Asymptotics in poisson order statistics
Citation for published version (APA):
Brands, J. J. A. M. (1986). Asymptotics in poisson order statistics. (Eindhoven University of Technology : Dept of Mathematics : memorandum; Vol. 8603). Technische Hogeschool Eindhoven.
Document status and date: Published: 01/01/1986
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Department of Mathematics and Computing Science
'1tA1~i\ 88-12 7Hly 1928 ASYMPfOTICS IN POISSON ORDER STATISTICS by J.J.A.M. BrandsReports on Applied and Numerical Analysis
Department of Mathematics and Computing Science Eindhoven University of Technology
P.O. Box 513 5600 MB Eindhoven The Netherlands
ASYMPfOTICS IN POISSON ORDER STATISTICS
by
JJAM. Brands
Department of Mathematics, Eindhoven University of Technology,
The Netherlands
ABSTRACT
In order statistics sums involving incomplete gamma functions are met. The asymptotic behaviour of such sums is studied, going beyond the results obtain-able by the centra11imit theorem.
1.
IntroductionSome colleagues"') of the author have posed the following problem: Determine the asymptotic behaviour for Il~CQ of the sums
(1) S(ll,m, n):=
L
Im(Jl,k)(1-/(Il,k))n ,k=O
(2) T(ll, m,n):=
L
(ll-k)/m (Jl, k)(l-I (ll,k))n ,k=O
where m and n are positive integers and f1
(3) 1(Il,k)::(k!r1
J
e-
Itledt {J..L>O, ke OVo)
o
This problem arose in the study of the expectation and variance of the order statistics in a random sample from the Poisson distribution with large mean ll. In section 2 we present the results. A brief description of the derivation is given in section 3. The details of the derivation are given in sections 4 to 8.
2. Results
The sums S and
Thave the following asymptotic behaviour:
(4) S (J,L, m , n)
=
A (m,n»).1112+
B (m,n)+
0 ().1-II2 ) (J,L~oo) , (5) T(J,L,m,n) = C(m,n»).1+D (m,n»).1112+
0(1) ().1~oo) ,where
00 (6) ACm,n)=..f2J
/",(x)rC-x)dx , (7) BCm,n)=-2/3J
x/",(x)r(-x)dx , (8) C(m,n)=-3B(m,n) , 00 (9) D (m,n)=..f2
J
fm(x)rC-x)( 2/3-x 2)dx ,and
x (10)f
(x):= 1t-1I2J
e-
s2 ds (XE /R) .Clearly f can be expressed in the errorfunction but fonnulas
(6)to
(10)do not become simpler in
doing so. Some coefficients are
(11) A (1,1)
=
2A (1,2) == 2A (2, 1) == 1t-l/2B (n,n)
=
0 (nE IN)B(1,2)=-B(2,1)=B(1,3)=-B(3,1)=
!
1t-1 D(1,1)=2D(l,2)=2D~2,1)=11t-1I2 .4
3. Sketch of the derivation
The results are obtained by taking the following steps.
(i)
The sums
areapproximated by sums over
I u-k 1$;).1 213with an error of
0 (e-CIL ) 1/3 (J,L~oo)where c is some positive constant
(ii) For
k E [).1-C ).12/3, ).1+C ).1213]the asymptotic behaviour of
J (J,L, k)for ).1~OO is determined.
Let
x
E /Rbe defined by
whereh(s):=-s+log(1+s) (s>-l) .
Roughly x = (j.t-k) (2~rll2. Then I (j.t,k) has a complete asymptotic expansion in powers of ~-1I2 wich is uniform with respectto x e JR, x
=
0 (~l!6) (j.t---+oo).I (j.t, k) =::;
I
(X)-2112 T11t-1I2e-
x2 ~-112+ ....
(~---+oo), where lis defined by (10), i.e.'VNe IN 'Vc>o
3
A>03
B>0 'V,.>A 'Vke INI k -
~
I S C~2J3
=:>Il(~'
k) - {/(x)+
e-x2f
q/(x)~-1!2}
Is
B~-(N+l)J2.
