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Limit Theorems for Randomly Indexed Sequences of Random Variables i=1Title Page
Krzysztof S. Kubacki i=2Contents
I NSTITUTE OF A PPLIED M ATHEMATICS , ACADEMY OF AGRICULTURE , P.O. B OX 158, 20-950 L UBLIN , P OLAND E-mail address:
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2000 Mathematics Subject Classification. Primary 60F05; Secondary 62E20, 60E15 Key words and phrases. Random limit theorems, randomly stopped martingales, random-sums limit theorems, rate of convergence, convergence of moments
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The author was supported in part by the Kosciuszko Foundation, NY, and the Committee for Scientific Research, Warsaw. A BSTRACT.
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This book is a survey of work on limit theorems for randomly indexed sequences of random variables.
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To my teacher, Professor Dominik Szynal, on the occasion of his 70th birthday Full Screen
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Contents Introduction
5
Chapter 1. On a problem of Tate 1.1. Early history 1.2. Characterizations of weak convergence and convergence in probability 1.3. Some characterizations of essential convergence in law and almost sure convergence 1.4. A characterization of stable convergence
9 9 13 13 19
Chapter 2. On a problem of Anscombe 2.1. The Anscombe theorem and its first generalizations 2.2. The Anscombe random condition 2.3. The Anscombe type theorem 2.4. Usefulness of the Anscombe random condition 2.5. An application to random sums of independent random variables 2.6. Applications to the stable convergence
23 23 27 30 33 42 52
Chapter 3. On a Robbins’ type theorem 3.1. The classical results 3.1.1. Some consequences of the First Robbins Theorem 3.2. On the limit behaviour of random sums of independent random variables 3.2.1. Introduction and notation 3.3. A Note on a Katz–Petrov Type Theorem 3.3.1. A Katz–Petrov type theorem. 3.3.2. Nonuniform estimates for random sums. 3.3.3. Proof of Theorem 3.6.
55 55 57 59 59 61 61 62 66
Chapter 4. Weak Convergence to Infinitely Divisible Laws 4.1. The case of finite variances 4.1.1. The main results 4.2. The case of not necessarily finite variances 4.2.1. The main results. 4.2.2. Proofs.
73 73 74 77 78 80
Bibliography
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Introduction The past fifty years or so have seen an intensive development of the studies devoted to limit laws of randomly indexed sequences of random variables. This development has been caused by numerous applications of the obtained theorems in queuing theory, reliability theory, renewal theory, sequential analysis and other sections of mathematics. Essentially, the studies of limit behaviour of randomly indexed sequences of random variables have been two pronged. Attempts have been made to find the limit laws for sequences of partial sums of independent or weakly dependent random variables in those cases where the sequence of random indices is either independent of or may depend on the summands. The first direction in research was originated by H. Robbins (1948). He observed that if {Xk , k ≥ 1 } is a sequence of independent and identically distributed random variables with EX1 = a, Var(X1 ) = c2 < ∞, and if {υn , n ≥ 1} is a sequence of positive, integer-valued random variables, independent of {Xk , k ≥ 1 } and such that Eυn = αn , Var(υn ) = β2n < ∞, then the limiting distribution of q (Sυn − ESυn )/ Var(Sυn ) p may not be normal or even infinitely divisible although (Sn − ESn )/ Var(Sn ) =⇒ N0,1 converges weakly to a normal random variable N0,1 with mean 0 and variance 1. To state his result exactly, put σ2n = Var(Sυn ), dn = |a|βn /σn , and hn (t) = E exp{it(Sυn − ESυn )/σn },
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gn (t) = E exp{it(υn − αn )/βn }. Go Back
Robbins has proved the following two statements. (A) If σn → ∞ and βn /σ2n −→ 0 as n → ∞ , then for all real t |hn (t) − gn (tdn ) exp{−t 2 (1 − dn2 )/2}| −→ 0
as n → ∞ .
. (B) Assume that a = EX1 = 0, and that (υn − αn )/βn =⇒ Z converges weakly to a random variable Z with characteristic function g(·), and distribution function G(·). If αn −→ ∞ and βn /αn −→ r as n → ∞ , where 0 < r < ∞, then for all real t (as n → ∞ ) √ . hn (t) = E exp(˙ıtSυn /c αn ) −→ g(˙ırt 2 /2) exp(−t 2 /2) Z ∞
= 0
e−t
2 y/2
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dG1 (y)
= E exp(−t 2 (rZ + 1)/2), . where G1 (y) = G((y − 1)/r) = P[rZ + 1 < y]. Observe that if P[υn = n] = P[υn = 3n] = 1/2, then αn = 2n, β2n = n2 , and σ2n = 2nc2 + n2 a2 . Moreover, P[(υn − αn )/βn = −1] = P[(υn − αn )/βn = 1] = 1/2,
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. . . so gn (t) = E exp{it(υn − αn )/βn } = (e−it + eit )/2 = cos(t). If a = EX1 6= 0, then dn = |a|βn /σn → 1, βn /σ2n → 0, and statement (A) yields hn (t) −→ cos(t), which proves that (Sυn − ESυn )/σn =⇒ Z converges weakly to a random variable Z equal ±1 with probability 1/2 each. Clearly, the distribution of the random variable Z is not infinitely divisible (see, e.g., [108, Theorem 5.1.1]). . On the other hand, if a = EX1 = 0, then σ2n = 2nc2 = c2 αn , and βn /αn → 1/2. It follows from statement (B) that 1 1 hn (t) −→ E exp(−t 2 ( Z + 1)/2) = (exp(−3t 2 /4) + exp(−t 2 /4)), 2 2 which implies that Sυn /σn =⇒ X converges weakly to a random variable X (X = I[Z = 1]N0,3/2 + I[Z = −1]N0,1/2 ) being a mixture of two independent normal random variables N0,3/2 and N0,1/2 . Clearly, the distribution of the random variable X is not infinitely divisible (see, e.g., Lukacs (1979), p. 376). Robbins type theorems will be presented in Chapter 3. They will include results due to J. Rosi´nski (1975), Z. Rychlik (1976), K.S. Kubacki, D. Szynal (1987, 1988) and V.M. Kruglov (1988), among others. When υn depends on the {Xk , k ≥ 1 } it is not possible in general to obtain the distribution of Sυn explicitly. However, substantial research has been conducted in the context of the preservation of classical limit theory (laws of large numbers and central limit theorems) under random time changes. This line of investigations was originated by F.J. Anscombe (1952) and R.F. Tate (1952). R.F. Tate has considered the following problem. Suppose that
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Yn −→ Y in some sense as n → ∞ , and υn −→ ∞ in some sense as n → ∞ .
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When can one conclude that Yυn −→ Y in some sense as n → ∞ ? His results were generalized by W. Richter (1965), D. Szynal, W. Zi˛eba (1986) and W. Zi˛eba (1988), among others, and will be presented in Chapter 1 (where ‘in some sense’ means one of the classical convergence modes, (in distribution, in probability or almost surely) or stable convergence introduced by A. Rényi (1958), or essential convergence in law introduced by D. Szynal and W. Zi˛eba (1974)). F.J. Anscombe has investigated the case where (1)
Yn =⇒ µ as n → ∞
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(converges weakly).
He observed that if (2)
∀ε > 0 ∃δ > 0 : lim sup P[ max |Yi −Yn | ≥ ε] ≤ ε, n→∞
|i−n|≤δn
then for every sequence {Nn , n ≥ 1} of positive, integer-valued random variables such that (3)
Nn P −→ 1, an
where an → ∞ are constants,
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we have (4)
YNn =⇒ µ as n → ∞ .
It ought to be noticed that no assumption is made about the independence of the sequences {Yn , n ≥ 1} and {Nn , n ≥ 1}. We will always denote by {υn , n ≥ 1} a sequence of positive, integer-valued random variables independent of the underlining sequence {Yn , n ≥ 1}. Condition (2), called the Anscombe condition, plays very important role in proofs of limit theorems for randomly indexed sequences of random variables. D.J. Aldous (1978) has pointed out that condition (2) is also a necessary one for (4) when (3) holds. A more general result than that of Anscombe (1952) and Aldous (1978) has been obtained by M. Csörg˝o, Z. Rychlik (1980, 1981) and K.S. Kubacki, D. Szynal (1985, 1986), among others, and will be presented in Chapter 2. Given the results of Kubacki and Szynal (1986) it is hardly surprising that these two branches of the theory are closely connected. In fact, this connection allowed them to extended some well-known results of the earlier studies to a larger class of random indices, see Kubacki and Szynal (1985, 1987b, 1988a). The procedure used there is simple but in addition to the Anscombe type theorem a limit theorem of Robbins type has been required. The main idea there focus on the following problem. Suppose that we do not know whether (1) holds (or suppose even that the sequence {Yn , n ≥ 1} does not weakly converge), but we do know that there is a sequence {υn , n ≥ 1} of positive, integer-valued random variables, independent of the underlining sequence {Yn , n ≥ 1} such that Yυn =⇒ µ converges weakly to an interesting us measure µ, e.g. gaussian measure. When in such cases can one conclude that YNn =⇒ µ ?
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CHAPTER 1 Title Page
On a problem of Tate In this chapter we give some characterizations of almost sure convergence, essential convergence in law and stable convergence in terms of the convergence of randomly indexed sequences. 1.1.
Early history
Let {Yn , n ≥ 1} be a sequence of random variables defined on a probability space (Ω, F, P), and let {Nn , n ≥ 1} be a sequence of positive, integer-valued random variables defined on the same probability space. We make no assumption about the independence of the sequences {Yn , n ≥ 1} and {Nn , n ≥ 1}. We will always denote by {υn , n ≥ 1} a sequence of positive integer-valued random variables independent of the underlining sequence {Yn , n ≥ 1}. Suppose that Yn −→ Y
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in some sense (as n → ∞ ),
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When can we conclude that YNn −→ Y in some sense (as n → ∞ ) ? The first elementary results of the above kind belong to R.F. Tate [180].
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T HEOREM 1.1. [180] Suppose that Yn =⇒Y as n → ∞ .
(1.1)
If {υn , n ≥ 1} is a sequence of positive, integer-valued random variables, independent of {Yn , n ≥ 1} and such that Full Screen
P
(1.2)
υn −→ ∞ as
n → ∞,
Yυn =⇒Y
n → ∞.
then1 (1.3)
as
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P ROOF. Let ϕX denote the characteristic function of the random variable X. By the independence assumption we have ∞
ϕYvn (t) =
∑ E(exp{itYk }|υn = k)P[υn = k] k=1 ∞
=
∑ ϕYk (t)P[υn = k]. k=1
1Condition (1.2) is equivalent to the condition υ =⇒ ∞ used throughout [180]. n
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Now, choose k0 so large that |ϕYk (t) − ϕY (t)| ≤ ε for
k ≥ k0
and then n0 so large that
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P[υn ≤ k0 ] ≤ ε for
n ≥ n0 .
We then obtain k0
|ϕYυn (t) − ϕY (t)| ≤
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∑ |ϕYk (t) − ϕY (t)|P[υn = k] k=1 ∞
+
∑
|ϕYk (t) − ϕY (t)|P[υn = k]
k=k0 +1 k0
∞
≤ 2 ∑ P[υn = k] + ε k=1
∑ P[υn = k]
k0 +1
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≤ 2P[υn ≤ k0 ] + ε ≤ 3ε,
which in view of the arbitrariness of ε proves the conclusion (1.3).
√ Theorem 1.1 has been proved also by A. Rényi [132] in the special case where Yn = (X1 + . . . + Xn )/ n for each n and X1 , X2 , . . . are independent and identically distributed with EX1 = 0 and EX12 = 1; see also A. Rényi [134], p. 472. We will formulate Rényi’s result and prove it (by a different method) in the section on Robbins Theorem. T HEOREM 1.2. [180] Suppose that {υn , n ≥ 1} is as in Theorem 1.1. If (1.4)
P
Yn −→Y
as
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n → ∞,
then (1.5)
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P
Yυn −→Y
as
n → ∞.
P
P ROOF. Observe that condition (1.4) is equivalent to Yn −Y −→ 0, which further is equivalent to Yn −Y =⇒ 0. By Theorem 1.1 we have Yυn −Y =⇒ 0
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as n → ∞ ,
which proves the assertion. The next results are due to W. Richter [138] (R.F. Tate [180] proved the same result, but for sequences {υn , n ≥ 1} only).
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T HEOREM 1.3. [138] Let {Yn , n ≥ 1} be a sequence of random variables such that a.s.
Yn −−→Y
(1.6)
as
n → ∞. Contents
Suppose further that {Nn , n ≥ 1} is a sequence of positive, integer-valued random variables such that a.s.
Nn −−→∞ as
(1.7)
n → ∞.
Then a.s.
YNn −−→Y
(1.8)
as
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n → ∞.
P ROOF. Let A = {ω : Yn (ω) 6−→ Y (ω)}, B = {ω : Nn (ω) 6−→ ∞} and C = {ω : YNn (ω) 6−→ Y (ω) }. Then C ⊂ A ∪ B , which proves the assertion. T HEOREM 1.4. [138] Let {Yn , n ≥ 1} be as in Theorem 1.3. Suppose that {Nn , n ≥ 1} is a sequence of positive, integer-valued random variables such that P
Nn −→ ∞ as
(1.9)
n → ∞.
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Then P
YNn −→Y
(1.10)
as
n → ∞. Go Back
P ROOF. In the proof of Theorem 1.4 we shall use the following well known result (see e.g. [24], Theorem 20.5): the sequence {ξn , n ≥ 1} of random variables converges in probability to a random variable ξ if and only if each subsequence of {ξn , n ≥ 1} contains a further subsequence which converges to ξ almost surely. P P Since Nn −→ ∞, then Nnk −→ ∞ for every subsequence {nk , k ≥ 1}. Furthermore, from each subsequence we can always select a further subsequence {nk( j) , j ≥ 1} such that
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a.s.
Nnk( j) −−→∞ as j → ∞. a.s.
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Finally, since Yn −−→Y, it follows from Theorem 1.3 that a.s.
YNnk( j) −−→Y as j → ∞. Theorem 1.3 can not be converted even in the case where {Nn , n ≥ 1} is a sequence of positive, integer-valued random variables independent of {Yn , n ≥ 1}. We have (1.11)
a.s.
a.s.
Yn −−→Y and Yυn −−→Y
need not imply
a.s.
υn −−→∞.
Indeed, let Yn = Y and P[υn = 1] = P[υn = 2] = 12 . Then Yn = Yυn = Y for all n, but υn 6−→ ∞.
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Furthermore, a.s.
a.s.
Yυn −−→Y and υn −−→∞
(1.12)
need not imply
a.s.
Yn −−→Y. Contents
To see this, let Yn = Y for n = 2k, k ≥ 1, and Yn = Y + 1 for n = 2k − 1, k ≥ 1. Let n if n = 2k, υn = n − 1 if n = 2k − 1, for k ≥ 1.
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a.s.
Then Yυn = Y for all n, and υn −−→∞. However, the sequence {Yn , n ≥ 1} oscillates between Y and Y + 1. The following example shows that the conclusion of Theorem 1.4 can not be sharpened, i.e. that a.s.
P
Yn −−→Y and Nn −→ ∞
(1.13)
need not imply
a.s.
YNn −−→Y.
a.s.
E XAMPLE 1.5. Let P[Yn = n1 ] = 1. Clearly Yn −−→0. Note that for every sequence {Nn , n ≥ 1} of positive, integer-valued random variables we have 1 YNn = , Nn which converges almost surely (respectively, in probability) to zero as n → ∞ if and only if Nn −→ ∞ almost surely (respectively, in probability).
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Also P
a.s.
Yn −→Y and Nn −−→∞
(1.14)
need not imply
P
YNn −→Y.
E XAMPLE 1.6. [138] Let Ω = [0, 1], F = the σ-field of measurable subsets of Ω, and let P be the Lebesque measure on [0, 1]. Set 1 if j2−m ≤ ω < ( j + 1)2−m , Yn (ω) = 0 otherwise,
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where n = 2m + j, 0 ≤ j ≤ 2m − 1, and let Nn = min{k ≥ 2n : Yk > 0 }. Then P
(i) Yn −→ 0 but not almost surely. a.s.
(ii) Nn −−→∞ as n → ∞ . (iii) YNn = 1 a.s. for all n. The proof of these facts is left to the reader.
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1.2.
Characterizations of weak convergence and convergence in probability
The following results are implied by Theorem 1.1 and Theorem 1.2. Contents
T HEOREM 1.7. The following conditions are equivalent: (i) Yn =⇒Y as n → ∞ ; (ii) Yυn =⇒Y as n → ∞ for every sequence {υn , n ≥ 1} of positive, integer-valued random variables, independent of {Yn , n ≥ 1} and such P
that υn −→ ∞ as n → ∞ . P ROOF. The implication (ii) =⇒ (i) is obvious: put υn = n. The reverse implication, (i) =⇒ (ii), follows by Theorem 1.1.
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T HEOREM 1.8. The following conditions are equivalent: P (i) Yn −→Y as n → ∞ ; P (ii) Yυn −→Y as n → ∞ for every sequence {υn , n ≥ 1} of positive, integer-valued random variables, independent of {Yn , n ≥ 1} and such P
that υn −→ ∞ as n → ∞ .
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Moreover, the following results are implied by Theorem 1.3. T HEOREM 1.9. The following conditions are equivalent: a.s.
(i) Yn −−→Y as n → ∞ ; a.s. (ii) Yυn −−→Y as n → ∞ for every sequence {υn , n ≥ 1} of positive, integer-valued random variables, independent of {Yn , n ≥ 1} and such
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a.s.
that υn −−→∞ as n → ∞ . Full Screen
T HEOREM 1.10. The following conditions are equivalent: a.s.
(i) Yn −−→Y as n → ∞ ; a.s. a.s. (ii) YNn −−→Y as n → ∞ for every sequence {Nn , n ≥ 1} of positive, integer-valued random variables such that Nn −−→∞ . In the next section we shall see that the similar statement to that of Theorem 1.10 for weak convergence or convergence in probability is false. So Theorem 1.7 (respectively, Theorem 1.8) is the only one analogy of Theorem 1.10 for weak convergence (respectively, convergence in probability). 1.3.
Some characterizations of essential convergence in law and almost sure convergence
The concept of essential convergence in law has been introduced by D. Szynal and W. Zi˛eba [176].
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D EFINITION 1.11. [176] A sequence {Yn , n ≥ 1} of random variables is said to converge essentially in law to a random variable Y (written E.D.
Yn −−−→Y ) if for every continuity point x of the distribution function FY (·) of Y we have2 P[ lim inf{Yn < x} ] = P[ lim sup{Yn < x} ] = FY (x). n→∞
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n→∞
It is well known that E.D.
L EMMA 1.12. If Yn −−−→Y, then Yn =⇒Y.
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P ROOF. Note by Fatou’s lemma that for every continuity point x of FY P[Y < x ] = P[ lim inf{Yn < x} ] ≤ lim inf P[Yn < x ] n→∞
n→∞
≤ lim sup P[Yn < x ] ≤ P[ lim sup{Yn < x} ] = P[Y ≤ x ], n→∞
n→∞
so P[Yn < x ] −→ P[Y < x ] as n → ∞ for every continuity point x of FY .
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E.D.
The following example shows that, in general, Yn =⇒Y does not imply Yn −−−→Y. E XAMPLE 1.13. [176] Let X, X1 , X2 , . . . be independent and identically distributed nondegenerate random variables. Obviously, Xn =⇒ X. Now, if x is a continuity point of FX such that 0 < FX (x) = p < 1, then lim P[
n→∞
\
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{Xk < x} ] = 0 6= P[ X < x ] = p > 0.
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E.D.
Thus Xn − 6 −−→X. a.s.
E.D.
L EMMA 1.14. If Yn −−→Y, then Yn −−−→Y. Close
P ROOF. Note that lim supn→∞ {Yn < x} ⊆ {Y ≤ x}, so P[ lim sup{Yn < x} ] ≤ P[Y ≤ x ]. n→∞
Conversely, {Y < x} ⊆ lim infn→∞ {Yn < x}, so
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P[Y < x ] ≤ P[ lim inf{Yn < x} ]. n→∞
2The function F (·) defined by F (x) = P[Y < x ] for all real x is said to be the distribution function of Y. The definition F (x) = P[Y ≤ x ] is also customary. This induces Y Y Y
only minor modifications in its properties, e.g. this function is continuous from the right, while P[Y < x ] is continuous from the left.
