AN ALMOST SURE LIMIT THEOREM FOR MOVING
AVERAGES OF RANDOM VARIABLES BETWEEN THE
STRONG LAW OF LARGE NUMBERS AND THE
ERD
OS-R
ENYI LAW
HARTMUT LANZINGER
Abstract. We prove a strong law of large numbers for moving aver- ages of the form (logn);pPn+(logk =n+1n)p]Xkwhen the moment condition
EexpftjX1j1=pgis imposed (with somep>1). It will turn out that due to the extreme terms among the Xk these means do not satisfy a strong law in the classical sense but we can identify its upper and lower limit.
1. Introduction
In this paper we intend to close a gap appearing in the theory of strong limit theorems for moving averages of random variables. To start with we recall the classical strong law of large numbers due to Kolmogoro (1930, 1933):
Theorem KLet
X
,(X
k)1k =1 be independent, identically distributed random variables.Then the sequence 1
n
n
X
k =1
X
k converges almost surely asn
!1 if and only if EjX
j<
1.In this case the a.s. limit equals
=EX
.One of the many ways to generalize this result involves so called moving averagesof random variables. In this context moving averages are means of the form
b
;1n Pn+bnk =n+1X
k with a monotonically increasing sequence (b
n)1n=1,b
n 2 N for alln
2 N such thatb
n ! 1 (n
! 1). For moving av- erages there are well{known analytical results that relate strong laws for these means to strong laws for certain classes of weighted means (partic- ularly summability methods). See e.g. Chow (1973) for such results con- cerning Euler methods, Bingham and Tenenbaum (1986) and Bingham and Goldie (1988)for corresponding theorems on Riesz means. Following this ap- proach one can prove strong laws for certain summability methods if strong laws for suitable moving averages are given. The reader may consult Lai (1974), Chow (1973), Bingham and Tenenbaum (1986), Bingham and Mae- jima (1985) as well as Bingham and Stadtmuller (1990) for a large varietyThe results of this paper form part of the author's dissertation written at the University of Ulm, Germany, under the guidance of Professor U. Stadtm uller.
URL address of the journal: http://www.emath.fr/ps/
Received by the journal February 25, 1997. Revised February 18, 1998. Accepted for publication September 17, 1998.
c
Societe de Mathematiques Appliquees et Industrielles. Typeset by LATEX.
of results of that kind. Strong laws for very general classes of summabil- ity methods were obtained by Kiesel (1993). Of course this list does not aim at completeness. For a more complete account the reader is referred to Bingham (1985, 1988). The behavior of the sequence
b
;1n Pn+bk =n+1nX
k1 diers fromb
;1n Pbk =1nX
k1 n=1n=1
as far as the covariance structures of the se- quences are concerned. This covariance structure is crucial for almost sure convergence whereas from the point of view of convergence in probability there is no dierence at all between both sequences. What can be said in general is that one has to impose the higher moment conditions the more slowly (
b
n)1n=1grows in order to obtain a strong law of large numbers. This can e.g. be seen from the following strong law for moving averages which is implicit in Chow (1973) and then is stated again in a more general framework by Bingham and Tenenbaum (1986):Theorem C{BT Let(
X
k)1k =1 be a sequence of independent, identically dis- tributed random variables andp >
1.Then 1
n
1=pn+n 1=p
]
X
k =n+1
X
k ! f:
s:
if and only ifEj
X
jp<
1 and EX
=:
Similar results hold in situations with more general moment conditions such asE
(jX
j)<
1 with functionsmore general than powers but with poly- nomial growth. For a theorem of this kind cf. Bingham and Goldie (1988).This particular theorem applies to functions like
(x
) =x
p for somep >
1 but not to (x
) =e
txfor at >
0. It is not very surprising that some condi- tion on the growth ofis needed in order for this result to hold since Shepp (1964) proved the following:Theorem S Let
X
,(X
k)1k =1 be independent, identically distributed random variables with EX
= 0 andM
(t
) =Ee
tX<
1 for allt
in a neighborhood of 0.Dene
m
(x
) = supt2R
f
xt
;logM
(t
)g forx >
0. Ifm
() = 1=c
holds forc
and then it follows:limsup
n!1
c
log1n
n+clogn]
X
k =n+1
X
k = a:
s:
If
c
varies in a suitable non{degenerate interval then the distribution ofX
is uniquely determined by the limit.Erdos and A. Renyi (1970) later proved a similar result, obviously unaware of Shepp's work.
