PROPERTIES OF CONVEX HULL GENERATED BY INHOMOGENEOUS POISSON POINT PROCESS Текст научной статьи по специальности «Науки о Земле и смежные экологические науки»

ISSN 2074-1871 Уфимский математический журнал. Том 12. № 3 (2020). С. 83-98.

PROPERTIES OF CONVEX HULL GENERATED BY INHOMOGENEOUS POISSON POINT PROCESS

I.M.KHAMDAMOV

Abstract. The paper is devoted to the limit distribution study of the exterior of a convex hull generated by independent observations of two-dimensional random points having Poisson distributions above the parabola. Following P. Groeneboom fl], we note that near the boundary of support, the Binomial point process is almost indistinguishable from the Poisson point process. Therefore, the approximation of a Binomial point process to a Poisson process is not considered here; it is believed that it is sufficient to study the functional of the convex hull generated by the Poisson point process. Using the modified P. Groeneboom technique, the so-called strong mixing and martingale properties of the vertex Markovian jump stationary process, the asymptotic expressions are obtained for the expectation and variance of the external part of the area of the convex hull inside the parabola. This is a continuation of results by H. Carnal in [2], where an asymptotic expression was found only for mean values of basic functionals of a convex hull. The asymptotic expression for the variance of the area of a convex hull was later obtained by J. Pardon [3] as no regularity conditions were imposed on the boundary of the support of a uniform distribution. The asymptotic expressions obtained here are used in the proofs of the central limit theorem for the area of the convex hull. Similar results were established in the studies by A.J. Cabo and P. Groeneboom [4] for the case as the initial distribution in a convex polygon is uniform.

Keywords: convex hull, random points, Poisson point process.

Mathematics Subject Classification: 60F05, 60D05

1. Introduction

Many researchers studied the distribution of boundary functionals of convex hulls. From an analytical point of view, these functionals have a very complicated nature. For many years, due to the lack of appropriate research methods, the main progress in this direction was limited by studying the expectation of a number of vertices, area and perimeter of a random polygon, see [2], [5], [6], [7], [8]). A great progress in this direction for the first time was achieved in [1]. Here the approximation methods for binomial point processes with homogeneous Poisson point processes were successfully applied. Then the powerful properties were found, namely, martingalitv, strong mixing, and stationaritv of functionals of the vertex process of a convex hull generated by the Poisson point process. The limit distribution for a number of vertices of a convex hull was obtained in the case, when the support of the initial uniform distribution is either a convex polygon or an ellipse. Then the method was developed for limit theorems for the area and perimeter of a convex hull in a polygon [4] and for the area in a disk [9]. These methods were applied in [10] to prove the limit theorems for the number of vertices, perimeter, and area of the convex hull, for the case when the tails of the initial distribution had an exponential form, including, in particular, the normal distribution. At the same time, using the methods given in [1], by the direct probabilistic method [11] the limiting joint probability

I.M.Khamdamov, Properties of convex hull generated by inhomogeneous Poisson point process.

© I.M.Khamdamov, 2020.

Поступила Ц января 2020 г.

distribution was obtained for the number of vertices, area and perimeter of a random convex hull generated by uniformly distributed points inside a polygon. In [12], in a one-dimensional case, it was proved that the contribution of the extreme terms of the variational series to the sum was significant when the distribution tail of initial random variable behaved as a regularly varying function. In a multidimensional case, the convex hull is the most correct generalization of the extreme terms of variational series and, therefore, the present study can be considered as a continuation of earlier research by authors.

2. Formulation of problem and main results

Let the distribution support A be the unit disk centered at a point (0,1), Assume that random points (ri, ai) are given in the polar coordinate system with the origin at (0,1) in a disk A, where ri and ai are independent and ai is uniformly distributed in [—k,-n] and

P (Ti > 1 - x)

0 < x < 1, P ^ 1,

(2.1)

where L(x) is the slowly varying function in sense of Karamata represented in the following form

L(u) = exp

e(t)

t

dt

e(t) ^ 0,

t oo.