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It follows that for every continuity point x of FY FY (x) = P[Y < x ] ≤ P[ lim inf{Yn < x} ] n→∞
≤ P[ lim sup{Yn < x} ] ≤ P[Y ≤ x ] = FY (x),
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n→∞ E.D.
i.e., that Yn −−−→Y.
E.D.
a.s.
P
The following example shows that, in general, Yn −−−→Y does not imply Yn −−→Y or even Yn −→Y. E XAMPLE 1.15. [176] Let Ω = [0, 1] and suppose that Yn (ω) ≡ Y (ω) ≡ ω for each n, where the random variable Y is uniformly distributed on Ω. Define a uniformly distributed on Ω random variable X by the formula ( ω+a if 0 ≤ ω < 1 − a, X(ω) = ω + a − 1 if 1 − a ≤ ω ≤ 1; 0 < a < 1. Then
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P[ lim inf{Yn < x} ] = P[ lim sup{Yn < x} ] = P[ X < x ], n→∞
n→∞
P
but Yn−→ 6 X as P[ ω : |Yn (ω) − X(ω)| ≥ ε ] = 1 for all ε < a and each n. Go Back
E.D.
P
Moreover, Yn −→Y does not imply Yn −−−→Y. E XAMPLE 1.16. [176] Let Ω = [0, 1], F = the σ-field of measurable subsets of Ω, and let P be the Lebesque measure on [0, 1]. Write IA for the indicator function of the set A. Define Y1 = 1,Y2 = I[0,1/2] ,Y3 = I[1/2,1] , and, in general, if bm = 2m , define Ybm +k = I[k/2m ,(k+1)/2m ] for k
= 0, 1, . . . , 2m − 1;
P
m = 0, 1, 2, . . . Then Yn −→ 0 (since P[Yn P[
6 0 ] = 2−n −→ 0). = [
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But
{Yk ∈ [1/2, 1]} ] = 1,
k≥n
Close
E.D.
which implies that Yn − 6 −−→0. Thus, the following implications hold true:
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a.s.
(1.15)
P
( Yn −−→Y ) −−−→ ( Yn −→Y ) y y E.D.
( Yn −−−→Y ) −−−→ ( Yn =⇒Y )
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Furthermore, D. Partyka and D. Szynal [122] have proved that: a.s.
E.D.
T HEOREM 1.17. [122] If Yn −−→Y then Yn −−−→Y 0 for every random variable Y 0 with the same distribution as Y. Contents
P ROOF. Obvious, use Lemma 1.14. E.D.
a.s.
T HEOREM 1.18. [122] If Yn −−−→Y then there exists a random variable Y 0 with the same distribution as Y and such that Yn −−→Y 0 . E.D.
It follows from Theorems 1.17 and 1.18 that Yn −−−→Y if and only if there exists a random variable Y 0 with the same distribution as Y such a.s.
E.D.
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a.s.
P
that Yn −−→Y 0 . Therefore, Yn −−−→Y and Yn −→Y at the same time if and only if Yn −−→Y [182]. Due to D. Szynal and W. Zi˛eba [177] and W. Zi˛eba [183] we have the following characterization of the essential convergence in law: T HEOREM 1.19. [177] The following conditions are equivalent: E.D.
(i) Yn −−−→Y as n → ∞ ; a.s. (ii) YNn =⇒Y as n → ∞ for every sequence {Nn , n ≥ 1} of positive integer-valued random variables such that Nn −−→∞. E.D.
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a.s.
P ROOF. Suppose that Yn −−−→Y . Then there exists a random variable Y 0 with3 L(Y 0 ) = L(Y ) such that Yn −−→Y 0 . Hence and by Theoa.s.
a.s.
rem 1.10, for every sequence {Nn , n ≥ 1} of positive, integer-valued random variables such that Nn −−→∞ we have YNn −−→Y 0 . Since PY 0 = PY , L(Y 0 ) = L(Y ), this implies that YNn =⇒Y . Conversely, suppose that (ii) holds. Let x be a continuity point of the distribution function FY of Y . Define S inf{k ≥ n : Yk < x} if ω ∈ ∞ k=n [Yk < x], τn,x (ω) = S n if ω 6∈ ∞ k=n [Yk < x ].
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a.s.
Then τn,x −−→∞ as n → ∞ and, therefore, Yτn,x =⇒Y as n → ∞ . Hence P[Yτn,x < x] −→ P[Y < x ] as n → ∞ , and since P(Yτn,x < x) = P(
∞ [
[Yk < x] ),
k=n
then P(
∞ [
[Yk < x] ) −→ P[Y < x] as n → ∞ .
k=n 3 L(Y 0 ) = L(Y ) means that Y 0 and Y have the same distribution.
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Similarly one can get P(
∞ \
[Yk < x] ) −→ P[Y < x] as n → ∞ ,
k=n
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E.D.
which proves that Yn −−−→Y as n → ∞ .
R EMARK 1.20. Condition (ii) of Theorem 1.19 is equivalent to the following condition: P (iii) YNn =⇒Y as n → ∞ for every sequence {Nn , n ≥ 1} of positive, integer-valued random variables such that Nn −→ ∞. P ROOF. Obviously, (iii) =⇒ (ii). It remains only to prove the reverse implication. Suppose that (ii) holds true, and that (iii) does not hold. We will obtain a contradiction. P P Let Nn −→ ∞ and YNn 6=⇒ Y. Then there exist ε > 0 and a subsequence {YNn(k) , k ≥ 1 } such that Nn(k) −→ ∞ and L(YNn(k) ,Y ) > ε
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for all k ≥ 1,
where L(X,Y ) denotes the Lévy metric on the space of real valued random variables4. We can choose a subsequence {Nn(k j ) , j ≥ 1 } such that a.s.
Nn(k j ) −−→∞ as j → ∞. But YNn(k j ) 6=⇒ Y as j → ∞, and a contradiction is obtained. Thus (ii) implies (iii).
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The following result immediately follows from Theorem 1.18, Theorem 1.19 and Remark 1.20, and has been also proved by E. Rychlik [144].
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P
C OROLLARY 1.21. [144] If YNn =⇒Y for every sequence {Nn , n ≥ 1} of positive, integer-valued random variables such that Nn −→ ∞, then a.s.
there exists a random variable X such that L(X) = L(Y ) and Yn −−→X. Full Screen
Now, for the moment, we specialize the random variable Y to a constant. C OROLLARY 1.22. [183] Let C be a constant. The following conditions are equivalent: a.s.
(i) Yn −−→C as n → ∞ ; P P (ii) YNn −→C as n → ∞ for every sequence {Nn , n ≥ 1} of positive, integer-valued random variables such that Nn −→ ∞. E.D.
a.s.
P ROOF. The assertion follows by Theorem 1.19, Remark 1.20 and the fact that Yn −−−→C if and only if Yn −−→C (see Theorems 1.17 and 1.18). Corollary 1.22 suggest the following more general result (a similar result has been proved also by A. Dvoretzky [53] under the additional assumption that Nn , n ≥ 1, are stopping times). 4L(X,Y ) = inf{h > 0 : F (x) ≤ F (x + h) + h, F (x) ≤ F (x + h) + h for all x } X Y Y X
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T HEOREM 1.23. [177] that The following conditions are equivalent: a.s.
(i) Yn −−→Y as n → ∞ ; P P (ii) YNn −→Y as n → ∞ for every sequence {Nn , n ≥ 1} of positive, integer-valued random variables such that Nn −→ ∞.
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P
P ROOF. The implication (i) =⇒ (ii) follows by Theorem 1.4. Suppose now that (ii) holds true. Then, obviously, Yn −→Y. E.D.
On the other hand, by Theorem 1.19 and Remark 1.20 we see that Yn −−−→Y, which further implies that there exists a random variable X a.s.
P
a.s.
P
with L(X) = L(Y ) such that Yn −−→X. Hence Yn −→ X. But also But also Yn −→Y, so that X = Y a.s., which implies that Yn −−→Y.
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C OROLLARY 1.24. Let {Yn , n ≥ 1} be a sequence of random variables such that Y1 ≤ Y2 ≤ . . . a.s. The following conditions are equivalent: (i) Yn =⇒Y as n → ∞ ; E.D.
(ii) Yn −−−→Y as n → ∞ ; P (iii) YNn =⇒Y as n → ∞ for every sequence {Nn , n ≥ 1} of positive, integer-valued random variables such that Nn −→ ∞. P ROOF. The equivalence of (ii) and (iii) follows by Theorem 1.19 and Remark 1.20. Furthermore, it is obvious that for every point y (of continuity of FY ) we have P(
∞ \
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[Yk < y] ) − P[Y < y ] ≤ P[Yn < y ] − P[Y < y ]
k=n
≤ P(
∞ [
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[Yk < y] ) − P[Y < y ].
k=n
Therefore (ii) implies (i). Suppose now that (i) holds. Since Y1 ≤ Y2 ≤ . . . a.s., then the sequence of events { [Yn < y] , n ≥ 1} decreases with n. Hence lim P(
n→∞
∞ \
[Yk < y] ) = lim P(
k=n
n→∞
∞ [
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[Yk < y] )
k=n Close
and by (i) we conclude that for every continuity point y of FY lim P(
n→∞
∞ [
[Yk < y] ) = lim P[Yn < y ] = P[Y < y ].
k=n
n→∞
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Thus, for every continuity point y of FY we have P( lim inf [Yn < y] ) = P( lim sup [Yn < y] ) = P[Y < y ], n→∞
E.D.
which means that Yn −−−→Y.
n→∞
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1.4.
A characterization of stable convergence
Let {Yn , n ≥ 1} be a sequence of random variables defined on a probability space (Ω, F, P). We write Yn =⇒Y (stably) if Yn =⇒Y and, for every set B ∈ F with P(B) > 0, there exists a distribution function FB such that for every continuity point y of FB we have
Contents
lim P[Yn < y|B ] = FB (y),
(1.16)
n→∞
where P(D|B) = P(D ∩ B)/P(B). In the special case where FB (y) = FY (y) for all B, we write Yn =⇒Y (mixing). These concepts are originally due to A. Rényi ([131, 133, 135]). A survey of stable and mixing limit theorems, and applications of these concepts, may be found e.g. in [?]. It is well known that: P If Yn −→Y, then Yn =⇒Y (stably). Indeed, for every set B ∈ F, with P(B) > 0 and every continuity point y of FB we have lim P[Yn < y|B] = lim P[ (Yn −Y ) +Y < y|B] = P[Y < y|B],
n→∞
n→∞
since
P
B Yn −Y −→ 0,
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where PB (A) = P(A|B).
Furthermore,
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P
Yn =⇒Y (stably) need not imply Yn −→Y. 2 E XAMPLE 1.25. [131] √ Let X, X1 , X2 , . . . be independent and identically distributed random variables with EX = 0 and EX = 1. Let Sn = X1 + . . . + Xn . Then Sn / n =⇒ N0,1 (mixing), but do not converge in probability.
T HEOREM 1.26. [183] The following conditions are equivalent: (i) Yn =⇒Y (stably) as n → ∞ ; P (ii) YNn =⇒Y (stably) as n → ∞ for every sequence {Nn , n ≥ 1} of positive, integer-valued random variables such that (a) Nn −→ ∞ and (b) the σ-field σ(Nn , n ≥ 1) generated by the random variables Nn , n ≥ 1, is countable5. [183], most P ROOF. The implication (ii) =⇒ (i) is obvious: put Nn = n. Suppose (i), and let {Nn , n ≥ 1} be a sequence of positive, integer-valued random P variables suchS that Nn −→ ∞ and the σ-field σ(Nn , n ≥ 1 ) is at most countably infinite: σ(Nn , n ≥ 1) = {Bi , i ≥ 1}, say, where Bi ∈ F, Bi ∩B j = 0/ for i 6= j, and ∞ i=1 Bi = Ω. For every ε > 0 there exists a positive integer L such that ∞
(1.17)
∑
i=L+1 5By “countable” we mean always “finite or countably infinite”.
P(Bi ) < ε/8.
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Furthermore, by (i), we have for every i ≥ 1 and every continuity point x of the distribution function FB∩Bi |P[Yn < x, B ∩ Bi ] − FB∩Bi (x)P(B ∩ Bi )| −→ 0 as n → ∞ . Therefore, there exists a positive integer M such that for all k > M
Contents
max |P[Yk < x, B ∩ Bi ] − FB∩Bi (x)P(B ∩ Bi )| < ε/(8L)
(1.18)
1≤i≤L P
Of course, since Nn −→ ∞, we have for sufficiently large n
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P[Nn < M] ≤ ε/8.
(1.19) For all i ≥ 1 and ω ∈ Bi let
Nn (ω) = kn,i , Dn = {i : 1 ≤ i ≤ L, kn,i ≥ M}. We note that ∞
P[Y < x, B] =
∑ FB∩Bi (x)P(B ∩ Bi)
i=1
=
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∑ FB∩Bi (x)P(B ∩ Bi) + ∑ FB∩Bi (x)P(B ∩ Bi), i6∈Dn
i∈Dn
where by (1.17) and (1.19) we have
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∑ FB∩Bi (x)P(B ∩ Bi)
i6∈Dn
=
∑
FB∩Bi (x)P(B ∩ Bi ) Full Screen
i:1≤i≤L, kn,i 0 we have |P[Y < x, B] − P[YNn < x, B]| ≤ |P[Y < x, B] −
∑ P[Ykn,i < x, B ∩ Bi]| + ε/4
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i∈Dn
≤|
FB∩Bi (x)P(B ∩ Bi ) − P[Ykn,i < x, B ∩ Bi ] | + ε/2
∑
i∈Dn
≤
∑ FB∩Bi (x)P(B ∩ Bi) − P[Ykn,i < x, B ∩ Bi] + ε/2
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i∈Dn
≤ Lε/(8L) + ε/2 < ε, which proves that YNn =⇒Y (stably).
By the same method one can prove
[?]
T HEOREM 1.27. The following conditions are equivalent: (i) Yn =⇒Y (mixing) as n → ∞ ; P (ii) YNn =⇒Y (mixing) as n → ∞ for every sequence {Nn , n ≥ 1} of positive, integer-valued random variables such that Nn −→ ∞ and the σ-field σ(Nn , n ≥ 1) is at most countable. The following results are quite useful:
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C OROLLARY 1.28. (cf. Corollary 1.22) Let C be a constant. The following conditions are equivalent: P (i) Yn −→C as n → ∞ ; P P (ii) YNn −→C as n → ∞ for every sequence {Nn , n ≥ 1} of positive, integer-valued random variables such that Nn −→ ∞ and the σ-field σ(Nn , n ≥ 1) is at most countable. C OROLLARY 1.29. Let {Xk , k ≥ 1 } be a sequence of independent random variables with zero means and finite variances. Let Sn = X1 + . . . + Xn , s2n = Var(Sn ). Then the following conditions are equivalent: (i) Sn /sn =⇒ N0,1 as n → ∞ ;
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P
(ii) SNn /sNn =⇒ N0,1 (mixing) as n → ∞ for every sequence {Nn , n ≥ 1} of positive, integer-valued random variables such that Nn −→ ∞ and the σ-field σ(Nn , n ≥ 1) is at most countable. Quit
R EMARK 1.30. It follows from Corollary 1.29 that, under (i), for any positive random variable λ with discrete distribution: S[λn] /s[λn] =⇒ N0,1 (mixing).
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CHAPTER 2 Title Page
On a problem of Anscombe In this chapter we give a generalization of the Anscombe’s theorem on the asymptotic behaviour of randomly indexed sequences of random variables. As an application of our result we give an extension of the random central limit theorem of J. R. Blum, D. L. Hanson, J. I. Rosenblatt [25] and J. A. Mogyoródi [114] to sequences of independent but nonidentically distributed random variables. 2.1.
The Anscombe theorem and its first generalizations
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Let {Yn , n ≥ 1} be a sequence of random variables defined on a probability space (Ω, F, P). Suppose that there exists a probability measure µ such that Yn =⇒ µ
(2.1)
as n → ∞ .
Let {Nn , n ≥ 1} be a sequence of positive, integer-valued random variables defined on the same probability space (Ω, F, P). The well known Anscombe’s theorem [7] gives conditions on sequences {Yn , n ≥ 1} and {Nn , n ≥ 1} under which YNn =⇒ µ as n → ∞ .
(2.2)
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They are as follows. T HEOREM 2.1. [7] Let {Yn , n ≥ 1} be a sequence of random variables satisfying (2.1). If {Yn , n ≥ 1} satisfies also the so-called “Anscombe condition”: for every ε > 0 there exists δ > 0 such that Go Back
(A)
lim sup P[ max |Yi −Yn | ≥ ε] ≤ ε, n→∞
|i−n|≤δn
and {Nn , n ≥ 1} is a sequence of positive, integer-valued random variables such that Nn P (2.3) −→ 1 as n → ∞ , an where {an , n ≥ 1} is a sequence of positive integers with an → ∞, then condition (2.2) holds. D. J. Aldous [?] has pointed out that condition (A) is also a necessary one for (2.2) when (2.3) holds. T HEOREM 2.2. [?] The following conditions are equivalent: (i) the sequence {Yn , n ≥ 1} satisfies (2.1) and the Anscombe condition (A); (ii) YNn =⇒ µ for every sequence {Nn , n ≥ 1} of positive, integer-valued random variables satisfying (2.3). Let {Xk , k ≥ 1 } be a sequence of independent random variables with zero means and finite variances. Let Sn = X1 + .√ . . + Xn , s2n = Var(Sn ), 2 and Yn = Sn /sn , n ≥ 1. From the above results one can only anticipate that if 0 < Var(Xk ) = σ < ∞ (k ≥ 1), Sn /σ n =⇒ N(0, 1), and if √ P Nn /an −→ 1 as n → ∞ , where {an , n ≥ 1} is a sequence of positive integers with an → ∞, then SNn /σ Nn =⇒ N(0, 1). A more general and stronger result than that of [7] and [?] has been given by M. Csörg˝o and Z. Rychlik [46].
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T HEOREM 2.3. [46] Let {kn , n ≥ 1} be a non-decreasing sequence of positive numbers. The following conditions are equivalent: (i) the sequence {Yn , n ≥ 1} satisfies (2.1) and the so called “generalized Anscombe condition” with norming sequence {kn , n ≥ 1} : for every ε > 0 there exists δ > 0 such that (A0 )
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lim sup P[
max
|ki2 −kn2 |≤δkn2
n→∞
|Yi −Yn | ≥ ε] ≤ ε;
(ii) YNn =⇒ µ for every sequence {Nn , n ≥ 1} of positive, integer-valued random variables satisfying P kN2 n /ka2n −→ 1
(2.4)
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as n → ∞ ,
where {an , n ≥ 1} is a sequence of positive integers with an → ∞. It is easy to see that in a special case where kn = n (n ≥ 1), both (A0 ) and (2.4) reduces to (A) and (2.3), respectively. P ROOF OF T HEOREM 2.3. (i) =⇒ (ii). Let C be the set of bounded and continuous real-valued functions defined on a real line. Let C0 be the subset of C consisting of functions f which satisfy (for all real x and y) | f (x)| ≤
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1 , | f (x) − f (y)| ≤ |x − y| . 2
It is well known ([23], the proof of Theorem 2.1, or [24], the proof of Theorem 25.8) that if E f (Yn ) −→ E f (Y ) for all f ∈ C0 then Yn =⇒Y. Given f ∈ C0 and ε > 0 choose δ as in (A0 ). Then
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E| f (YNn ) − f (Yan )| ≤ ε + P[|YNn −Yan | > ε] ≤ ε + P[|kN2 n − ka2n | > δka2n ] + P[
max
|ki2 −ka2n |≤δka2n
|Yi −Yan | ≥ ε],
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and so by (A0 ) and (2.4), lim sup E| f (YNn ) − f (Yan )| ≤ 2ε,
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and (ii) follows. (ii) =⇒ (i). For the converse, it is clear that Yn =⇒ µ, so suppose that (A0 ) fails. Then there exist ε > 0 and a subsequence b1 < b2 < . . . of positive integers such that
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n→∞
(2.5)
P[ max |Yi −Ybn | ≥ ε] > ε f or all n ≥ 1, i
where the maximum is taken either over all i such that kb2n ≤ ki2 ≤ (1 + 1/n)kb2n or over all i such that kb2n (1 − 1/n) ≤ ki2 ≤ kb2n . We shall only consider the first case as the second one can be treated in the same way.