1=1(iii) The sums
L
are approximated by integralsJ ...
dk with an error which, forI
,.-k
1!>,.:113 I,.-k
1!>,.213every positive number r, is 0 (j.t-r) (j.t---+oo). Then these integrals are transformed into integrals over x and then approximated by integrals
J
dx with errors of the kind4. The truncation of the sum
The function l(j.t, k) interpreted as a function of the real variable ke [0,00) is decreasing on [0,00). This statement follows from
11
~1(~,k)=(f'(k+l)r2
J
dtf
e-I-'ttkl log (tt-1)d't<0I) 11
Let kS~;.t2J3:=a. Then substituting t=k(1+s) we have l-I(~, k):;; 1-1 (j.t,a)=(r(a +1)r1 e-aa o+1
J
eoh(s)ds ,a-I 11-1
where h(s):=-s+log(1+s) . The function h is concave and negative on (-1,00) • whence
h(s):;;h(~-l_l)
+
(s _~a-l+
1) hi (~-1_l) (s~~-l_l).It follows that
0< e-a a o+1
(rca
+
1))-1f
eoh(s)dsSe-aao+l(f'(a+l)rl eah (0-1 Il-l) (a h' ~-1-1)rla-11l-l
:m 113 113
0<
L
<1l.1l3/2 e-TIl=
O(e41 ) (Jl~(X»Qs;k$a
k-11l-1
Let
iell+J.L2I3 •Then
/(1l,k)=e-kkk+I(Hk+l)rlJ
ekh(s)ds.Since
h(s)S-1!2s2on
(-1, 0]-1
we have
k-11l-l 1(Jl, k):S;e-kkk+1(r(k+l»-1J
e-1I2ks2 ds -1 :S;e-kkk+1(r(k+l)r1J
e-112ks2 ds l-k-11l 00 :S;ekkk+l (Hk+ 1)r1J
e-1I2k (l-k-11l)s ds l-k-1!1Hence, for both of the sums
0<
L
:s; 1l-1I6L
k e-Yzkm (!1113+1r
2 :S;2Ile-l/3m!1113 (~8) .1&!!1+!12!3 1&!!1+112!3
Hence. in both cases
where c is a positive number.
S. The incomplete gamma function
As we have seen already the substitution
(12) t=k(1+s)
in the integral representation of
I (f..1.,k)gives
k-1 11-1 (13) l{J!,k)=(k!r1e-ke+1J
ekh(s)ds • -Iwhere
(14) h(s):=-s+log(l+s)
(s>-I).(15) Y :=k 112, h (s) 11/2 sgn (s) (s >-1). Then we get
(16)
where
To study the transfonnation (15) we introduce first in (13) (18) 't=lh(s)11I2sgn(s).
Then
(19)
..J2
't=S(l-2/3s+2/4s2-2/5s3+ .... )1/2 (lskI) ,=s-l!3s2+7/36s3+. . . (Is 1 <1) •
the radius of convergence being one since It (s)1 s 2 has no zeros inside the unit cirele. By the Bumann-Lagrange theorem we can expand s as a powerseries in 't with a positive radius of con-vergence, say p.
We calculate (20)
(21)
The transfomation (18) changes the integral (13) into
k-1f2x
(22) I (Il,k)=(k !fl e-kkk+1
J
We shall study the asymptotic behaviour of this integral for k*"700 and fixed XE IR.. Therefore we
shall denote the right hand of (22) by i(x,k).
In order to use (21) we must truncate the interval of integration. Now it is easily shown that
where
c
is a positive number.Inside the circle I 't I~ 1I2p the powerseries in (21) is also an asymptotic series, i.e. for every NE IN we have
(23)
Hence
(24) I-(x,k)
=
(k ,.)-Ie-kkk+lJ
e
_k't2(£:+ - ' t .... 4+ +
CN-IT-N-1+O ( N»d'"-r .1't1S;1I2p, 'td-'12x 3
Now we change the lower bound into -oc and, evenually, the upper bound into r1l2x, thereby making an error of the kind
0
(e-Ckl13) (k-too) uniformly inxE JR.Then we substitute 't=k-I12y and we get
(25) i(x,k)=(k!r1
e-
kk"+l!2 {e-irJ2+4/3k-l!2y+~
k-1y 2+ ... -<X>k1f3
+0 (e-C ) (k-too) unifonn]y inXE JR.