We assume that Xi = r^ sin a^ 1 — Yi = ri cos ai and we denote by Cn the convex hull generated by these random points (X1, Y1); (X2,Y2);...; (Xn, Yn), The svmbols un, sn and ln stand for the numbers of vertices, area and perimeter of Cn, respectively. Under such assumptions, asymptotic mean values for the functional of the convex hull were obtained in [2].

In this paper, we study the functional sn as the random points are distributed in the same way as in (2,1), Following [1],[11], [13]—[15], it is sufficient to study the functional of the convex hull in the case when it is generated by the Poisson point process. That is, random points with distribution (2,1) are easily approximated by inhomogeneous Poisson point processes nn (•), defined inside a parabola

„2

Rn

{

x

(x,y): wn ^ y

}

with intensity

^n(A) = < (bn)

1- wS L(

2bn

0

dxdy as A c R„

as AC Rn

where bn is the least root of equation

nh

-(e+1 )

L (bn) = 1

For all a G R we define the vertices of process Wn(a) = (Xn (a), Yn(a)) as points (Xi,Yi), for which Yi — aXi is minimal. In view of the definition of Wn(a), that this is a non-stationarv Markovian jump process.

Before proceeding to the main results of the paper, we need to introduce some notations.

Let 0 = a0 < a1 < ■ ■ ■ < at ^ a be the time of the proeess jumps {W(c), c ^ a}, denote by

2

A0 the area of the domain enveloped by lines y = Y(0), y = Ai is the area of the domain enveloped by the lines

y = a— (x - X(ai-i)) + Y(a¿-i), y = ai (x - X(a¿-i)) + Y(a¿-i),

y

x 2b,

2

b

1

n

2

X

for each 1 ^ i ^ k.

Let A [a] be the the area of the domain enveloped by the curves

y = ak (x - X(ak)) + Y(ak), y = a (x - X(ak)) + Y(ak), y =27-.

2on

If to assume that A(0,a) = A0 + Ai + ■ ■ ■ + Ak + A[a], then, by from the property of the independence of increments of the Poisson point process, A(0, a) behaves as the sum of a random number of independent random variables, see [13].

As P. Groeneboom noted, to simplify the calculations, L(x) = 1 can be taken, since all the calculations can easily be transferred to the case of an arbitrary slowly varying function. Then

„2 \ /3-i

( {y- 20 dxdy as A c ^

A°n(A) = I 2-KyJbn i \ 2W (2.2)

[ 0 as AC Rn.

2

bn = n

The following theorem holds true. Theorem 2.1. At n ^ ro, the identity holds

EA(0, a) = abnX^, DA(0, a) = abHX^,

where

xni) - Ci, X(n2) ^ C2 (ft) + C303) + cA(13),

r (B) = 2ft (B (1;ft) + B (I;ft)) V~2n \ "+1 r /2ft + 3N ^) V2n(2/3 + 1) \B (ft + 1; 2 + lj .

as n ^ ro. Here t X^ and X(2) are constants; they coincide with the corresponding constants given in [2].

Theorem 2.2. As n ^ ro,

A(0, a) - abnX(n) d

abn Xn)

>N (0,1),

where 4> denotes the convergence in distribution and N(0,1) is a standard normally distributed random variable.

3. Properties of vertex processes of a convex hull

In this section we prove a series of preliminary lemmata, which will be employed then in the proof of Theorems 2.1 and 2.2.

The first lemma provides forms of distributions Wn(a) in various situations.

Lemma 3.1. Let

a2bn

s = y — ax +--—.

