_{t}, z_{d,l}(t) ∈ γ_{t}, z_{d,r}(t) − z_{d,l}(t) = λz with λ > 0 and |z_{d,r}(t) − z_{d,r}(t)| = d(z, t). For z ∈ ℝ^{2} let z^{⊥} denote the vector z rotated by +π/2, and define K_{t}^{+}(z) and K_{t}^{−}(z) as the half planes whose boundary is the line going through the points z_{d,r}(t) and z_{d,l}(t) and which are in the direction z^{⊥} and −z^{⊥} of this line respectively. For z ∈ ℝ^{2} and 0 < |z| ≤ d(z, t) define the (unique) points z_{t,r}^{+}, z_{t,l}^{+} ∈ K_{t}^{+}(z) ∩ γ_{t} and z_{t,r}^{−}, z_{t,l}^{−} ∈ K_{t}^{−}(z) ∩ γ_{t} such that z_{t,r}^{+} − z_{t,l}^{+} = z and z_{t,r}^{−} − z_{t,l}^{−} = z.

For z_{1}, z_{2} ∈ γ_{R} define the function

F_{R}(z_{1}, z_{2}) = \frac{4c^{2}}{R^{4}} \frac{|z_{1}||z_{2}| cos(φ(z_{1}) − ψ(z_{1})) cos(φ(z_{2}) − ψ(z_{2}))}{| sin(ψ(z_{1}) − ψ(z_{2}))|}

and for some z ∈ ℝ^{2}, 0 < |z| ≤ d(z, R) the functions

f_{R}^{+}(z) = F_{R}(z_{R,l}^{+}, z_{R,r}^{+}) , f_{R}^{−}(z) = F_{R}(z_{R,l}^{−}, z_{R,r}^{−}) .

For A ⊂ ℤ^{2} and B ⊂ ℤ^{2} define their sum

A + B = {x + y, x ∈ A, y ∈ B}

and

A^{(2)}(R) = RA + (−RA) .

We claim that

∫_{A^{(2)}(R)} [f_{R}^{+}(z) + f_{R}^{−}(z)] dz = 16c^{2} |A|^{2} .  (2.3)

Put

h_{R}(z) = |𝕆_{R} ∩ (𝕆_{R} − z)| , z ∈ ℝ^{2} .  (2.4)

We then also claim that

lim_{R→∞} { ∑_{m∈ℤ^{2}\{0}} h_{R}(m) − ∫_{A^{(2)}(R)} [f_{R}^{+}(z) + f_{R}^{−}(z)] dz} = 0 .  (2.5)

Relations (2.2), (2.3) and (2.5) together imply (2.1) hence also the Theorem. To prove (2.3) we introduce the maps

G^{±}: A^{(2)}(R) \ {0} ↦ γ_{R} × γ_{R} ,
G^{+}(z) = (z_{R,l}^{+}, z_{R,r}^{+}), G^{−}(z) = (z_{R,l}^{−}, z_{R,r}^{−})

Observe that the set Int G^{+}(A^{(2)}(R) \ {0}) ∩ Int G^{−}(A^{(2)}(R) \ {0}) is empty, the maps G^{±} are diffeomorphisms on Int A^{(2)}(R)\{0}, and G^{+}(A^{(2)}(R)\{0})∪G^{−}(A^{(2)}(R)\{0}) is γ_{R} × γ_{R} \ {(z, z), z ∈ γ_{r}}.

The inverses of the maps G^{±}(z) have Jacobians |sin (ψ(z_{R,r}^{±}) − ψ(z_{R,l}^{±}))|. To see this we make the following observation: Let [z_{1}, z_{1} +dz_{1}] and [z_{2}, z_{2} +dz_{2}] be two small curves on γ_{R} starting from some points z_{1} and z_{2} respectively. Then the inverse of the map G^{±} maps the set [z_{1}, z_{1}+dz_{1}]×[z_{2}, z_{2}+dz_{2}] approximately to z_{2}−z_{1}+Δ(dz_{1}, dz_{2}), where Δ(dz_{1}, dz_{2}) is a parallelogram with one vertex at the origin, whose sides are the vectors dz_{1} and dz_{2}. The area of this parallelogram is |dz_{1}||dz_{2}|| sin(ψ(z_{1}) − ψ(z_{2}))|. We can approximate the area of the image of the above domain by the inverse map (G^{±})^{−1} with the area of this parallelogram. Since this approximation gives only an error of order o(|dz_{1}||dz_{2}|) the Jacobian has the form we have stated.

The above relations imply that

∫_{A^{(2)}(R)} [f_{R}^{+}(z) + f_{R}^{−}(z)] dz

= ∫_{γ_{R}×γ_{R}} \frac{4c^{2}}{R^{4}} |z_{1}||z_{2}| cos(ψ(z_{1}) − φ(z_{1})) cos(ψ(z_{2}) − φ(z_{2})) dz_{1} dz_{2}

= [\frac{2c}{R^{2}} ∫_{γ_{R}} |z| cos(ψ(z) − φ(z)) dz]^{2}

= [2c ∫_{γ_{1}} |z| cos(ψ(z) − φ(z)) dz]^{2} = 16c^{2} |A|^{2} ,

hence relation (2.3) holds. To prove relation (2.5) we need some geometrical facts formulated in relations (2.9) and (2.10) and a lemma about the value of h_{R}(z). They will be proved in the next Section.