Note that if the moment generating function of
X
exists in an open neigh- borhood of 0 then there is ac
0 2Rsuch that for everyc > c
0 we may nd some 2 Rwithm
() = 1=c
. Hence this condition is fullled at least for allc
2 (c
0 1). We may in particular letc
vary in an interval of positive length.We will denote such results as Erdos{Renyi{Shepp laws or Erdos{Renyi laws in the following. In contrast to the classical strong laws of large numbers we do not have invariance of the limit in this case. These topics also attracted a number of authors. We want to mention only Csorgo and J. Steinebach (1981), Steinebach (1978), Kiesel and Stadtmuller (1996), Deheuvels and Devroye (1987).
Comparing the above theorems one discovers that none of them applies to moving averages like (log
n
);2Pn+logk =n+12n]X
k under the moment conditionE(expfj
X
j1=2g<
1. The function (x
) = expfx
1=2g is not of polynomial growth so the result of Bingham and Goldie does not apply. But on the other hand we do not have any information about the moment generating function Ee
tX that may not exist for anyt
6= 0 in spite of the moment condition. So Theorem S does not apply either. Thus we can ask the ques- tion whether anything can be said about the behavior of moving averages like (logn
);2Pn+logk =n+12n]X
k under the above moment condition, if a classical strong law or an Erdos{Renyi type law holds or if a third possibility applies.The answer to this question (in a more general form) is the main objective of this work. There are only few papers dealing with such averages. We only want to mention de Acosta and Kuelbs (1983) who partially examined such means in a very general setting (i.e. for random variables taking values in a separable Banach space).
2. Main results
From now on we assume without further comment that
X
, (X
k)1k =1 are in- dependent, identically distributed random variables. We further setg
p(x
) = sgnx
jx
j1=p forp
1 andx
2R.Theorem 2.1. For some
p >
1 we deneg
(x
) =g
p(x
) anda
n = (logn
)p. Further lett
1t
22(0 1].Then the following are equivalent:
(i) E
e
tg (X)<
1fort
2(;t
1t
2),Ee
tg (X)=1fort
62;t
1t
2] andEX
= .(ii) liminf
n!1
a
1nn+a
n
X
k =n+1
X
k =;t
1p1 a:
s:
and limsupn!1
a
1nn+a
n
X
k =n+1
X
k =+ 1t
p2 a:
s:
Remark 2.2. Note that there might be dierent triples (
t
1t
2) such that the respective values of +t
;p2 and ;t
;p1 agree. Thus the statement of part (ii))(i) of the assertion is to be read as follows:If liminfn!1
a
;1n Pn+ank =n+1X
k and limsupn!1a
;1n Pn+ank =n+1X
k both are - nite almost surely then Ee
tg (X)<
1in a neighborhood of 0. HenceEjX
j<
1.
Setting E
X
= we can now write liminfn!1a
;1n Pn+ak =n+1nX
k as well as limsupn!1a
;1n Pn+ank =n+1X
k in the form ;t
;p1 and +t
;p2 , respectively, then the moment condition (i) holds.Similar remarks also apply to all other results of this kind stated here.