Then

p w)e =dkexp (- £ Bip+(v - 0" dxiv

P ( Wn(a) E (dx,dy))

exp

(-HB{0v— S dxdv•

2 \ P—

P( Wn(a) = Wn(0)/Wn(0) = (x, y))

(

exp

\

V2b„s p V2bny p

/ ( s- £) iu —1 (»- i) du

x—abn x

/

Proof. Let v = a(u — x) + y be a straight line passing through points (x, y) with an angular coefficient a and D(a, x, y) be the domain enveloped bv curves v = a(u — x) + y and v = 2^-. It is easy to see that if ui and u2 are the roots of equation

u

2f = a(u — x)+

then Mi,2 = abn ±\J2bns. Let us find A°n (D(a, x, y)); we recall that A°n was introduced in (2,2), We note that

u

y + a(u — x) — —— = y — ax +

a2 bn (u — a bn)2

2 bn

2

Then

U2

(D(a,x, y))

2nVbn

u2

y + a(u — x) —

2 n

du

(u — a bn)2

2b n

2nVbn

/2bns p

Is--— | du

J \ 2 bn)

U1

-V2bns

V2s2

(l — u2)P du

'K

sP+2

Let

d = \J (Ax)2 + (Ay)2, v = au + v = au + c+b

be two lines parallel to v = a(u — x) + y and passing at a distance d from above and below, respectively.

By D-(a ,x, y) and D+(a ,x, y) we denote the domains enveloped respectively by bounded curves

v = au + c _,

u

—— and

2b n

v = au + c+,

u

2bn

Assume that

Ax,y = (x,x + Ax) x (y,y + Ay)

It follows from the definition of Wn(a) that if the number of points k(A) in A of inhomogeneous Poisson point processes nn (■) satisfies the inequality

P ( Wn(a) e Ax,y) ^ P (n(Ax,y) ^ 1, w (A-(a,x, y)) = 0) . (3.1)

On the other hand, it is easy to see that

P ( Wn(a) e Ax,y) ^ P (n(Ax,y) = 1,w (D+(a, x, y) — Ax,y) = 0) . (3.2)

Taking into consideration a Poisson process property, namely, the independence of increments, using inequalities (3.1) and (3.2) as well as (2.2) as d ^ 0, we arrive at the first identity of the lemma.

We proceed to the proof of the second identity. It is easy to confirm understand that by the property of the independence of increments, the chain of identities hold:

P ( W(a) = W(0)/W(0) = (x, y)) =P (w (D(a,x, y) — D(0,x, y)) = 0/w (D(0,x, y)) = 0)

2

P

i

2

=P (k (D(a,x,y) - D(0,x,y)) = 0) =P (n (D*(a,x,y))) = 0) = exp (-An (D*(a,x,y))), where D*(a,x,y) is a set of points in the domain enveloped by lines

, , u2

v = a(u - x) + y, v = y, v = .

2 On

By the definition of a measure of An(-), we obtain

abn +y/2 bns /

An (D*(a,x,y))=^-^\ J is - (U - ahn) 1 du -

2rK\fbn

2b„

v P V2bny R

* - / (' - £)P*

/2 bn s a V2b^y

2r^Vn\ I {s - 22n] du -

\x-ab „

('-3du-1 (v-i) duy

This completes the proof. Assume that

Rn (a) = Xn (a) - abn, Sn (a), = Yn (a) -

□

Xn (a) + Rn (a)

It is obvious that and therefore

2bn 2bn

Tn(0) = Wn(0) a.s. P (Tn(0) E (dr,ds)) = P (Wn(0) E (dr,ds))

Tn (a) = (Rn (a) ,Sn (a))

Lemma 3.2. Tn(a) forms a Markovian jump process and P (Tn(0) E (dr,ds))

ft I

exp

2nVbn

P(Tn(a) = (n, si)/Tn(0) = (ra, so))

{-£*(>+} C - ¿f

drd s,

(

exp

1

V2*

\

J+2

(1 - t2)Pd t-s

P+2 o

(1 - t2)Pdt

n

r0

\/2b"sl \/2bn s0

2y

where ri = r0 - abn, si = s0 - ar0 + ^^, and

P(Tn(a) E (dri,dsi)/Tn(0) = (ro, so)) = P (Tn(a) E (dri,dsi))

if

ab,n - \j2b,ns i > \J2b,n so,

and

P(Tn(a) E (dn,ds2)/Tn(0) = (n, si))

1

exp

(~—(p+2

V V^A

(1 - t2)Pdt

sl —s2 + a bn a^/ 2b„S2 ^/2bns2

1

i

1

i

- Si

ß+2

(1 - t2)ßdt

s1-s2 + ab„ a^/ 2 bnsi ^/2bris-±

( s2 —| dr2ds2.