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Let {Gi , 1 ≤ i ≤ M} be disjoint and open subsets such that 0 < µ(Gi ) = µ(Gi ), µ( M i=1 Gi ) > 1 − ε/2, and diameter(Gi ) < ε/2. Thus, by 0 (2.5), there exists a set G ∈ {Gi , 1 ≤ i ≤ M} and a subsequence {bn , n ≥ 1} of {bn , n ≥ 1} such that S
P[Yb0n ∈ G, max |Yi −Yb0n | ≥ ε] ≥
(2.6)
i
ε , 2M
Contents
where the maximum is taken over all i such that kb20 ≤ ki2 ≤ (1 + 1/n)kb20 . Define Nm = min(Cm1 , Cm2 ), where n
n
Cm1 = max{i : ki2 ≤ (1 + 1/m)kb20m } , Cm2 = min{i ≥ b0m : Yi 6∈ G} .
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a.s.
Then kN2 m /kb20 −−→1 as m → ∞ and so (2.4) holds and hence YNm =⇒ µ as m → ∞. But, by (2.6), we have m
P[YNm 6∈ G] = P[Yb0m 6∈ G] + P[Yb0m ∈ G,YNm 6∈ G] ε , ≥ P[Yb0m 6∈ G] + 2M
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i.e., there exists ε > 0 for which lim sup P[YNm 6∈ G] ≥ µ(Ω \ G) + m→∞
ε , 2M
which proves that YNm 6=⇒ µ. This is a contradiction to (ii), and it proves that the sequence {Yn , n ≥ 1} must satisfy
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(A0 ).
R EMARK 2.4. Let {Xk , k ≥ 1 } be a sequence of independent random variables with zero means and finite variances. Let Sn = X1 + . . . + Xn , = Var(Sn ), and Yn = Sn /sn , n ≥ 1. Then the sequence {Yn , n ≥ 1} of random variables satisfies (A0 ) with norming sequence {kn , n ≥ 1}, where kn2 = s2n , n ≥ 1. s2n
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P ROOF. For every n and 0 < δ < 1, define stopping times T1 = min{k : s2k ≥ (1 − δ)s2n } and T2 = max{k : s2k ≤ (1 + δ)s2n }.
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Note that, by Kolmogorov’s inequality, we have for all ε > 0 P[ max |Si | ≥ ε] ≤ s2T2 ε−2 ≤ (1 + δ)s2n ε−2 T1 ≤i≤T2
and P[ max |Si − ST1 | ≥ ε] ≤ (s2T2 − s2T1 )ε−2 ≤ 2δs2n ε−2 . T1 ≤i≤T2
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Hence, for every ε > 0 and 0 < δ ≤ 1/3, we get P[
max
|s2i −s2n |≤δs2n
|Yi −Yn | ≥ ε ] ≤ P[
max
|s2i −s2n |≤δs2n
+P[
|Si − Sn | ≥ εsn /2 ]
max
|s2i −s2n |≤δs2n
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|Si (si − sn )|/si sn ≥ ε/2 ]
≤ 8δ(4 + δ)ε−2 ,
which proves the conclusion.
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It follows from Remark 2.4 that if Sn /sn =⇒ µ, then SNn /sNn =⇒ µ for every sequence {Nn , n ≥ 1} of positive, integer-valued random P
variables such that s2Nn /s2an −→ 1, where {an , n ≥ 1} is a sequence of positive integers with an → ∞. We see that in Theorem 2.1, and as well as in Theorem 2.3, the assumption Yn =⇒ µ (cf. condition (2.1)) is essential one to prove that YNn =⇒ µ. However, in practical situations we very often do not know whether (2.1) holds and even sometimes we know that the sequence {Yn , n ≥ 1} does not weakly converge. When in such cases we can state that a sequence {YNn , n ≥ 1} (with random indices) converges weakly to an interesting us measure µ, e.g. gaussian measure ?
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E XAMPLE 2.5. [149] Let {Xk , k ≥ 1 } be a sequence of independent random variables defined by P[X22n = 22 and Xk , for k
n 6 22 (n ≥ 1, k =
n−1
n−1
] = P[X22n = −22
]=
1 (n ≥ 1), 2
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≥ 1), has the normal distribution function N(0, 1) with mean zero and variance one. Put n
Sn =
n
∑ Xk , s2n =
∑ Var(Xk ) , Yn = Sn/sn ,
k=1
k=1
n ≥ 1. Full Screen
Then Y22n −1 =⇒ N0,1
(as n → ∞ ),
and Y22n =⇒ X
√ where N0,1 denotes a normal random variable with mean 0 and variance function 1, and the random variable X = (Z + N0,1 )/ 2 , where1 Z is independent of N0,1 and 1 P[Z = −1] = P[Z = 1] = . 2 0 Let {Nn , n ≥ 1} be a sequence of positive integer-valued random variables such that (2.7)
n 1 P[ Nn0 = 22 − 1 ] = 1 − , n
n
P[ Nn0 = 22 ] =
1 n
(n ≥ 1).
1The random variable X has the characteristic function given by the formula ϕ (t) = cos(t/√2) exp(−t 2 /4). X
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Theorem 2.3 does not allow to confirm the weak convergence of the randomly indexed sequence {YNn0 , n ≥ 1} although it is known [149] that YNn 0 =⇒ N0,1 (see also Example 2.17 lather on in this chapter). This example suggest that in the studies of the limit behaviour of randomly indexed sequences of random variables the assumption (2.1) should be replaced by a weaker one, e.g., that Yτn =⇒ µ for a sequence {τn , n ≥ 1} of positive, integer-valued random variables such that
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P
τn −→ ∞, or its particular case: Yan =⇒ µ for a sequence {an , n ≥ 1} of positive integers tending to infinity. When in such case (when we know that the sequence {Yn , n ≥ 1} does not weakly converge) we can state that a randomly indexed sequence {YNn , n ≥ 1} converges weakly ? The next section gives an answer to this question and contains, as particular cases, the results quoted above. 2.2.
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The Anscombe random condition
Let {Yn , n ≥ 1} be a sequence of random variables defined on a probability space (Ω, F, P), and let {αn , n ≥ 1} be a non-decreasing sequence of positive random variables (0 < αn ≤ αn+1 a.s., n ≥ 1) defined on the same probability space (Ω, F, P). Furthermore, let {τn , n ≥ 1} be a P
sequence of positive, integer-valued random variables defined on (Ω, F, P) and such that τn −→ ∞. Page 27 of 94
D EFINITION 2.6. (Anscombe random condition) A sequence {Yn , n ≥ 1} is said to satisfy the Anscombe random condition with norming sequence {αn , n ≥ 1} of positive random variables and filtering sequence {τn , n ≥ 1} of positive, integer-valued random variables if for every ε > 0 there exists δ > 0 such that (A∗ )
lim sup P[ n→∞
max
|α2i −α2τn |≤δα2τn
|Yi −Yτn | ≥ ε] ≤ ε.
One can easily see that in the special case where αn = kn and τn = n (n ≥ 1), (A∗ ) reduces to (A0 ), and hence, when kn2 = n (n ≥ 1), (A∗ ) reduces to (A). Moreover, we notice that if a sequence {Yn , n ≥ 1} satisfies (A), then for any sequence {an , n ≥ 1} of positive integers with an → ∞ we have ∀ε > 0 ∃δ > 0 : lim sup P[ max n→∞
|i−an |≤δ an
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|Yi −Yan | ≥ ε] ≤ ε,
√ with i.e. it satisfies (A∗ ) with norming sequence { n , n ≥ 1} (i.e. α2n = n) and any filtering sequence {an , n ≥ 1} of positive integers such that an → ∞. The analogous remark refers to (A0 ). Namely, if a sequence {Yn , n ≥ 1} satisfies (A0 ) with norming sequence {kn , n ≥ 1}, then it satisfies (A∗ ) with the same norming sequence {kn , n ≥ 1} (i.e. αn = kn ) and any filtering sequence {an , n ≥ 1} of positive integers such that an → ∞. The following lemma generalize these remarks. L EMMA 2.7. If a sequence {Yn , n ≥ 1} satisfies (A0 ) with the norming sequence {kn , n ≥ 1}, then it satisfies (A∗ ) with the same norming sequence {kn , n ≥ 1} and any filtering sequence {τn , n ≥ 1} independent of {Yn , n ≥ 1}.
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P ROOF. Let M > 0 be fixed. By the independence assumption P[
max
|ki2 −kτ2n |≤δkτ2n ∞
+
|Yi −Yτn | ≥ ε] ≤ P[τn ≤ M] Contents
P[τn = r]P[
∑
max
|ki2 −kr2 |≤δkr2
r=M+1
|Yi −Yr | ≥ ε]. JJ
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But lim P[τn ≤ M] = 0,
n→∞ P
since τn −→ ∞. Choosing then M so large that for all r > M P[
max
|ki2 −kr2 |≤δkr2
|Yi −Yr | ≥ ε] ≤ ε ,
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we obtain the desired result.
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There arises, obviously, a question about conditions which the filtering sequence {τn , n ≥ 1} should satisfy in the case where we reject in Lemma 2.7 the assumption that {τn , n ≥ 1} is independent of {Yn , n ≥ 1}. The following lemma gives an answer to this question: L EMMA 2.8. If a sequence {Yn , n ≥ 1} satisfies (A0 ) with the norming sequence {kn , n ≥ 1}, then it satisfies (A∗ ) with the same norming sequence {kn , n ≥ 1} and any filtering sequence {Nn , n ≥ 1} satisfying (2.4). P ROOF. By the assumption and the remark after Definition 2.6 we conclude that the sequence {Yn , n ≥ 1} satisfies (A∗ ) with the norming sequence {kn , n ≥ 1} and any filtering sequence {an , n ≥ 1} of positive integers such that an → ∞. Let {Nn , n ≥ 1} be a sequence of positive, integer-valued random variables satisfying (2.4), i.e. (2.8)
P
kN2 n /ka2n −→ 1,
where {an , n ≥ 1} is a sequence of positive integers with an → ∞. Put Bn = [ |kN2 n − ka2n | ≤ ηka2n ] (n ≥ 1),
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where η is a fixed positive number. Then for every ε > 0 and δ > 0 we have P[
max
2 |≤δk2 |ki2 −kN Nn n
+P[
|Yi −YNn | ≥ ε] ≤ P(Bcn )
2 |≤δk2 |ki2 −kN Nn n
+P[|YNn −Yan | ≥ ε/2, Bn ] |Yi −Yan | ≥ ε/2] ≤ P(Bcn ) + P[ max |ki2 −ka2n |≤δ∗ ka2n
+P[
max
|ki2 −ka2n |≤ηka2n
Contents
|Yi −Yan | ≥ ε/2, Bn ]
max
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I
|Yi −Yan | ≥ ε/2],
where δ∗ = δ(1 + η) + η. Since the sequence {Yn , n ≥ 1} satisfies condition (A∗ ) with the norming sequence {kn , n ≥ 1} and filtering sequence {an , n ≥ 1} and since, by (2.8), P(Bcn ) → 0, the last inequality proves that the sequence {Yn , n ≥ 1} satisfies (A∗ ) with norming sequence {kn , n ≥ 1} and filtering sequence {Nn , n ≥ 1}. R EMARK 2.9. Further it will be shown that condition (2.4) is in a sense necessary one for (A0 ) to imply (A∗ ). It will be proved in Subsec-
Page 29 of 94
P
tion 2.4 that it is not enough to assume kN2 n /kτ2an −→ 1 as n → ∞ . L EMMA 2.10. If a sequence {Yn , n ≥ 1} satisfies (A∗ ) with the norming sequence {αn , n ≥ 1} and filtering sequence {n, n ≥ 1} (i.e. τn = n a.s., n ≥ 1), then it satisfies (A∗ ) with the same norming sequence {αn , n ≥ 1} and any filtering sequence {υn , n ≥ 1} independent of {(αk ,Yk ), k ≥ 1}.
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The proof of this lemma runs similarly as the proof of Lemma 2.7, so it is omitted. The following lemma generalizes Lemma 2.8. L EMMA 2.11. [99] If a sequence {Yn , n ≥ 1} satisfies (A∗ ) with the norming sequence {αn , n ≥ 1} and filtering sequence {τn , n ≥ 1}, then it satisfies (A∗ ) with the same norming sequence {αn , n ≥ 1} and any filtering sequence {Nn , n ≥ 1} such that
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P
α2Nn /α2τan −→ 1 as n → ∞ ,
(2.9)
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where {an , n ≥ 1} is a sequence of positive integers with an → ∞. P ROOF. Put Bn = [|α2Nn − α2τan | ≤ ηα2τan ] (n ≥ 1), where η is a fixed positive number. Then for every ε > 0 and δ > 0, we have P[
max
|α2i −α2Nn |≤δα2Nn
|Yi −YNn | ≥ ε] ≤ P(Bcn )
+P[
max
|Yi −Yτan | ≥ ε/2]
+P[
max
|Yi −Yτan | ≥ ε/2],
|α2i −α2τan |≤δ∗ α2τan |α2i −α2τan |≤ηα2τan
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where δ∗ = δ(1 + η) + η. The assumption and this inequality imply the desired result. 2.3.
The Anscombe type theorem Contents
The following theorem generalizes Theorem 2.3. T HEOREM 2.12. [99] Let {αn , n ≥ 1} be a non-decreasing sequence of positive random variables and let {τn , n ≥ 1} be a sequence of P
positive, integer-valued random variables such that τn −→ ∞. The following conditions are equivalent: (i) Yτn =⇒ µ and the sequence {Yn , n ≥ 1}satisfies Anscombe random condition (A∗ ) with norming sequence {αn , n ≥ 1} and filtering sequence {τn , n ≥ 1}; (ii) YNn =⇒ µ for every {Nn , n ≥ 1} satisfying (2.10)
P
α2Nn /α2τan −→ 1 as n → ∞ ,
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I
where {an , n ≥ 1} is a sequence of positive integers with an → ∞. P ROOF. Let ε > 0 and a closed set A ⊂ R be given. Then, for every δ > 0, we have P[YNn ∈ A] ≤ P[Yτan ∈ Aε ] + P[YNn ∈ A,Yτan 6∈ Aε ]
Page 30 of 94
≤ P[Yτan ∈ Aε ] + P[|YNn −Yτan | ≥ ε] ≤ P[Yτan ∈ Aε ] + P[|α2Nn − α2τan | > δα2τan ] +P[
max
|α2i −α2τan |≤δα2τan
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|Yi −Yτan | ≥ ε],
where Aε = {x : ρ(x, A) ≤ ε}, ρ(x, A) = inf{|x − y| : y ∈ A}. Since ε > 0 can be chosen arbitrarily small, we see by (2.10), (A∗ ) and the assumption Yτn =⇒ µ that for every closed set A ⊂ R we have lim sup P[YNn ∈ A] ≤ µ(A) ,
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n→∞
i.e. YNn =⇒ µ. If (ii) holds, then putting Nn = τn (n ≥ 1), we conclude that Yτn =⇒ µ. Suppose that (A∗ ) with the norming sequence {αn , n ≥ 1} and filtering sequence {τn , n ≥ 1} fails. Then there exist an ε > 0 and a subsequence n1 < n2 < . . . of positive integers (n j → ∞ as j → ∞) such that for all j≥1 (2.11)
P[ max |Yi −Yτn( j) | ≥ ε] > ε, i∈Bn( j)
where Bn = {i : α2τn ≤ α2i ≤ (1 + 1/ j)α2τn } or Bn = {i : (1 − 1/ j)α2τn ≤ α2i ≤ α2τn }.
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We shall only consider the first case as the second one can be treated similarly. Let {Gi , 1 ≤ i ≤ M} be disjoint and open subsets such that 0 < µ(Gi ) = µ(Gi ) , µ(
M [
Contents
Gi ) > 1 − ε/2 ,
i=1
and the diameter(Gi ) < ε/2. Thus, by (2.11), there exists a set G ∈ {Gi , 1 ≤ i ≤ M} and a subsequence {m( j) , j ≥ 1} of the sequence {n( j) , j ≥ 1} such that for all j ≥ 1 ε |Yi −Yτm( j) | ≥ ε ] ≥ (2.12) P[Yτm( j) ∈ G , max 2M ≤α2i ≤(1+1/ j)α2τ α2τ
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I
m( j)
m( j)
Let N j = min(C1j ,C2j ), where C1j = max{i : α2i ≤ (1 + 1/ j)α2τm( j) },
C2j = min{i ≥ τm( j) : Yi 6∈ G}.
Since the sequence {αn , n ≥ 1} is non-decreasing, we have (for j ≥ 1) α2N j ≤ (1 + 1/ j)α2τm( j) .
α2τm( j) ≤ αC2 1 ,
α2τm( j) ≤ αC2 2 ,
j
j
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Hence P
α2N /α2τm( j) −→ 1 j
as j → ∞,
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which proves that the sequence {N j , j ≥ 1} satisfies (2.10). But, by (2.12), we have P[YN j 6∈ G] = P[Yτm( j) 6∈ G] + P[YN j 6∈ G,Yτm( j) ∈ G] ≥ P[Yτm( j) 6∈ G] +P[Yτm( j) ∈ G,
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α2τ
m( j)
max
≤α2i ≤(1+1/ j)α2τ
|Yi −Yτm( j) | ≥ ε]
m( j)
≥ P[Yτm( j) 6∈ G] + ε/(2M),
Close
which proves that YN j 6=⇒ µ as j → ∞. This is a contradiction to (ii), and it proves that the sequence {Yn , n ≥ 1} must satisfy (A∗ ) with norming sequence {αn , n ≥ 1}and filtering sequence {τn , n ≥ 1}. It is easy to see that putting αn = kn , τn = n (n ≥ 1), Theorem 2.12 reduces to Theorem 2.3. It is so since in this case (A∗ ) reduces to (A0 ), whereas (2.10) reduces to (2.4). C OROLLARY 2.13. Let {kn , n ≥ 1} be a non-decreasing sequence of positive numbers and let {bn , n ≥ 1} be a sequence of positive integers with bn → ∞. The following conditions are equivalent: (i) Ybn =⇒ µ and the sequence {Yn , n ≥ 1} satisfies (A∗ ) with norming sequence {kn , n ≥ 1} and filtering sequence {bn , n ≥ 1} ;
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(ii) YNn =⇒ µ for every {Nn , n ≥ 1} satisfying P
kN2 n /ka2n −→ 1 as n → ∞ ,
(2.13)
where {an , n ≥ 1} is a sequence of positive integers with an → ∞ and {an , n ≥ 1} ⊆ {bn , n ≥ 1}.
Contents
C OROLLARY 2.14. Let {kn , n ≥ 1} be a non-decreasing sequence of positive numbers and let {τn , n ≥ 1} be a sequence of positive, integerP
valued random variables such that τn −→ ∞. The following conditions are equivalent: (i) Yτn =⇒ µ and the sequence {Yn , n ≥ 1} satisfies (A∗ ) with norming sequence {kn , n ≥ 1} and filtering sequence {τn , n ≥ 1}; (ii) YNn =⇒ µ for every {Nn , n ≥ 1} satisfying
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I
P
kN2 n /kτ2an −→ 1 as n → ∞ ,
(2.14)
where {an , n ≥ 1} is a sequence of positive integers with an → ∞. In the particular case, where kn2 = n, τn = [λn] (n ≥ 1), where λ is a positive a.s. finite random variable (P[0 < λ < ∞] = 1), we have C OROLLARY 2.15. The following conditions are equivalent: √ (i) Y[λn] =⇒ µ and the sequence {Yn , n ≥ 1} satisfies (A∗ ) with norming sequence { n , n ≥ 1} and filtering sequence {[λn], n ≥ 1}; (ii) YNn =⇒ µ for every {Nn , n ≥ 1} satisfying
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P
Nn /an −→ λ as n → ∞ ,
(2.15)
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where {an , n ≥ 1} is a sequence of positive integers with an → ∞. C OROLLARY 2.16. Let {αn , n ≥ 1} be a non-decreasing sequence of positive random variables and let {bn , n ≥ 1} be a sequence of positive integers with bn → ∞. The following conditions are equivalent: (i) Ybn =⇒ µ and the sequence {Yn , n ≥ 1} satisfies (A∗ ) with norming sequence {αn , n ≥ 1} and filtering sequence {bn , n ≥ 1} ; (ii) YNn =⇒ µ for every {Nn , n ≥ 1} satisfying
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P
α2Nn /α2an −→ 1 as n → ∞ , where {an , n ≥ 1} is a sequence of positive integers with an → ∞ and {an , n ≥ 1} ⊆ {bn , n ≥ 1}.