Lemma 1. There is some ε > 0 such that the function h_{R}(z) defined in (2.4) satisfies the following estimates:

a) For 1 ≤ |z| < εR, h_{R}(z) < \frac{const.}{R |z|} .

b) For all 0 < η ≤ ε and ηR ≤ |z| ≤ (1 − η)d(z, R)

R^{2} {h_{R}(z) − [f_{R}^{+}(z) + f_{R}^{−}(z)]} → 0 as R → ∞ ,

and the convergence is uniform in z.

c) Let us fix some positive constant B > 0. Then the following inequalities hold:

c1) h_{R}(z) < \frac{const.}{R^{3/2}\sqrt{d(z, R) − |z|}} if (1 − ε)d(z, R) ≤ |z| ≤ d(z, R) − \frac{B}{R}.

c2) h_{R}(z) < \frac{const.}{R} if d(z, R) − \frac{B}{R} ≤ |z| ≤ d(z, R) + \frac{B}{R}.

d) h_{R}(z) = 0 if |z| > d(z, R) + \frac{B}{R}, and B is larger than c times the diameter of the set A.

To prove relation (2.5) let us introduce the sets

D_{1}^{ε}(R) = {z ∈ ℝ^{2}, 0 < |z| < εR}
D_{2}^{ε}(R) = (1 − ε)A^{(2)}(R) \ D_{1}^{ε}(R)
D_{3}^{ε}(R) = (1 − \frac{B}{R^{2}} ) A^{(2)}(R) \ (1 − ε)A^{(2)}(R)
D_{4}^{ε}(R) = (1 + \frac{B}{R^{2}} ) A^{(2)}(R) \ (1 − \frac{B}{R^{2}} ) A^{(2)}(R) .

Define the discrete measure μ on the positive half-line [0, ∞],

μ([0, x]) = {the number of lattice points in the set xA^{(2)}} ,

where A^{(2)} denotes A^{(2)}(R) with R = 1. Define also the signed measure

ν([0, x]) = μ([0, x]) − x^{2}Area (A^{(2)}) .

If A is an oval, then A^{(2)} is again an oval. (See [3]). (This means that the boundary of A^{(2)} is again strictly convex, smooth, and has positive curvature at all points.) Hence the results known for ovals can be applied to A^{(2)}. In particular, we can state because of a result of Colin de Verdière [5] that

|ν([0, x])| < const. x^{2/3} for all x > 1 .  (2.6)

Let us also remark that the normals of γ_{R} in the points z_{d,r}(R) and z_{d,l}(R) satisfy the relation

ψ(z_{d,r}(R)) = ψ(z_{d,l}(R)) + π  (2.7)

for all z ∈ ℝ^{2} \ {0}, i.e. the normals in the points z_{d,r}(R) and z_{d,l}(R) point in opposite directions. The half-line λz, λ > 0, intersects the boundary of A^{(2)}(R) at distance d(z, R) from the origin. Hence Part c1) of Lemma 1 bounds the value of h_{R}(z) for z ∈ D_{3}^{ε}(R) and Part c2) bounds the value of h_{R}(z) for z ∈ D_{4}^{ε}(R).

By Part a) of Lemma 1

∑_{m∈ℤ^{2}∩D_{1}^{ε}(R)\\{0\}} h_{R}(m) < const. \frac{1}{R} ∑_{\begin{matrix}m∈ℤ^{2} \\ 0<|m|<εR \end{matrix}} \frac{1}{|m|} < const. ε.

Since R^{2} (f_{R}^{+}(Rz) + f_{R}^{−}(Rz)) is uniformly continuous in the set \frac{1}{R}D_{2}^{ε}(R), hence

∑_{m∈ℤ^{2}∩D_{2}^{ε}(R)} h_{R}(m) − ∫_{D_{2}^{ε}(R)} [f_{R}^{+}(z) + f_{R}^{−}(z)] dz → 0 as R → ∞

by Part b) of Lemma 1. Put

H_{R}(t) = sup_{z∈t∂A^{(2)}} h_{R}(z) ,

where ∂A^{(2)} denotes the boundary of A^{(2)}. By using Part c1) of Lemma 1, integrating by parts and applying (2.6) we can write

∑_{m∈ℤ^{2}∩D_{3}^{ε}(R)} h_{R}(m) ≤ ∫_{(1−ε)R}^{(R− \frac{B}{R})} H_{R}(x) μ(dx)

< \frac{const.}{R^{3/2}} ∫_{(1−ε)R}^{(R− \frac{B}{R})} \frac{x}{\sqrt{R − x}} dx + \frac{const.}{R^{3/2}} ∫_{(1−ε)R}^{(R− \frac{B}{R})} \frac{ν( dx)}{\sqrt{R − x}}

< const. \sqrt{ε} + \frac{const.}{R^{3/2}} [\frac{ν([0, x])}{\sqrt{R − x}}]_{(1−ε)R}^{(R− \frac{B}{R})}

+ const. ∫_{(1−ε)R}^{(R− \frac{B}{R})} \frac{ν([0, x])}{R^{3/2} (R − x)^{3/2}} dx

< const. (\sqrt{ε} + R^{−1/3} + R^{−5/6} ∫_{(1−ε)R}^{(R− \frac{B}{R})} \frac{1}{(R − x)^{3/2}} dx)

< const. [R^{−1/3} + \sqrt{ε}] .

Similarly, by Part c2) of Lemma 1

∑_{m∈ℤ^{2}∩D_{4}^{ε}(R)} h_{R}(m) ≤ const. ∫_{(R− \frac{B}{R})}^{(R+ \frac{B}{R})} \frac{1}{R} μ( dx)

≤ const. ∫_{(R− \frac{B}{R})}^{(R+ \frac{B}{R})} \frac{x}{R} dx + \frac{const.}{R} ν ([R − \frac{B}{R} , R + \frac{B}{R} ])

< const. [\frac{1}{R} + R^{−1/3}] < const. R^{−1/3} .