V 2 2bJ 2 2

2 \ ß-1

Here

(r-i, si) e D = |(r, s) : s^ , abn - V^bns2 ^ V2bns i,

s 2 +

a2 n

2

+ ar2 ^ si ^ s2 -

a2

2

+ a ri.

Proof. Statement 1 of the lemma is implied immediately by the definition of Tn(a) and Lemma 3,1,

In what follows, we consider three possible cases regarding the position of the vertex process Wn(a).

a

Case 2: There are jumps as the time changes from a to b, but the sets D(a,x\, y\) and D(b, x2, y2), determining the jump state Wn(-) at the time point a and b are disjoint (Statement

3);

Case 3: As the time changes from a to b, there are jumps and the sets D(a,xi, y\) and D(b,x2, y2) determining the jump state Wn(■) at the moment of time a and b intersect (Statement 4),

All cases can be treated in a similar way, this is why we deal with Case 1 only. We are going to calculate the probability

P(a, b) = P ( W(b) = W(a)/W(a) = (x, y));

here the methods from second part of the proof of Lemma 3,1 are used. We have:

bbn+/ 2b„S(b)

P(a, b) = exp I —

+ y 2<Jn^yJ)

(-wC ( / (

2^\fb~n

abn+^J 2b„S(a)

b(u — x) + y — —— du

vT\13

2 b n)

(

a(u — x) + y — —— ) du 2 On

where

S(a) = y — ax +

a2 bn 2 ,

S (b) = y — bx +

2 n

We let r(a) = x — abn, r(b) = x — bbn, then

, . u2 a2 bn (u — abn)2

a(u — x)+y — 2^ = y — ax + ~y

2 n

= s(a) —

(u — abn)2

2b n

Substituting the variables, we get

P( , ) = exp

( f /2b„s(b)

1

2ix^fb~n

( s(b)—k)du — i {s(a)—w) du

yj 2b„s(a)

,2\ I3

r(b)

r{a)

2

if \\

i i

■ exp

1

V

s(b)3+2 J (1 - t2)3dt- s(a)3+2 J (1 - t2)13 dt

r(b) r(a)

y V 2 bns(b) v/2b„s(a) J

=P (T(b) = (r{b), s{b)) /T(a) = (r(a), s(a))).

/

Eedenoting

ro = x, so = y, n = x - abn,

, a2bn „ , , b2bn

si = y - ax +--—, r2 = x - bbn, s2 = y — bx +

2 ' 2 " 2 * 2 ' we obtain

th th \ , (b -a)2bn r2 = n - (b - a)bn, s2 = si - (b - a)ri +--^-.

Hence,

P (T(b) = (r2, s2) /T(a) = (n, Si)) = P (T(b - a) = (S2) /T(0) = (n, si)) this proves Statement 2 of the lemma is obtained. The proof is complete, □

We consider the following a-algebras generated by process Tn(a)\

^ = a {Tn (c) : c^ 0} , ^ = a [T,n(c) : O a} . Lemma 3.3. For every A E and B E 0n+, the inequality

|P (A n B) - P(A)P(B) | ^ Tn(a)

holds, where

mM < 4exp f 'B (» ))■

In particular, if

— 2/3 s 1

a>-n-n, an = ^2bn logn, e*n = (logn)-2//+T) , 0 <6< 1,

n

then

Tn(a) ^ 4exp (-c (logn)i+3 j .