Close
The following example elucidates the usefulness of these considerations: E XAMPLE 2.17. (cf. Example 2.5) Let {Xk , k ≥ 1 }, Sn , Yn , and Nn0 be as in Example 2.5. It is easy to see that the sequence {Yn , n ≥ 1} satisfies (A0 ) with norming sequence {kn , n ≥ 1}, where kn = sn (cf. Remark 2.4). Hence and by Lemma 2.7, the sequence {Yn , n ≥ 1} satisfies also (A∗ ) with the norming sequence {sn , n ≥ 1} and any filtering sequence {an , n ≥ 1} of positive integers with an → ∞. Thus, by P
n
Corollary 2.13, YNn =⇒ N0,1 for every {Nn , n ≥ 1} satisfying s2Nn /s2an −→ 1, where {an , n ≥ 1} ⊆ {22 − 1, n ≥ 1} and an → ∞; and YNn =⇒ X for P
n
every {Nn , n ≥ 1} satisfying s2Nn /s2a0 −→ 1, where {a0n , n ≥ 1} ⊆ {22 , n ≥ 1} and an 0 → ∞. n
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For the sequence {Nn0 , n ≥ 1} defined by (2.7) we have n
P[ |s2Nn0 /s222n −1 − 1| ≥ ε ] = P[ |s2Nn0 /s222n −1 − 1| ≥ ε, Nn0 = 22 ] n
≤ P[ Nn0 = 22 ] =
1 −→ 0 n
Contents
for all ε > 0,
so P
s2Nn0 /s222n −1 −→ 1
as n → ∞ .
Hence we conclude that in this case YNn0 =⇒ N0,1 . Let us further notice that if A is an event independent of {Xk , k ≥ 1 } and 2n 2 − 1 on A υn = n 22 on Ac , then P[Yυn < x ] −→ Φ(x)P(A) + P[ X < x ]P(Ac ) as n → ∞ for every real x. Therefore,
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1 Yυn =⇒ N0,1 I(A) + √ (Z + N0,1 )I(Ac ), 2
(2.16)
where N0,1 , I(A), and Z are independent. Moreover, the sequence {Yn , n ≥ 1} satisfies (A∗ ) with norming sequence {sn , n ≥ 1} and filtering
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sequence {υn , n ≥ 1}. Indeed, by the construction υn −→ ∞ and υn is for every n independent of {Yk , k ≥ 1}. Since the sequence {Yn , n ≥ 1} satisfies (A0 ) with norming sequence {sn , n ≥ 1}, Lemma 2.7 confirms the desired result. Hence, and by (2.16) and Corollary 2.14, we have 1 YNn =⇒ N0,1 I(A) + √ (Z + N0,1 )I(Ac ), as n → ∞ 2
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P
for every {Nn , n ≥ 1} such that s2Nn /s2υan −→ 1 (n → ∞), where {an , n ≥ 1} is a sequence of positive integers with an → ∞. This fact implies the n statements after (2.7), i.e. putting A = Ω or, equivalently, υn = 22 − 1 (n ≥ 1), we obtain YNn =⇒ N0,1 for every sequence {Nn , n ≥ 1} satisfying P
n
Close
n
s2Nn /s2an −→ 1, where {an , n ≥ 1} ⊆ {22 − 1, n ≥ 1} and an → ∞; putting however A = 0/ or, equivalently, υn = 22 (n ≥ 1), we obtain YNn =⇒ X P
n
for every sequence {Nn , n ≥ 1} satisfying s2Nn /s2a0 −→ 1, where {a0n , n ≥ 1} ⊆ {22 , n ≥ 1}, and a0n → ∞. n
2.4.
Usefulness of the Anscombe random condition
Now we shall give two examples of sequences {Yn , n ≥ 1} which fulfil (A∗ ) whereas they satisfy neither (A) nor (A0 ). At the end of this section we shall give an example proving Remark 2.9.
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n
E XAMPLE 2.18. Let {Xk , k ≥ 1 } be a sequence of random variables defined as follows: the random variables Xk , k 6= 22 (n ≥ 0, k ≥ 1), are independent and have the normal distribution function N(0, 1), and n
22 −1
X2 = −X1 ,
X22n = −
∑ n−1
j=22
Contents
(n ≥ 1).
Xj +1
√ . Let us put Sn = ∑nj=1 X j , Yn = Sn / n (n ≥ 1). Then a.s. (n ≥ 0),
Y22n = 0
(2.17)
(n → ∞).
Y22n −1 =⇒ N0,1
Indeed, for every n ≥ 0 we have S22n = 0 a.s., which proves that Y22n = 0 a.s. For the proof of the second property in (2.17) we put Sn = Sn∗ + Sn∗∗ (n ≥ 1), where n
Sn∗ =
(2.18)
∗
∑
n
n
Xi =
i=1
Sn∗∗ =
∑ Xi ,
i=1 i∈N∗
∗∗
∑
n
=
Xi
i=1
∑
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I
Xi ,
i=1 i∈N∗∗ Page 34 of 94
and n
N∗ = { j ∈ N : j 6= 22 , n ≥ 0} ,
(2.19)
N∗∗ = N \ N∗ ,
and note that, for every n ≥ 1, Sn∗ is the sum of independent random variables, S∗2n = S∗2n 2
a.s., S∗∗2n
S1∗∗ = 0
(2.20)
2
−i
= −S∗ n−1 22
2
−1 n
a.s. for n ≥ 0, while n−1
a.s. for 1 ≤ i ≤ 22 − 22
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(n ≥ 1).
−1
Moreover, by the Kolmogorov’s inequality we have, for every ε > 0, P[|S∗∗2n 2
−1
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q n 22 − 1| ≥ ε] = P[|
n−1 22 −1
∑
Xi | ≥ ε
q
n
22 − 1]
i=1 i∈N∗
≤2
2n−1
2 2n ε (2 − 1) −→ 0
Close
( n → ∞ ).
Hence, and by taking into account that E exp{˙ıtS∗2n 2
−1
Quit
n q 22 −1 q n n 2 / 2 − 1} = E exp{˙ıt ∑ X j 22 − 1}
j=1 j∈N∗
n n = exp{−t 2 (22 − n − 1) 2(22 − 1)} −→ exp{−t 2 /2}
( n → ∞ ),
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we obtain Y22n −1 = S∗2n 2
−1
q n q n 22 − 1 + S∗∗2n 22 − 1 =⇒ N0,1
( n → ∞ ),
−1
2
which ends the proof of (2.17). Now we shall prove that the sequence {Yn , n ≥ 1 } does not satisfy the Anscombe condition (A). Indeed, if the sequence {Yn , n ≥ 1} fulfilled (A), than in view of the remarks after Definition 2.6 it would fulfil (A∗ ) with norming sequence √ n { n, n ≥ 1} and any filtering sequence {an , n ≥ 1} of positive integers with an → ∞ ( n → ∞ ), e.g. with an = 22 − 1 (n ≥ 1) for which Y22n −1 =⇒ N0,1
P
n
|i−32 |≤δ32
+ P[
max n
|i−32 |≤δ32
≤ 2P[
n
max
n
J
I
( n → ∞ ),
√ 2n 2n |S ||3 − i|/ i 3 ≥ ε/2] i n
[(1−δ)32 ]≤i≤[(1+δ)32 ]
II
n
√ which is a contradiction to (2.17). Thus, the sequence {Yn , n ≥ 1} does not satisfy (A∗ ) with the norming sequence { n, n ≥ 1} and filtering n sequence {22 − 1, n ≥ 1}, and than, by the earlier considerations, we conclude that the sequence {Yn , n ≥ 1} does not satisfy (A). √ n And now we shall prove that the sequence {Yn , n ≥ 1} satisfies (A∗ ) with the norming sequence { n, n ≥ 1} and filtering sequence {32 , n ≥ 1}. To this end we note that for every ε > 0 and for every δ > 0 we have q n n n (2.21) P[ max |Yi −Y32 | ≥ ε] ≤ P[ max |Si − S32 | ≥ ε 32 /2] n n n n |i−32 |≤δ32
JJ
( n → ∞ ).
Hence, by Corollary 2.13, YNn =⇒ N0,1 ( n → ∞ ) for every sequence {Nn , n ≥ 1} such that Nn /(22 − 1) −→ 1 ( n → ∞ ). So then, for Nn = 22 a.s. (n ≥ 1), we would have Y22n =⇒ N0,1
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Page 35 of 94
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q n |Si − S[(1−δ)32n ] | ≥ ε 32 /4] + P[
max
n n [(1−δ)32 ]≤i≤[(1+δ)32 ]
q n |Si | ≥ ε [(1 − δ)32 ]/2δ],
where [x] denotes the integral part of the real number x. Further, by (2.20) and the Kolmogorov’s inequality, the right-hand side of (2.21) is less than or equal to q n ∗ ∗ 2P[ max |Si − S[(1−δ)32n ] | ≥ ε 32 /4] n n [(1−δ)32 ]≤i≤[(1+δ)32 ]
n n n ≤ 32{[(1 + δ)32 ] − [(1 − δ)32 ]} ε 32 −→ 64δ/ε2
( n → ∞ ),
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n
n
since, for [(1 − δ)32 ] ≤ i ≤ [(1 + δ)32 ], we have ∗∗ Si∗∗ − S[(1−δ)3 2n ] =
i
∑
Xj = 0
a.s.,
n j=[(1−δ)32 ]+1 ∗∗ j∈N
Contents
while the second term on the right-hand side of (2.21) is less than or equal to q n ∗ P[ max |Si | ≥ ε [(1 − δ)32 ]/4δ] n n
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II
J
I
[(1−δ)32 ]≤i≤[(1+δ)32 ]
+ P[
max n
n
[(1−δ)32 ]≤i≤[(1+δ)32 ]
2
≤ 16δ [(1 + δ)3
2n
2n
|Si∗∗ | ≥ ε
q n [(1 − δ)32 ]/4δ]
]/ε [(1 − δ)3 ] + P[|S2∗2n −1 | ≥ ε 2 2n 2n 2 2
q n [(1 − δ)32 ]/4δ] n
≤ 16δ {[(1 + δ)3 ] + 2 }/ε [(1 − δ)32 ] −→ 16δ2 (1 + δ)/ε2 (1 − δ) ( n → ∞ ),
Page 36 of 94
where 16δ2 (1 + δ)/ε2 (1 − δ) ≤ 64δ/ε2 for 0 < δ ≤ 1/2. Hence, for every ε > 0 and 0 < δ ≤ 1/2, we have lim sup P[ n→∞
max n
|i−32 |≤δ32
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|Yi −Y32n | ≥ ε] ≤ 128δ/ε2 , n
√ n which proves that the sequence {Yn , n ≥ 1} satisfies (A∗ ) with the norming sequence { n, n ≥ 1} and filtering sequence {32 , n ≥ 1}. Let us further notice that Y32n =⇒ N0,1 ( n → ∞ ). Indeed, since for every n ≥ 1 we have q q n n Y32n = S3∗2n / 32 + S3∗∗2n / 32 , where, by the Kolmogorov’s inequality, for every ε > 0, q q n n n n ∗∗ ∗ 2 P[|S32n / 3 | ≤ ε] = P[|S22n −1 | ≥ ε 32 ] ≤ 22 /ε2 32 −→ 0
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(n → ∞)
and q n ∗ E exp{˙ıtS32n 32 } = E exp{˙ıt
32
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n
∑
q n Xj 32 }
j=1 j∈N∗
n n = exp{−t 2 (32 − n − 1) 2 · 32 } −→ exp{−t 2 /2}
( n → ∞ ),
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we have Y32n =⇒ N0,1 ( n → ∞ ). Hence, by Corollary 2.13, YNn =⇒ N0,1 ( n → ∞ ) for every {Nn , n ≥ 1} satisfying Nn /an −→ 1 ( n → ∞ ), where n {an , n ≥ 1} ⊂ {32 , n ≥ 1} (an → ∞, n → ∞ ). n
E XAMPLE 2.19. Let {Xk , k ≥ 1 } be a sequence of random variables defined as follows: the random variables Xk , k 6= 22 (n ≥ 0, k ≥ 1), are n independent and have the normal distribution function N(0, 1) for k 6= 32 (n ≥ 1) and n−1
P[X32n = 32
] = P[X32n = −32
n−1
]=
1 2
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(n ≥ 1), JJ
II
J
I
while n
22 −1
X2 = −X1 ,
X22n = −
∑ n−1
j=22
Xj
(n ≥ 1).
+1
Let us put n
kn2 =
∑ Var(X j )
m
for n 6= 22 (m ≥ 0), Page 37 of 94
j=1
and n
k222n
= k222n −1
for n ≥ 0,
Sn =
∑ X j,
. and Yn = Sn /kn
(n ≥ 1). Go Back
j=1
Then Y22n = 0
(2.22)
a.s. (n ≥ 0),
Y22n −1 =⇒ N0,1
( n → ∞ ).
Indeed, for every n ≥ 0 we have S22n = 0 a.s., which proves that Y22n = 0 a.s. (n ≥ 0). For the proof of the second property in (2.22) we notice n that k222n −1 ∼ 22 for sufficiently large n, and Y22n = S2∗2n −1 /k22n −1 + S2∗∗2n −1 /k22n −1 where
Sn∗
and
Sn∗∗
(n ≥ 1),
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are defined as in (2.18). Furthermore, by (2.20), we have 2n
Y22n −1 = (S∗2n 2
−1
− S∗ n−1 22
−1
2 −1 ∗ ) k22n −1 = ∑ X j k22n −1 2n−1
j=2
2n
2 −1
=
∑ n−1
j=22
X j k22n −1 +1
for every n ≥ 1,
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where
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n
22 −1
E exp{˙ıt
∑ n−1
j=22
X j k22n −1 } +1
= cos(t · 3
2n−2
2n
/k22n −1 ) exp{−t 2 (2 − 2
2n−1
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II
J
I
− 1)/2k22n } 2
−1
−→ exp{−t 2 /2}
( n → ∞ ).
Hence, Y22n −1 =⇒ N0,1 ( n → ∞ ), which completes the proof of (2.22). From (2.22) and that
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k22n = k22n 2
2
−1
(n ≥ 1), Full Screen
we conclude that the sequence {Yn , n ≥ 1} satisfies neither the Anscombe condition (A) nor the generalized Anscombe condition (A0 ) with the norming sequence {kn , n ≥ 1}. The proofs of these facts run similarly as in Example 2.18. Suppose for example that the sequence {Yn , n ≥ 1} satisfies the generalized Anscombe condition (A0 ) with norming sequence {kn , n ≥ 1}. Then in view of the remarks after Definition 2.6 it would fulfil (A∗ ) with (the same) norming sequence {kn , n ≥ 1} and any filtering sequence n {an , n ≥ 1} of positive integers with an → ∞ ( n → ∞ ), e.g. with an = 22 −1 (n ≥ 1) for which Y22n −1 =⇒ N0,1 ( n → ∞ ). Hence by Corollary 2.13 P
we would have YNn =⇒ N0,1 ( n → ∞ ) for every sequence {Nn , n ≥ 1} such that kN2 n /k222n −1 −→ 1 ( n → ∞ ). So then, since it follows from the assumptions that k222n = k222n −1 for all n ≥ 1, we would conclude that Y22n =⇒ N0,1 ( n → ∞ ). But this is a contradiction to (2.22) which states that Y22n = 0 a.s. for all n ≥ 0. Thus, the sequence {Yn , n ≥ 1} does not satisfy (A∗ ) with the norming sequence {kn , n ≥ 1} and filtering sequence n {22 − 1, n ≥ 1} and then, by the earlier considerations, we conclude that the sequence {Yn , n ≥ 1} does not satisfy the generalized Anscombe condition (A0 ) with the norming sequence {kn , n ≥ 1}, as well.
Close
Quit
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n
Now we shall prove that the sequence {Yn , n ≥ 1} satisfies (A∗ ) with the norming sequence {kn , n ≥ 1} and filtering sequence {32 , n ≥ 1}. To this end we note that, for every ε > 0 and for every 0 < δ < 1, we have (2.23) P[
max
|ki2 −k22n |≤δk22n 3
|Yi −Y32n | ≥ ε]
Contents
3
≤ P[
max
|ki2 −k22n |≤δk22n 3
+ P[
3
max
|ki2 −k22n |≤δk22n 3 3
≤ P[
|Si − S32n | ≥ εk32n /2]
|Si ||k322n − ki2 |/ki k322n ≥ ε/2]
max
k22n ≤ki2 ≤(1+δ)k22n 3 3
+ P[
k22n ≤ki2 ≤(1+δ)k22n
|Si∗ ≥ εk32n /4δ]
J
I
3
+ P[
max
k22n ≤ki2 ≤(1+δ)k22n 3 3
n
II
|Si∗ − S3∗2n | ≥ εk32n /2]
max
3
JJ
|Si∗∗ ≥ εk32n /4δ], Page 39 of 94
n
since k322n −1 ∼ 32 and k322n ∼ 2 · 32 for sufficiently large n, whence {i ∈ N : |ki2 − k322n | ≤ δk322n } = {i ∈ N : k322n ≤ ki2 ≤ (1 + δ)k322n } for sufficiently large n, and since for
i ∈ {k322n
≤ ki2
≤ (1 + δ)k322n }
Go Back
we have i
Si∗∗ − S3∗∗2n =
∑n
j=32
Xj = 0
a.s. Full Screen
+1
j∈N∗∗
for sufficiently large n. As Sn∗ is the sum of independent random variables with finite variances, then, by the Kolmogorov’s inequality, the first term on the right-hand side of (2.23) is less than or equal to 4δ/ε2 , the second one is less than or equal to 16δ2 (1 + δ)k322n ε2 k322n ≤ 32δ/ε2 , and the third term on the right-hand side of (2.23) is equal to
Quit
P[|S2∗2n −1 | ≥ εk32n /4δ] ≤ 16δ2 k222n −1 /ε2 k32n −→ 0
( n → ∞ ).
Hence, for every ε > 0 and for every 0 < δ < 1, we have lim sup P[ n→∞
Close
max
|ki2 −k22n |≤δk22n 3 3
|Yi −Y32n | ≥ ε] ≤ 36δ/ε2 ,
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n
which proves that the sequence {Yn , n ≥ 1} satisfies (A∗ ) with the norming sequence {kn , n ≥ 1} and filtering sequence {32 , n ≥ 1}. n Similarly it can be proved that the sequence {Yn , n ≥ 1} satisfies (A∗ ) with the norming sequence {kn , n ≥ 1} and filtering sequence {32 − 1, n ≥ 1}. Let us further notice that Y32n −1 =⇒ N0,1 ,
(2.24)
Y32n =⇒ X ,
( n → ∞ ),
and k322n /k322n −1 −→ 2
Contents
( n → ∞ ),
JJ
II
J
I
where X is a random variable with the characteristic function
√ 2 ϕX (t) = cos(t/ 2)e−t /4 .
