U-max-Statistics and Limit Theorems for Perimeters and Areas of Random Polygons

0.26 MB·English

by  E. V. Koroleva

U MAX-STATISTICS AND LIMIT THEOREMS FOR
−
PERIMETERS AND AREAS OF RANDOM POLYGONS
E. V. KOROLEVA, YA. YU. NIKITIN
Abstract. Recently W. Lao and M. Mayer [6], [7], [9] considered U-max - statistics,
where insteadof sum appearsthe maximum overthe same setof indices. Such statistics
often appear in stochastic geometry. The examples are given by the largest distance
betweenrandompointsinaball,themaximaldiameterofarandompolygon,thelargest
scalar product within a sample of points, etc. Their limit distribution is related to the
distribution of extreme values.
Among the interesting results obtained in [6], [7], [9] are limit theorems for the max-
imal perimeter and the maximal area of random triangles inscribed in a circumference.
3
In the present paper we generalize these theorems to convex m-polygons, m 3, with
1 ≥
randomverticesonthe circumference. Next, a similarproblemis solvedfor the minimal
0
2 perimeter and the minimal area of circumscribed m-polygons which has not been previ-
ously considered in literature. Finally, we discuss the obtained results when m .
n →∞
a
J

7

] 1. Introduction
R
P 1.1. One-sample U-statistics and parametric functionals. Let ξ ,...,ξ be inde-
1 n
h. pendent random elements taking values in a measurable space (X,A) and having identical
t distribution P. Let = P be some class of probability distributions on (X,A) and let
a
P { }
m θ(P) be a functional on .
P
[ The functional θ(P) is called regular [5], if θ(P) can be represented as

1
v
θ(P) = ... h(x ,...,x )P(dx )...P(dx ) (1)
0 1 m 1 m
6 ZX ZX
3
1 with some real-valued symmetric Borel function h(x ,...,x ) which is called the kernel,
. 1 m
1 while the integer number m 1 is called the degree of functional θ(P).
0 ≥
Halmos and Hoeffding [3], [4] began to study the class of unbiased estimates of θ(P)
3
1 called U-statistics, which are defined as follows. Consider a kernel h(x ,...,x ) of para-
1 m
:
v metric functional (1). Then the U-statistic of degree m is defined as
i
X 1
n −
r U = h(ξ ,...,ξ ),
a n m i1 im
(cid:18) (cid:19) J
X
where n m and J = (i ,...,i ) : 1 i < ... < i n is a set of increasing
1 m 1 m
≥ { ≤ ≤ }
permutations of indices i ,...,i .
1 m
It turns out that numerous statistical estimates and test statistics belong to the class
of U-statistics. This entailed the intensive development of the theory, see [5] and [8].
2000 Mathematics Subject Classification. 60F05, 62G32, 62H11.
Key words and phrases. U-max statistics, Weibull distribution, random perimeter, random area, in-
scribed polygon.
1
1.2. U-max-statistics. U-max-statistics appear for the description of extreme counter-
parts of U-statistics, they are given by the formula:
H = maxh(ξ ,...,ξ ).
n i1 im
J
U-min-statisticsaredefinedsimilarly, theycanbereducedtoU-max-statisticsbychanging
the sign of the kernel. Here are some examples of U-max and U-min-statistics.
1. Largest interpoint distance max ξ ξ , where ξ ,ξ ,... are i.i.d. points in the
i j 1 2
1 i<j nk − k
d-dimensional unit ball Bd,d 2.≤ ≤
≥
2. Largest scalar product max ξ ,ξ , where ξ ,ξ ,... are i.i.d. points in the ball
i j 1 2
1 i<j nh i
Bd, d 2. ≤ ≤
≥
3. Smallest spherical distance: min β , where β denotes the smaller of two central
i,j i,j
1 i<j n
≤ ≤
angles between U and U . It is assumed that the vertices U ,...,U are i.i.d. points on
i j 1 n
the unit sphere Sd 1,d 2.