Proof We introduce events:

Gi ={u : Sn(0) ^ Op} , G2 ={u : Sn(a) ^ Op} , G = Gi n G2.

Since the sets Gi and G2 are disjoint, due to the property of the independence of increments, inhomogeneous Poisson point processes for each A E Gi E and B E G2 E 0n+ we have

P (A nB/G) = P (A/G) P (B/G)

Then it is easy to see that

P (A n B) ^ P (A n B n G) =P (G) P (A n B/G)

=P ( G) P (A/G) P (B/G) _P (A nG) P (B nG) = P ( G) ^P (A nG) P (B nG) = (P( A) - P (G)) (P(B) - P (G))

^P(A)P(B) - 2P (G) . Here G is a completement to the event G. On the other hand, P(A nB) ^ P(A nB n G) + P (G) =P(G)P (A nB/G) + P (G)

=P( G)P (A/G) P (B/G) + P (G) P(A n G)P(B n G)

P( G)

+ P(G)

P (g) _

=P(A n G)P(B nG) + ^-J-P(A n G)P(B n G) + P (G)

P( G)

^P(A)P(B) + 2P (G) .

Then it is easy to see that

P (G) = P (G1U G) ^ P (Gl) + P (G2) = 2P (Gl) =2P^Sn(0) ^ .

By last three inequalities we obtain

|P (A nB) - P(A)P(B) I ^4P(Sn(0) > = 4h°n(Y > ^ ^

8 J " n " 8

- TZ f 4B )}

Hence, as a > we have

On

|P (A nB) - P(A)P(B )| ^ 4exp{-c (logn)1+. The proof is complete, □

This Lemma implies immediately that Lemma 3.4. For all £ > 0, a> 0 the inequality

<x

£ <(a)

nXa) < <x>

n=1

holds true.

We define

V2bns p_i -J2bns_r

M(k) (t;R2) =i (u - r)k( s-dui uk( s-M^D du, V; ; 2nTn j ) V 2bn) 2kTK J V 2bn ) ,

r 0

where t = (r, s ),

Lemma 3.5. The processes

a

A(a) - 2 f (V2bns(b) - r(b)j2 db

2

0

and

a

. , ,_ , 2

12 /

A2(a) -J A(b) (y2b ns (b) - r(b)j' db 0

a

0„ = a {T(c) = (r(c), s(c)) : 0 ^ c ^ a} . Proof. We first note that according the proof of Lemma 3,2 we have

Q / x \3-i

P ( W(a + h) E (dx2,dy2) /W(a) = (xi, yi)) = y2 - -f dx2dy2 + o(h). (3.3)

Let

D(0(a,h,r, s) = |(u, v) : x ^ u ^ abn + \J2bns(a), a(u - x) + y ^ v ^ (a + h)(u - x D(i)( , h, , ) = ( u, ) : a n n ( a) < u ^ (a + h)bn + y/2b n s (a + h),

u

—r~ ^ v ^ (a + h)(u - x) + y >,

2 n

D( a, h, , ) = D(0)(a,h,r, s) UD(i)(a,h,r, s), where

( \ I a bn ( , M ( , M , (a + h)2 bn

s(a) = y - ax +--—, s(a + h) = y - (a + h)x +

22 It is easy to see that the areas are

ab n+V 2b ns

SD(°)(a,h,r,s) = J h(u - x)du = h (V2bnS + abn - xj = (^2b,ns - r^j ,

X

(a+h) b„ +^J2 bn s(a+h)

SD(1) (a,h,r,s) = j ^(a + h)(u -x)+y - du = 0(h,2).

ab n + ^j2 b„ s(a)

Further, for each k, for small h > 0 we get E{A(a,a + h), in D(0)(a,h,r, s), there are k vertices W (-)/W (a) = (x, y)} = 0(hk+i), E {A(a, a + h), in D(°\a, h, r, s), there are no vertices W(-)/W(a) = (x, y)} = SD(0)(a,h,r,S)P \in D(0)( a, h, r, s) no vertices W(-)/W(a) = (x, y)J

= h -ry exp {-An (D(a,h,r, s))} + o(h)

h 2

= ^ -r^j +o(h).

