(#2013)
INFIN ITELY DIVIS JBLE LA\\S PART IALLY ATTRACTED TO A POI SSON LAW
t
.."
SANDOR CSORGO and ROSSITZA DODUKEKOVA
Groshev, in 1941, derived a necessary and sufficient condition for a
fi~ed
infinitely divisible law to belong to the domain of par-
tial attraction of some Poisson law. The condition is expressed
through the spectral function of the Levy-Khinchin canonical
representation for the characteristic function of the infinitely
divisible law in question. The mean A of the attracting law does
not play any role in Groshev's condition. In the present paper
we investigate the problem for any fixed
A
>
a
and derive cha-
racterizations involving A in terms of the measure functions
ofa unique
probabilistic
representation for an infinitely
divisible law, obtained within the framework of a new "probabilistie approach"
to
sums of independent and identically
distributed random variables. Here, as in [2] , where an arbitrary law attracted to
a
Poisson is characterized, the intricate
behaviour of the normalizing constants shows up. Depenciing on
whether A > 1 or A < 1 there are two qualitatively different
ways of choosing them and hence the attracting Poisson law
arises in two different ways. A concrete construction illustrates these results and shows also
bilities may
that if A
=1
then both possi-
Gccur~
* s. Csorgo was partially supported by the Hungarian National
Foundation for Scientific Research, Grants 1808/86 and 457188
and R•. Dodimekova was partially supported by the Ministry of
Culture, Science and Education of Bulgaria, Contract No. 16848
- F3.
.
-
1.
I~TRODUCTION,
2 -
THE RESULTS,
A~D
DISCUSSIO~
Let F(x) = P{X s x}, xeR, be the distribution function of a
given random variable X, and let X1tX Zt ••• be independent copies of X. We say, that F is partially attracted by the Poisson
distribution with mean A, if there exist a subsequence en'} of
the sequence en) of the positive integers and normalizing and
centering constants An' > 0 and Cn,eR such that
1
n'
-A {1; X. - C ,) -+ D YA as n' -+
n
n' j=l J
(1.1)
GO
where -+D denotes convergence in distribution and YA is a Poisson random variable ~ith mean A > 0, that is
k
P{Y>..
A
= k} = IT
e
->..
k = Ot1,2t . . . .
The distribution functions satisfying (1.1) form the domain of
partial attraction of the limiting Poisson law. We write for this
Dp (Poisson (A)). The domain of partial attraction of
any infinitely divisible law G is defined in a way analogous to
set
(1.1). Khinchin proved in 1937
that for every such G the set
Dp(G) is not void (cf. [4]). Then, of course, Dp (Poisson (>..))
is not void. It is also known ([4]) that if (1.1) is true,
then en') can not be the whole {n}; otherwise the limiting
distribution should be stable which is not true for the Poisson
distribution. (That is exactly the reason that we talk about
partial
attraction~)
There are many reasons for the interest in the characterization of the distributions partially attracted to a Poisson
law. In,1941; Groshev (5) characterized the distribution functions in the set
U
).>0
Dp (Poisson (>..)). He proved that
- 3 -
Fe
U D (Poisson (A))
1.>0 p
if and only if
x2
1
(1. 2)
dF(hx)
Z
lim inf ~>£ 1+x
h-+
1
dF (hx)
=
0
for any
£
>
o.
CD
IX-11<£
This condition, being in terms of F, does not show immediately in
what sense F is restricted
role of the mean
A
just as it does not show at all the
of the attracting law •.In the same paper,
Groshev also considered a special case of the problem when the
underlying F is infinitely divisible and proved that
Fe
U D (Poisson (A))
p
1.>0
if and only if
.2
1
(1. 3)
.1E..:.lJ >£
lim inf .
h-+ CD
where G(.) is the
1
~ dG(huJ
1+ u
lu-11<£
dG(hu)
l~vy-Khinchin
=
0,
measure function of the canoni-
cal form of the characteristic function of F. As in the general
case., the condition above completely disguises the role of the
parameter A. Note that although (1.3) and (1.2) are very much
alike, the result in the special case is not a direct implication from the result in the general case. '1:.0 obtain (1-.3) Groshev uses the special form of the general criterion of convergence to an infinitely divisible law obtained for the case of
an infinitely divisible underlying distribution function (see
[ 4] ) •
In [2], operating within the framework of the recently
- 4 -
developed "probabilistic approach" to
SUl':".S
of independent
and identically distributed random variables ([ 3] , [1))
racterised an arbitrary distribution function F
to D
p
~'e
cha.-
which belongs
(Poisson (A)). Starting out from the general convergence
theorems of the probabilistic approach we delineate the portion
of the sample that contributes the limiting Poisson behaviour
of the sums in (1.1) and describe the effect of extreme values.
This leads to the analytical condition (1.4) below for
F£Dp(poisson (A)) expressed by the quantile function
Q(s)
= in~ {x ~
~
F (x)
s},
a<
s < 1, and the corresponding "trun-
cated variance" function, which by setting uAv
l-s l-s
J
(uAV-uv)dQ(u)dQ(v) ,
2
a (s) = J
s
s
= min(u,v)
a< s <
is
1/2.