The above relations together with Part d) of Lemma 1 imply that

∑_{m∈ℤ^{2}\\{0\}} h_{R}(m) − ∫_{D_{2}^{ε}(R)} [f_{R}^{+}(z) + f_{R}^{−}(z)] dz = O (\sqrt{ε} + R^{−1/3}) .  (2.8)

We claim that

I_{1} = ∫_{D_{1}^{ε}(R)} [f_{R}^{+}(z) + f_{R}^{−}(z)] dz = O(ε)  (2.8′)

and

I_{2} = ∫_{A^{(2)}(R)\(1−ε)A^{(2)}(R)} [f_{R}^{+}(z) + f_{R}^{−}(z)] dz = O (\sqrt{ε}) .  (2.8′′)

Since ε > 0 can be chosen arbitrary small the above relations imply (2.5).

In Section 3 we shall prove the following statements. There is some ε > 0 such that if |z| = uR with some 0 < u < ε, then the normals of γ_{R} satisfy the inequality

|ψ(z_{R,r}^{±}) − ψ(z_{R,l}^{±})| > const. u ,  (2.9)

8 ZHEMING CHENG, JOEL L. LEBOWITZ AND PÉTER MAJOR

and if |z| = (1 − u)d(z, R) with some 0 < u < ε, then

|ψ(z_{R,r}^{±}) − ψ(z_{R,l}^{±}) − π| > const. \sqrt{u} .  (2.10)

In the first case we get that

|f_{R}^{+}(z)| + |f_{R}^{−}(z)| < const. \frac{1}{R^{2}u}  (2.11)

and in the second case

|f_{R}^{+}(z)| + |f_{R}^{−}(z)| < const. \frac{1}{R^{2}\sqrt{u}} .  (2.12)

Thus we get, by integrating in a polar coordinate system and applying the estimate (2.11), that

I_{1} < const. \frac{1}{R^{2}} ∫_{0}^{ε} R^{2}u\frac{1}{u} du < const. ε .

Relation (2.12) implies that for (1 − ε)R ≤ t ≤ R

∮_{t∂A^{(2)}} [f_{R}^{+}(z) + f_{R}^{−}(z)] dz ≤ const. ∮_{t∂A^{(2)}}(1 − t)^{−1/2}R^{−3/2} dz

≤ const. \frac{1}{\sqrt{(1 − t)R}} .

Integrating first on the curves γ_{t}, (1 − ε)R < t < R, we get that

I_{2} < const. ∫_{0}^{εR} R^{−1/2}u^{−1/2} du < const. \sqrt{ε} .

Hence we proved the Theorem with the help of Lemma 1 and formulas (2.9) and (2.10).

3. Proof of Lemma 1 We shall need a result about convex sets in the proof. To formulate it we introduce some notations. Let us fix some vector z ∈ ℝ^{2}, z ≠ 0. Introduce the coordinate system whose coordinate axis x is in the direction \frac{z}{|z|} and the coordinate axis y is its rotation

with +\frac{π}{2} , in the direction \frac{z^{⊥}}{|z^{⊥}|}. In this new coordinate system let (x_{0}(t), y_{0}(t)) and

(x_{1}(t), y_{1}(t)) be the points of tangency of the curve γ_{t} with the line parallel to the vector z in the half-spaces K_{t}^{+}(z) and K_{t}^{−}(z) respectively. For y_{1}(t) < u < y_{0}(t) the line y = u intersects the curve γ_{t} in the points x_{l,t}(u) and x_{r,t}(u), x_{l,t}(u) < x_{r,t}(u). We shall prove the following

Lemma 2. There are some positive constants A > 0, 0 < B_{1} < B_{2} depending only on the curve γ such that

B_{1}\sqrt{t(y_{0}(t) − u)} < x_{r,t}(u) − x_{0}(t) < B_{2}\sqrt{t(y_{0}(t) − u)} ,
B_{1}\sqrt{t(y_{0}(t) − u)} < x_{0}(t) − x_{l,t}(u) < B_{2}\sqrt{t(y_{0}(t) − u)}
\frac{B_{1}\sqrt{t}}{\sqrt{y_{0}(t) − u}} < −\frac{d}{du}x_{r,t}(u) < \frac{B_{2}\sqrt{t}}{\sqrt{y_{0}(t) − u}} ,

\frac{B_{1}\sqrt{t}}{\sqrt{y_{0}(t) − u}} < \frac{d}{du}x_{l,t}(u) < \frac{B_{2}\sqrt{t}}{\sqrt{y_{0}(t) − u}} ,

if y_{0}(t) − tA < u < y_{0}(t), and

B_{1}\sqrt{t(u − y_{1}(t))} < x_{r,t}(u) − x_{1}(t) < B_{2}\sqrt{t(u − y_{1}(t))} ,
B_{1}\sqrt{t(u − y_{1}(t))} < x_{1}(t) − x_{l,t}(u) < B_{2}\sqrt{t(u − y_{1}(t))}
\frac{B_{1}\sqrt{t}}{\sqrt{u − y_{1}(t)}} < \frac{d}{du}x_{r,t}(u) < \frac{B_{2}\sqrt{t}}{\sqrt{u − y_{1}(t)}} ,

\frac{B_{1}\sqrt{t}}{\sqrt{u − y_{1}(t)}} < −\frac{d}{du}x_{l,t}(u) < \frac{B_{2}\sqrt{t}}{\sqrt{u − y_{1}(t)}} ,

if y_{1}(t) < u < y_{1}(t) + tA.