This yield the first statement of the Lemma. Using (3.4), we obtain

E {A2(a,a + h)/W (a) = (x, y)} = 0(h2). (3.5)

Then, by (3.3) and (3.5), we have E(A2(0, a + h) - A2(0, a)/W(a) = (x, y)) =E(A2(a, a + h)/W(a) = (x, y))

+ 2A(0, a)E(A(a, a + h)/W(a) = (x, y)) + o(h)

=hA(a) (V2 bns -r^J +o(h). The proof is complete. □

(3.4)

4. Proof of Theorems 2.1 and 2.2

In this section, the asymptotic behavior of the moments A(0, a) is found as n ^ ^ for a a

Using Lemma 3.5, we have

a

1

EA(0, a)=- E( y/2bns(b) - r(b)) db

0

aE -r

x V 2b„so , 1

I -ro)2exp(-^B{f+1;i

0 _

{So - k)

2 \ P_1 0

s 0 - — ) dro x /1 \ 1

wJ s'+1 exA - 'SB(B + i;1))ds j(1 - r)M - Sf-1 *

0 ^ ' _1 aKP T T

-J1J2.

Let us find each factor J1 and J2. We see that

00 2^+3

J = /^1 exp{-& (f + i;2)}d(^r(W,)

= 2

1

J r(1 - r2)l_1dr = 0, _1

we get:

1 1 1

2

J2 = (1 - r)2(1 - r2)l_1dr = (1 - r2)l_1dr - 2 r(1 - r2)l3_1 dr

Hence,

_1 _1 _1 1 1 1

+ J r2(1 - r2)l3_1 dr = 2 J(1 - r2)l_1dr + 2 J r2(1 - r2)l_1 dr

_1 0 0 1 1

J t_2 (1 - t)ll_1dt + J t2 (1 - t)l_1dt = B(+ b( 3;,).

N

A(1) 2, (B (2;f) +B (I;ff)) / V/2VT \ ^ _ /2, + 3\ n ~ T2n(2f3 + 1) \B (f + 1; 2)) \2f + 1)

It is easy to see that

E A2(0, a)

j s(b) - r(b)) db

db ^ EA(6 - (k + 1)h, b - kh) ^2bnS(b) - r

k=0

dbjr E(E(D*(b - (k + 1)h,b - kh, x0, y0)

L— n V V

Mty - r(b)j2 /0(b -kh, tt)^ +o(h)

k=0

(4.1)

db ^ E

k=0

» s(b) - r

E(D*(b - (k + 1)h, b - kh, x0, y0)/W(b - kh) = (x0, y0))) + o(h)

where

D*(a - h, a, x0, y0) ^|(u, v) : abn - \J2bns(a) ^ u ^ x0

(a - h)(u - x0) + y0 ^ v ^ a(u -

Then the area of the domain D*(a - h, a, x0, y0) is

xq

S

D* (a—h,a,XQ,yo)

((a - h)(u - x0) + y0 - a(u - x0) - y0) du

ab n-\J 2b „ s(a)

XQ

h(u - x0)du

ab n — \j 2 b„s(a)

--'-z i abn - Vbns(a) - x0^j

h 2 h 2

2

s(a) + r(a)Sj

2

Hence,

E

s(b) - r(b)) E (A*(b - (k + 1)h, b - kh)/0(b - kh, tt))

)

h e

2

'h

is(b) - r

E

t s(b - kh) + r(b - kh)) /0(b - kh, tt)

is(b) - r

t s(b - kh) + r(b - kh)) /0(b - kh, tt)

(4.2)

h

-E 2

( s(b) - r(b)j2 (y2bns(b - kh) + r(b - kh))^ .