Remark 2.3. Observe that the upper limit occuring in part (ii) of the as- sertion equals limsupn!1
X
n=a
nwhich follows easily from the usual Borel{Cantelli argument. That means that the moving averages contain terms of the size of the norming constants again and again and that the moving av- erages become at most as large as these terms. This suggests that precisely these terms determine the non-classical behavior of these means which will be shown later. In the classical case such terms cannot occur because the moment conditions imposed on
X
orKX
for arbitraryK >
0 are equivalent.Remark 2.4. The special case
t
1 =t
2 =1 of Theorem 2.1 yields a strong law in the classical sense.A similar proof also yields
Theorem 2.5. For some
p >
1 we denea
n= (logn
)p. Then the following are equivalent:(i) E
e
tjXj1=p<
1 for somet >
0, EX
=. (ii)lim
n!1
c
1nn+c
n
X
k =n+1
X
k =for every monotonically increasing sequence(
c
n)1n=1 withc
n1 andc
na
n ! 1 (n
!1):
In Theorem 2.1 the upper and lower limits might dier from E
X
. So we are in a situation that at least resembles the one of the law of the iterated logarithm. Like there one can also ask for the set of limit points of the sequence of moving averages in our setting. This question is answered by the next result. We denote here and later on the set of limit points of a given real sequence (x
n)1n=1 byC(fx
ng).Theorem 2.6. For some
p >
1 we deneg
(x
) =g
p(x
) anda
n = (logn
)p. Further lett
1t
22(0 1].Then the following are equivalent:
(i) E
e
tg (X)<
1fort
2(;t
1t
2),Ee
tg (X)=1fort
62;t
1t
2] andEX
= .(ii) C
( 1
a
nn+a
n
X
k =n+1
X
k)!
=
;
t
1p1 + 1t
p2 a:
s:
We now reconsider an aspect observed above. We have seen that the non- vanishing upper limit in Theorem 2.1 was a consequence of the largest terms occuring. This leads to the idea that it should be possible to prove a strong law in the sense of almost sure convergence to the mean under the moment conditions imposed above for moving averages slightly modied by excluding some terms with large modulus. This is done in Theorem 2.7.
Results of this kind were already shown by Mori (1976, 1977) for the classi- cal strong law of large numbers and by Grin (1988, 1988a) for the law of the iterated logarithm with some ideas being due to Feller (1968). In these classical theorems the moment conditions Ej
X
j<
1 orEX
2<
1, respec- tively, can be weakened by removing extremal terms. The methods we use in the sequel partially rely on techniques developed by Mori and Grin.Theorem 2.7. For some
p >
1 we deneg
(x
) =g
p(x
) anda
n = (logn
) . Further lett
1t
2>
0 be given.We assume E
e
tg (X)<
1 for allt
2;t
1t
2] and EX
=.Finally let (
r
n)1n=1 be a monotonically increasing sequence of positive inte- gers such thatr
n!1 (n
!1). Assume that there exists an>
1 and a>
0 withliminf
n!1
a
nr
n(logn
)>
0:
(2.1) Letn+a
n
X
k =n+1
X
k denote the sum n+anXk =n+1
X
k with ther
nlargest and ther
nsmallest terms excluded.Then 1
a
nn+an
X
k =n+1
X
k ! a.s. (n
!1):
Remark 2.8. Condition (2.1) is satised e.g. by
r
n = (logn
) with 2 (0p
).Theorem 2.9 shows that the condition
r
n!1 (n
!1) of Theorem 2.7 is necessary. So there is nor
2Nsuch that the moving averages without ther
largest and ther
smallest terms converge to the mean almost surely unlessE
e
tjXj1=psgnX<
1 for allt
2Rin which case we have almost sure conver- gence toEX
for the original moving averages already without removing any terms.Theorem 2.9. For some
p >
1 we deneg
(x
) =g
p(x
) anda
n = (logn
)p. Further letr
2Nas well ast
0t
1t
2>
0 be given.We assume E
e
tg (X)<
1 fort
2;t
1t
0] and Ee
t2g (X)=1. Finally letn+a
n
X
k =n+1
X
k denote the sum n+anXk =n+1
X
k with ther
;1 largest sum- mands removed. Thenlimsup
n!1
a
1nn+an
X
k =n+1
X
kr
p1t
p2:
Remark 2.10. The proof of Theorem 2.9 shows that the r-1 smallest sum- mands can be removed, too.