Proof of Lemma 2. Let us first restrict our attention to the case t = 1. It is more convenient to work with the inverse of the function x_{r,1}(u). Let (x, g(x)) be a small part of the curve γ in the neighborhood of the point (x_{0}(1), y_{0}(1)), and let ρ(x) be the curvature of the curve γ in the point (x, g(x)). We can write in an interval [x_{0}(1), x_{0}(1)+ η], η > 0,

ρ(x) = −\frac{[1 + g′(x)^{2}]^{3/2}}{g′′(x)} .

Put z(x) = g′(x). Since z(x_{0}(1)) = 0, the last relation implies that

∫_{0}^{z(x)} \frac{dt}{(1 + t^{2})^{3/2}} = − ∫_{x_{0}(1)}^{x} \frac{1}{ρ(t)} dt

or

\frac{z(x)}{\sqrt{1 + z^{2}(x)}} = −P (x)

with P (x) = ∫_{x_{0}(1)}^{x} \frac{1}{ρ(t)} dt. Since the curvature of γ is separated both from zero and

infinity, there are some constants K_{2} > K_{1} > 0 such that

K_{1}(x − x_{0}(1)) < P (x) < K_{2}(x − x_{0}(1))

Since z(x) = \frac{−P (x)}{\sqrt{1 − P^{2}(x)}} the last relation implies that

C_{1}(x − x_{0}(1)) < −g′(x) < C_{2}(x − x_{0}(1))

with some C_{2} > C_{1} > 0 in an interval x ∈ [x_{0}(1), x_{0}(1) + η). Since g(x_{0}(1)) = y_{0}(1) we get by integrating that

−\frac{C_{2}}{2} (x − x_{0}(1))^{2} < g(x) − y_{0}(1) < −\frac{C_{1}}{2} (x − x_{0}(1))^{2}.

These formulas imply that x_{r,1}(u), the inverse of g(x), satisfies Lemma 1 for t = 1 in an interval [y_{0}(1) − A, y_{0}(1)]. The remaining statements of Lemma 1 for t = 1 can be proved in the same way. The case of general t > 0 follows from the case t = 1 because of the homogeneity properties of γ_{t}. □

Now we turn to the proof of relations (2.9) and (2.10). Here again we can restrict our attention to the case R = 1. Let us recall that the curvature ρ(x) of γ in a point x ∈ γ and the angle of the normal ψ(x) in this point satisfy the relation

\frac{dψ(x)}{ds(x)} = ρ(x) ,  (3.1)

where s(x) is the length of the arc (x_{0}, x) of γ with some fixed x_{0} ∈ γ.

Under the conditions imposed for formula (2.9) the length of the arc (z_{1,l}^{±}, z_{1,r}^{±}) is greater than u. Since ρ(x) is bounded from below by a positive constant, we get formula (2.9) by integrating (3.1).

The proof of (2.10) is similar. Here we can apply formula (2.7). Because of this formula it is enough to show that

|ψ(z_{1,l}^{±}) − ψ(z_{d,l}(1))| > const. \sqrt{u}
|ψ(z_{1,r}^{±}) − ψ(z_{d,r}(1))| > const. \sqrt{u} (3.2)

under the conditions imposed for (2.10).

Given two vectors z_{1} and z_{2} let ∠(z_{1}, z_{2}) denote the angle between them. For z ∈ γ_{1} let n(z) denote the normal vector to the curve γ_{1} at z. We make the following observation: For any z ∈ ℝ^{2}\{0} consider the end points z_{d,l} = z_{d,l}(1) and z_{d,r} = z_{d,r}(1) of the interval with maximal length in A in the direction of z. The normal of γ_{1} in these points cannot be almost orthogonal to z. More explicitly, there is some η > 0 such that

−\frac{π}{2} + η < ∠ (−z, n(z_{d,l})) < \frac{π}{2} − η .  (3.3)

(This statement is equivalent to the following one: If z is a boundary point of the oval A^{(2)}, then the vector z cannot be almost orthogonal to the normal of the boundary of A^{(2)} in this point. The equivalence of these two statements follows from the following argument. The vector z_{d,l} − z_{d,r} is on the boundary of A^{(2)}, and it is parallel to z. The normal of A^{(2)} in this point is parallel to n(z_{d,l}). The proof of the second statement is simpler.)

We claim that under the conditions imposed for formula (2.10) the distance of the parallel lines going through the points z_{1,l}^{±} and z_{1,r}^{±} and through the points z_{d,l}(1) and

z_{d,r}(1) is bigger than const. \sqrt{u}. Put m_{l} = m_{l}(z) = \frac{1}{cos ∠(z, n(z_{d,l}))} and m_{r} = m_{r}(z) =

\frac{1}{cos ∠(z, n(z_{d,r}))}. Then m_{r} = −m_{l} by (2.7), and by (3.3) there is some ∞ > K > 0

such that −K < m_{r}(z) < K for all z ∈ ℝ^{2} \ {0}.