Then, by Corollary
n P 2.13, YNn =⇒ N0,1 ( n → ∞ ) for every {Yn , n ≥ 1 } satisfying kN2 n /ka2n −→ 1 (n → ∞), where {an , n ≥ 1} ⊂ {32 − 1 , n ≥ 1} n P and YNn =⇒ X (n → ∞) for every {Yn , n ≥ 1 } satisfying kN2 n /ka2n −→ 1 (n → ∞), where {an , n ≥ 1} ⊂ {32 , n ≥ 1} (an → ∞,
(an → ∞, n → ∞ ), n → ∞). n Let us still notice that if A is an event independent of Xk , k 6= 22 (n ≥ 0, k ≥ 1), and ( n 32 − 1 on A, (2.25) υn = n 32 on Ac ,
Page 40 of 94
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then P[Yυn < x] → Φ(x)P(A) + P[X < x]P(Ac ), n → ∞ , i.e. Yυn =⇒ N0,1 I(A) + XI(Ac )
(2.26)
( n → ∞ ), Full Screen
and the sequence {Yn , n ≥ 1} satisfies the Anscombe random condition (A∗ ) with the norming sequence {kn , n ≥ 1} and filtering sequence {υn , n ≥ 1}. Thus, by Corollary 2.14, YNn =⇒ N0,1 I(A) + XI(Ac ) ( n → ∞ ),
Close
P
for every {Nn , n ≥ 1} satisfying kN2 n /kυ2 an −→ 1 ( n → ∞ ), where {an , n ≥ 1} is a sequence of positive integers with an → ∞ ( n → ∞ ). That fact n implies the statements after (2.24), i.e. putting A = Ω or, equivalently, υn = 32 − 1 a.s. (n ≥ 1), we obtain YNn =⇒ N0,1 ( n → ∞ ) for every P
n
{Nn , n ≥ 1} satisfying kN2 n /ka2n −→ 1 ( n → ∞ ), where {an , n ≥ 1} ⊂ {32 − 1 , n ≥ 1} (an → ∞, n → ∞ ); putting however A = 0/ or, equivalently, n
P
n
υn = 32 a.s. (n ≥ 1), we obtain YNn =⇒ X ( n → ∞ ) for every {Nn , n ≥ 1} satisfying kN2 n /ka2n −→ 1 ( n → ∞ ), where {an , n ≥ 1} ⊂ {32 , n ≥ 1} (an → ∞, n → ∞ ). The following example proves Remark 2.9.
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E XAMPLE 2.20. [50] Let U, Z1 , Z2 , . . . be independent random variables such that U has a uniform distribution on the interval (0, 1) and, for each n ≥ 1, Zn has a normal distribution, N(0, 1), with mean zero and variance one. Let n
Yn = n−1/2 ∑ Zi
Contents
(n ≥ 1),
i=1
n
J(ω) = {[2 U(ω)] + 1, 1 ≤ n < ∞}, JJ
and
II
Yn0 = Yn I[n 6∈ J] (n ≥ 1). J I √ The sequence {Yn0 , n ≥ 1} satisfies (A) [50] and than, by Lemma 2.8, it satisfies as well (A∗ ) with norming sequence { n, n ≥ 1} and any filtering podac dowod sequence {Nn , n ≥ 1} of positive integer-valued random variables satisfying (2.3) (i.e. (2.4) with kn2 = n, n ≥ 1). Let us put Nn = [2nU], n ≥ 1. The sequence {Nn , n ≥ 1} does not satisfy (2.4) but satisfies (2.14). Indeed, putting e.g. τn = [2nU] + 1, n ≥ 1, Page 41 of 94 we have P
Nn /τn −→ 1
(2.14’)
( n → ∞ ),
2 ∗ i.e. condition √ (2.14) with kn = n (an = n, n ≥ 1). Now we shall prove that the sequence {Yn , n ≥ 1} does not satisfy (A ) with the norming sequence { n, n ≥ 1} and filtering sequence {Nn , n ≥ 1}. This fact will prove Remark 2.9. It follows from Theorem 1.1 that
YNn =⇒ N0,1
(2.27)
( n → ∞ ),
where Nn = [2nU] (n ≥ 1),
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P
since Yn =⇒ N0,1 ( n → ∞ ), Nn −→ ∞ ( n → ∞ ) and, for every n ≥ 1, the random variable Nn is independent of Zi (i ≥ 1). Furthermore, by the construction of the set J(ω) we have Close
P[YN0 n
−n
6= YNn ] = P[Nn ∈ J] ≤ P[U < (n − 1)2 ] 2n
+ ∑ P[(k − 1)2−n ≤ U < k2−n ; [2nU] ∈ {[2nU] + 1, 1 ≤ n < ∞}]
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k=n 2n
≤ on (1) + ∑ P[(k − 1)2−n ≤ U < k2−n ; k − 2 ≤ 2mU < k − 1 k=n
for some m ≥ 1]
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2n
≤ on (1) + ∑
n
∑
Title Page
P[(k − 1)2−n
k=n m=log2 2n (1−2/n)
≤ U < k2−n ; (k − 2)2−m ≤ U < (k − 1)2−m ] n
≤ on (1) +
∑n
Contents
P[(n − 2)2−m ≤ U < (2n − 1)2−m ]
m=log2 2 (1−2/n) n
= on (1) + (2n − n + 1)
∑n
2−m
JJ
II
J
I
m=log2 2 (1−2/n)
= on (1) +
2−n+1 (2n − n + 1) 2 · −→ 0 n 1 − 2n
( n → ∞ ),
since on (1) = P[U < (n − 1)2−n ] → 0 (n → ∞). (It should be noticed that the last equality above follows from the well known property: s
∑
m=r
1 − ( 12 )s−r 2−m = 2−r 1 + 2−1 + . . . + 2−s+r = 2−r 1
Page 42 of 94
2
used with s = n and r = log2 2n (1 − 2/n). Since s − r = log2 n/(n − 2), in a straightforward manner we obtain n
∑
2−m = 2
m=r
2−n 2−n+1 2 1 − 2− log2 n/(n−2) = · , 1 − 2/n 1 − 2/n n
and thus the desired equality follows.) P This fact and (2.27) imply (by the law-equivalence lemma: If Yn −Yn0 −→ 0 or P[Yn 6= Yn0 ] → 0, then the sequences L(Yn ) and L(Yn0 ) of laws are equivalent) that Yn0 =⇒ N0,1 (n → ∞). Since (2.14’) holds, and Yτ0n = 0
where τn = [2nU] + 1 (n ≥ 1), √ we conclude that the sequence {Yn0 , n ≥ 1} does not satisfy (A∗ ) with the norming sequence { n, n ≥ 1} and filtering sequence {Nn , n ≥ 1}, where Nn = [2nU], n ≥ 1. (2.28)
2.5.
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a.s. (n ≥ 1),
Close
An application to random sums of independent random variables Quit
We now show that the given results allow to extend the random central limit theorem of J. R. Blum, D. L. Hanson, J. I. Rosenblatt [25] and J. A. Mogyoródi [114] to sequences of independent, nonidentically distributed random variables. T HEOREM 2.21. [98] Let {Xk , k ≥ 1} be a sequence of independent random variables with EXk = 0, EXk2 = σ2k < ∞, k ≥ 1, such that . (2.29) Yn = Sn /sn =⇒ µ, n → ∞ ,
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where Sn = ∑nk=1 Xk , s2n = ∑nk=1 σ2k , n ≥ 1, and s2n → ∞ as n → ∞. Suppose that λ is a positive random variable (P[0 < λ < ∞] = 1), and that P
{υn , n ≥ 1} is a sequence of positive integer-valued random variables independent of {λ, Xk , k ≥ 1} with υn −→ ∞, n → ∞ , such that for any given ε > 0 and some constant r > 0 # " 2 s [(λ+c)υn ] − λr ≥ ε = 0. (2.30) lim lim sup P s2υn 0≤c→∞ n→∞ If {Nn , n ≥ 1} is a sequence of positive integer-valued random variables such that P
s2Nn /s2υn −→ λr ,
(2.31)
JJ
II
J
I
n → ∞,
then YNn = SNn /sNn =⇒ µ,
(2.32)
Contents
n → ∞.
R EMARK 2.22. Note that condition (2.30) implies that for any given ε > 0 (2.300 )
lim lim sup P[|s2[(λ+c)υn ] /s2[λυn ] − 1| ≥ ε] = 0.
Page 43 of 94
0 0 and N > 0, we have P[|s2[(λ+c)υn ] /s2[λυn ] − 1| ≥ ε] ≤ P[|s2[(λ+c)υn ] /s2υn − λr |(s2υn /s2[λυn ] ) ≥ ε/2]
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+ P[λr |s2υn /s2[λυn ] − 1/λr | ≥ ε/2] ≤ P[s2υn /s2[λυn ] > M] + P[|s2[(λ+c)υn ] /s2υn − λr | ≥ ε/(2M)]
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+ P[λr > N] + P[|s2υn /s2[λυn ] − 1/λr | ≥ ε/(2N)]. Since the condition (2.30) with c = 0 implies that
Close
lim sup P[s2υn /s2[λυn ] > M] = P[λr < 1/M] n→∞
and that lim sup P[|s2υn /s2[λυn ] − 1/λr | ≥ ε/(2N)] = 0,
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n→∞
then for any M > 0 and N > 0 lim lim sup P[|s2[(λ+c)υn ] /s2[λυn ] − 1| ≥ ε] ≤ P[λr < 1/M] + P[λr > N].
Letting now M → ∞ and N → ∞
0 0 and 0 < δ < 1/2 we have (cf. Remark 2.9) lim sup P max |Yi −Ym | ≥ ε ≤ 8δ(4 + δ)/ε2 . |s2i −s2m |≤δs2m
m→∞
Page 46 of 94
Hence and by Lemma 2.23, for every ε > 0, ∞
lim lim sup P 0 M]
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m n
m n
n
We see that for any given M > 0 n→∞
n→∞
+ lim sup P[|s[λm υn ] /s[λυn ] − 1| ≥ ε/(2M)] n→∞
≤ µ{x : |x| > M} + lim sup P[|s2[λm υn ] /s2[λυn ] − 1| ≥ ε/(2M)]. n→∞
Full Screen
Moreover, since for every n ≥ 1 1 ≤ s2[λm υn ] /s2[λυn ] ≤ s2[(λ+2−m )υn ] /s2[λυn ]
a.s., Close
then, by (2.300 ), we get lim lim sup P[|s2[λm υn ] /s2[λυn ] − 1| ≥ ε/(2M)] = 0.
m→∞
n→∞
Hence
Quit
lim lim sup P[|Y[λm υn ] (1 − s[λm υn ] /s[λυn ] )| ≥ ε/2] ≤ µ{x : |x| > M},
m→∞
n→∞
which implies (letting M → ∞) that (2.39)
lim lim sup P[|Y[λm υn ] (1 − s[λm υn ] /s[λυn ] )| ≥ ε/2] = 0.
m→∞
n→∞
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Now we see, using (2.30) and (2.300 ), that for any given δ > 0 (2.40) lim lim sup P |S[λm υn ] − S[λυn ] | ≥ εs[λυn ] /2 m→∞ n→∞ ≤ lim lim sup P m2−m ≤ λr < m, |s2[λυn ] /s2υn − λr | ≤ 2−m , |s2[λm υn ] /s2λυn − 1| ≤ δ, |S[λm υn ] − S[λυn ] | ≥ εs[λυn ] /2 m→∞ n→∞ |Si − S[λυn ] | ≥ εs[λυn ] /2 ≤ lim lim sup P m2−m ≤ λr < m, |s2[λυn ] /s2υn − λr | ≤ 2−m , max m→∞
|s2i −s2[λυ ] |≤δs2[λυ
n→∞
n]
n
≤ lim lim sup P m2−m ≤ λr < m, (λr − 2−m )s2υn ≤ s2[λυn ] ≤ (λr + 2−m )s2υn , m→∞ n→∞ p |Si − S[λυn ] | ≥ εsυn λr − 2−m /2 max (1−δ)(λr −2−m )s2υn ≤s2i ≤(1+δ)(λr +2−m )s2υn m2m −1
≤ lim lim sup m→∞
n→∞
∑
JJ
II
J
I
P k2−m ≤ λr < (k + 1)2−m , |s2[λυn ] /s2υn − λr | ≤ 2−m ,
k=m
max
(1−δ)(λr −2−m )s2υn ≤s2i ≤(1+δ)(λr +2−m )s2υn m2m −1
≤ lim lim sup m→∞
n→∞
∑
|Si − S[λυn ] | ≥ εsυn
p (k − 1)2−m /2
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k=m
max
p |Si − S j | ≥ εsυn (k − 1)2−m /2
m2m −1 ∞
= lim lim sup n→∞
Page 48 of 94
P k2−m ≤ λr < (k + 1)2−m ,
(1−δ)(k−1)2−m s2υn ≤s2i ,s2j ≤(1+δ)(k+1)2−m s2υn
m→∞
Contents
∑ ∑ P[υn = b]P k2−m ≤ λr < (k + 1)2−m,
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k=m b=1
max
(1−δ)(k−1)2−m s2b ≤s2i ,s2j ≤(1+δ)(k+1)2−m s2b
|Si − S j | ≥ εsb
p (k − 1)2−m /2 Close
m2m −1 ∞
≤ lim lim sup m→∞
n→∞
∑ ∑ P[υn = b]P k2−m ≤ λr < (k + 1)2−m,
k=m b=1
max
(1−δ)(k−1)2−m s2b ≤s2i ≤(1+δ)(k+1)2−m s2b m2m −1
≤ 2 lim
m→∞
∑
|Si − ST 1 | ≥ εsb b
p (k − 1)2−m /4 ∞
P k2−m ≤ λr < (k + 1)2−m lim sup n→∞
k=m
∑ P[υn = b]× b=1
p P max |Si − ST 1 | ≥ εsb (k − 1)2−m /4 k2−m ≤ λr < (k + 1)2−m , Tb1 ≤i≤Tb2
b
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where Tb1 = Tb1 (δ, k, m) = min{i : (1 − δ)(k − 1)2−m s2b ≤ s2i }, Tb2 = Tb2 (δ, k, m) = max{i : s2i ≤ (1 + δ)(k + 1)2−m s2b }.
Contents
But by Lemma 2.23 and the Kolmogorov’s inequality we get p lim sup P max |Si − ST 1 | ≥ εsb (k − 1)2−m /4 k2−m ≤ λr < (k + 1)2−m p = lim sup P max |Si − ST 1 | ≥ εsb (k − 1)2−m /4 b→∞
Tb1 ≤i≤Tb2
b
≤ 16 lim sup E(ST 2 − ST 1 )2 /(ε2 s2b (k − 1)2−m ) b→∞
II
J
I
b
Tb1 ≤i≤Tb2
b→∞
JJ
b
b
Page 49 of 94
≤ 16 lim sup {(1 + δ)(k + 1)2−m s2b − (1 − δ)(k − 1)2−m s2b }/(ε2 s2b (k − 1)2−m ) b→∞ 2
≤ 16(2kδ + δ + 3)/(ε (k − 1)) ≤ 64δ/ε2 + 48/((m − 1)ε2 )
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for m ≤ k ≤ m2m − 1, m ≥ 3. Therefore, using the Toeplitz lemma [107, p. 238], we get (for m ≥ 3) ∞
lim sup n→∞
∑ P[υn = b]P b=1
max |Si − ST 1 | ≥ εsb
Tb1 ≤i≤Tb2
b
p (k − 1)2−m /4 k2−m ≤ λr
Full Screen
< (k + 1)2−m ≤ 64δ/ε2 + 48/((m − 1)ε2 ) for m ≤ k ≤ m2m − 1, Close
and hence, and by (2.40), we have lim lim sup P |S[λm υn ] − S[λυn ] | ≥ εs[λυn ] /2 ≤ 128δ/ε2 .
m→∞
n→∞
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Letting now δ → 0, we have lim lim sup P |S[λm υn ] − S[λυn ] | ≥ εs[λυn ] /2 = 0,
m→∞
n→∞
which together with (2.39) implies (2.38). The proof of Lemma 2.27 is complete.
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P ROOF OF L EMMA 2.28. Note that for any given ε > 0 and δ > 0, we have (2.41) P max |Yi −Y[λυn ] | ≥ ε |s2i −s2[λυ ] |≤δs2[λυ n
n]
Contents
≤P
n]
n
+P
−1 |S[λυn ] ||s−1 i − s[λυn ] | ≥ ε/2
max
|s2i −s2[λυ ] |≤δs2[λυ n
≤P
|Si − S[λυn ] |/si ≥ ε/2
max
|s2i −s2[λυ ] |≤δs2[λυ
max
|s2i −s2[λυ ] |≤δs2[λυ
JJ
II
J
I
n]
p |Si − S[λυn ] | ≥ ε 1 − δs[λυn ] /2
n]
n
p + P |Y[λυn ] | ≥ ε 1 − δ/2δ . By Lemma 2.27 we know that p p lim sup P |Y[λυn ] | ≥ ε 1 − δ/2δ = µ{x : |x| ≥ ε 1 − δ/2δ}.
Page 50 of 94
p lim lim sup P |Y[λυn ] | ≥ ε 1 − δ/2δ = 0.
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n→∞
Letting δ → 0, we get (2.42)
δ→0
n→∞
Following the considerations of the proof of Lemma 2.27 we can show that for any given ε > 0, δ > 0 and m ≥ 3 p lim sup P max |Si − S[λυn ] | ≥ ε 1 − δs[λυn ] /2 n→∞
|s2i −s2[λυ ] |≤δs2[λυ n
Full Screen
n]
≤ P[λr < m2−m ] + P[λr > m] + 128δ/ε2 (1 − δ) + 96/ε2 (1 − δ)(m − 1). Letting m → ∞, we get for every ε > 0 and 0 < δ ≤ 1/2 p (2.43) lim sup P max |Si − S[λυn ] | ≥ ε 1 − δs[λυn ] /2 n→∞
Close
|s2i −s2[λυ ] |≤δs2[λυ n
n]
≤ 256δ/ε2 −→ 0 Combining (2.41), (2.42) and (2.43) we get the desired result: for every ε > 0 there exists a δ > 0 such that |Yi −Y[λυn ] | ≥ ε ≤ ε. lim sup P max n→∞
|s2i −s2[λυ ] |≤δs2[λυ n
n]
as δ → 0.
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P ROOF OF T HEOREM 2.21. It follows from (2.30) and (2.31) that P
s2Nn /s2[λυn ] −→ 1
(2.44)
as
n → ∞.
By (2.35) and Lemma 2.28 we have Y[λυn ] =⇒ µ and the sequence {Yn , n ≥ 1} satisfies the Anscombe random condition (A∗ ) with the norming sequence {sn , n ≥ 1} and filtering sequence {[λυn ], n ≥ 1}. Thus, by Corollary 2.14, we obtain YNn =⇒ µ as n → ∞ for every sequence {Nn , n ≥ 1} satisfying (2.44), which is the statement of Theorem 2.21. C OROLLARY 2.29. Let {Xk , k ≥ 1} be a sequence of independent and identically distributed random variables with EX1 = 0, EX12 = σ2 > 0. If {Nn , n ≥ 1} is a sequence of positive integer-valued random variables such that P
Nn /υn −→ λ as
(2.45)
Contents
JJ
II
J
I
n → ∞,
where λ is a positive random variable (P[0 < λ < ∞] = 1) and {υn , n ≥ 1} is a sequence of positive integer-valued random variables independent P
of {λ, Xk , k ≥ 1} with υn −→ ∞ as n → ∞, then
√ SNn /σ Nn =⇒ N0,1 .
P ROOF. It is easy to see that in this case the condition (2.30) holds true with r = 1, while the condition (2.31) is a consequence of (2.45).
Page 51 of 94
We note that in the case when λ is a positive random variable having a discrete distribution, then the conditions (2.30) and (2.31) in Theorem 2.21 can be replaced by the weaker condition (2.44). T HEOREM 2.30. Let {Xk , k ≥ 1} be a sequence of independent random variables with EXk = 0, EXk2 = σ2k < ∞, k ≥ 1, such that . (2.290 ) Yn = Sn /sn =⇒ µ, n → ∞ , where Sn = ∑nk=1 Xk , s2n = ∑nk=1 σ2k , n ≥ 1, and s2n → ∞ as n → ∞. Let {Nn , n ≥ 1} be a sequence of positive integer-valued random variables such that P
(2.440 )
s2Nn /s2[λυn ] −→ 1,
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n → ∞,
where λ is a positive random variable having a discrete distribution and {υn , n ≥ 1} is a sequence of positive integer-valued random variables Close
P
independent of {λ, Xk , k ≥ 1} with υn −→ ∞. Then (2.320 )
YNn = SNn /sNn =⇒ µ,
n → ∞. Quit
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2.6.