x
Since
J
e-i 0 (k-NI2 y N) dy=
0 (k-NI2 ) (k-too) unifonnly in XE JR, we have(26) i(x,k)=g(k)
-.&
unifonnly in XE JR, where Hence xJ
e-
i
CJz+4/3k- 1I2y+' .. +CN_1k-(N-l)l2y N-l) dy N-l(28) i(x,k)=g (k)
L
ct.fJ(x)k-l12+O(k-N/2) (k-too) unifonnly in XE JR,1=0
where the
c/s
are given by (21) andx
-7-Integration by parts gives
(30) fi(x)=e-x2p[(x}+
l~li
r{1~1
)(21trl12f
(x) (Ie IN) .where
fis
defined by (10) and(31)
l
(/-1>1( ) " 2-312 -112r( 1+] ) '"
PI X
.=-
n2
4-is=O
Letting
x~oo
we find thatfi(oo)=O if I is odd andfi(oo)=r{l~l
)(2nr112 if I is even.Since limI(Il,k)=l we get the asymptotic series for (g(k»-l by letting x~oo in (28). We have I1-tOO
00
(32) 1:::: g(k) :Eclfi(oo)k-1I2 (k~oo) .
1=0
Comparing (28) (30) and (32) we see that the factor with whichf (x) occurs in (28) has the same asymptotic expansion as the factor
g
(k) in (32). Hence(33)
_ 00 2
I(x,k»::::f(x}+g(k) L,cle-x PI(x)k-1I2 (k~oo)uniformlyinxe fR.
1=1
(Le. after truncation the (absolute) error is smaller than c k-(N+l)12 with c independent of x).
According to the results of section 4 we restrict ourselves to values of ke [1l-Jl2l3 ,1l+1l2l3]. Then
x=o (1l1l6) CIl~oo).
In order to get an asymptotic series for I (Il,k) for Il~OO we have to express k as a function of Il and x. From (17) we have kh(j.tk-l_1)=-x 2, sgnx==sgnCll-k). Putting k==Il(1-'I') and ll-l12x=z we get (34) 'I'+(1-'I')10g(1-'I')=z2 whence (35) Then (36)
:E . )"
=z2 (I'l'l <1),sgnz=sgn'l'. i=2 (t-l 1(37) ljI= 'LdjZi (I Z I <r) .
i=l
Hence
(38) k=Jl- idiIJ.1-il2XI (Ix
I<r~).
i=l
Clearly, for IJ. sufficiently large. x is within the range of convergence since x=O (J.l.1I6) (J.l.~oo). A few coefficients dj are calculated.
(39)
We need also expansions for k-1I2 • k-1 and k-312:
Clearly, there is a positive number ro such that for all ae 1R the function (l-'VYl. has
a
power-series expansion in Z which converges for I Z I <ro; if00
(40) {l-'V)u= 'La/a)zj (I z I <ro) j=D
then
(41) kU=JluO-'V)u= ia/a)IJ.U
-il2 Xj (Ix I<ro IJ.1I2) . j=O
We calculate
(42) k-l12 = IJ.-1I2 +2-1 12 IJ.-1 x+..LJ..l-3J2x 2+ 13
12
IJ.-2x 3+ ...12 36
(43)
(44)
An asymptotic expansion for g (k) can be determined from (32).
(45) g(k) ;:: ig[k-lJ2
(k~oo)
1=0
Some coefficients gl are
(46) gO=l,g2=-1I12andg1=Oif 1 = odd.
9
-(47)
1(~,k)==I
(x)+e-x2L
ql(X)~-1I2 ~~oo)
1=1
uniformly in x
=
O(~1I6) ~~oo).The ql's are polynominals. We calculate
(48) (49)
q 1 (x)=_21I2Tl1t-112
q2(X)=_~1t-1I2X
12
6. The replacement of the sum
byan integral
We have alreadyII
(50)
~1(~,k)=(r(k+l»-2
J
dtJ
e-H:(t'1/log(t't-1)d'to
0By induction one can prove easily that
(51) 1 II
~1(~,k)=(r(k+l))-I-1
J
dtJ ...
dk°
0J
e l l - 1 4 - " ' 4 ( t't) ... 'tl )kL( t, 'tl, . . . ,'tl )d 1:)'" d t[ , owhere the functions L(tO,tl, ... ,t[) are defined by
(52) L(to)=1
With methods simular to those used in the treatment of I (~,k) in section 5 we can prove easily that
(54)
Furthermore, if we restrict ourselves to values of k such that I ~-k 1::;~213, then the integrals in II. 11.+211.213 11.+211.213
(51) can be replaced by
J J. . . J
11.-211.213 11.+211.213 11.-211.213113
with an error of the kind 0 (e -ell ) ~ ~oo).