Let us fix a new coordinate system with the origin in a point of the line going through the points z_{l,r} and z_{d,r}, with the x axis in the direction of the vector z and y axis in the direction z^{⊥}, and let us work in it. For ε > v > 0 let z_{l}^{+}(v) = (z_{l}^{+}(v), v) and z_{r}^{+}(v) = (z_{r}^{+}(v), v), z_{l}^{+}(v) < z_{r}^{+}(v), be the two intersections of γ and the line y = v, and put z(v) = z_{l}^{+}(v) − z_{r}^{+}(v). It follows from Lemma 2 (with its application in the coordinate system with coordinate axes parallel to the normal and to the tangent vector of the curve γ_{1} in the points z_{d,l}(1) and z_{d,r}(1) respectively) that there is some C > 0 such that z_{l}^{+}(v) < z_{d,l}(1) + vm_{l} + Cv^{2} and z_{r}^{+}(v) > z_{d,r}(1) + vm_{l} − Cv^{2}. The above relations imply that |z(v)| = z_{r}^{+}(v) − z_{l}^{+}(v) > d(z, 1) − 2Cv^{2}. Since we imposed the condition |z| = d(z, 1) − ud(z, 1), this relation implies that the distance between the parallel lines going through the points z_{1,l}^{+} and z_{1,r}^{+} and through the points z_{d,l}(1) and z_{d,r}(1) is greater than const. \sqrt{u}. The same statement holds if z_{1,l}^{+} and z_{1,r}^{+} are replaced by z_{1,l}^{−} and z_{1,r}^{−}. Hence the arcs z_{1,l}^{±}, z_{d,l} and the arcs z_{1,r}^{±}, z_{d,r} are longer than const. \sqrt{u}. Hence relation (3.1) and the strict positivity of ρ(x) imply formula (3.2) and hence formula (2.10) too.

Now we turn to the proof of Lemma 1. We introduce the abbreviation R^{+} instead of (R + \frac{c}{R}) and R^{−} instead of (R − \frac{c}{R}).

Proof of Part a). Let us introduce the coordinate system with x axis parallel to z and y axis parallel to z^{⊥}. We shall estimate, by means of Lemma 2, the length of the intersection of the set 𝕆_{R} ∩ (𝕆_{R} − z) with the lines y = u for different u-s.

Define u_{R^{+}}(z) as the (unique) solution of the equation

x_{r,R^{+}}(u) − x_{l,R^{+}}(u) = |z|, u > u_{d,R^{+}} ,

where u_{d,R^{±}} is defined by the formula

x_{r,R^{±}} (u_{d,R^{±}}) − x_{l,R^{±}} (u_{l,R^{±}}) = max_{u} [x_{r,R^{±}}(u) − x_{l,R^{±}}(u)] ,

12 ZHEMING CHENG, JOEL L. LEBOWITZ AND PÉTER MAJOR

that is, it is the level u at which the horizontal line y = u has the longest intersection with the set R^{±}A.

We claim that there is some K > 0 such that

x_{r,R^{+}}(u) − x_{l,R^{+}}(u) < |z| + \frac{const.}{|z|} if u_{R^{+}}(z) − \frac{K}{R} ≤ u ≤ u_{R^{+}}(z) ,  (3.4)
x_{r,R^{−}}(u) − x_{l,R^{−}}(u) > |z| if u_{d,R^{−}} ≤ u ≤ u_{R^{+}}(z) − \frac{K}{R} ,  (3.4′)

and

\begin{matrix}x_{r,R^{+}}(u) − x_{r,R^{−}}(u) & < \frac{1}{2} \\ x_{l,R^{+}}(u) − x_{l,R^{−}}(u) & < \frac{1}{2} \end{matrix}\} if u_{d,R^{−}} ≤ u ≤ u_{R^{+}}(z) − \frac{K}{R} .  (3.4′′)

First we show that relations (3.4)—(3.4′′) imply that

|𝕆_{R} ∩ (𝕆_{R} − z) ∩ K_{R^{−}}^{+}(z)| < \frac{const.}{R |z|} .  (3.5)

To see this we show that the intersection of 𝕆_{R} ∩ (𝕆_{R} − z) with the line y = u has a length smaller than \frac{const.}{|z|} if u_{R^{+}}(z) − \frac{K}{R} ≤ u ≤ u_{R^{+}}(z), and it is empty if u > u_{R^{+}}(z) or u_{d,R^{−}} ≤ u ≤ u_{R^{+}}(z) − \frac{K}{R}. The above relations imply (3.5).

The above intersections are contained in the interval [x_{l,R^{+}}(u), x_{r,R^{+}}(u) − |z|] whose length is less than \frac{const.}{|z|} if u_{R^{+}}(z)− \frac{K}{R} ≤ u ≤ u_{R^{+}}(z) by (3.4). The distance x_{r,R^{+}}(u)− x_{l,R^{+}}(u) is less than |z| for u > u_{R^{+}}(z), because it is a (convex) monotone decreasing function of u in the interval [u_{d,R^{+}}, y_{0}(R^{+})] and it equals |z| for u_{R^{+}}(z). This implies that the intersection 𝕆_{R} ∩ (𝕆_{R} − z) ∩ {(x, u), x ∈ ℝ^{1}} is empty for u > u_{R^{+}}(z). To see that it is empty for u_{d,R^{−}} ≤ u ≤ u_{R^{+}}(z) − \frac{K}{R} too, observe first that the intersection of 𝕆_{R} with the line y = u consists of two intervals, [x_{l,R^{+}}(u), x_{l,R^{−}}(u)] and [x_{r,R^{−}}(u), x_{r,R^{+}}(u)]. The intersections

[x_{l,R^{+}}(u), x_{l,R^{−}}(u)] ∩ ([x_{l,R^{+}}(u), x_{l,R^{−}}(u)] − |z|)

and

[x_{r,R^{−}}(u), x_{l,R^{+}}(u)] ∩ ([x_{r,R^{−}}(u), x_{l,R^{+}}(u)] − |z|)

are empty because of (3.4′′) and the condition |z| > 1. The intersection

[x_{l,R^{+}}(u), x_{l,R^{−}}(u)] ∩ ([x_{r,R^{−}}(u), x_{l,R^{+}}(u)] − |z|)

is empty because of (3.4′), and the intersection

[x_{r,R^{−}}(u), x_{l,R^{+}}(u)] ∩ ([x_{l,R^{+}}(u), x_{l,R^{−}}(u)] − |z|)

is always empty. In such a way we deduced (3.5) from (3.4)—(3.4′′).