Applications to the stable convergence
Recall that Yn =⇒ µ(stably) if Yn =⇒ µ and for every B ∈ F with P(B) > 0 there exists a probability measure µB such that Yn =⇒ µB as n → ∞ under the conditional measure P( |B). In the special case where µB = µ for all B ∈ F with P(B) > 0 we say Yn =⇒ µ(mixing). As a simple consequence of Theorem 2.3 we get T HEOREM 2.31. [150] The following conditions are equivalent: (i) Yn =⇒ µ(stably) and satisfies (A0 ); (ii) YNn =⇒ µ(stably) for every sequence {Nn , n ≥ 1} satisfying (2.4).
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P ROOF. The implication (ii) =⇒ (i) is obvious: put Nn = n to get Yn =⇒ µ(stably), and use Theorem 2.3 to get {Yn , n ≥ 1} satisfies (A0 ). To prove the reverse implication fix B ∈ F with P(B) > 0. Fix ε > 0. Given ε0 = εP(B) choose δ > 0 such that lim sup P[
max
|ki2 −kn2 |≤δkn2
n→∞
|Yi −Yn | ≥ ε0 ] ≤ ε0
Then lim sup P[ n→∞
max
|ki2 −kn2 |≤δkn2
≤ lim sup P[ n→∞
max
|Yi −Yn | ≥ ε]/P(B)
max
|Yi −Yn | ≥ ε0 ]/P(B) ≤ ε,
|ki2 −kn2 |≤δkn2
≤ lim sup P[ n→∞
|Yi −Yn | ≥ ε|B]
|ki2 −kn2 |≤δkn2
which proves that the sequence {Yn , n ≥ 1} satisfies (A0 ) under the conditional measure P(·|B). Moreover, by (i), the sequence {Yn , n ≥ 1} converges weakly (to a measure µB ) under the conditional measure P(·|B). Thus, by Theorem 2.3, YNn =⇒ µB under the conditional measure P(·|B) for every sequence {Nn , n ≥ 1} satisfying (2.4). Hence, YNn =⇒ µ(stably) for every {Nn , n ≥ 1} satisfying (2.4).
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R EMARK 2.32. We note that if Yn =⇒ µ(mixing) not only stably, then in the part (ii) of Theorem 2.31 we obtain YNn =⇒ µ(mixing) for every {Nn , n ≥ 1} satisfying (2.4). Close
C OROLLARY 2.33. [150] (cf. Corollary 1.29) Let {Xk , k ≥ 1 } be a sequence of independent random variables with zero means and finite variances. Let Sn = X1 + . . . + Xn , s2n = Var(Sn ). Then the following conditions are equivalent: (i) Sn /sn =⇒ N0,1 ; (ii) SNn /sNn =⇒ N0,1 (mixing) for every sequence {Nn , n ≥ 1} of positive, integer-valued random variables such that P
s2Nn /s2an −→ 1, where an → ∞ are positive integers. In what follows we shall need the following simple consequence of Theorem 2.12.
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T HEOREM 2.34. [99] Let {αn , n ≥ 1} be a non-decreasing sequence of positive random variables and let {τn , n ≥ 1} be a sequence of P
positive, integer-valued random variables such that τn −→ ∞. The following conditions are equivalent: (i) Yυn =⇒ µ (stably) and the sequence {Yn , n ≥ 1} satisfies Anscombe random condition (A? ) with norming sequence {αn , n ≥ 1} and filtering sequence {τn , n ≥ 1}; (ii) YNn =⇒ µ (stably) for every {Nn , n ≥ 1} satisfying (2.10), i.e. P
α2Nn /α2τan −→ 1,
where an → ∞ are constants.
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Theorem 2.34 together with Remark 1.1.30 yields the following generalization of Corollary 2.33. C OROLLARY 2.35. Let {Xk , k ≥ 1 } be a sequence of independent random variables with zero means and finite variances. Let Sn = X1 + . . . + Xn , s2n = Var(Sn ). Then the following conditions are equivalent: (i) Sn /sn =⇒ N0,1 ; (ii) SNn /sNn =⇒ N0,1 (mixing) for every {Nn , n ≥ 1} such that P
s2Nn /s2[λan ] −→ 1 as n → ∞ ,
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where λ is a positive random variable with discrete distribution and {an , n ≥ 1} is a sequence of positive integers with an → ∞.
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CHAPTER 3 Title Page
On a Robbins’ type theorem 3.1.
The classical results
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Let {Xk , k ≥ 1 } be a sequence of indepenent, identically distributed random variables, with a common distribution function F. For each k ≥ 1, set . a = EXk =
Z +∞
. c2 = Var(Xk ) =
Z +∞
X˜k = Xk − a,
x dF(x), −∞
x2 dF(x) − a2 ,
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0 < c2 < +∞.
−∞
Let {υn , n ≥ 1} be a sequence of positive, integer-valued random variables, independent of {Xk , k ≥ 1 }. Assume that the distribution function of υn is well defined by the values pnk , k ≥ 1, where ∞
pnk = P[υn = k],
k ≥ 1;
pnk = 1.
∑
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k=1
Put . αn = Eυn =
∞
k pnk ,
∑
β2n
. = Var(υn ) =
k=1
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∞
∑
k
2
pnk − α2n ,
k=1
. gn (t) = E exp{˙ıt(υn − αn )/βn } =
∞
∑
exp{˙ıt(k − αn )/βn }pnk .
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k=1
. . Under these assumptions on υn , the distribution functions of Sυn = X1 + . . . + Xυn and S˜υn = X˜1 + . . . + X˜υn depend on υn , and ∞
ESυn =
∑
∞
E(I[υn =k] Sk ) =
k=1
∞
∑ (ESk ) pnk = ∑ ak pnk = aαn , k=1
k=1
ES˜υn = E(Sυn − aυn ) = ESυn − aαn = 0, Cov(S˜υn , υn ) = E(υn S˜υn ) − (Eυn )(ES˜υn ) = E(υn S˜υn ) ∞
=
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∞
∑ E(I[υn=k] kS˜k ) =
∑ k(ES˜k ) pnk = 0,
k=1
k=1
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. σ2n = Var(Sυn ) = Var(S˜υn + aυn ) = Var(S˜υn ) + Var(aυn ) + 2 Cov(S˜υn , aυn ) = Var(S˜υn ) + Var(aυn ) Title Page
∞
= E(S˜υn )2 + a2 Var(υn ) =
∑ (ES˜k2) pnk + a2β2n k=1
∞
=
∑ c2k pnk + a2β2n = c2αn + a2β2n ,
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k=1
. fn (t) = E exp{˙ıt(Sυn − ESυn )/σn } = E exp{˙ıt(Sυn − aαn )/σn } ∞
=
∑ k=1 ∞
=
∑
exp{−˙ıtaαn /σn } (E exp{˙ıtX1 /σn })k pnk exp{˙ıta(k − αn )/σn } (E exp{˙ıt(X1 − a)/σn })k pnk
k=1
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υn ˜ = E exp{˙ıta(υn − αn )/σn } ϕ(t/σ , n) where
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. ˜ ˜ = Eeı˙t X1 = Eeı˙t(X1 −a) = ϕ(t)
Z +∞
eı˙tx dF(x + a).
−∞
Due to Robbins [140] we have the following version of the central limit theorem.
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T HEOREM 3.1 (First Robbins Theorem). If σ2n −→ ∞ and
(3.1)
βn = o(σ2n ) as n → ∞ , Full Screen
then for all real t lim fn (t) = lim gn (t sgn(a) dn ) exp{−t 2 (1 − dn2 )/2},
(3.2)
n→∞
n→∞
. where dn = (a2 β2n /σ2n )1/2 = |a|βn /σn = a sgn(a)βn /σn , 0 ≤ dn ≤ 1. 1
P ROOF.
2
Close
It follows from (3.1) that
2 E (υn − αn )/σ2n = β2n /σ4n −→ 0 ( n → ∞ ), so in view of the Chebyshev’s inequality, for every ε > 0, (3.3) P |υn − αn | ≥ εσ2n −→ 0 ( n → ∞ ). 1 sgn(a) = −1, 0 or 1 according as a < 0, a = 0 or a > 0. Obviously, aβ /σ = sgn(a) d . n n n 2 The proof presented is a slightly modified version of the original Robbins’ proof.
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Put αn . ˜ ψn (t) = gn (t sgn(a)dn ) ϕ(t/σ n) αn ˜ . = E exp{˙ıta(υn − αn )/σn } ϕ(t/σ n)
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Then αn υn ˜ ˜ | fn (t) − ψn (t)| = Eeı˙ta(υn −αn )/σn ϕ(t/σ − ϕ(t/σ n) n) υn −αn ˜ ≤ E ϕ(t/σ − 1 ≤ 2P[|υn − αn )| > bσ2n ] n) υn −αn ˜ + E ϕ(t/σ − 1 I[|υn − αn )| ≤ bσ2n ], n) whence (for any given b > 0) (3.4)
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lim | fn (t) − ψn (t)|
n→∞
υn − αn ) υn −αn ˜ − 1 I[| ≤ lim E ϕ(t/σ | ≤ b]. n) n→∞ σ2n
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Since EX1 = a, Var(X1 ) = c2 < ∞ and |eı˙tx − 1 − ı˙tx| ≤ x2 /2,
x ∈ IR (Burrill, p. 334),
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it follows from (3.1) that (3.5)
˜ |ϕ(t/σ n ) − 1| = |
Z +∞ −∞ 2 2
(eı˙tx/σn − 1 − ı˙tx/σn )dF(x + a)| Full Screen
≤ t c /2σ2n −→ 0,
n → ∞.
3.1.1. Some consequences of the First Robbins Theorem. Note that if a = EX1 = 0, then dn = 0 for all n ≥ 1, and the assumption (3.1) reduces to the following one (3.6)
c2 αn −→ ∞ and βn = o(αn )
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as n → ∞ . Quit
Hence and by the First Robbins Theorem we have C OROLLARY 3.2. If a = EX1 = 0 and (3.6) is satisfied, then lim fn (t) = lim E exp{˙ıtSυn /
n→∞
n→∞
p 2 c2 αn } = e−t /2 .
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R EMARK 3.3. In the next subsection we will show that that the assumptions of Corollary 3.2 can be weakened, i.e. instead of (3.6) it will be ? sufficient to assume that P
αn −→ ∞ and
(3.7)
υn /αn −→ 1
as n → ∞ .
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It is easy to see that (3.6) implies (3.7). Indeed, by (3.6) and the Chebyshev’s inequality, for every ε > 0 we have P [|υn − αn | > εαn ] ≤ β2n /ε2 α2n −→ 0
as n → ∞ .
But in (3.7) we do not assume that β2n = Var(υn ) exists (is finite). From the First Robbins Theorem we also immediately get the following results.
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C OROLLARY 3.4. If a = EX1 6= 0 and σn −→ ∞ and
(3.8)
as n → ∞ ,
βn = o(σn )
then lim fn (t) = lim E exp{˙ıt(Sυn − aαn )/σn } = e−t
(3.9)
n→∞
2 /2
n→∞
. Page 58 of 94
P ROOF. Note that (3.8) implies βn = o(σ2n ), i.e. (3.1) holds. It follows from (3.8) that dn = |a|βn /σn −→ 0
as n → ∞ . Go Back
Moreover, denoting Zn := (υn − αn )dn /βn , n ≥ 1, we see that for every n ≥ 1
EZn = 0 and
EZn2 = dn2 −→ 0
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( n → ∞ ).
P
The Chebyshev’s inequality implies that Zn −→ 0. Hence ∞
E exp{˙ıtZn } =
exp{˙ıt(k − αn )dn /βn }pnk
∑
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k=1
= gn (tdn ) −→ 1
as n → ∞ .
Thus, using Theorem 3.1, we get
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lim fn (t) = lim gn (tdn ) exp{−t
n→∞
n→∞
= e−t which proves the desired result (3.9).
2 /2
2
(1 − dn2 )/2}
,
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3.2.
On the limit behaviour of random sums of independent random variables
3.2.1. Introduction and notation. Let (Xnk )n,k∈N be a doubly infinite array (DIA) of random variables such that for every n, the random variables Xnk , k ≥ 1, are independent, let Fnk be the distribution function of Xnk , and let Snk= ∑kj=1 Xn j . We put Z +∞
ank = EXnk =
x dFnk (x),
−∞
σ2nk = Var(Xnk ) =
k
Lnk =
Z +∞ −∞
x2 dFnk (x) − a2nk ,
k
∑
2 = an j , Vnk
∑
j=1
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σ2n j , b2nk = max σ2n j ,
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1≤ j≤k
j=1
while for Ynk = Xnk − ank , k ≥ 1, n ≥ 1, we write ϕnk (t) = E exp{itYnk } =
n
Z +∞ −∞
exp{itx} dFnk (x + ank ),
fnk (t) = ∏ ϕn j (t). j=1
Now let {Nn , n ≥ 1} be a sequence of positive integer-valued random variables such that Nn is for every n independent of Xnk , k ≥ 1. We assume that the distribution function of Nn is determined by the values
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∞
pnk = P[Nn = k], k ≥ 1;
∑
pnk = 1.
k=1 n Under these assumptions on Nn , the distribution function of SnNn = ∑N k=1 Xnk depends on Nn , and
∞
ESnNn =
∞
Lnk pnk = An ,
∑
ELnNn =
k=1 2 = EVnN n
∑ Vnk2 pnk = ρn ,
∑
Lnk pnk ,
k=1
∞
∞
Var(LnNn ) =
k=1
∑
2 Lnk pnk − A2n = ∆2n ,
∑
∞
2 2 Vnk pnk + ∑ Lnk pnk − A2n = ρn + ∆2n = σ2n .
k=1
Close
k=1
Furthermore, let H be a bounded, non-decreasing function such that H(−∞) = 0, 0 ≤ H(x) ≤ 1 for all x, and H(+∞) = 1. We write (exp{itx} − 1 − itx)/x2 for x 6= 0, gt (x) = for x = 0, −t 2 /2 Z (3.10)
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k=1 ∞
Var(SnNn ) =
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+∞
f (t) = exp −∞
gt (x) dH(x) .
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Then f (t) is a characteristic function (ch. f.) of an infinitely divisible distribution with zero mean and unit variance (see, e.g., [108, Theorem 5.5.3]). Contents
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3.3.
A Note on a Katz–Petrov Type Theorem
The note contains a generalization of the well-known Katz–Petrov theorem [123] on the rate of convergence in the central limit theorem. As an application of our result we give nonuniform estimates of the rate of convergence in the random central limit theorem. The results obtained extend and strengthen some results of [4, 21, 31, 101, 120, 121, 146] and [155]. 3.3.1. A Katz–Petrov type theorem. Let {Xk , k ≥ 1} be a sequence of independent random variables variables) such that EXk = 0 and EXk2 = σ2k < ∞ for each k ≥ 1. Put n
Sn =
∑ Xk ,
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n
s2n =
k=1
∑ σ2k . k=1
Let G be the class of functions, defined for all real u, that satisfy the following conditions: (A) g is nonnegative, even, and nondecreasing in the interval (0, ∞); (B) u/g(u) does not decrease in (0, ∞). Note that G contains the functions g(u) = |u|δ , 0 ≤ δ ≤ 1; g(u) = |u|I[|u| < c]+cI[|u| ≥ c], c > 0; and g(u) = c(1−δ)/2 |u|δ I[|u| < c1/2 ]+|u|I[c1/2 ≤ |u| < c] + cI[|u| ≥ c], 0 ≤ δ ≤ 1, c > 0. The Katz–Petrov theorem can be stated (in a general form) as follows (cf. [123] or [124, p. 141]). T HEOREM 3.5. Let {Xk , k ≥ 1} be a sequence of independent random variables such that EXk = 0, EXk2 = σ2k < ∞, k ≥ 1, and let {gn , n ≤ 1} be a sequence of functions from G such that EXk2 gn (Xk ) < ∞ for all 1 ≤ k ≤ n, n ≥ 1. Then there exists a positive universal constant C such that, for all n ≥ 1 and all real x, (3.11)
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n ∆n,x = |P[Sn < xsn ] − Φ(x)| ≤ C ∑ EXk2 gn (Xk ) [s2n gn (sn )] .
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k=1
We are going to prove the following result. Close
T HEOREM 3.6. Let {Xk , k ≥ 1} and {gn , n ≥ 1} be as in Theorem 1. Then there exists a positive universal constant C such that, for all n ≥ 1 and all real x, (3.12)
n ∆n,x ≤ C ∑ EXk2 gn (Xk ) [s2n (1 + |x|)2 gn (sn (1 + |x|))] . k=1
Theorem 3.6 generalizes some results of [21], and at the same time strengthens the main result of [4]. As an application of Theorem 3.6 we give nonuniform estimates of the rate of convergence in the random central limit theorem. But one can easily see that Theorem 3.6 enables us to strengthen some results of [16] and [118], also. We do not, however, go into details here.
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3.3.2. Nonuniform estimates for random sums. It is well known that there are sequences {Xk , k ≥ 1} of independent random variables such that {Sn /sn , n ≥ 1} does not weakly converge, whereas for every x Rn,x = |P[SNn < xsNn ] − Φ(x)| → 0
as n → ∞ ,
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for a sequence {Nn , n ≥ 1} of positive integer–valued random variables independent of {Xk , k ≥ 1} (cf. e.g. Example 2.5 and Example 2.17 in Chapter 2). How fast does in such cases Rn,x tend to 0 ? From Theorem 3.6 we have immediately T HEOREM 3.7. Let {Xk , k ≥ 1} and {gn , n ≥ 1} be as in Theorem 3.5, and let {Nn , n ≥ 1} be a sequence of positive integer–valued random variables independent of {Xk , k ≥ 1}. Then n Nn o (3.13) Rn,x ≤ CE ∑ Xk2 gNn (Xk ) s2Nn (1 + |x|)2 gNn (sNn (1 + |x|)) .
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k=1
The following consequences of Theorem 3.7 are sometimes useful: T HEOREM 3.8 (cf. [118], Corollary 6). Let {Xk , k ≥ 1} be a sequence of independent random variables such that EXk = 0, EXk2 = σ2k , E|Xk |2+δ = β2+δ < ∞, k ≥ 1, for some 0 < δ ≤ 1. Then k Rn,x ≤ CE
n Nn
∑ β2+δ k
k=1
2+δ o sNn (1 + |x|)2+δ ,
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and, for all p > 1/(2 + δ), Z
+∞
p dx Rn,x
−∞
1/p
≤ Cp E
n Nn
∑
k=1
2+δ β2+δ k /sNn
P ROOF. This follows from (3.13) with gn (u) = |u|δ , n ≥ 1.
o
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.
T HEOREM 3.9 (cf. [101], Theorem 1). Let {Xk , k ≥ 1} be a sequence of independent random variables such that EXk = 0, EXk2 = σ2k < ∞, k ≥ 1. Then (3.14) Rn,x ≤ C EΛNn (sNn (1 + |x|)) + ELNn (sNn (1 + |x|)) /(1 + |x|)2 = E{ΨNn (sNn (1 + |x|))}/(1 + |x|)2 , and, for all p > 1/2, (3.15)
Z
+∞
−∞
p Rn,x dx
1/p
≤ C p EΨNn (sNn ) ,
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where 1 Λk (u) = 2 sk Lk (u) =
k
Z
y2 dFX j (y) ,
∑
j=1
1 2 sk u
k
Z
∑
j=1
Ψk (u) = Cu−1
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|y|≥u
|y|3 dFX j (y) ,
|y| 1/(2 + δ), (3.17)
Z
+∞
−∞
p Rn,x dx
1/p
2+δ ∗ ≤ C p E B2+δ , Nn ΨNn (sNn ) sNn
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Nn 2+δ where B2+δ Nn = ∑k=1 βk .
P ROOF. Using (3.13) with
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gn (u) = (sn (1 + |x|))(1−δ)/2 |u|δ I[|u| < (sn (1 + |x|))1/2 ] + |u|I[sn (1 + |x|))1/2 ≤ |u| < sn (1 + |x|)] + sn (1 + |x|)I[|u| ≥ sn (1 + |x|)] , n ≥ 1 ,
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we find that Rn,x ≤
n C 2+δ −(1−δ)/2 E (B2+δ Nn /sNn )(sNn (1 + |x|)) 2+δ (1 + |x|) +
+
1
Nn
|y|≥(sNn (1+|x|))1/2
Nn
Z
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o |y|2+δ dFXk (y)
∑
s2+δ Nn k=1
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|y|2+δ dFXk (y)
∑
s2+δ Nn k=1 1
Z
|y|≥sNn (1+|x|)
2+δ ∗ = E (B2+δ /s )Ψ (s (1 + |x|)) (1 + |x|)2+δ , N Nn n Nn Nn where Ψ∗k (u) = C
n k −(2+δ) −(1−δ)/2 u + Bk ∑
Z
j=1
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|y|≥u1/2
−(2+δ)
+ Bk
|y|2+δ dFX j (y)
k
Z
∑
j=1
o |y|2+δ dFX j (y) .