−
≥
4. Largest perimeter max peri(U ,U ,U ) and largest area max area(U ,U ,U)
i j l i j l
1 i<j<l n 1 i<j<l n
≤ ≤ ≤ ≤
among all inscribed triangles, whose vertices are formed by triplets of points taken from
a sample U ,...,U of independent and uniformly distributed points on the unit circum-
1 n
ference.
Lao and Mayer [6], [7], [9] initiated the study of U-max-statistics and proved for them
the basic limit theorem. They used some modification of a statement on Poisson conver-
gence from the monograph of Barbour, Holst and Janson [1, p.35].
Lao-Mayer Theorem [6]. Let ξ ,...,ξ be i.i.d. random elements in some measurable
1 n
space (X,A) and let h be a symmetric Borel function, h : Xm R. Put
→
H = maxh(ξ ,...,ξ ),
n i1 im
J
and denote for any z R
∈
n
p = P h(ξ ,...,ξ ) > z , λ = p ,
n,z 1 m n,z n,z
{ } m
(cid:18) (cid:19)
τ (r) = P h(ξ ,...,ξ ) > z, h(ξ ,ξ ,...,ξ ) > z /p .
n,z 1 m 1+m r 2+m r 2m r n,z
{ − − − }
Then for all n m and for each z R we have:
≥ ∈
P H z exp λ
n n,z
| { ≤ }− {− }| ≤
m 1
n n m − m n m
(1 exp λ ) p − + − τ (r) . (2)
n,z n,z n,z
≤ − {− } m − m r m r
( )
(cid:20)(cid:18) (cid:19) (cid:18) (cid:19)(cid:21) r=1 (cid:18) (cid:19)(cid:18) − (cid:19)
X
Clearly, the result can be reformulated for the minimal value of the kernel by replacing
h with h.
−
Remark 1. (Lao-Mayer [6]). If the sample size n tends to infinity, then the error in (2) is
asymptotically of order
m 1
−
O p nm 1 + τ (r)nm r ,
n,z − n,z −
!
r=1
X
where for m > 1 the sum is dominating, see [1].
2
Silverman and Brown [11] formulated the conditions which ensure that the general
theorem used in [6] provides a non-trivial Weibull limit law.
Silverman-Brown Theorem [11]. In the setting of Lao-Mayer theorem, if for some
sequence of transformations z : T R, T R, the conditions:
n
→ ⊂
lim λ = λ > 0, (3)
n,zn(t) t
n
→∞
lim n2m 1p τ (m 1) = 0, (4)
n − n,zn(t) n,zn(t) −
→∞
hold for each t T, then
∈
lim P H z (t) = exp λ (5)
n n t
n { ≤ } {− }
→∞
for each t T.
∈
Remark 2. (Lao-Mayer [6]). If m > 2, then the condition (4) can be replaced by the
weaker condition:
lim n2m rp τ (r) = 0 (6)
− n,zn(t) n,zn(t)
n
→∞
for all r 1,...,m 1 .
∈ { − }
Remark 3. (Lao-Mayer [6]). Condition (3) implies p = O(n m), and therefore (5) is
n,z −
valid with the rate of convergence
m 1
−
O n 1 + n2m rp τ (r) .
− − n,zn(t) n,zn(t)
!
r=1
X
For the perimeter and the area of inscribed triangles (see example 4) Lao and Mayer
in [6], [7] and [9] obtained the following results.
Theorem A (Perimeter of inscribed triangle). Let U ,U ,... be independent and uni-
1 2
formlydistributed points on the unit circumference S, and letperi(U ,U ,U ) be the perime-
i j l
ter of triangle formed by the triplet of points U ,U ,U . Set H = max peri(U ,U ,U ).
i j l n i j l
1 i<j<l n
≤ ≤
Then for each t > 0
2t
lim P n3(3√3 H ) t = 1 exp .
n
n { − ≤ } − −9π
→∞ (cid:26) (cid:27)
1
The rate of convergence is O n .