The condition of [3] is the following:
2
(1 .4)
lim in f
E
a (-) +
n
=
a
1 for
n..;-"
<
E
S A/2.
This implies that depending on whether A>l or A<l there a.re two
principally different ways of choosing the normalizing constants
An' in (1.1). When A > 1, then 6 1 <
positive 6
and 6 • When A
2
1
The case A
=
<
1,
YnT ~~~/n') <
then
vnr ~(l/n')
6 2 for two
";-'0
as n' ..;--.
n'
1 is a real borderline and both possibilities may
occur as shown by the Construction in [3].
Just as the canonical representation of the characteristic
function of an infinitely divisible law in (1.S) below is a
central point in the classical limit theory, the probabilistic
..
representatio~
of an infinitely divisible law in (1.9) below
plays an essential role within the framework of the probabilis-
- 5 -
tic approach. It is desirable then, in case if in (1.1) the
underlying law F is infinitely divisible,' to have the necessary
and sufficient conditions for F£Dp(Poisson
in terDS of the
(A))
components of the probabilistic representation of F, what is the
aim of the present paper.
~ote
that here as in (5], again, a di-
rect implication from the general result (1.4) is not possible.
To formulate the main results we need to introduce the probabilistic representation of an infinitely divisible law. Let
~(t),
t~R,
be the characteristic function of an arbitrary infi-
nitely divisible random variable Y in its Levy form,
(1 .5)
It'
2 2
a t
(t) = exp{ite - -Z-
0
1 ( e itx - 1
+
_llG
- ~)dL(X) +
1+x
00
Here 0 and a
+
f ( e itx - 1 - ~)dR(X) .
1+x
0
~
0 are constants and L(x) and R(x) are two non-
decreasing uniquely determined left-continuous and right-continuous functions on (_llG, 0) and (0, 00), respectively, with
L(_oo)
=
R(+oo) = 0 such that
o
f
-£
£
x 2 dL(x) + f x 2 dR(x) < 00
for any
£
> O.
0
Introduce the functions
(' .6)
" 1 (u) = inf (x < 0: L(x) > u), 0 < u < 00,
(1. 7)
'2(u) = inf{x < 0: -R(-x) > u), 0 <
U
<
00,
which obviously are non-decreasing, non-positive and right-continuous, and also the constant
'f j (u)
l+'f~(u)
du -
du} •
J
Further, let Nj (.), j • ',2 be two independent standard left-
- 6 -
continuous Poisson processes with jump points s~j), n
=
1,2, ... ,
given by
•
I
I
n=l
(S~j) <
o
u),
$
u < .,
j
=
1,2,
where I(.) is the indicator function and with independent exponential random variables Efj), E~j) , ... , all having mean 1,
= E1
(j) + ••• + E n
(j) '
Sn(j)
n....c;.,
1 J' - 1, .
2
Introduce also the random variables
v(j) = (_l)j+l {
(1. 8)
J
S(j)
1
sO)
1
(u-N. (u) )dli'. (u) +
J
J
J
ud"j
1
j
(U)+lf j
(1)}
= 1,Z.
Let Z be a standard normal variable independent of
(N 1 (.), NZ(.)). Theorem 3 in [3] states that the random variable
V(l) + oz + V(2) + e - y(f , 'fZ) has characteristic function
1
~(t) of (1.5), that is we have the distributional equality
(1. 9)
From now on we assume that the given distribution function
F is infinetely divisible. In fact,without loss of generality
we can assume that
(1.10)
X = V(l)
.
+
oZ
+ V(2)
:= V(lf , lfZ'O)
1
which by (1.9) means of course that
The simple reason is the following. The condition (1.1) is satisfied if and only if it is also satisfied for independent
copie~
of
~-
C, with the same An' and with centering constants
Bn , • Cn , -n' C, for any real C. This remark and (1.10) above
- 7 -
(~"~2,o)£Dp(Poisson
allow us to write further
(A)) instead
of FeDp(poisson (A)), just as it is adopted in [1].
For
a
n and for h > 0 we introduce
positive integer
the functions
z
00
2
a,(n;h)
Z •
aZ(n,h)
=
n{
=
a, (n;h)
If,(u)du
J
+ a
Z
J
+
Z
'I' Z (u)
duJ ,
hln
h/n
2
+
'1'2, (h/n)
+
2
If 2 (h/n)
and a~(rn ) = ra~(n; '/r), r = ',2, .•. , which are all ~ell defined due to the property that
-
J 'I'~(u)du <
00,
J
£
£
> 0;
j
= ',2
obtained from the square - integrability of Land R.
Let
A
> 0 be fixed and introduce the notation
r, (A) = min{r:
positive integer, Ar >
r
l}.
THEOREM ,. We have ('I',,'I'Z,a)eDp(Poisson (A)) if and onLy
i f for each r
r,(A) there exists a genuine subsequence {n'l
~
of the positive integers such that
s
1
(1.11)
aZ(rn')
'l'1(~)
• 0, if s > 0,
. r-a r'
1
(1.12)
if 0 < s < >. ,
0 , if· S
as n' • -, where a r
>
0 is some constant,
a (n' ;h)
(1.13)
lim lim sup
h·-
n'·-
> ). ,
1
a 2 (rn')
&
and
o.