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|y|≥u
−(1−δ)/2
Furthermore, one can easily see that, for every k ≥ 1, Ψ∗k is bounded by C(σ1 + 2), nonincreasing on [σ1 , ∞], and that limu→∞ Ψ∗k (u) = 0. The proof of Theorem 3.11 is complete as (3.17) is an immediate consequence of (3.16). R EMARK 3.12. One can easily see that in the special case where Nn = n almost surely, n ≥ 1, Theorem 3.11 strengthens Corollary 2 to Theorem 4 of [21] and at the same time generalizes to nonidentically distributed random variables the Corollary of [120]. 2+δ 2+δ 2+δ T HEOREM 3.13. Let {Xk , k ≥ 1} be as in Theorem 3.11, and suppose that B2+δ Nn /sNn = O E(BNn /sNn ) with probability 1 as n → ∞, and that Z 1 Nn 2+δ |y| dFXk (y) → 0 as n → ∞ , E ∑ B2+δ k=1 Nn |y|≥sNn /ϕn
where {ϕn , n ≥ 1} is a sequence of positive numbers going to infinity. Then 2+δ sup(1 + |x|)2+δ Rn,x = o (E(B2+δ Nn /sNn )) as x
n→∞.
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P ROOF. Only some modifications are necessary in the proof of Theorem 3.11 to make it applicable in this case: put gn (u) = sn (1 + |x|)/ϕn )1−δ |u|δ I[|u| < sn (1 + |x|)/ϕn ] + |u|I[sn (1 + |x|)/ϕn ≤ |u| < sn (1 + |x|)] + sn (1 + |x|)I[|u| ≥ sn (1 + |x|)] ,
n≥1.
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The following corollaries generalize some results of [31] and [121]. C OROLLARY 3.14. Let {Xk , k ≥ 1} be a sequence of independent and identically distributed distributed ) random variables such that EX1 = 0, EX12 = 1, and let gn ∈ G be such that EX12 gn (X1 ) < ∞, n ≥ 1. Then there exists a positive universal constant C such that rn,x = |P[SNn < xsNn ] − Φ(x)| √ ≤ CE X12 gNn (X1 )/[(1 + |x|)2 gNn ( Nn (1 + |x|))] . R EMARK 3.15 (cf. [121]). Obviously, if gn = g, n ≥ 1, where g ∈ G is such that EX12 g(X1 ) < ∞ and limu→∞ g(u) = ∞, and if for some C > 0 P[Nn < C/(εn (1 + |x|))] = O 1/g ((1 + |x|)/εn )1/2 ,
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where {εn , n ≥ 1} is a sequence of positive numbers with εn → 0 as n → ∞, then (1 + |x|)2 rn,x = O 1/g (C1 (1 + |x|)/εn )1/2 , Go Back
where C1 = min(C, 1). C OROLLARY 3.16. Let {Xk , k ≥ 1} be a sequence of independent and identically distributed random variables such that EX1 = 0, EX12 = 1. Then there exists a bounded and nonincreasing function Ψ on (0, ∞) with limu→∞ Ψ(u) = 0 such that √ rn,x ≤ EΨ( Nn (1 + |x|))/(1 + |x|)2 , and, for all p > 1/2, Z
+∞
−∞
p rn,x dx
1/p
≤ C p EΨ
√ Nn .
C OROLLARY 3.17. Let {Xk , k ≥ 1} be a sequence of independent and identically distributed random variables such that EX1 = 0, EX12 = 1, E|X1 |2+δ < ∞ for some 0 < δ < 1. Then there exists a bounded and nonincreasing function Ψ1 on [1, ∞) with limu→∞ Ψ1 (u) = 0 such that n o √ δ/2 (3.18) rn,x ≤ E Ψ1 ( Nn (1 + |x|))/[Nn (1 + |x|)2+δ ] , and, for all p > 1/(2 + δ), (3.19)
Z
+∞
−∞
p rn,x dx
1/p
√ δ/2 ≤ C p E Ψ1 Nn /Nn .
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−δ/2
−δ/2
C OROLLARY 3.18. Let {Xk , k ≥ 1} be as in Corollary 3.17, and suppose that Nn = O(ENn P Nn −→ ∞ (in probability) as n → ∞. Then −δ/2 sup(1 + |x|)2+δ rn,x = o (ENn ) ,
Title Page
) with probability 1 as n → ∞, and that
Contents
x
and for all p > 1/(2 + δ), Z
+∞
−∞
p rn,x dx
1/p
−δ/2
= o (ENn
).
P ROOF. This immediately follows from (3.18), (3.19), and the assumptions of the corollary.
C OROLLARY 3.19. Let {Xk , k ≥ 1} be a sequence of independent and identically distributed random variables such that EX1 = 0, EX12 = 1, E|X1 |3 < ∞. Then −1/2 sup(1 + |x|)3 rn,x = O(ENn ) ,
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x
and for all p > 1/3, Z
+∞
−∞
p rn,x dx
1/p
−1/2
= O(ENn
).
R EMARK 3.20. We note that [155] gives a similar order of approximation for the weak convergence in the random central limit theorem. But the order of approximation deduced for the weak convergence cannot in general be transferred to the associated strong convergence in distribution (see e.g. [28]). 3.3.3.
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Proof of Theorem 3.6. The proof is based on the following auxiliary result.
P ROPOSITION 3.21. Let Xk , 1 ≤ k ≤ n, be independent random variables such that EXk = 0, EXk2 = σ2k < ∞, 1 ≤ k ≤ n, and let g = gn,x ∈ G be such that EXk2 g(Xk ) < ∞ for all 1 ≤ k ≤ n. Then there exists a positive universal constant C1 such that n (3.20) ∆n,x ≤ C1 ∑ EXk2 g(Xk ) [s2n x∗2 g(sn (1 + |x|))] ,
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k=1
where x∗ = max(1; |x|).
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P ROOF. In the proof we use the ideas of [4, 21] and [166] (see also [124, pp. 141–144]), therefore we indicate only the changes that should be made in our case. Observe first that for all x ∆n,x ≤ 1, x2 ∆n,x ≤ 1 ([124, p. 151]), whence ∆n,x ≤ x∗−2 for all x. Moreover, if for some x n
∑ EXk2g(Xk )/[s2ng(sn(1 + |x|))] ≥ 10−1 k=1
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then
n ∆n,x ≤ x∗−2 ≤ 10 ∑ EXk2 g(Xk ) [s2n x∗2 g(sn (1 + |x|))] , k=1 Contents
i.e. (3.20) is trivially true in this case with C1 = 10. It remains to show (3.20) for the case n (3.21) ∑ EXk2g(Xk ) [s2ng(sn(1 + |x|))] < 10−1 k=1
with arbitrarily fixed x. For j = 1, . . . , n, define the truncated r.v. by Xn j = X j if |X j | < sn (1 + |x|), and Xn j = 0 otherwise. Let
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an j = EXn j = EX j I[|X j | < sn (1 + |x|)] , c2n j = EXn2j − E 2 Xn j = EX j2 I[|X j | < sn (1 + |x|)] − a2n j , n
An =
n
∑ an j ,
j=1
Vn2
=
∑ c2n j .
j=1
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Then it follows from elementary calculations that (3.22)
|an j | ≤ EX j2 g(X j )/[sn x∗ g(sn (1 + |x|))] , j = 1, . . . , n ,
(3.23)
0 ≤ s2n −Vn2 ≤ 2 ∑ EX j2 g(X j )/g(sn (1 + |x|)) .
n
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j=1
Furthermore, we note that n
n
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Vn2 = s2n − ∑ EX j2 I[|X j | ≥ sn (1 + |x|)] − ∑ a2n j j=1
j=1
n ≥ s2n 1 − ∑ EX j2 g(X j )/[s2n g(sn (1 + |x|))] −
j=1 n
∑ E|X j |I[|X j | ≥ sn(1 + |x|)]/sn
2
j=1
9 n 2 ≥ s2n − ∑ EX j2 g(X j ) [s2n (1 + |x|)g(sn (1 + |x|))] %Bigr) 10 j=1 ≥
89 2 s , by (3.21). 100 n
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Hence 89 2 s ≤ Vn2 ≤ s2n . 100 n
(3.24)
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As in the above–mentioned papers, we obtain (3.25)
h n i ∆n,x ≤ P ∑ (Xn j − an j ) < yVn − Φ(y)
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j=1
n
+ |Φ(y) − Φ(x)| + ∑ P[|X j | ≥ sn (1 + |x|)] j=1
= Tn,1 + Tn,2 + Tn,3 , say , where Page 68 of 94
y = (xsn − An )/Vn .
(3.26) By Chebyshev’s inequality we have
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n
(3.27)
Tn,3 =
∑ P[|X j | ≥ sn(1 + |x|)]
j=1 n
≤
∑ EX j2g(X j )/[s2n(1 + |x|)2g(sn(1 + |x|))]
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j=1 n
≤
∑ EX j2g(X j )/[s2nx∗2g(sn(1 + |x|))] .
Close
j=1
Next we observe that (3.28)
Tn,2 ≤ |Φ(y) − Φ(xsn /Vn )| + |Φ(xsn /Vn ) − Φ(x)| −1/2
≤ (2π)
xsZn /Vn 2 Zy −t 2 /2 e dt + e−t /2 dt . xsn /Vn
x
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Furthermore, by (3.23) and (3.24), we have Z xsn /Vn 2 2 −t /2 (3.29) e dt ≤ |x|e−x /2 (sn /Vn − 1) x
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−x2 /2
= |x|e (s2n −Vn2 )/[Vn (sn +Vn )] 89 −1/2 n 2 |x|e−x /2 ≤2 EX j2 g(X j )/[s2n g(sn (1 + |x|))] 100 j=1 89 −1/2 n 2 |x|x∗2 e−x /2 EX j2 g(X j )/[s2n x∗2 g(sn (1 + |x|))] =2 100 j=1 89 −1/2 n (3/e)3/2 EX j2 g(X j )/[s2n x∗2 g(sn (1 + |x|))] , ≤2 100 j=1
∑
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∑
∑
and, by (3.22) and (3.24), putting z = min(|x|sn /Vn ; |y|),
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Zy 2 −t 2 /2 e dt ≤ e−z /2 |An /Vn |
(3.30)
xsn /Vn −z2 /2
≤e
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n
∑
EX j2 g(X j )/[Vn sn x∗ g(sn (1 + |x|))]
j=1
89 −1/2 n 2 x∗ e−z /2 ∑ EX j2 g(X j )/[s2n x∗2 g(sn (1 + |x|))] 100 j=1 89 −1/2 n ≤ ∑ EX j2g(X j )/[s2nx∗2g(sn(1 + |x|))] 100 j=1
≤
−z2 /2
since x∗ e
Close
≤ 1 (this inequality is obvious for |x| ≤ 1; if |x| > 1, then we observe that |y| = |xsn /Vn ||1 − An /(xsn )| ≥ |x||1 − An /(xsn )| ≥
by (3.21) and (3.22), which implies that z ≥
9 10 x∗ ).
Combining (3.28)–(3.30) we find that n
(3.31)
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Tn,2 ≤ C2 ∑ EX j2 g(X j )/[s2n x∗2 g(sn (1 + |x|))] , j=1
9 |x| 10
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where C2 = (1 + 2(3/e)3/2 )/(1, 78π)1/2 < 1.4035. Finally, applying Bikelis’ theorem [21, Theorem 2] to the sequence {(Xn j − an j ), 1 ≤ j ≤ n} of truncated random variables, we obtain h n i Tn, j = P ∑ (Xn j − an j ) < yVn − Φ(y) (3.32)
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j=1 n
≤ C3 ∑ E|Xn j − an j |3 /[Vn3 (1 + |y|)3 ] . j=1
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But for j = 1, . . . , n, E|Xn j − an j |3 ≤ 4(E|Xn j |3 + |an j |3 ) ≤ 8E|Xn j |3 ≤ 8sn (1 + |x|)EX j2 g(X j )/g(sn (1 + |x|)) . Furthermore, using (3.26), we have |x| = |(yVn + An )/sn | ≤ |y| + |An /sn |
as Vn /sn ≤ 1 , Page 70 of 94
n
≤ |y| + ∑
EX j2 g(X j )/[s2n x∗ g(sn (1 + |x|))]
by (3.22),
j=1
≤ |y| + 10−1 ,
by (3.21),
whence x∗2 (1 + |x|)/(1 + |y|)3 ≤ x∗2 /(1 + |y|)2 + x∗2 /[10(1 + |y|)3 ] ≤
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11 . 10
Consequently,
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n
Tn,1 ≤ C3 ∑ E|Xn j − an j |3 /[Vn3 (1 + |y|)3 ] j=1
n
Close
≤ 8C3 sn (1 + |x|) ∑ EX j2 g(X j )/[Vn3 (1 + |y|)3 g(sn (1 + |x|))] j=1
89 −3/2 n ≤8 C3 (x∗2 (1 + |x|)/(1 + |y|)3 ) ∑ EX j2 g(X j )/[s2n x∗2 g(sn (1 + |x|))] 100 j=1 89 −3/2 n ≤ 8.8 C3 ∑ EX j2 g(X j )/[s2n x∗2 g(sn (1 + |x|))], 100 j=1 and combining this result with (3.25), (3.27) and (3.31) we get the desired conclusion (3.20) with C1 = 1 +C2 + 8.8(89/100)−3/2C3 .
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Note. Proposition 3.21 strengthens Theorem 2.1 of [4]. P ROOF OF T HEOREM 3.6. It follows from Proposition 3.21 that for all x n
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∆n,x ≤ C1 ∑ EX j2 gn (X j )/[s2n gn (sn (1 + |x|))] , j=1 n
x2 ∆n,x ≤ C1 ∑ EX j2 gn (X j )/[s2n gn (1 + |x|))] . j=1
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Consequently, n
(1 + x2 )∆n,x ≤ 2C1 ∑ EX j2 gn (X j )/[s2n gn (1 + |x|))] , j=1
and it is easily seen that n
∆n,x ≤ 4C1 ∑ EX j2 gn (X j )/[s2n (1 + |x|)2 gn (sn (1 + |x|))] . j=1
The proof of Theorem 3.6 is complete.
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R EMARK 3.22. We note that Theorem 3.6 can also be deduced in a different way from Theorem 4 of [21] (cf. e.g. [165]). Go Back
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Page 72 of 94
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CHAPTER 4 Title Page
Weak Convergence to Infinitely Divisible Laws Let (Xnk ) (k ≥ 1, n ≥ 1) be a doubly infinite array (DIA) of random variables (r.v.’s) defined on a common probability space (Ω, F, P). Assume that (Xnk ) is adapted to an array (Fnk ) (k ≥ 0, n ≥ 1) of row-wise increasing sub-σ-fields of F, i.e. Xnk is Fnk -measurable and Fn,k−1 ⊆ Fnk . Fn0 / Ω}. need not be the trivial σ-field {0, Now let {Nn , n ≥ 1} be a seqence of positive integer-valued r.vs defined on the same probability space (Ω, F, P). Let us denote Nn
SnNn =
∑ Xnk ,
Nn
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ϕnk (t) = E(exp(˙ıtXnk )|Fn,k−1 ), fnNn (t) = ∏ ϕnk (t), k=1
k=1
Recently, several papers have appeared which are devoted to the study of the limit distribution of SnNn as n → ∞ . There have been two basic problems on the filed. One is when a limit theorem is already given and the question is about the optimal conditions ensuring that the same theorem, or a mixtured version of it, remain true with random indices (see, e.g. [169], [141], [47], and their references). The other problem is to prove directly random-sums limit theorems, and to determine the class of possible limit distributions. Of course, the conditions ensuring the random limit theorem ought to be reduced to the classical ones when Nn = n. However, there are (Xnk ) such that SnNn satisfies the random limit theorem whereas Snn does not weakly converge (see, e.g. [149, Example 1]). The latter problem has been considered by many authors. The case when Nn , Xn1 , Xn2 , . . . are independent r.v.’s has been investigated in [169], [141] and [149], while the case when Nn are stopping times has been investigated in [148], [20] and [80] (see also references in these papers). A general case, when Nn need not be (for every n) independent of Xnk or a stopping time, has been considered in [97] under the assumption that Xnk have finite conditional variances and in [90, 92] without this assumption. In the sequel we will abuse notation slighlty in the interes of brevity denoting E(Y |Fn,k−1 ) by Ek−1Y , where Y is some variable (e.g., = Xnk ) taken from the n-th row1. Throughout, I(A) denotes the indicator function of the set A, and the various kinds of convergence, with probability 1 a.s.
L
P
Page 73 of 94
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r (almost sure ), in Lr -norm, in probability, and weak (in distribution) are denoted by −−→ , −→ , −→ and =⇒ , respectively. All equalities and inequalities between r.v.’s are considered in the sense “with probability one”, and all limits are taken as n → ∞ , unless stated otherwise.
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4.1.
The case of finite variances
(Xnk ) is called a martingale doubly infinite array (MDIA) if E|Xnk | < ∞, and E(Xnk |Fn,k−1 ) = 0 almost surely (a.s.) for all k ≥ 1 and n ≥ 1. Let us denote 2 σ2nk = E(Xnk |Fn,k−1 ), VN2n =
Nn
∑ σ2nk , k=1
1Hence, we will denote P(A |F nk n,k−1 ) by Pk−1 (Ank ).
bNn = max σ2nk . 1≤k≤Nn
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4.1.1. The main results. To begin with considerations we see that if Nn is Fn0 -measurable and Fn0 is the trivial σ-field, then there exists P a seqence {kn , n ≥ 1} of positive integer numbers such that P(Nn = kn ) = 1 for all n ≥ 1. So, if Nn −→ ∞, then kn −→ ∞. On the other hand, if P
Nn , Xn1 , Xn2 , . . . are independent random variables, then E fnNn (t) = E exp(˙ıtSnNn ). Thus2 for a fixed real t, fnNn (t) −→ A(t) as n → ∞ , if and only if E exp(˙ıtSnNn ) −→ A(t) as n → ∞ , since | fnNn (t)| ≤ 1. The following routine lemma will be useful throughout this chapter. L EMMA 4.1.. ([97]) Let (Xnk ) be a DIA of r.vs adapted to an array (Fnk ) of row-wise increasing sub-σ-fields, and let {Nn , n ≥ 1} be a sequence of positive integer-valued random variable such that
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P
fnNn (t) −→ A(t) as n → ∞ ,
(4.1) for some t, implies (4.2)
∗ ∗ E exp(˙ıtSnN )/( fnN (t))−1 −→ 1 as n → ∞ , n n
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where A(t) is a complex number such that |A(t)| 6= 0, and ∗ = SnN n
Nn
∑ Xnk∗ , k=1
Nn
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∗ ∗ fnN (t) = ∏ Ek−1 exp(˙ıtXnk ), n k=1
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∗ Xnk
. = Xnk I[| fnk (t)| ≥ |A(t)|//2].
Then, under (4.1), E exp(˙ıtSnNn ) −→ A(t) as n → ∞ .
Close
. P ROOF. For each fixed n, let Ank = {| fnk (t)| ≥ |A(t)|//2}. Note that An0 = Ω ⊇ An1 ⊇ An2 . . . and Acnk ∩ Ans = 0/ for k ≤ s. Furthermore, Ω = Anm + Acnm = Anm + (Acnm ∩ An,m−1 ) + Acn,m−1 = . . . = Anm + (Acn,m ∩ An,m−1 ) + . . . + (Acn2 ∩ An1 ) + (Acn1 ∩ An0 ).