3. Auxiliary results First we want to introduce some notation.
x
2
X
k =x
1
means X
x
1 k x
2 k 2Z
.
s
x =s
x]is dened the same way for an arbitrary real sequence (s
n)1n=0 and an arbitraryx
0.The variable
C
is supposed to represent a positive constant that may change within one sqeuence of inequalities.For
p >
1 we denote byq
the numberq >
1 withp
;1+q
;1= 1.We use the notations
Lx
= maxf1 logx
g,LLx
=L
(Lx
) and recur- sively deneL
0x
=x
and, for 2 N:L
x
=L
(L
;1x
) (forx
0, respectively). We further write for>
0:L
x
= (Lx
).First we state a result that can be found in Lai (1974a).
Lemma 3.1. Let(
Y
k)1k =1 and(Z
k)1k =1 be two sequences of random variables on a probability space ( FP
) such that (Y
1Y
2::: Y
n) andZ
n are inde- pendent for alln
2N. Further leta b
2RandZ
n;p!b
asn
!1.Then the following holds:
a) If limsup
n!1
(
Y
n+Z
n)a
+b
thenlimsupn!1
Y
na
a:
s:
b) If liminfn!1
(
Y
n+Z
n)a
+b
thenliminfn!1
Y
na
a:
s:
We will also frequently use the well{known inequalities1 +
x
+x
22
e
;jxje
x 1 +x
+x
22
e
jxj (3.1)e
x 1 +x
+x
22
H
(x
) (3.2)where
H
(x
) = maxf1e
xg or simply1 +
x
e
x (3.3)for
x
2R.Next we state some easy technical lemmas.
Lemma 3.2. Let the functions
h
k: (0 1)!(0 1) (k
= 1 2) be monoton- ically increasing. Leth
1(x
)!1 (x
!1) as well ash
2(x
)(
h
1(x
))2 !0 (x
!1).
Then we have for every
y
2R:
1 +
y h
1(x
)
h
2 (x)
exp
yh
2(x
)h
1(x
)
(
x
!1):
Proof. It suces to provelog
1 +
y h
1(x
)
h
2 (x)
;
yh
2(x
)h
1(x
) !0 (x
!1):
But this follows easily by Taylor expansion of the logarithm.The next lemma essentially contains, as a corollary, the Poisson approxima- tion for sums of Bernoulli distributed random variables.
Lemma 3.3. Let(
p
n)1n=1and(r
n)1n=1 be two sequences of numbers such thatp
n20 1] andnp
n!0(n
!1) as well asr
n2Nandr
2n=n
!0(n
!1).Then
n
X
=rn
n
p
n(1;p
n)n; (np
n)rnr
n! (n
!1):
Proof. We know
n
X
=r
n
n
p
n(1;p
n)n; ;nr
np
rnn (1;p
n)n;rn
n
rnr
n!1;
r
nn
rn
p
rnn (1;p
n)n:
From Lemma 3.2 we obtain1;
r
nn
rn
exp
;
r
n2n
!1 (
n
!1) as well as(1;
p
n)ne
;npn !1 (n
!1):
Furthermorer
n! (np
n)rnn
X
=rn
n
p
n(1;p
n)n; Xn=rn
r
n!(n
;r
n)!(
n
;)!!p
; rnn (1;p
n)n;n
X
=rn
(
n
;r
n)!(
n
;)!(;r
n)!p
; rn n(1;p
n)n;= n;rXn
=0
n
;r
n
p
n(1;p
n)n; rn; = 1:
Remark 3.4. The special case
r
n =r
= const can e.g. be found in Mori (1976).Lemma 3.5. Let
Y
be a random variable dened on a probability space ( FP
). Further leta
2Randt
1t
2>
0. Letg
(x
) =g
p(x
) forx
2R. ThenE
e
tg (Y)<
1 for allt
2(;t
1t
2) if and only ifE
e
tg (Y;a)<
1 for allt
2(;t
1t
2):
Proof. It suces to prove one direction. So assume E
e
tg (Y)<
1 for allt
2(;t
1t
2).Fix
t
2 (0t
2). Ifa
0 theng
(Y
;a
)g
(Y
) and the assertion follows immediately.If
a <
0 the assertion follows fromY
;a
;a
forY
0 and forY >
0Y
;a
=Y
+ja
jY
1=p+ja
j1=ppi.e.