−2
(cid:16) (cid:17)
As a comment to this result, we note that among all triangles inscribed in the unit
circumference, the regular triangle has the maximal value of perimeter equal to 3√3.
It is a classical and well-known result, see [12]. Clearly, the maximal perimeter H of
n
random triangle tends to this value. The theorem gives the required normalization for
this convergence, describes the limit distribution and establishes the rate of convergence.
Theorem A is proved by means of the following Lemma A.
Lemma A. Let U ,U ,U be independent and uniformly distributed points on the unit
1 2 3
circumference S. Then
4
lim s 1P peri(U ,U ,U ) 3√3 s = .
− 1 2 3
s +0 { ≥ − } 3π
→
Now we proceed to random areas.
3
Theorem B (Area of inscribed triangle). Let U ,U ,... be independent and uniformly
1 2
distributed points on the unit circumference S. Set G = max area(U ,U ,U ), where
n i j l
1 i<j<l n
≤ ≤
area(U ,U ,U ) is the area of random triangle formed by the triplet of points U ,U ,U .
i j l i j l
Then for each t > 0
3√3 2t
lim P n3 G t = 1 exp .
n
n ( 4 − ! ≤ ) − −9π
→∞ (cid:26) (cid:27)
1
The rate of convergence is O n .
−2
(cid:16) (cid:17)
Commenting this result, we note that the area of the triangle inscribed into the unit
circumference has the maximal value 3√3 when its vertices are the vertices of regular
4
triangle [12]. The following Lemma B plays an important role in the proof of Theorem B.
Lemma B. Let U ,U ,U be independent and uniformly distributed points on the unit
1 2 3
circumference S. Then
3√3 4
lim s 1P area(U ,U ,U ) s = .
− 1 2 3
s +0 ( ≥ 4 − ) 3π
→
In this paper we consider the limit behavior of more general U-max - statistics of this
type related to m-polygons, m 3. In the sequel C ,C ,... denote positive constants
1 2
≥
depending only on m.
2. Inscribed polygon
2.1. Perimeter of inscribed polygon. The result ofthis section is thegeneralization of
Theorem A for inscribed triangles to the case of convex m-polygons, m 3. We underline
≥
that the proof of this theorem in [6] turned out to be inapplicable for random perimeters
and areas of m-polygons with m > 3. Therefore we had to use some new ideas.
Theorem 1. Let U ,U ,... be independent and uniformly distributed points on the unit
1 2
circumference S, and let
m = max peri(U ,...,U ) (7)
Pn 1 i1<...<im n i1 im
≤ ≤
be the maximal perimeter among the perimeters of all convex m-polygons, generated
m
P
by m points from U ,...,U , m 3. Then for each t > 0 we have
1 n
≥
m−1
π t
lim P nm2m−1 2msin m t = 1 exp 2 ,
n→∞ m −Pn ≤ − (−K1m)
n (cid:16) (cid:17) o
where K1m = m32 πsin mπ m2−1 Γ m2+1 . The rate of convergence is O n−m1−1 .
(cid:16) (cid:17)
Next Lemma is(cid:0)crucial(cid:1)in the p(cid:0)roof(cid:1)of Theorem 1.
Lemma 2.1.Let U ,...,U be m independent and uniformly distributed points on the
1 m
unit circumference. Consider the convex inscribed m-polygon with such vertices and with
the perimeter m = peri(U ,...,U ). We have the following limit relation:
1 m
P
lim s−m2−1P m 2msin π s = Γ(m+1), (8)
s→+0 nP ≥ 4 m − o K1m
where the constant K is from Theorem 1.
1m
Proof of Lemma 2.1. For m = 3 Lemma 2.1 coincides with the result of Lao and
Mayer [6]. The perimeter m is maximal [12] for the regular m-polygon (its side is
P
2sin π) and this maximal value equals 2msin π. Consider for i = 1,...,m 1, m 3,
m m − ≥
the central angle β = ∠U OU . By rotational symmetry, these angles are independent
i 1 i+1
and uniformly distributed on the interval [0,2π]. Let show that the points U can be taken
i
in order of increasing angles, so that our m-polygon corresponds to the following figure.