To formulate the next theorem we need the function
-
00
x{ f
h/x
=
'f
2 (u)du
1
1j
+ a
-
2
+
x >
o.
Obviously,
(1.14)
b
2
(n;h) = a 12 (n;h)
b 2 (n ; 1)
+ h
{lf2 (~)
1 n
2
2
a (n;1) = a 2 (n) •
2
=
TH E0 REM 2. The t h r e e con d i t i
0
r. s
(1. 1 2 ) a r. d (1. 1 3 )
(1. 1 1 ),
are satisfied along Some subsequence {nl) of the positive intean d h ence ( %1'%2,a
gers~
UI
(1.15)
W
)
E
Dp (P olsson
.
( A)) ~.~ fan don 1~_y ~.~ f
2
b (x;t) + b2(x;A+t)+'f~(t/X)
lim inf
2
xt oo
b (X;A-t)
o<
= 1 for every
t
S 'A/2.
As a matter of fact, instead of (1.15) we will use the
condition
(1.16)
b
2
2
(n;t) + b (n;A+t) +
2
lim inf
'i'~(*)
b (n; A-£)
n~C11
o<
=
t
~
1 for every
)../2,
which is equivalent, as we show just now, to (1.15). Note first
that if h < s, then
2
2
b (x;h) - b (x;s) = x{
+
h {lf~ (~) +.
~ xc.!x - ~)
x
2 h
If~ (~)}
(lf2, ('!) +
x
2 s
= h{'i',Ci) - 'l Ci)
- s
six
six
2
2
J 'i',(u)du + f If (u)du)
h/x
h/x 2
{lf~ (~)
+
If~ (i)}
If2(.!)}+ h{lf2(~) + If2(~)} _ s
2x
'x
2x
+
2 h
2 s
h{ lf 2(x). - 'i'2(i)
s) +
{lf 2 <-x
l
If 2 <-xs )}
2
~ 0,
2
which means that b (x;h) is non-increasing as a function of h.
Hence
- 9 -
2 (x;c)
b2
b (x;>"-c)
~
1 for x
> 0 and 0 < c
~
A/2,
2
and since b (x jA +C ) ~ 0,
2
2
b (x; A - t)
b (x; >- - E:)
(1.15)
means
that
for
~
0,
0 < E: ~ A/2 there exists a
every
subsequence of real positive numbers {x m) = {xm(t)), such that
x ~.. and
In
'1'2 (~)
2
2
b (x ; c)
m
b (x ; A + E:)
~
1,
In
--""2-- - -
b (xm;>--t)
1 x
~ 0,
2
TIl
b (Xrn;A-t)
~ O.
2
Because of the monotonicity of the functions b 2 (x,t) and'1'l(c/x)
~n E:,
none
of the three ratios above increases in c. Using
this, one can easily show that the above conditions hold if and
only if there exists a subsequence {n') of the positive integers,
independent of
£,
such that
2
c)
b (n'i
= 1,
b Z (n I ; A- £)
o<
c ~
A/2,
(1.18)
b 2 (n l ;}.+c)
lim
= 0,
2 l
nl~ ... b
(n jA-E)
o<
c
~
A/2,
(1.19)
lim
o
£
~
>'/2.
(1 . 17)
1 im
n'~'"
n'~'"
'~(t/n')
=
b 2 (n' j >. - £ )
0,
<
The conditions (1.17)-(1.19) are obviously equivalent to
(1.20)
lim
n'~'"
2
b (n'jc)
+
b
2
b (n';A+c)
2
(n' j A- £)
+
'i(c/n')
=
1, 0 < e: :: A/2 ,
and the latter implies, of course, (1.16). Conversely, (1.16)
implies (1.17)-(1.19), again by the monotonicity of the func.
b 2 an d '12 W1t
. h respect to
t10ns
E.
Some results, obtained .in the proofs of our theorems, will
be used to establish the
followin~ Coroll~ry chaTacteTizin~
- ,0 -
the
two different ways of behaviour of the normalizing con-
stants in (1.1) by comparing AnI and aZ(nl).
COROLLARY. Suppose (1.1). If l > 1, then there exist t~o
6,
positive constants
a 2 (n I
0, S
0z
<
such that
)
AnI
S oZ'
a 2 (n I)
If l
< "
then
A ,
~
0 as n l
~ ~
n
The proofs are in the next section. In Section 3, we illustrate the above results by a concrete construction of ~2' that
is, a concrete constraction of an infinitely divisible random
variable in Dp (Poisson (l)). This construction works for all
1
> 0 and
i~ will turn out that the case when 1
=,
is really a
borderline case in the sense that both possibilities of the
Corollary show up.