g
(Y
;a
)g
(Y
) +g
(ja
j).The assertion for negative values of
t
follows similarly.Because we not only want to deduce a limit theorem from a moment condi- tion but also vice versa we need a result that allows us to do this step. In that respect the following proposition is extremely useful. We use a notation matching the situation in later sections.
Proposition 3.6. Let
g
:R!Rbe a strictly increasing function such thatg
(x
)!1 (x
!1) andg
(x
)!;1 (x
!;1):
For
a
n=g
;1(logn
) leta
nn
K <
1 for alln
2N:
Letlimsup
n!1
a
1nn+a
n
X
k =n+1
X
ks
0 as well asliminf
n!1
a
1nn+an
X
k =n+1
X
k ;s
1 with suitable constantss
0>
0 ands
1>
0.Then Ej
X
j<
1. If EX
= 0 then we have alsoE
e
g(1sX)<
1 for alls > s
0 and alls <
;s
1.Proof. Set
Y
n= 1a
nn+an;1
X
k =n+1
X
k andZ
n=X
n+ana
n . Note thatZ
n;p!0 (n
!1). Then Lemma 3.1 immediately implies liminf
n!1
Y
n;s
1:
Hence we obtainlimsup
n!1
X
n+ana
n = limsupn!11
a
nn+a
n
X
k =n+1
X
k;Y
n!
s
0+s
1<
1:
Thereforelimsup
n!1
X
nn
K
(s
0+s
1):
Thus for everyx > K
(s
0+s
1) we haveP
X
nn > x
i:
o:
= 0
:
Because of independence of the eventsf
X
n> nx
gwe obtain from the Borel{Cantelli lemma using the notation
Y
=X x
:1
>
X1n=1
P
(X
n> nx
) =X1n=1
P
(Y > n
)= X1
k =1 k
X
n=1
P
(Y
2(k k
+ 1])X1k =1 Z
fY2(k k +1]g
(
Y
;1)dP
Z
fY>0g
(
Y
;1)dP
=EY
+;P
(Y >
0):
Thus we have proved E
Y
+<
1 and thereforeEX
+<
1. EX
;<
1 can be shown similarly. From now on we assume thatEX
= 0.Now we set
Y
n=X
n+1a
n andZ
n= 1a
nX
k =n+2
X
k. Since EX
exists we obtain for arbirarily small>
0:P
(Z
n>
) =P
a
n
;1
X
k =1
X
k> a
n!
!0 (
n
!1) by the weak law of large numbers. Hence Lemma 3.1 giveslimsup
n!1
Y
ns
0:
So for everys > s
0:P
(X
n> sa
ni:
o:
) = 0:
By the Borel{Cantelli lemma1
>
X1n=1
P
(X
n> sa
n) =X1n=1
P
e
g(Xsn)> n
= X1
k =1 k
X
n=1
P
e
g(Xsn) 2(k k
+ 1]1
X
k =1
Z
e
g(Xs)2(k k +1]
e
g(Xs);1dP
Ee
g(Xs);1:
So we have proved Ee
g(Xs)<
1 for alls > s
0.The second assertionE
e
g(Xs)<
1for alls <
;s
1 follows applying the same argument to (;X
k)1k =1.4. Proofs of Theorems 2.1 and 2.6
First we prove the following result which contains one part of the main theorem.