Fig. 1
Consider the permutation of angles with increasing indices β β ... β .
1 2 m 1
Let prove for any 0 < z < 2msin π the equality of probabilities{ ≤ ≤ ≤ − }
m
P m z,β β ... β = P m z,β β ... β ,
{P ≥ i1 ≤ i2 ≤ ≤ im−1} {P ≥ 1 ≤ 2 ≤ ≤ m−1}
where i ,...,i is any permutation of indices 1,...,m and β is the central angle
{ 1 m} { } is
defined by the points U and U ,s = 1,...,m 1.
Let C be the (mi1 1)-dimise+n1sional cube w−ith the edge 2π. Note that the sides of
m 1
− −
m-polygon are calculated by the law of cosines:
β β β
l l 1 m 1
UlUl+1 = 2sin − − ,1 l m 1; UmU1 = 2sin − .
| | 2 ≤ ≤ − | | 2
We have
P m z,β β ... β =
{P ≥ i1 ≤ i2 ≤ ≤ im−1}
m 1
−
= P U U + U U z,β β ... β =
{ | ik ik+1| | im i1| ≥ i1 ≤ i2 ≤ ≤ im−1}
k=1
X
m 1
1 −
= 1 U U + U U z,β β ... β dβ ...dβ .
(2π)m 1 { | ik ik+1| | im i1| ≥ i1 ≤ i2 ≤ ≤ im−1} i1 im−1
− ZCm−1 k=1
X
After the change of variables β = β ,...,β = β the last integral becomes
i1 1 im−1 m 1
−
1
1 U U +...+ U U z,β ... β dβ ...dβ =
1 2 m 1 1 m 1 1 m 1
(2π)m−1 ZCm−1 {| | | | ≥ ≤ ≤ − } −
= P m z,β β ... β ,
1 2 m 1
{P ≥ ≤ ≤ ≤ − }
as required. (cid:3)
5
Our arguments imply that
π π
P m 2msin s = (m 1)!P m 2msin s, β β ... β . (9)
1 2 m 1
{P ≥ m − } − {P ≥ m − ≤ ≤ ≤ − }
The condition β β ... β provides that the points U ,...,U stand in
1 2 m 1 1 m
≤ ≤ ≤ −
increasing order just as in Fig.1.
Now let show that for small s > 0 and under the condition
π
m 2msin s (10)
P ≥ m −
all central angles of our m-polygon differ very little from 2π. Take the smallest and the
m
largest side of our polygon with opposite central angles 2π 2ψ and 2π + 2ϕ, ϕ >
m − m
ψ > 0. One can displace these sides so that they are adjacent. Their lengths are equal to
2sin(π +ϕ) and 2sin(π ψ). Now let make the so-called ”symmetrization” [12] replacing
m m−
our two sides by two equal sides having the common vertex on the circle in the center of
the arc subtending the sum of the angles. Then the length of new two sides will increase
[12, Probl. 55b] and is equal to 4sin(π +ϕ ψ). The increment of the perimeter is just
m −
π ϕ ψ ϕ+ψ
∆ = 4sin + − 1 cos , ϕ ψ > 0.
m 2 − 2 −
(cid:18) (cid:19)(cid:18) (cid:19)
Supposetheinitialperimeter ofourm-polygonwas2msin π σ withσ s.Consequently
m− ≤
π π
2msin σ +∆ 2msin ,
m − ≤ m
whence it follows that ∆ s and therefore
≤
ϕ+ψ s
1 cos . (11)
− 2 ≤ 4sin π + ϕ ψ
−
m 2
Clearly, the sum of the largest and smallest cent(cid:0)ral angles(cid:1)4π +2ϕ 2ψ is smaller than
m −
2π. Hence 0 < π + ϕ ψ < π and we have from (11) the inequality
−
m 2 2
ϕ+ψ s
1 cos = C s,
− 2 ≤ 4sin π 1
m
which implies that for small s > 0
ϕ+ψ 2arccos(1 C s) < C √s.