2. PROOFS
It is well known (see [4]) that for the Poisson law with mean
1
>
0 the constant
zero, e ., l/2,
0
2
L(x)
in the representation (l.S) equals
= 0,
x > 0, R(x)
= -1
for 0 < x < , and
Rex) • 0 for ~ ~ 1. Simple calculations show, that in this case
the function difined by (1.6) is identically equal to zero and
the one in (1.7), "'hich \\'e denote by
by
-1, if
[ 0, if
O<U<l
'f).
in the sequel, is given
- 11 -
Furthermore we need the random variable V(O'~A' 0) defined
by (1.10). Easy calculations in (1.8) give
(z. 1)
where, for simplic ity, ,,;e put N(A)
Let Vj
Proof of Theorem 1.
('f
1
NZ(A).
=
,1f
2
,a), j
= 1,2, ... , be inde-
pendent copies of V(1f1'~2,a) and suppose (1.1). This means, according to the remark following (1.10) that we have
n'
1
--- {~ V.(1f ,If ,a) An' j =1 J 1 Z
B ,)
n
for some An' > 0 and Bn ,. Let
n' r
{ ~ (k
ja:1 m=1
VJ~m))
-
~D
YA,
{vjm));=l
rB ,) ~D Y ,
n
r"
n'
~ ~,
be independent copies
'
n' ~ ~
=
1 , Z, ...
Introducing
= V~m)
J
and Cn* ,
means
1
An'
=
rB n'
+
,
1 S
m S r,
j
rA An I and using (2.1) we see that the above
rn'
*
!. - C ,) ~D V(O'~rA'O).
j =1 J
n
--- {t
We are in the
5
i tua tion of part (v) .of Theorem 12
necessarily implies
in [1] , ,·:hich
- 12 -
lim lim sup
h+~
n'+=
(Z.4)
a,(rn';h)
A
=
n'
o.
Also,
1im sup
n' . . . =
I If j (r~ f) 1<
aZ(rn')
"", for s > 0, j = " Z,
since otherwise we would have aZ(rn')/A n , . . . 0 as n' . . . ~, which
by part (iv) of Theorem 12 in [1] is impossible because the
limit in (2.3) is non-zero for 1
~
s <
r~.
Therefore, again by
part (v) of the same Theorem 12, there exist an
{n"}
C
{n'}
and a constant 0 < a r < "" such that
Putting this together with (2.2), (2.3) and (2.4), we obtain
(1.'1),
(1.'Z) and (1.13) along {n"}
since a 1 (rn";h)
= v'T a 1 (n";h/r) and,obviously,
lim lim sup
n"........
h.. . =
= lim
lim sup
h.. . =
n"+'"
a 1 (n";h/r)
\
aZ(rn")
a,(n";h)
a 2 ( rn " )
This completes the proof of necessity.
Kow we show sufficiency. Writing n~ for the k-th term
of the sequence (n') in (1.11), (1.12) and (1.13) and setting
.,
,
'jkC.) = '1'j(./nk )/a 2 ( rn k) := 'j
j
=
1,2,; k
=
of [1] we get
,
(n k ) ·
,
(.)/a2(rn k ),
1,2, ..• , by the distributional equality (2.42)
- 13 -
1
(Z .5)
where
1
l!n('l"Z)
=
1
n( J ud'Z(u) - J udli'l (u)}.
lIn
lIn
The general convergence theorem for inf ini te ly di vis ib Ie 1a. ws ,
that is, Theorem 11 in [1] states
that under the conditions
(1 • 11 ), ( 1 • 12) and (1. 13) ,
V(li'1k,li'2k'
as
k
~ ala2(rn~)) ~D
V(O,arli'A'O)
By (Z.5) and (Z.1) we get
~~.
1
as
k
~
c
., which proves sufficiency.
The following four lemmas are required for the proof of
Theorem Z.
LEMMA 1. Let A > 1 be a fi:ed number and {nIl be a subsequence such that
(2.6)
1
a 2 (n ')
1
(Z.7)
8
as n'
(2.8)
s
'1(nr)
2
(n ')
s
"z (nr)
~
0, 5
+
> 0,
{-a,
0, s
a > 0 is Borne
alCnt;h)
lim lim sup aZln I) •
h~·
n'~·
~ -~
~here
0 <
5
< A,
> A,
constant~
o.
and
- 14 -
Then~
as n'
-+
DO,
O<h<l\,
(2.9)
and
•
h > 1\.
(2.10)
Proof. By (1.14) we have
(2. 11)
b 2 (n';h) = a ~ (n ' ; h)
2
b (n' ; 1 )
a ~ (n' ; 1)
+h{
If 2 (~)
1n
a~ en' ; 1)
+
If2(h )
2 J1T
2
)
az(n t ;1)
First we investigate the behaviour of the right-hand side of
(2.11). Let 0 < h <
S
and consider
sIn
a 12 (n;h) - a 21 (n;s) = n { J
hln
2
If 1 (u)
du
sIn
+
f
2
If 2 eu)
du} .
hln
Then
n(!n - nh) {W2(!)