Proposition 4.1. For some
p >
1 we deneg
(x
) =g
p(x
) anda
n= (Ln
)p. Further lett
1t
2>
0 be given such that Ee
;t1g (X)<
1 and Ee
t2g (X)<
1. In particular, this implies EjX
j<
1 and we may assume EX
= 0.Then
;
t
1p1 liminfn!1a
1nn+an
X
k =n+1
X
k limsupn!1
a
1nn+an
X
k =n+1
X
kt
1p2:
(4.1) Proof. It suces to prove the inequality for the upper limit. Fixs
22(0t
2).We decompose
X
k intoX
k0 =X
k1fXk s;p
2 a
k
g and
X
k00=X
k;X
k0. Then1
X
k =1
P
(X
k006= 0) = X1k =1
P
e
s2g (X)> k
= X1 X1
P
e
s2g (X)2(n n
+ 1]Ee
s2g (X)<
1:
Thus
P
(X
k006= 0 i.o.) = 0 and therefore limn!1
a
1nn+an
X
k =n+1
X
k00= 0:
(4.2)Since 1
a
n+an=a
n ! 1 asn
! 1 it suces to consider onlyn
large enough that for some ~s
2 2(s
2t
2):s
2a
n+ana
n
1=q
s
~2< t
2:
Now observe thatE
X
k0 0 for allk
2N,EX
2<
1andE(X
2e
~s2g (X))<
1. Then (3.2) yields fork
=n
+ 1::: n
+a
n andt
=s
p2a
;1=qn :E
e
tXk0 1 +t
22E((
X
0)2kH
(tX
k0))1 +
t
2 2
E
X
2+E((X
0)2ke
tXk0 1fXk0 0g)1 +
t
2 2
E
X
2+E(X
2expfts
;p=q2a
1=qkg
(X
)g)1 +
s
2p2 2a
2=qn
E
X
2+E(X
2e
~s2g (X))expnO
(a
;2=qn )o where we have used (3.3) in the nal step. Hence we obtain for anyx >
0:P
n+an
X
k =n+1
X
k0> xa
n!
e
;txan n+anYk =n+1 E
e
tXk0expn;
xs
p2logn
+O
(a
1;2=qn )oexpf;
xs
p2logn
+o
(logn
)gwhich yields a convergent series if
x > s
;p2 . Sinces
2 2(0t
2) was arbitrary this proveslimsup
n!1
a
1nn+a
n
X
k =n+1
X
k0t
1p2:
(4.3) Now the assertion follows from (4.2) and (4.3).The same proof yields the following variant of Proposition 4.1:
Proposition 4.2. For some
p >
1 we deneg
(x
) =g
p(x
) anda
n= (Ln
)p. Let 1.Let
t
1t
2>
0 be given such that Ee
;t1g (X)<
1 and Ee
t2g (X)<
1. In particular, this implies EjX
j<
1 and we may assume EX
= 0.Then
;
t
1p1 liminfn!1a
1nn
]+a
n
X
k =n
]+1
X
k limsupn!1
a
1nn
]+a
n
X
k =n
]+1
X
kt
1p2:
(4.4)Proof. (Theorem 2.1) We rst assume that (i) holds. Then the expected value
=EX
exists. NowEe
tg (X; )<
1fort
2(;t
1t
2) andEe
tg (X; )=1 fort
62;t
1t
2] by Lemma 3.5. Therefore we may without loss of generality assume that = 0 since otherwiseX
k may be replaced byX
k;.Proposition 4.1 immediately yields limsup
n!1
a
1nn+a
n
X
k =n+1
X
kt
1p2 a:
s:
Similarly we can prove the corresponding statement for the lower limit. On the other hand we know from Kolmogorov's 0{1 law that the upper and lower limit are constant almost surely. If now
limsup
n!1
a
1nn+a
n
X
k =n+1
X
kt
1p a:
s:
with some
t > t
2 then we obtain from Proposition 3.6 that Ee
g (sX) =E
e
s1=pg (X)<
1 for alls
2 (0t