1 2
≤ −
Consequently all random central angles differ from the angle 2π by no more than O(√s.)
m
Let estimate the deviation of the angles β from β ,k = 2,...,m 1, under the
k k 1
− −
condition (10). We introduce the auxiliary random angles α = 0,α ,...,α ,α = 0
0 1 m 1 m
−
such that
2πk
β = +α , k = 1,...,m. (12)
k k
m
The random angles α ,...,α are independent and each α is uniformly distributed on
1 m 1 k
2πk,2π 2πk . In terms of −α we have
− m − m k
(cid:2) (cid:3) m π α α
m = 2 sin + k − k−1 ,
P m 2
k=1 (cid:18) (cid:19)
X 6
and the inequality (10) takes the form
m
π α α π
k k 1
2 sin + − − 2msin s. (13)
m 2 ≥ m −
k=1 (cid:18) (cid:19)
X
The argument given above shows that any central angle differs from the expected value
2π by no more than O(√s.) Hence we have under (10)
m
max α α C √s. (14)
k k 1 3
1 k m| − − | ≤
≤ ≤
and consequently
max α C √s. (15)
k 4
1 k m 1| | ≤
≤ ≤ −
Let return to the formula (13). As the differences α α are small, we can expand
k k 1
| − − |
in (13) the sine function in Taylor series with the remainder term. Hence we obtain for
some small random angles η ,k = 1,...,m 1, that
k
−
m
π 1
2 sin + (α α ) =
k k 1
m 2 − −
k=1 (cid:18) (cid:19)
X
m m
π 1 π 1 π
= 2msin sin (α α )2 + cos +η (α α )3,
k k 1 k k k 1
m − 4 m − − 24 m − −
Xk=1 Xk=1 (cid:16) (cid:17)
and therefore (13) is equivalent to
m m
1 π 4s
(α α )2 cos +η (α α )3 . (16)
k − k−1 − 6sin π m k k − k−1 ≤ sin π
Xk=1 m Xk=1 (cid:16) (cid:17) m
Clearly
m m
π
cos +η (α α )3 max α α (α α )2,
k k k 1 k k 1 k k 1
| m − − | ≤ 1 k m| − − | − −
Xk=1 (cid:16) (cid:17) ≤ ≤ Xk=1
hence, using (16) and (14), we may write the inequalities
π
P m 2msin s, β β ... β
1 2 m 1
P ≥ m − ≤ ≤ ≤ − ≤
n m o
24s
P (α α )2
≤ (Xk=1 k − k−1 ≤ 6sin mπ −max1≤k≤m|αk −αk−1|) ≤
m
4s
P (α α )2 1 ,
≤ ( k − k−1 ≤ sin π )
k=1 m
X
where s s as s 0. Quite analogously
1
→ →
m
π 4s
P m 2msin s, β β ... β P (α α )2 2 ,
P ≥ m − 1 ≤ 2 ≤ ≤ m−1 ≥ ( k − k−1 ≤ sin π )
n o Xk=1 m
where s s as s 0.
2
→ →
Now let introduce the quadratic form Q(α) = Q(α ,...,α ) by
1 m 1
−
m m 1 m 1
1 − −
Q(α) = (α α )2 = α2 α α .
2 k − k−1 k − k k−1
k=1 k=1 k=1
X 7 X X
We have by (9) for small s > s ,s s,s s as s 0 :
1 2 1 2
→ → →
2s 1 π 2s
P Q(α) 2 P m 2msin s P Q(α) 1 . (17)
≤ sin π ≤ (m 1)! P ≥ m − ≤ ≤ sin π
(cid:26) m(cid:27) − n o (cid:26) m(cid:27)
Now we proceed to the calculation of the probability in the right-hand side of (17). The
quadratic form Q(α) has the following matrix of size (m 1) (m 1):
− × −
1 1/2 0 0 0 ... 0
−
1/2 1 1/2 0 0 ... 0
 − − 
0 1/2 1 1/2 0 ... 0
− −
B =  0 0 1/2 1 1/2 ... 0 .