T, n
+
2
T2 (-ns )} <- a,2( n; h) - a 12( n;s )
W
and hence
(2.12)
2
a 1 (n;h)
2
a (n;1)
z
a~(n;s)
a~ (n; 1)
Let h > A. By (1.14), (2.6) and (2.7) the right-hand side of
(2. 12) goes .to zero as n' ..... and this gives
- 15 -
a
~
•
2
a~(n' ;h)
a1en';s)
---,;,--a ~ en' ; 1)
,
ai en ';l)
+
0 as n'
+
~,
A < h <
5,
which implies
a ~ en' ; h)
a ~ en' ; s)
(2.13) lim sup
= lim sup
=: ~2(s).
n'+m
a~en';l)
n t +m a~(n';l)
Since ~ 2 (5) does not increase, there exis ts 1 im ~ 2 (s)
s+m
Now (2.13) and (2.8) imply
~
=
=:
~ ~
O.
0 and this together with (2.6),
(2.7) and (2.11) prove (2.10). To prove (2.9), put first s = 1,
o < h < 1 in (2.12). By (~.6) and (2.7) and by A
> 1
t~e
inequalities (2.12) give
lim {
a~ en' ;h)
n'+ma~(n';l)
a ~ (n' ; 1 )
--=----} = (l-h)a 2 ,
a~ (n' ; 1)
while by putting h = 1, 1 < s < A they give
a 2 en' ; 1)
a~en';S)
lim {--.,.~--­ - 2
) = (s-1)a 2 •
n '+m a 2 (n' ; 1)
a 2 (n I ; 1)
These two limiting equalities together with
a ~ (n' ; 1)
a ~ (n' ; 1)
as n'
+
(2.14)
-,
=
1
2ji"T
a ~ (n ' ; 1)
2
a 2 en' ; 1)
+
~
1 - a 2,
imply that
a~ (n' ; h)
a~(n';l)
~
1 - ha 2 t as n'
Now by (2.6), (2.7) and
(2.9).
,2(1 ) }
1 fiT 1
- {'2(1
(2~14)
~
-
t
o<
h < A.
used in (2.11) we easily get
- 16 -
2. Let A
LE~~
~
1 be a fi=ed number and suppose (2.9),
(2.10) and that
If
1
1 (nr)
0, as n'
--~~ ~
(2.15)
~ ~
.
b (n' ; 1 )
Then there exist a subsequence (nil) C (n') and a constant
a > 0 such that (2.6), (2.7) and (2.8) are satisfied
~ith
n'
rep laced by nil.
Proof.
For h > A, (2.10) and (2.11) give
a ~ (n' ; h)
a~(n';l)
~
0, as n' ..
00
which implies of course (Z.8) along the original (n'). Furthermore, the definition of b 2 and (1.14) easily imply that for
h < s,
a ~ (n; s)
a~ (n; 1)
+
2
= b 2 Cn;h)
b 2 Cn;s)
b (n; 1) - b 2 (n; 1 )
s(
We substitute the right-band side of the above equality into
(2.12) and get
(2.16)
,,2(.!)
1 n }
2
.
a (n,l)
Z
2
b (n;s)
2
b (n; 1)
,2(.!)
,2 (~)
2 n
2' n } + s ( -2-(....;.;.'-8 (n; 1)
8
n,l)
2
2
- 17 -
By putting 0 < h < 1,5 ': 1 into (2.16), and by (2.9) we get
o
'I' 2 (1!-)
1 n'
~
'1'2 (_')
1 n'
0, as n'
_-~- -+
-2--a Z (n';1)
a~ en'
; 1)
-+
co,
for 0 < h < 1. From (2.15) and this (2.6) follcws for
o<
< 1. Since for s > 1 we have by (2.15)
S
'I'2(s )
1JiT
'1'2(1 )
1j}T
s
a~(nl;1)
0 as n'
--"'2-~- -+
a (n';1)
2
(2.6) is true for all s >
o.
-+
I
to
,
When s = 1 and- 0 < h < 1 in (2.16),
then by (2.9) we obtain
(2.17)
a ~ (n I ; 1)
0
-+
t
as n I
-+ ....
1 t we can find an (n") C (n') such that
Since
2 1
'i'z(nrr)
-+ a
2
2
~
0
t
a s r!"
-+
DO
,
a2(n";1)
which and (2.17) give
,2 (h )
2 jilT
--..r-~-
a~(n";1)
-+
a
2
t
as n"
-+
DO
t
for 0 < h < 1.
When putting h • 1 and 1 < s < A in (2.16) and using (Z.9) we
obtain that this relation holds true for 1 < h < A as well.
Using this and the relation
2 h
, 2 (n")
2
- s
a (n",l)
Z
b Z ( " h)
2 n;
.. 0 t as n"
-+ ... ,
b (n";l)
which according to (2.11) and (2.10) is true for h > A,
we get
-
18 -
2 h
(2.18)
a 5 n" -+-
'i'2(nrr)
if
O<h<A,
a~(n";1)
if
h > A
\~ e
DO.
cIa i m t hat
a
2
> O. Suppo sen 0 t. The n
2 1
'I' 2 (li")
a~(n";1)
-+ 0, as n" -+
DO
and together with (2.15) this leads to
2
a 1 (n";1)
a~(n"; 1)
=
1 -
'i' 2 (1 )
1 Ii""
a ~ (n" ; 1)
2 (1 )
2nrr
_--'I'
a~(n";1)
-+- 1 as nil -+•
DO
•
The condition above, the already proved (2.6) and (2.18) with
show that we are in the case (i) of Theorem 12 of (1] •
a = 0
Then part (v) of the theorem gives that along {n"} we should
have
lim limsup
h-+- nil-+--
a,2 (n";h)
a~(n";1)
e-
=,
which contradicts the already proved (2.8). Hence a 2 > 0 and
o
this completes the proof.