a

2

a

2

a

a

2

n

2

n

2

2

2

On the other hand, we readily confirm that

E (V2bnS(a) - r(a)j2 = E (V2bnS(a) + r(a)^J' Let h = b/m, then from the latter identity (4,1), (4,2) imply that

m—i

2

0 k=°

E A2(0, a) = J db ( V2bnS(b) — r{

E (A*(b — (k + 1)h, b — kh)/S(b — kh, m))^j + o(h)

a -t

1 f, m— i

0 k=0

2

Hence, asm ^ m, we get

db £e(( \/2 bnS (b) — r(b)j (v/2b nS (b — kh) + r(b — kh)) h + o(h)

a b

EA2(0,a) = 2Jdbj dcE j ^2bnS(b) — r(bf)2 (^26nS(c) + r(cf) ^

00

Then it follows from Lemma 3,5 that

a b

DA(0, a) =\Jdbj dcE^ (y2bnS(b) — r(bf)2 (^2Ks(c) + r(cf) ^

00

2

6E (V2bns(b) — r(b) 2

a b

1

00 a

db dcE[[ V2bnS(b) — r(b)) ( y/2bnS(c) + r(c)

— 4 f E( V2 bn s (b) — r(b)) db

0

b

E , X, 2„n

0

a b

s(c) + r(c)^ de + i E (y2bns(c) + r(c)j dc

2 I db j dc\ ns(b) — r(b)) (V2bns(c) + r(c

00

— E ^2bns(b) — r(b)j2 E ^2bnS(c) + r(c)^j^

a b

— \j dbJ d(b — °){E (^2bnS (b — c) — r(b — cf) 2 {y2b nS (0) + r

0 0

— E ^2bn s(b — c) — r(b — c)) 2 E (^M® + r(0)^j ^ .

2

2

2

Hence,

a

DA(0, a) =1 J (a - b) db (e ^2 b.ns (b) - r(b)^2 ^2b nS (0) + r

0

- E ^2bnS(b) - r(bf)2 E (y+ r(0)^j 2 ^

1

n

aen a

1

2 J +2

0 a

: h + h-

Let a > ^n bn/en, then we let

{

a2 n

A = { S(0) , S(a) <

a2 n

8

)

Gi nG2

It is easy to see that for each r > 0

E

^2b nS (a) - r(af) +

- E ^2b nS (a) - r(af) 2 E ^2 bn s (0) + r = 0 ((a2bn)23+T exp ( -C (a2bn)3+2-r ))

2=

Hence,

We are going to estimate ii. Let a ^ ^bn/en. We denote

( f2 __a2 b

Ki(a) = I (T0, S0) x (n, si) : Si > ^, i = 0,1, n = ro - a bn, Si = S0 - ar i--^

}

K2(a)

S0) x (Ti, Si) : Si > , i = 0,1, \J 2 bn s 0 > a bn - sj 2 bnS i, Si

2

n

a2 n a2 n a2 n a2 n

+--— + ari ^ s0 ^ si--— + ar-0, «0 - ar0 +--— ^ si ^ S0 - ari--—

2

K3(a) = \ (T0, S0) x (n, si) : Si > ——, i = 0,1, \J 2 bn s 0 < a bn - \J 2bnS i

2

K4(a) = |(T0, S0) x (n, si) : Si > ,i = 0,1, \J2bns0 ^ abn - \J2bnSi Given b ^ aen, by the steady-state character of the process T(a) we have

h =1 t (a - b) db ( E

a(1 - n)

^E ^2bnS(b) - r(bf) 2 + r

- E ^2bnS(b) - r(b)Sj2E (yZbnMfi) + r(0)^j^ db(E (V2bnS(b) - r(b)Sj2 + r(0)^2

2

8

2

3

2

E

(V2bns(b) - r(b)j2E (y+ r(0)^j

a(1 - n)

A[(b) + A'2(b) - A3(b) db

where

A[(b) =E (^2bns(b) - r(b))2 (V^M® + r(0))2; (T(0),T(b)) e K(b) A2M =E ({^2bns(b) - r(b))2 + r(0))2; (T(0),T(b)) e K(b)