 ... ... −... ... −... ... ... 
 
 0 ... ... ... 1/2 1 1/2 
 − − 
 0 ... ... ... 0 1/2 1 
 − 
 
This matrix is symmetric tridiagonal. The spectrum of such matrices is known, see [2,
p.137] or [13], so that all eigenvalues of the matrix B are
πk
λ = 1 cos , k = 1,...,m 1.
k
− m −
Then using the appropriate orthogonal transformation, we can replace the quadratic
m 1 m 1
form − α2 − α α by the quadratic form
k − k k−1
k=1 k=1
P P
m 1
− πk
1 cos Y2 (18)
− m k
k=1 (cid:18) (cid:19)
X
in new variables Y . Now set for brevity
k
π πk
l2 = 2s /sin 1 cos ,k = 1,...,m 1.
k 1 m − m −
(cid:18) (cid:19)
As the Jacobian of the orthogonal transformation is 1, we get by (17) that
π (m 1)!
P m 2msin s − dx ...dx =
nP ≥ m − o ≤ (2π)m−1 Z(cid:26)mkP=−11x2k−mkP=−11xkxk−1≤si2ns1mπ (cid:27) 1 m−1
(m 1)!
= − dy ...dy =
(2π)m−1 Z(cid:26)mkP=−11(1−cosπmk)yk26si2ns1mπ (cid:27) 1 m−1
(m 1)! (m 1)!
= − dy ...dy = − b Π (s ).
(2π)m−1 Z(cid:26)mkP=−11(cid:16)ylkk(cid:17)261(cid:27) 1 m−1 (2π)m−1 m−1 m−1 1
Hereb isavolumeof(m 1)-dimensionalball ofunitradius, andΠ (s )isaproduct
m 1 m 1 1
− − −
of semi-axes of the small ellipsoid
m−1 y2
(y ,...,y ) : k 6 1 ,
( 1 m−1 k=1 lk2 )
8 X
which is equal to
m 1 m 1 m−1 m 1 1
− − 2s1 s1 2 − πk −
Π (s ) = l = = sin .
m−1 1 k=1 k k=1 ssin mπ 1−cos mπ (cid:18)sin mπ (cid:19) k=1 (cid:18) (cid:18)2m(cid:19)(cid:19)
Y Y Y
To simplify this expression we use t(cid:0)he identi(cid:0)ty(cid:1)fr(cid:1)om [10, formula (6.1.2.3)]:
m 1
− πk √m
sin = .
2m 2m 1
k=1 (cid:18) (cid:19) −
Y
Hence the product of semi-axes of the ellipsoid is equal to
m−1 m−1
Π (s ) = s1 2 2m−1 = 2m−1s1 2 .
m−1 1 (cid:18)sin mπ (cid:19) √m √m sin π m2−1
m
The volume of (m 1)-dimensional ball of the unit radius(cid:0)is we(cid:1)ll-known and equals
−
m 1
(√π) −
b = . (19)
m−1 Γ m+1
2
Therefore the volume of the (m 1)-dimen(cid:0)sional(cid:1)ellipsoid is equal to
−
V (s ) = b Π (s ) = (√π)m−1 2m−1s1m2−1 ,
m−1 1 m−1 m−1 1 Γ m2+1 √m sin π m2−1
m
(cid:0) (cid:1)
from where we finally obtain: (cid:0) (cid:1)
P m 2msin π s (m−1)!V (s ) = (m−1)!(√π)m−1 2m−1s1m2−1 =
P ≥ m − ≤ (2π)m−1 m−1 1 (2π)m−1 Γ m2+1 √m sin π m2−1
n o m
(cid:0) (cid:1)
m−1 (cid:0) (cid:1)
s 2 Γ(m) m−1Γ(m+1)
= 1 = s 2 .