LEt~
3. Suppose (2.9), (2.10) and (2.15) are satisfied
aZong {n'} and A > 1. Then for every 0 <
(2.19)
2
b (n';A+E)
(2.20)
2
-+
0,
-+
a,
b (n';A-E)
,2 (£ )
(2.21)
2
I
lJiT
b (n' ;A-E)
as n'
-+
re.
E
S A/2,
- 19 -
Proof. Take any
b 2 (n I
;
t)
=
b 2 (n';A-c)
t,
0 <
b 2 (n ' ; t)
b 2 (n';l)
t
A/2. By (2.9),
$
[b 2 (n I ; Ab 2 (n ' ;1)
c)]
-1
-+
1
n' . . .
'
00
,
which is (2.19). Further by (2.9) and (2.10) for the ratio in
(2.20) we have
2
b (n' ;A+t)
b 2 (n' ;A-£)
=
r
2
b (n';A+t)
2
b (n ' ; 1 )
b~(n';A-E)]-1
b (n I
;
0, n'
-+
-+
00
1)
and finally (2.15) and (2.9) give
'I'2(c )
1 jiT
=
2
b (n';A-c)
\112 (c )
"1 Ii'"
b 2 (n ' ;1)
2
rLbz(n
b (n
l
I
;A-c) ] -1 -+ 0,
; 1)
n'
-+ ... ,
which is (2.21).
LHNA 4. Suppose that (2.19),
(2.20) and (2.21) hold true
aZong SOme subsequence {n'l with A > 2. Then (2.9), (2.10)
and (2.15) hoZd true aZong the same subsequence and with the
same A.
Proof. The condition A
A -
t
> 2
implies that for 0
<
t
~
1,
$
> 1. Then by (2.19),
1
s b
2
2
(n I
;
E) S
b (n' ; 1)
2
b (n' ;t)
2
-+
1, n '
-+ - ,
o
-+
1, n'
-+ - ,
1 <
b (n';A-t)
<
E
and
1 S
2
2
b ( '1)
b (n'·,1)
Zn;
S
2
ben' ;t)
b (n' ;A-1)
t
~ A/2,
which together give (2.9).
Furthermore, by (2.20) and (2.19) for 0
<
E ~
A/2 we have
A/2,
-
20 -
and since for every h >
2
b (n' ;h)
2
b (n ' ;1)
b
2
there exists a srr.;J11 c, such thnt
A
(n'; A+c)
0
-+
t 2 (n ;1)
::s;
,
'
".e have ( 2 • 1 0) a 1 s 0 sat i s fie d. Fin a 11 y ". e use (2. 2 1)
VI
ith
t
=
and get
If 2 (1
.2
b
lDT
J'
(n I
1)
;
2 (1 )
1
J1T
--..2----:.--=.::.----If
~
b (n I
;
0, n '
-+
-1)
A
-+
DC
"'hich is (2.15).
o
Proof of Theorem 2.
(1.11),
first
~e
prove necessity.
(1.12) and (1.13) for SOT'le
A> 0, a
fLxed r
and some {n ' }. Then Ke have
(2.22)
1
a 2 ( rn '. )
'Y(s)
0
1 rn'
..... ,
s > 0,
-0.
(2.23)
{
o<
r'
°,
s < rA,
s > rA,
as n ' . . . -, Khere a r > 0 is some constant, and
(2.24)
1 1m
·
l'1m sup
h........
n'........
a (rn I ; h)
a1 (rn') =
2
°
•
By Lemma 1,(2.22), (2. f3) and (2.24) give
(2.25)
b 2 (rn' ;h)
2
b (rn';l)
.....
f'
0,
0 < h < r). ,
h
>
TAo ,
and, as a special case of (2.22), we have also
(2.26) ,
Assu~e
~
r (A)
1
~
1
-
25
n' ...
DC.
By Lc-r;na.:>, cppJied along
b L (I n' ; jj )
b 2 (I II
of
I;
{rn'}
r i. -+ b)
.,,::it11.\ = rA,
Cf:G
h )
r,:,'" 1=(-r n-.. . . .
-+
1_
]in
(2.27)
2i -
2
b (rn'; rA-h)
n ' ... ",
for 0 < h < rA/Z. Since
b 2 (rn';h):: r a Z (n';h/r)
r{~Z(n"
=
b y put tin g
~O\"
£
= hI r
~)
'r
of
i n ( Z. 2 7) ". e
~(~2(h/r)
r
1 D'
0
SOF,e
of
~2(h/r,.))