p), thus particularly for ans
withs
1=p> t
2 contradicting the assumptions. So we have provedlimsup
n!1
a
1nn+a
n
X
k =n+1
X
k = 1t
p2 a:
s:
and the proof ofliminf
n!1
a
1nn+an
X
k =n+1
X
k =;1t
p1 a:
s:
is similar.Now we assume (ii). By Proposition 3.6 E
e
tg (X)<
1 for allt
in a neigh- borhood of 0. So EjX
j<
1 and we denote EX
=. If we replaceX
k byX
k ; again, if necessary, we may assume without loss of generality that = 0.Then Proposition 3.6 implies E
e
tg (X)<
1 for allt
2(;t
1t
2). If Ee
tg (X)<
1 for at > t
2 then as in the rst part of the proof limsupn!1
a
1nn+an
X
k =n+1
X
kt
1p a:
s:
would follow from Proposition 4.1 and this would contradict the assump- tions. The assumption E
e
tg (X)<
1 for somet <
;t
1 similarly leads to a contradiction.Remark 4.3. In the special case where an asymptotic for the tail of the distribution function of
X
is known Theorem 2.1 can also be proved using a large deviations result implicit in Nagaev (1979) (this can also be found with a short proof in Gantert (1996)).Proof. (Theorem 2.6) That (i) follows from (ii) is obvious from Theorem 2.1.
So we assume (i). Using Proposition 4.2 we can show exactly as in the proof of Theorem 2.1 that the moment condition (i) is equivalent to
liminf
n!1
a
1nn
]+a
n
X
k =n
]+1
X
k =;t
1p1 a:
s:
andlimsup
n!1
a
1nn
]+a
n
X
k =n
]+1
X
k =+ 1t
p2 a:
s:
Since
1 is arbitrary we nd for everys
2h;t
;p1 +t
;p2 i a subse- quence with an almost sure upper or lower limit that equalss
. The re- mainder follows from a standard argument: If we choose a countable dense subsetS
of h;t
;p1 +t
;p2 i it follows by excluding the exeptional sets corresponding to elements ofS
:C ( 1
a
nn+an
X
k =n+1
X
k)!
S
a.s.Since the set of limit points must be closed we obtain
C ( 1
a
nn+an
X
k =n+1
X
k)!
;
t
1p1 + 1t
p2 a:
s:
The reverse inclusionC ( 1
a
nn+an
X
k =n+1
X
k)!
;
t
1p1 + 1t
p2 a:
s:
is an obvious consequence of Theorem 2.1. This completes the proof.
5. Proofs of Theorems 2.7 and 2.9
Proposition 5.1. For some
p >
1 we deneg
(x
) =g
p(x
) anda
n =L
pn
. Further lett
1t
2>
0 be given.We assume E
e
tg (X)<
1 for allt
2;t
1t
2] and EX
= 0.Finally let (
u
n)1n=1 and(l
n)1n=1 be monotonically decreasing sequences of re- als with 0l
nu
n1,a
nu
n !1a
nl
n!1 (n
!1) andu
= limn!1
u
n andl
= limn!1
l
n. SetU
n= n+anXk =n+1
X
k1; ln
t p
1 anX
k
un
t p
2 an
:
Then;
l
1=qt
p1 liminfn!1a
1nU
nlimsupn!1
a
1nU
nu
1=qt
p2:
Proof. SetY
k(n)=X
k1f;lnt
;p
1 a
n X
k u
n t
;p
2 a
ng and
n = E