A3(b)

i ^2 |(ro,so): sq>^

}

P (T(0) e (dr-0,ds0)) +

( \

J P (T(b) e (dn,ds1)) (y2b^1 - n)2

yjVl,si): «1>22,J 2bso>bbn_j2bnsi | y

Using Statement 2 of Lemma 3,2, for sufficiently large n we have

ae„

= a(1 - 0£n)

A1

A1 (b) db

3

V2a bIff

1

■ exp

0 1 1 -

0 (ro,so)x(n,si)eKi(6) ( / ,

1

1

V2 b n s 1

1+

0

1

V2:

1

\

J+1 I A ,2\l J, J+ 2

(1 -12)1 d t- 4

V2 bn s 0

(1 -12 y dt

)

\\

--1

\ J2b:

n^i

V26nS0

II

J+2

■ exp

3 00

(

1

B[3 + 1;2I I dr-0ds0

V^dbn3 I l+2 I ,f r-0 -- / ds0s0 d

1

ro _ ^ bn j2bnso V 2«o

1--- | exp

so

2 n 1 0

,_ , . d\b\ — — 1 +—-¡==_

\J2b,ns0) J \ \ s0 J \s0j\ \J2b,ns0

_V 2bnso 0

2

)

- j exp ^

1+1 / / \ 1+2

s 0 2 / (S1\ 2

v^ll \ s 0

(1 - 2)l

V ^o

bn

^ (-T+i;B{3

(1 - ^ I)exp| - ^ +1;^)(1 - ¿J "

V2fenS0

/A

2

1

1

1

(X

-C2{ß)ab% ^ G\ab3 s0 exp (~G2Sfß+ 2 j ds0 ^ C3ab%.

o

n

ae

n

A =a(1 2d£n) J A2(b)db

o

x x 1

^ fdbf sO^dso f dro(1 + To)2 (1 - r2)ß-1 exp ^^B (ß + 1; 1

in

O O -1

1

1 ( ß+2

dr1ds 1(1 — r1)2 exp ^ — ^ ^sß+ 2 J (1 — u2)ß du

m

(r-i,si)eK|(Vo,so) t*(b)

— <?2 / (1 — u»)'du))( ^ )8-1 (1 —r2)ß-1

3 x 1 x , ß+1

ßlßf s 0ß+lds of dro(1 + ro)2 (1 — fO Y-1 j db * exp i — ^ B(ß + 1; 1

o 1 o

■J dnds1(1 — n)2 exp i — ^-i ( s *)ß+2 J (1 — t2 )ßdt

(r1,sl)^K**(b*,ro,so) ^ ^ t**(b*)

t** (b*)^s*

where

(1 — t2)ßdt\ |(s 1)ß-1 (1 — rlf-1 ~ G3(ß)ah

S1 bbn v 1 — S * , b

*

s* = —, b* = . n , t**(b*)= . \L +

s 0' 2 b*y/si 2^'

n

aen

n)

A3 = A3(b)db

o

3 x 1 x ß+1

ß2abn f ds oj dro j dbs2f+3 exp ( — ^ ^ß + 1;\))(1 — r20)ß-1

o -1 o

• J 83+2drldSl expf- ^^B (ß + 1;\^ (1 - ri)13-1

3

=C4(ß)ab2.

The proof of Theorem 2,1 is eomplete,

Theorem 2,2 is implied by Theorem 2,1, Lemma 3,4 and Theorem 17,2,2 in [16].

1

3

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Isakjan Mamasalievich Khamdamov,

Uzbekistan Academy of Sciences

V.I.Romanovskiy Institute of Mathematics,

Mirzo-Ulugbek 81,700125,

Tashkent, Uzbekistan

Tashkent University of

Information Technologies,

A.Timur street 108,700020,

Tashkent, Uzbekistan

E-mail: isakjan.khamdamovSmail.ru,

khamdamov.isakj anSgmail.com