2 -n')
b t a i n (1. Z0 ) and
aSSUIT,e (i.1b). T}1cn of cot;rse
and (1.19) along
=
<:11
Z( h ')
h{!2(~)
+!Zi-nT )
1 rn'
of
1
\,'C
}j (' nee
Lave (1.1i),
subsequence {n'}. Ta}:e rZ
(1. 16) •
(1.18)
r Z (A)
=
=
n i n {r: r p 0 sit i \' e i n t e g e r, r A > Z} and ,,' ri t e (1. 1 i), (1. 1 8) .; n d
(1.19)
in the form of (Z.Z7) with r
=
r Z and h
na 4 the latter inplies (2.Z5) and (2.26)
~ith
roa 2, used for the subsequence {r 2 n'}
A
find an {n"} C{n'} such that (2.6),
fied along {rZn") for A
stitute t
= s/T Z
=
and
=
rZc. Ey
r
=
r ZA
=
Lc~-
r Z' By Lcrn-
> Z "'e can
(2.7) and (2.8) are satis-
TZA and sOT:le a
>
O. Further "'e sub-
and get (1.11), (1.12) and (1.13) "'ith n' rep-
laced by n". Then, as sho"'n in the suffidenc)'
proof of
o
Theorem 1, (1f 1 , 't'Z' o)EDp(Poisson (A)).
Froof of the Corollary. Suppose
(1.1).Fo]10~in~
the neces-
sity proof of Theorem 1, one can see, that if r ~ Tl(A) then
for e"e,ry {n"} C tnt} there exists an {n"'} C {n"} such that "'e have
(2.6), (2.7) .c.nci (~.Ei) aleIi£; {Tn'~'} 'With LL
=
T~ and, as n lll
..... 00,
- 22 -
a (rn"')
2
a
~here
6,
-+
< 6 <
~
This is nothing but
(2.28)
a <
,,;here
6
1
=
< 6 2 = °2(r) < -. If
6 1 (r)
).
=
> 1 , then r 1 (A)
and '\<ie get the first statel7lent of the Corollary by putting
r
=
1
1 in (2.28). Let A < 1 and r> max(r l (A), 1- A) . Then
a ~ (n ' )
2
b (n';l)
=
2
An'
2
An'
~
2
2
2
b (n';1)
6 2 (r) - ......2-->---'-~
b (rn';1)
6~(r)
=
2
b (n' ;1)
=
2
2
b (n';1)
b Cn';).-1/r)
b (n' ;A-l/r)
b (n'; l/r)
2
r
rb (n';1/r)
2
-+
a'
since, using first Lerr,ma 1 along {rn"') and "'ith
Le ml7l a 3 a Ion g {r nil') and ". i t h A = r ). and c = 1
2
b Cn" t ;1)
b 2 (n"
-+
b 2 (n"';A-l/r)
0,
A_ 1 / r )
t ;
(1
A
=
rA and then
n d not i n g t h ~ t r >
1, as n'"
-+
as n' ~ ~,
-+ - ,
b 2 (n" , ; 1 / r)
o
and since {nil) "'as arbitrary.
3. A
CO~STRUCTI0N
Denote
t
°
J
= t
°
J
(c) = c 2
-2 j
,,'here c is a number, 0
j
<
= 0,
J , 2 , ••• ,
c:s 1, and
zj -1
b
°
J
= bJo (c)
Intr~duce
= (t.
J-
the function
1- t 0) -1 =
J
Z
j-l
c(1-Z- Z
' j
)
=
0,1,2, ••
I...... [
e-
-23-
k
1,2, ... ,
=
and the corresponding infinitely divisible random
V(O,~!,o),
given b)' (1.10) and (1.8). Since to
=
varj~ble
c/2 < 1, simple
calculations in (1.8) show that
t
=J
V(O,'i',O)
o
o
N ( s ) d ~. (s)
=
-
"'here b o = o.
Fix). > ~ We will illustrate the results of Theorem 1 and
of the corol]ary b.Y choosing X
nk =
(3.1)
and An
k
r~k
1
=
= V(O,'r,O),
{nl}
~ ~k}'
minil:! integer, t
k
-
=
{n k } k= 1 ' "'here
=
0,1,2, ... ,
= bk + l' k = 0, 1 , 2 , .•• ,
and discuss a bit the situation when n k above is replaced by
JTl , "'here
k
(3.2)
mk
·l.~kj
= max{Z:Z
integer. Z "
~k)'
k • 0.1.2 ••••
Let nk be defined by (3.1). Then by elementary considcrations we obtain that for all k large enough,
s
t k+ 1 < - < t k ,
nk
if
t}~+ 1 <~
nk
r~k > t k ,
if
r~k1 tk '
< tk,
(3.3)
tk
t
k
=
-nAk
s
< -n
< t k- 1 ,
< t _ ,
k
k 1
if
,
s < ).
1
A
A
=
if
s >
).
.
These together \dth the asymptotic equality
2~
-
-
(3.4)
imply that if s
~' (~
*
A,
)
{ - 1
~_k_ .. ~' (s)
A
A
nk
(3.5)
=
i f
s < ),',
if
s
'
0,
~ A,
as k • ~. Further ~e ~ant to know the behaviour of the ratios
function
t
2
0
a,(n k ; h) = n k J
h/n k
If
Relations (3.3) show that
2
(s)ds.