Let us recall the following notation. The point (x_{0}(R^{+}), y_{R}^{+}) is the point of tangency of the curve γ_{R^{+}} with the line parallel to z in the half-space K_{R^{+}}^{+}(z). To prove relations (3.4)—(3.4′′) let us first observe that because of the first two relations in Lemma 2

B_{1}R(y_{0}(R^{+}0) − u_{R^{+}}(z) < z^{2} < B_{2}R(y_{0}(R^{+}0) − u_{R^{+}}(z) .

In particular, for |z| ≥ 1 y_{0}(R^{+}) − u_{R^{+}}(z) > \frac{const.}{R} . Hence, by the third and fourth relations in Lemma 2

C_{1} \frac{R}{|z|} < −\frac{d}{du} (x_{r,R^{+}}(u) − x_{l,R^{+}}(u)) < C_{2} \frac{R}{|z|} for u_{R^{+}}(z) − \frac{K}{R} < u < u_{R^{+}}(z) (3.6)

with some C_{2} > C_{1} > 0. Since x_{r,R^{+}} (u_{R^{+}}(z)) − x_{l,R^{+}} (u_{R^{+}}(z)) = |z| we get relation (3.4) by integrating the right-hand side of (3.6) in the interval [u, u_{R^{+}}(z)].

Because of the left-hand side of (3.6) we can choose for any D > 0 a number C =

C(D) > 0 such that for u_{0} = u_{R^{+}}(z) − \frac{C}{R} x_{r,R^{+}}(u_{0}) − x_{l,R^{+}}(u_{0}) > |z| + \frac{D}{|z|}. We

rewrite this relation by turning from γ_{R^{+}} to γ_{R^{−}}. In this calculation we exploit that

|z| < const. R if z ∈ γ_{R^{+}}. Putting u_{1} = (1 − η)u_{0} = (1 − \frac{c}{R^{2}} ) (1 + \frac{c}{R^{2}} )^{−1} u_{0} we have

u_{1} > u_{R^{+}}(z) − \frac{K}{R} with an appropriate K > 0, η = O(R^{−2}) and

x_{r,R^{−}}(u_{1}) − x_{l,R^{−}}(u_{1}) = (1 − η) (x_{r,R^{+}}(u_{0}) − x_{l,R^{+}}(u_{0}))

> (1 − η) (|z| + \frac{D}{|z|}) > |z|

if D > 0 is sufficiently large. This means that relation (3.4′) holds for u_{1} > u_{R^{+}}(z)− \frac{K}{R}.

Because of the monotonicity of x_{r,R^{−}}(u_{0}) − x_{l,R^{−}}(u_{0}) for u_{d,R^{−}} < u < y_{0}(R^{−}) relation (3.4′) holds.

To prove relation (3.4′′) first we observe that x_{r,R^{+}}(ū) = (1+β)x_{r,R^{−}}(u) with 1+β =

(1 + \frac{c}{R^{2}} ) (1 − \frac{c}{R^{2}} )^{−1} and ū = (1 + β)u. Hence x_{r,R^{+}}(ū) = x_{r,R^{−}}(u) + O (\frac{1}{R}) and

|u − ū| < \frac{L}{R} with some L > 0. The derivative \frac{d}{du}x_{r,R^{+}}(u) is a monotone decreasing function of u, hence it follows from the third relation in Lemma 2 that

|\frac{d}{du}x_{r,R^{+}}(u)| < \frac{R}{3L}

if y_{1}(R^{+}) − \frac{K}{R} ≤ u ≤ y_{0}(R^{+}) − \frac{K}{R}, and K > 0 is chosen sufficiently large. Hence

|x_{r,R^{+}}(u) − x_{r,R^{+}}(ū)| < \frac{R}{3L}|u − ū| + O (\frac{1}{R}) < \frac{1}{2} .

The first relation of (3.4′′) is proved, and the second one can be proved in the same way.

We have proved relation (3.5). It can be proved in the same way if K_{R^{+}}^{+}(z) is replaced by K_{R^{+}}^{−}(z). These two relations together imply Part a).

Proof of Part b). We define two parallelograms P^{+}(z) and P^{−}(z). The parallelogram P^{+}(z) is bounded by two pairs of parallel lines, the lines of the first pair are going through the points (1 − \frac{c}{R^{2}} ) z_{R,l}^{+} and (1 + \frac{c}{R^{2}} ) z_{R,l}^{+} and they have normal ψ(z_{R,l}^{+}), the lines of the second pair are going through the points

(1 − \frac{c}{R^{2}} ) z_{R,r}^{+} − z and (1 + \frac{c}{R^{2}} ) z_{R,r}^{+} − z ,

and they have normal ψ(z_{R,r}^{+}). The parallel pairs of lines bounding P^{−}(z) are going

through the points (1 − \frac{c}{R^{2}} ) z_{R,l}^{−} and (1 + \frac{c}{R^{2}} ) z_{R,l}^{−} with normal z_{R,l}^{−}(ψ), and through

the points (1 − \frac{c}{R^{2}} ) z_{R,r}^{−} − z and (1 + \frac{c}{R^{2}} ) z_{R,r}^{−} − z with normal z_{R,r}^{−}(ψ). The par-

allelograms P^{+}(z) and P^{−}(z) have area f_{R}^{+}(z) and f_{R}^{−}(z) respectively, and they are disjoint if |z| > ηR. Since the difference of these parallelograms and the domains 𝕆_{R} ∩ (𝕆_{R} − z) ∩ K_{R}^{+}(z) and 𝕆_{R} ∩ (𝕆_{R} − z) ∩ K_{R}^{−}(z) have an area of order o (R^{−2}), the above relations imply Part b) of Lemma 1.