√m πsin π m2−1 Γ m2+1 1 K1m
m
In the same manner we get f(cid:0)rom (17(cid:1)) the o(cid:0)pposi(cid:1)te inequality
π m−1Γ(m+1)
P m 2msin s s 2 .
P ≥ m − ≥ 2 K
1m
n o
Lemma 2.1 immediately follows from these two inequalities. (cid:3)
Proof of Theorem 1. Consider the transformation
π π
2m
zn(t) = 2msin s = 2msin tn−m−1,
m − m −
and denote
n n!
λ = P m > z (t) = P m > z (t) ,
n,zn(t) m {Pn n } m!(n m)! {Pn n }
(cid:18) (cid:19) −
where m is the maximal perimeter from (7).
Pn
9
As s = tn−m2−m1, we note that nmsm2−1 = tm2−1. Since for fixed m and large n it holds
n! nm, then we have from Lemma 2.1 that
(n m)! ∼
−
nlim λn,zn(t) = m1! nlim nmsm2−1 ·s−m2−1P{Pnm > 2msin mπ −s} =
→∞ →∞
= m1!tm2−1 nlim tn−m2m−1 −m2−1 P{Pnm > 2msin mπ −tn−m2−m1} =
→∞
(cid:16) (cid:17) m−1
1 m−1Γ(m+1) t 2
= t 2 = := λt > 0.
m! K K
1m 1m
Thus, the condition (3) from Silverman-Brown Theorem holds true.
Now for any r 1,...,m 1 denote by m;i1,...,im the perimeter of the convex
∈ { − } P
inscribed m-polygon based on the random points U ,...,U ,i < ... < i , on the unit
i1 im 1 m
circumference. We must verify the condition (6):
lim n2m rP m;1,...,m > z (t), m;m r+1,...,2m r > z (t) = 0
− n − − n
n {P P }
→∞
for each r 1,...,m 1 .
∈ { − }
Lemma 2.2 (verification of the condition (6)). For each r 1,...,m 1 it holds
∈ { − }
lim n2m rP m;1,...,m > z (t), m;m r+1,...,2m r > z (t) = 0.
− n − − n
n {P P }
→∞
Proof of Lemma 2.2. Using same arguments as in the proof of Lemma 2.1, we may assume
that β < β < ... < β , so that the points U follow one after one in increasing order.
1 2 m 1 j
Then the condition m;−1,...,m > z (t) implies by (15) that for some constant C > 0
n 5
P
α < C √s,..., α < C √s.
1 5 m 1 5
| | | − |
The second condition m;m r+1,...,2m r > z (t) implies analogously that
− − n
P
α < C √s,..., α < C √s.
m r+1 5 2m r 1 5
| − | | − − |
The intersection of these two events is the event
2m r 1
− −
α < C √s .
j 5
{ }
j=1
\
All angles α are independent and have the uniform distribution on the intervals
j
2πj,2π 2πj of length 2π. Then the expression of interest can be estimated for each
− m − m
r = 1,...,m 1, as follows
(cid:0) − (cid:1)
n2m rP m;1,...,m > z (t), m;m r+1,...,2m r > z (t)
− n − − n
{P P } ≤
n2m−r C5√s/2π 2m−r−1 = O(n2m−r−(2mm−r−−11)m) = O(n−mm−−r1) = o(1).
≤
(cid:3)
(cid:0) (cid:1)
By Remark 3 it follows that the rate of convergence in the worst case r = m 1 is
−
O nm+1pn,zn(t)τn,zn(t)(m−1) = O nm+1n−mm−21 = O n−m1−1 .
(cid:16) (cid:17) (cid:16) (cid:17)
We see that(cid:0)this result coincides with(cid:1)the result of Lao and Mayer in Theorem A for
m = 3 but deteriorates when m grows.
10