~hen
h <
A
or h = A hut
1
r~k >
then
\
By simple calculations this implies that
.
~hen k •
~,
e-
(3.6)
and
2
2j - , [ 1-2 - 2j - 1J -1
+
Again by (3.3) when h > A or h = A but
and this implies
,
(3.8)
•
that when k • -,
a ~ (n k ; h) -
Zk
A
~
2
k-1
2Z
,
if h > A ,
- 25 -
and
k
r 22
j=l
(3.9)
j
-1 [ _ -1] -1
1-2 2
= a 2 (n k ;1)
1
+
j
'1,2(_1) and also (3.6)-(3.9)
n
k
show that we have to distinguish four cases.
Case 1: A>'. By (3.3), '1,2(_') = b 2 + and"then by (3.6)
k 1
n
k
and (3.4) used with h = 1 we get
(3.10)
From the above and using consecutively (3.8) and (3.4) we obtain that
~hen
k
-+ ....
,
0, for every h > A,
-+
and
a~ (n k )
AZ
n
k
These relations together with (3.5) illustrate Theorem' and
the first case of the corollary.
Case 2: A < ,. Since for every r
~
r,(A) we have '/r < A
and since
by using (3.6) with h = l/r and (3.3) with s = llr we obtain
2' (
a 2 rn k
)". _
X'
(>. + 1) -12 k .. 1
c
2
2
,k
Then (3.8) implies that when k
-+ -
-+ - ,
- 26'-
(3.11)
a ~ (n ; h)
k
• 0, for every h
2
a (rn )
2
k
Furtherr.lore, in this case by (3.3)
'1
> 1.
2
(_1 )
n
k
= b k2 ard then by (3.8)
•
"
and (3.4),
(3.12)
which implies, of course, that
..... 0, k ..... -
This, (3.5) and (3.11) together illustrate Theorem 1 and the
second case of the corollary.
Case 3: 1
an integer. By
==
1 and lJe choose 0 < e
< 1 so that l/e is r.ot
(3.6) and (3.3) used with h
s = 1 Ire 0 r res po ndin g 1y, "'.here r
~
==
l/r
< 1 and
r 1 (1) = 2, ",'e get
e.
This and (3.8) imply that when k ..... -
(3.13)
Z
a 1 (nk;h)
2
BZ(n k )
. . . 0, for every h > 1.
Furthermore, by (3.3), 'I' 2 (_1_)
n
(3. 14)
a 2l (n )
k
2
~
k
'---.. O. k
.....
bk + 1
b k2 + 1 and since by (3.7) and
(3.4)~
-
oJ
•
then in this case
(3.15)
1\2
nk '
k
~ 00.·
It is' seen now by (3.13) and (3.15) that we are in the
situation~f
Case 1. (The present case also illustrates part (iii) of Theorem 12
- 27 -
in[l].)
Case 4: A = 1 ar.d c = 1. Relations (3.6) and (3.3) used
with h
=
llr <
1
and s
2
=
aZ(rn k ) - (r-l)2
llr give
Zk+ 1
, k
~ ~,
and then this and (3.8) imply (3.13). By (3.3) , ~2( __
n1 )
by (3.4) and (3.9) used with A = 1 we get
=
k
bk~ and
which means that
(3.16)
Using again (3.9) and (3.4) we obtain
~
0,
The latter and (3.13) sho",' that ",oe are in the situation of
Case 2.
It is perhaps interesting to remark the following.
As we
~ee,
the role of the "natural" normalizing constant fOT
our problem is taken by a 2 (n), while in the general case, that
is, when the limit in (1.1) is an arbitrary (infinitely divisible) random variable this role may be taken either by
1 (n)
or by 3 2 {n), as Theorem 12 of [1) sho",·s. Relations (3.10),
(3.12) and (3.16) show, .that in cases 1, 2 and 4 of the above
3
-
26 -
show. The situation changes if
This second
~e
rcplace n k by mk of (3.2).
construction. illustrates the seme results in the
same way, but in all four cases, that is, for cyery 1 > 0,
~e
•
\
REFERE~CES
1.
CS~RGtl,
S. A probabilistic approach to
attraction~
2. CSORGtl, S.,
do~ains
of partial
Adv. in Appl. Math. To appear.
DODU~EKOVA,
R. The domain of partial attraction
of a Poisson law. 1989. Submitted.
3. CSORGB, S., HAEUSLER E. and ~~SOK, D.M. A probabilistic ap-
proach to the asymptotic distribution of sums of independent, identically distributed random variables. Adv. in Appl.
~:a th
., 9, 1988, 259 - 233.
4. GNEDEKKO, B.V. and KOLMOGOROV, A.N. Limit distributions for
sums of independent random variables.
Addison-~esley,.Read­
ing. Massachusetts, 1954.
5. GROSHEV, A.V. The domain of attraction of the Poisson
~aw.
Izvestija Akad. Nauk USSR, Sere Mat. 5,1941, 165-1721Russianl
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~EPARTHfNT
OF STAllSTJCS
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•