Proof of Part c). Let us work in the coordinate system with origin z_{d,l}(R^{+}), with xcoordinate axis in the direction −n(z_{d,l}(R^{+})), the normal of γ_{R^{+}} in the point z_{d,l}(R^{+}) showing inside the domain A_{R^{+}}, and with y-axis in the direction z_{e} = −n(z_{d,l}(R^{+}))^{⊥}, the tangent of γ_{R^{+}} in this point which is obtained when the x axis is rotated with angle +π/2. Let y_{R^{+}}^{±}(u) be the y coordinate of the intersection of the set γ_{R^{+}} ∩ K_{R^{+}}^{±}(z) with the line x = u and y_{R^{−}}^{±}(u) the y coordinate of the intersection of the set γ_{R^{−}} ∩ K_{R^{−}}^{±}(z) with this line. We shall estimate the length of the intersection of 𝕆_{R} with the line x = u. We shall prove that the Lebesgue measure of this intersection satisfies the inequality

λ (𝕆_{R} ∩ \{(u, y), y ∈ ℝ^{(1)}\}) < const. max (\frac{1}{\sqrt{Ru}}, 1) if 0 < u < ηR  (3.7)

with some η > 0.

By Lemma 2

|y_{R^{+}}^{±}(u)| < const. \sqrt{Ru} if 0 < u < AR ,

and this inequality implies (3.7) in the case u < \frac{K}{R} with some K > 0. We shall show, using again Lemma 2, arguing similarly as in the proof of relation (3.4) in the proof of Part a) that

|y_{R^{+}}^{±}(u) − y_{R^{−}}^{±}(u)| < \frac{const.}{\sqrt{Ru}} , if \frac{K}{R} < u < ηR ,  (3.8)

which relation completes the proof of (3.7). To prove (3.8) we have to express y_{R^{−}}^{±}(·) by

y_{R^{+}}^{±}(·) and to exploit that by Lemma 2 |\frac{d}{dv} y_{R^{−}}^{±}(v)| < const. \begin{matrix}√ & \overline{R} \\ √ & \overline{u} \end{matrix} if \frac{K}{R} < u < v < u+ \frac{K}{R}.

Let v_{R^{+}}^{+} = (v_{R^{+},1}^{+}, v_{R^{+},2}^{+}) and v_{R^{+}}^{−} = (v_{R^{+},1}^{−}, v_{R^{+},2}^{−}) be the points of intersection of γ_{R^{+}} and γ_{R^{+}} − z in the half planes K_{R^{+}}^{+}(z) and K_{R^{+}}^{−}(z) respectively. We shall prove that there is some constant K > 0 such that

𝕆_{R} ∩ (𝕆_{R} − z) ∩ K_{R^{+}}^{+}(z) ∩ \{(v, y), y ∈ ℝ^{1}\} = ∅ if v ∉ [v_{R^{+},1}^{+} − \frac{K}{R} , v_{R^{+},1}^{+} + \frac{K}{R} ] (3.9)

and

𝕆_{R}∩(𝕆_{R}−z)∩K_{R^{+}}^{−}(z)∩\{(v, y), y ∈ ℝ^{1}\} = ∅ if v ∉ [v_{R^{+},1}^{−} − \frac{K}{R} , v_{R^{+},1}^{−} + \frac{K}{R} ] . (3.9′)

We also claim that under the conditions of Part c1) or Part c2) of Lemma 1

v_{R^{+},1}^{±} > const. (d(z, R) − |z|) .  (3.10)

Relations (3.7), (3.9), (3.9′) and (3.10) together imply Part c) of Lemma 1. To prove relation (3.10) let us consider the projection of the vectors v_{R^{+}}^{±} and (d(z, R) − |z|)\frac{z}{|z|} −

v_{R^{+}}^{±} to the direction of the vector −n(z_{d,l}(R^{+})). The sum of these two vectors, which is the projection of d(z, R) − |z| to −n(z_{d,l}(R^{+})) is longer than const. (d(z, R) − |z|) because of relation (3.3). On the other hand, the proportion of the length of these two vectors is separated both from zero and infinity because of relation (3.3) which implies this relation if the projection is done in the orthogonal direction (d(z, R) − |z|)^{⊥} and Lemma 2.

To prove relations (3.9) and (3.9′) we introduce the following notation. Let s_{R^{±}}^{±}(u) be the y coordinate of the intersection of the set (γ_{R^{±}} −z)∩K_{R^{±}}^{±}(z) with the line x = u. Since s_{R^{+}}^{+}(v_{R^{+},2}^{+}) = y_{R^{+}}^{+}(v_{R^{+},2}^{+}) we get by expressing s_{R^{−}}^{+}(·) through s_{R^{+}}^{+}(·), exploiting the lower bound on the derivative of the function s_{R^{+}}^{+}(·) given by Lemma 2 and arguing

similarly to the proof of relation (3.7) that s_{R^{−}}^{+}(v) > x_{R^{+}}^{+}(v) or v < 0 if v < v_{R^{+},1}^{±} − \frac{K}{R}

with some sufficiently large K > 0. If v < 0, then the set 𝕆_{R} ∩ {(v, y), y ∈ ℝ^{1}} is empty. Hence

𝕆_{R} ∩ \{(v, y), y ∈ ℝ^{1}\} = ∅ if v < v_{R^{+},1}^{+} − \frac{K}{R} .