Then it follows from (51) that for I ~-k 1::;~213 (55)
From (52) and (53) it follows that
[
J..I.+2J.12/3l 1+1
Ml+l ~ M/log 213J· (Ie IN 0), J..I.-2J..1.
whence, by (54) and (55), we have for all ke IN
(56) I dlll(J.I.k)1 =O(J.I-II3) (J.I-+oo) (Ie INo).
dk
Now we apply the Euler-Maclaurin sumfonnula: For every fixed re IN we have
q q (57) ~f(k)= Jf(x)dx+1I2fCq)+1I2f(P) k=p p +
i
Bu,
[f21-1)(q)_f21-1)(P~
1=1 (2l).j
+o[
/1f2')(X)11
(q~p)
.Takingf(k)= Iffl(J.I,k) (1-/(J.I,k»n,p=r
J..I.-J.l2l~.
q =lJ..l.+J..I.213J '
we find, for every re IN, using (54), (56) and (57)(58) ~ Im(J..I..k) (1-/(J.l, k)t
=
I k-I1I$I1213
Simularly
11+11213
(59)
L
(u-k)l m(J.I,k)(1-/(J..I.,k)t =J
(J.I-k)lmCJ..I., k) (l-/(J.I, k»n dkI k-I1I$I1213 1141113
7. The asymptotic behaviour of Sand T
According to section 6 the sums ~ can be replaced by integrals with small errors which II1-k 1$11213
are, for every r>O • 0 (J.I-r) (J.I-+oo). Then these integrals are transfonned into integrals
B(jJ.) dk
J ...
-dx where A (J..I.) and B (J.l) are asymptotically equivalent with1I2~
J..I.1I6 (J..I.-+oo)11
-By means of (38) and (47)
we
get complete asymptotic expansions for the two intcgrands. (60) 1m {J.t.k)(1-1 {J.t.k»11 :=
~
S/(X)Il-I/2 (J.t-,)oo) •/=-1
(61) (J.t-k)lm{J.t.k)(l-I{J.t.k)t:
=
~
t/(X)Il-//2 (J.t-,)oo) , 1=-2both unifonnly in XE
m. ,
x=
O{J.tl/6) (J.t -,) 00). The functions s/(x) and tl(X)are
absolutely integr-able over (-00,00). Now we proceedas
follows. Let N E IN. Then1L+1L213
t
Im{J.t,k)(1-I{J.t,k)t : dx IL-J! N B (p.) [ N 1 lJ
=L
/J-l12f
s/(x)dx+O 11--2 '2+'6 1=-1 -A(p.)It is easily seen that the functions s/(x) are of the fonn
2 2 2
p(x.e-x ,f(x»
e-
X+q(x,e-X .f(x» f (x)(1-f
(x» ,
where p and q are polynomials. HenceJ
Isl(x)ldxandJ
ISI(x)ldxareO(e-CILII3) (J.t-,)oo).x<-A(p.) x>B (p.)
It follows that S (/J,m, n) has a complete asymptotic expansion
(62) S(Il,m,n)=
~
Jl-1I2J
sl(x)dx (Jl-,)oo)/=-1 - 0 0
A simular argument holds for T(Il.m,n).
(63) T(Jl.m,n)
=
~
/J-lJ2J
t[(x)dx (Jl-,)oo). 1=-2 We calculate (64) S-l (x) ={2r(x)rC-x)
(65) so(X)=2/31t-1I2 [ nr
(x)r-1C-x)-m r -1(X)r(-xJ e-x2-2/3xfmCx)r(-x) (66) L2(X) =2 x r(x)rC-x)
(67) t-l (x) = 23123-111-112[nr(X)r-l(-x)-mr-l(x)r(-x~
xex2-21I2x2r(x)r(-x) 2The tenn with factor
e-
x in the right hand of (65) gives 0 upon integration.2
Integration by parts of the tenn with factor xe-x in the right hand of (67) gives 00
00
D(1,2)=D(2, 1)=213A (1,2)-112"2
f
x 2(j(x)f2(_X)+
f2(x) f (-x»dx = 00=2/3A (1,2)-112"2