By changing the role of γ_{R} and γ_{R} − z we get that

𝕆_{R} ∩ \{(v, y), y ∈ ℝ^{1}\} = ∅ if v > v_{R^{+},1}^{+} + \frac{K}{R} .

The last two relations together imply (3.9). The proof of (3.9′) is similar.

In such a way we have proved Part c) of Lemma 1. The proof of Part d) is trivial, since in this case even the set R^{+}A ∩ (R^{+}A − z) is empty. □

4. Some concluding remarks

In this Section we discuss the conjecture about the Poissonian distribution of a randomly placed circle suggested by the computer study of Cheng and Lebowitz [4] and briefly explain what kind of approach is suggested by the present paper.

It is relatively easy to show that the Poissonian limit for the number of lattice points in 𝕆_{R}(c, α) would follow from the following generalization of formula (2.1):

lim_{R→∞} Eξ_{R}(α)(ξ_{R}(α) − 1) ⋯ (ξ_{R}(α) − k + 1) = |𝕆_{R}|^{k} = [4c |A|]^{k} for all k ≥ 1 . (4.1)

Actually relation (4.1) is equivalent to the statement that all moments of the random variable ξ_{R}(a) converge to the corresponding moments of a Poissonian random variable with parameter |𝕆_{R}| = 4c |A|. Some modification of the argument leading to the proof of formula (2.2) gives that

Eξ_{R}(α)(ξ_{R}(α) − 1) ⋯ (ξ_{R}(α) − k + 1)

=  ∑_{m_{1}∈ℤ^{2}\{0},...,m_{k−1}∈ℤ^{2}\{0} the points m_{1},...,m_{k−1} are different.}|𝕆_{R} ∩ (𝕆_{R} − m_{1}) ∩ ⋯ ∩ (𝕆_{R} − m_{k−1})| . (4.2)

It is relatively simple to prove the following identity:

∫ |𝕆_{R} ∩ (𝕆_{R} − z_{1}) ∩ ⋯ ∩ (𝕆_{R} − z_{k−1})| dz_{1} … dz_{k−1} = [4c |A|]^{k} .  (4.3)

Hence to prove the Poissonian limit it would be enough to show that for large R the replacement of the sum in (4.2) by the integral in (4.3) causes a negligible error for all k = 1, 2, …. Actually, this fact was proved for k = 1 and 2 in this paper. But the proof for larger k is much harder. In our proof we exploited the independence caused by the random shift α ∈ [0, 1]^{2}. But this independence is not sufficient for the proof of (4.3) if k ≥ 3. To prove formula (4.3) in this case some deep number theoretical statement would be needed which states that certain functions of the k-tuples of lattice points (m_{1}, …, m_{k}) are almost uniformly distributed. We could give an explicit formulation of this statement, but since this would require complicated notations and would lead to a problem that we cannot handle we omit it.

We also discuss briefly the higher dimensional version of the problem in this paper. Let A be a convex set with a nice boundary, α ∈ ℝ^{d} a vector, uniformly distributed

in the d-dimensional unit cube, c > 0 a fixed number and

𝕆_{R}(c, α) = [(R + \frac{c}{R^{d−1}} ) A − α] \ [(R − \frac{c}{R^{d−1}} ) A − α] .

The volume of the set 𝕆_{R}(c, α) is 2cd Volume (A) + O (\frac{1}{R}). We are interested in the

number of lattice points ξ_{R}(α) in the randomly shifted set 𝕆_{R}(c, α). It can be proved that the first two moments of ξ_{R} tend to the first two moments of a Poissonian random variable with parameter 2d Volume (A). This can be proved by methods similar to those of the present paper. Moreover, in this case a stronger result can be proved. It can be shown that in the d-dimensional case

lim_{R→∞} Eξ_{R}(α)(ξ_{R}(α)−1) ⋯ (ξ_{R}(α)−k +1) = [2cd Volume (A)]^{k} for 1 ≤ k ≤ d . (4.4)

This means that the first d moments of the random variable ξ_{R} tend to that of a Poissonian random variable with parameter 2cd Volume (A) as R → ∞. To explain why relation (4.5) holds let us remark that relations (4.2) and (4.3) remain valid for all d ≥ 2 if the area is replaced by Volume, ℤ^{2} \ {0} by ℤ^{d} \ {0} and 4c |A| by 2cd Volume (A). We have to show that by replacing the sum in (4.2) by the integral in (4.3) we commit a negligible error. Let us also observe that for k ≤ d the expression

h(z_{1}, …, z_{k−1}) = Volume (𝕆_{R} ∩ (𝕆_{R} − z_{1}) ∩ ⋯ ∩ (𝕆_{R} − z_{k−1}))

changes very little if the arguments in this expression are changing with an order of constant. Hence the same technique works for k ≤ d, d > 2, as for the case d = 2 in the second Section of the present paper. Actually some technical difficulties have to be overcome if we want to carry out this program. We do not go into the details.

Acknowledgement: The authors would like to thank E. Makai Jr. for some useful explanations on convex geometry. J. L. Lebowitz and P. Major would like to thank the IHES in Bures–Sur–Yvette where this work was completed. The work of Zh. Cheng and J. L. Lebowitz was supported in part by the NSF Grant DMR 89 18902.

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