2Other cases will be discussed later on
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Then ∗ | fnN (t)| = n
Nn
∏ |Ek−1 exp(˙ıtXnk∗ )| =
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k=1 Nn
= ∏ |I(Ank )Ek−1 exp(˙ıtXnk ) + I(Acnk )| = k=1 Nn
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= [I(Ank )|ϕnk (t)| + I(Acnk )] k=1 Nn −1 s Nn
∏
= ∏ |ϕnk (t)|I(AnNn ) + k=1
Nn
×
∏
∑ ∏ |ϕn j (t)|I(An j ∩ Ans ∩ Acn,s+1) ×
s=1 j=1
I(Acn j ∩ Ans ∩ Acn,s+1 ) + I(Acn1 )
j=s+1
≥
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|A(t)| I(AnNn ) + 2
Nn −1 s
∑ ∏ |ϕn j (t)|I(Ans ∩ Acn,s+1) ×
s=1 j=1
|At | c I(An1 ) 2 ≥ |A(t)|//2, a.s. for all n,
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× I(Ans ∩ Acn,s+1 ) +
and, by (4.1),
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P
Nn [
! {Xnk 6= Ynk }
Nn [
=P
k=1
! Acnk
= P(AcnNn ) −→ 0 as n → ∞ .
k=1 Close
Moreover, by (4.1), P
Nn [
! ∗ { fnk (t) 6= fnk (t)}
=P
Nn [ k [
! Acn j
k=1 j=1
k=1
as n → ∞ , and then (4.10 )
P
∗ fnN (t) −→ At as n → ∞ . n
∗ ) −→ A(t) as n → ∞ . Thus, it suffices to show that E exp(˙ıtSnN n
= P(AcnNn ) −→ 0 Quit
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Now we have ∗ ∗ ∗ |E exp(˙ıtSnN ) − A(t)| ≤ |E exp(˙ıtSnN ){1 − A(t)( fnN (t))−1 |+ n n n ∗ ∗ + |A(t)| · |E exp(˙ıtSnN )( fnN (t))−1 − 1| ≤ n n
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∗ ≤ (2//|A(t)|)E| fnN (t) − A(t)| + n ∗ ∗ + |A(t)| · |E exp(˙ıtSnN )( fnN (t))−1 − 1| n n
−→ 0 as n → ∞ , by (4.10 ), (4.2) and the dominated convergence theorem [107, p. 125].
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R EMARK 4.2.. Suppose that fnk (t) = fk (t) for all n and k. (Put e.g. Xnk = Xk //Vn (k ≥ 1, n ≥ 1), where Vn2 = ∑nj=1 E(X j2 |F j−1 ), and a.s.
P
fn (t) = ∏nk=1 ϕk (t//Vn ), n ≥ 1.) Note that if fn (t)−−→A(t) as n → ∞, where |A(t)| 6= 0, and if Nn −→ ∞ (as n → ∞), then (4.1) holds (see e.g. P
[40, Lemma 2]). Taking into account that in our case fn (t) −→ A(t) is eqivalent to the almost sure convergence (since P(lim inf Ankn ) = 1), we n→∞
P
P
see that (4.1) takes place if for instance fnkn (t) −→ A(t), kn → ∞, and Nn −→ ∞.
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Now we are going to extend one of the results of [27] and [149]. We will assume that (Xnk ) is a MDIA such that there exists a finite constant C for which lim P(VN2n > C) = 0,
(4.3)
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n→∞
and P
b2nNn −→ 0 as n → ∞ .
(4.4)
P ROPOSITION 4.3.. ([97]) Let {Nn } be a sequence of positive integer-valued random variables and suppose (Xnk ) is a MDIA such that (4.3) and (4.4) hold. Then Z +∞ −2 P ı˙tx fnNn (t) −→ exp e − 1 − ı˙tx x dG(x) as n → ∞ , −∞
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where G is a bounded nondecreasing function, if and only if for all continuity points a, b of G Nn
(4.5)
P
∑ Ek−1Xnk2 I(a < Xnk ≤ b) −→ G(b) − G(a) as n → ∞ . k=1
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P ROOF. We may assume that (4.6)
VN2n ≤ C a.s. for all n ∈ N.
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4.2.
The case of not necessarily finite variances
The aim of the present paper is to consider the general case without any assumptions about the existence of moments of the r.v.’s Xnk . We put Z +∞ ı˙tx 1 + x2 (4.7) A(t) = exp ı˙tγ + exp(˙ıtx) − 1 − dK(x) , 1 + x2 x2 −∞ where γ is a fixed real number, and K is a bounded and nondecreasing real function such that K(−∞) = 0. Then A is the characteristic function (ch.f.) of an infinitely divisible distribution in the Lévy - Khintchine representation (see, e.g. [108]). The following routine lemma will be useful throughout this paper. L EMMA 4.4 ([97]). Let (Xnk ) be a DIA of r.v.’s adapted to an array (Fnk ) of row-wise increasing sub-σ-fields, and let {Nn , n ≥ 1} be a sequence of positive integer-valued r.v.’s such that Nn
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P
fnNn = ∏ Ek−1 exp(˙ıtXnk ) −→ A(t),
(4.8)
k=1
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for some t, implies ? ? −1 E exp(˙ıtSnN )( f ) −→ 1, nN n n
(4.9) where ? SnN = n
Nn
∑ Xnk? , k=1
Nn
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? ? fnN (t) = ∏ Ek−1 exp(˙ıtXnk ) n k=1
and ? Xnk = Xnk I(| fnk (t)| ≥ |A(t)|/2).
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Then, under (4.8), E exp(˙ıtSnNn ) −→ A(t). R EMARK 4.5. One can easily prove that (4.8) implies (4.9) for example in the following cases: / Ω}, Fnk = σ{Xn1 , Xn2 , . . . , Xnk }; a: Nn is for every n independent of {Xnk , k ≥ 1}, and Fn0 = {0, b: Nn is for every n a stopping time w.r.t. {Fnk , k ≥ 1};
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c: Xnk = Xk for every k and all n, and Nn −→ ∞; d: P(kn ≤ Nn ≤ rn ) −→ 1 and maxkn ≤k≤rn α(Fnk , Ank ) −→ 0, where / Ω, {Nn = k}, {Nn 6= k}} , kn ≤ rn are positive integers, and Ank = {0, α(A1 , A2 ) = sup{|P(B ∩C) − P(B)P(C)| : B ∈ A1 , C ∈ A2 }. ? )( f ? (t))−1 − 1, k ≥ 1, forms a uniformly bounded The proof of Remark 4.5 is based on the fact that for every n the sequence Zkn = exp(˙ıtSnk nk n martingale w.r.t. {Fnk , k ≥ 1} such that EZk = 0, and is not deatiled here.
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4.2.1.
The main results. Let (τnk ) (k ≥ 1, n ≥ 1) be a DIA of nonnegative r.v.’s such that |Ek−1 Xnk I(|Xnk | ≤ τnk )| < ∞ for every k and all n.
We put
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ank = Ek−1 Xnk I(|Xnk | ≤ τnk ), Ynk = Xnk − ank . What we consider to be the basic random limit theorem for partial sums of dependent r.v.’s, can now be formulated as follows. T HEOREM 4.6.. Let (Xnk ) and {Nn , n ≥ 1} be as in Lemma 4.4, and assume that (4.10)
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P
max Pk−1 (|Ynk | ≥ ε) −→ 0, for all ε > 0,
1≤k≤Nn
(4.11)
2 Ynk P ∑ Ek−1 1 +Y 2 −→ 0, k=1 nk
(4.12)
nk I(Ynk ≤ y) −→ K(y) ∑ Ek−1 1 +Y 2
Nn
Nn
k=1
Y2
P
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nk
for every point y of continuity of K and for y = ±∞, and Nn
(4.13)
Ynk P ∑ ank + Ek−1 1 +Y 2 −→ γ. k=1 nk
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Then SnNn weakly converges to an infinitely divisible distribution, whose ch.f. A is given by (4.7). Theorem 4.6 extends the main result of [97], Theorem 1, to partial sums of dependent r.v.’s such that their frist moments need not exist. In the special case when the DIA (Xnk ) is “conditionaly infinitesimal”, i.e. (4.14)
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P
max Pk−1 (|Xnk | ≥ ε) −→ 0, for all ε > 0,
1≤k≤Nn
and τnk = τ, where τ is a positive constant, Theorem 4.6 yields
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T HEOREM 4.7.. Let (Xnk ) and {Nn , n ≥ 1} be as in Lemma 4.4, and assume that (4.12), (4.13) and (4.14) are satisfied. Then SnNn weakly converges to an infinitely divisible distribution, whose ch.f. is given by (4.7). One cen see that in the special case, when {Nn , n ≥ 1} is a sequense of contans and for every n the r.v.’s Xn1 , Xn2 , . . . are independent, (4.12) and (4.13) with τnk = τ are just the conditions that are necessary and sufficient in order that SnNn =⇒ X, where X has ch.f. A (see, e. g. [107, p.309]). In the case when for every n the r.v.’s Nn , Xn1 , Xn2 , . . . are independent Theorem 4.7 reduces to Theorem 6.1 of [141]. Moreover, under the assumption that {Nn , n ≥ 1} is a sequence of stopping times Theorem 4.7 is the 1-dimensional analogy of Theorem 1 from [20]. Of course, in all these cases (4.8) implies (4.9).
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As simple conseqences of Theorem 4.6 we also get: C OROLLARY 4.8.. Let (Xnk ) and {Nn , n ≥ 1} be as in Lemma 4.4, and assume that (4.10), (4.11) and (4.13) are satisfied, and that Nn
(4.15)
∑ k=1
Ek−1
2 Ynk P −→ D, 2 1 +Ynk
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where D is a positive number, and (for every real y) Nn
(4.16)
2 Ynk D P D E ∑ k−1 1 +Y 2 I(Ynk ≤ y) −→ 2 + π arc tg(y). k=1 nk
Then SnNn weakly converges to the Cauchy distribution with ch.f. A(t) = exp(˙ıtγ − D|t|).
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C OROLLARY 4.9.. Let (Xnk ) and {Nn , n ≥ 1} be as in Lemma 4.4, and assume that (4.10), (4.11), (4.13) and (4.15) are satisfied with γ = D, and that Nn
(4.17)
2 Ynk P ∑ Ek−1 1 +Y 2 I(|Ynk − 1| ≥ ε) −→ 0, for all ε > 0. k=1 nk
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Then SnNn weakly converges to the Poisson distribution with parameter λ = 2D. C OROLLARY 4.10.. Let (Xnk ) and {Nn , n ≥ 1} be as in Lemma 4.4, and assume that (4.11), (4.13) and (4.15) are satisfied, and that Nn
(4.18)
∑
Ek−1
k=1
2 Ynk P I(|Ynk | ≥ ε) −→ 0, for all ε > 0. 2 1 +Ynk
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Then {SnNn } weakly converges to the normal distribution N(γ, D) with mean γ and variance D. It is easy to verify that (4.18) is equivalent to Nn
(4.19)
P
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∑ Pk−1(|Ynk | ≥ ε) −→ 0, for all ε > 0. k=1
Now we see that in the special case, when {Nn , n ≥ 1} is a seqence of contans and τnk = 0, Corollary 4.10 reduces to Theorem 2.3 of [52]. From Corollary 4.10 one can deduce the following random version of the “normal convergence criterion” (cf. [107, p. 316]). C OROLLARY 4.11.. Let (Xnk ) and {Nn , n ≥ 1} be as in Lemma 4.4, and assume that Nn
(4.20)
P
∑ Pk−1(|Xnk | ≥ ε) −→ 0, for all ε > 0. k=1
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Nn
(4.21)
∑
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P
Ek−1 Xnk I(|Xnk | ≤ τ) −→ γ,
k=1 Nn
(4.22)
∑
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P 2 Ek−1 Xnk I(|Xnk | ≤ τ) − E2k−1 Xnk I(|Xnk | ≤ τ) −→ D,
k=1
where τ > 0 is finite and arbitrarily fixed. Then SnNn =⇒ N(γ, D). In the special case, when {Nn , n ≥ 1} is a seqence of stopping times, Corollary 4.11 reduces to Theorem 2.1 of [80]. 4.2.2.
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Proofs. In order to prove Theorem 4.6 we need the following auxiliary results.
L EMMA 4.12.. (4.10) holds if and only if dnNn = max Ek−1
(4.23)
1≤k≤Nn
2 Ynk 2 1 +Ynk
P
−→ 0.
Furthermore, (4.10) implies that
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(4.24)
bnNn
P Y nk −→ 0. = max Ek−1 2 1≤k≤Nn 1 +Ynk Go Back
Proof. Obvious. P ROPOSITION 4.13.. Let (Xnk ) be a DIA of r.v.’s adapted to an array (Fnk ) of row-wise increasing sub-σ-fields, and let {Nn , n ≥ 1} be a sequence of positive integer-valued r.v.’s satisfying (4.10), (4.11), (4.12) and (4.13). Then (4.8) holds.
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Proof. For every t let us define ht (x) =
( 1+x2 ı˙tx exp(˙ıtx) − 1 − 1+x 2 x2 1 2 2t
for x 6= 0, for x = 0.
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The function ht is continuous and bounded (i.e. there exists a constant L = Lt > 0 such that |ht (x)| ≤ Lt for all x). Thus, (4.25)
max |Ek−1 exp(˙ıtYnk ) − 1| = 2 Y Y nk nk ≤ = max ı˙tEk−1 + Ek−1 ht (Ynk ) 2 2 1≤k≤Nn 1 +Ynk 1 +Ynk
1≤k≤Nn
P
≤ |t|b2nNn + Lt dnNn −→ 0, as (4.23) and (4.24) hold.
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Furthermore, from the Taylor series it is seen that Nn
Nn
k=1
k=1
∑ | log(1 + znk ) − znk | ≤ ∑ |znk |2 .
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This with znk = Ek−1 exp(˙ıtYnk ) − 1, by (4.25), yields for fixed t Nn
Nn
k=1 Nn
k=1
|Atn | = | ∑ log Ek−1 exp(˙ıtYnk ) − ∑ [Ek−1 exp(˙ıtYnk ) − 1]| 2 Ynk ≤ ∑ |Ek−1 exp(˙ıtYnk ) − 1| ≤ t ∑ Ek−1 2 1 +Ynk k=1 k=1 2
+(2|t|Lt bnNn + Lt2 dnNn )
2
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Nn
Nn
2 Ynk P ∑ Ek−1 1 +Y 2 −→ 0, k=1 nk Page 81 of 94
as (4.10), (4.11) and (4.12) with y = +∞ hold. Since Nn
Nn 2 Ynk Ynk log fnNn (t) = ı˙t ∑ ank + Ek−1 + + Atn , E h (Y ) t k−1 nk ∑ 2 2 1 +Y 1 +Y k=1 k=1 nk nk
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then for the proof of (4.8) it is sufficient to show that (4.26)
Rtn
Nn
Y2 P = ∑ Ek−1 ht (Ynk ) nk 2 −→ 1 +Ynk k=1
Z +∞ −∞
ht (x) d K(x) .
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For arbitrary positive ε1 , ε2 , and ε3 , choose an integer m sufficiently large and subdivision x0 < x1 < . . . < xm , all continuity points of K, so that
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Z +∞ m |B(m)| = ∑ ht (x j−1 ) K(x j ) − K(x j−1 ) − ht (x) dK(x) < ε1 , j=1 −∞ Quit
max |x j − x j−1 | < ε2 ,
1≤ j≤m
|ht (x j ) − ht (x j−1 )| < ε3 ,
and (4.27)
K(x0 ) + K(+∞) − K(xm ) < ε3 .
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(We recall that K is a bounded and nondecreasing function such that K(−∞) = 0, and that x0 → −∞ and xm → +∞ as m → ∞.) Then, it follows that Nn m Y2 |Cn (m)| = ∑ ∑ Ek−1 ht (Ynk ) nk 2 I(x j−1 < Ynk ≤ x j )− k=1 j=1 1 +Ynk Nn m 2 Ynk − ∑ ∑ ht (x j−1 )Ek−1 I(x < Y ≤ x ) j ≤ j−1 nk 2 1 +Ynk k=1 j=1 Nn
≤ ε3 ∑ Ek−1 k=1
2 Ynk , 2 1 +Ynk
Nn 2 Y |Dn (m)| = Rtn − ∑ Ek−1 ht (Ynk ) nk 2 I(x0 < Ynk ≤ xm ) ≤ 1 +Y k=1 nk
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Nn
Nn 2 2 Ynk Ynk I(Y ≤ x ) + E 0 nk ∑ k−1 1 +Y 2 − 2 1 +Ynk k=1 k=1 nk ! Nn 2 Y − ∑ Ek−1 nk 2 I(Ynk ≤ xm ) , 1 +Ynk k=1
≤ Lt
∑ Ek−1
where the last factor on the majorant side converges in probability to K(x0 ) + K(+∞) − K(xm ), which is less than ε3 by (4.27). Thus, by the relations given above, we obtain Rtn
Nn
Y2 = ∑ Ek−1 nk 2 = 1 +Ynk k=1 (
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Z +∞ −∞
ht (x) d K(x) + B(m)+
) 2 Ynk I(x j−1 < Ynk ≤ x j ) − K(x j ) − K(x j−1 ) + + ∑ ht (x j−1 ) ∑ Ek−1 2 1 +Ynk j=1 k=1 m
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Nn
+Cn (m) + Dn (m), which, by (4.12), gives the desired result (4.26). The proof of Theorem 4.6 is easily based on Proposition 4.13 and Lemma 4.4 and is not detailed here. In order to prove Theorem 4.7 we need the following auxiliary results.
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L EMMA 4.14.. Under (4.10) and (4.12) with y = +∞ (4.11) is equivalent to Nn
P
∑ |Ek−1 exp(˙ıtYnk ) − 1|2 −→ 0.
(4.28)
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PROOF. From the proof of Proposition 4.13 it follows that (4.11) implies (4.28). The reverse implication can be proved similarly. L EMMA 4.15.. Suppose that (4.14) holds, and that τnk = τ, where τ is a positive constant. Then P
max |ank | −→ 0
(4.29)
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1≤k≤Nn
and there is a constant M depending only on τ and t such that for every k and all n (4.30)
|Ek−1 exp(˙ıtYnk ) − 1| ≤ MEk−1
2 Ynk . 2 1 +Ynk
Furthermore, in this case (4.10) holds. PROOF. From (4.14) it follows that for every 0 < ε < τ
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P
max |ank | < ε + τ max Pk−1 (|Xnk | ≥ ε) −→ ε,
1≤k≤Nn
1≤k≤Nn
which states that (4.14) implies (4.29). Furthermore, one can see that (4.14) combined with (4.29) implies (4.10). Thus, for the proof of Lemma 4.15 we wish to show that (4.30) holds. To this end we note that by a routine technique of subsequences we can assume that τ (4.31) |ank | ≤ for every k and all n. 2 Obviously, there is no loss of generality, because if (Xnk ) does not satisfy (4.31), then we can set X˜nk = Xnk I(|ank | ≤ 2τ ). Then (X˜nk ) will form a DIA of r.vs adapted to (Fnk ) and, by (4.29), τ n ˜ P(∪N max |ank | > ) −→ 0 k=1 {Xnk 6= Xnk }) ≤ P(1≤k≤N 2 n and Nn τ Nn P(∪k=1 { fnk (t) 6= ∏ E j−1 exp(˙ıt X˜n j )}) ≤ P( max |ank | > ) −→ 0. 2 1≤k≤Nn j=1 Furthermore, (4.14) holds with Xnk replaced by X˜nk , and τ . Ek−1 X˜nk I(|X˜nk | ≤ τ) = ank I(|ank | ≤ ) = a˜nk , 2 τ . Y˜nk = X˜nk − a˜nk = Ynk I(|ank | ≤ ), 2
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Ynk Y˜nk = Ek−1 I(|ank | ≤ 2 2 1 + Y˜nk 1 +Ynk Y˜ 2 Y2 Ek−1 nk 2 = Ek−1 nk 2 I(|ank | ≤ 1 + Y˜ 1 +Y Ek−1
nk
nk
τ ), 2
τ Ek−1 exp(˙ıtY˜nk ) − 1 = (Ek−1 exp(˙ıtYnk ) − 1) I(|ank | ≤ ), 2 ˜ from which we conclude that (4.10)–(4.13) and (4.28)–(4.30) hold with Ynk replacing Ynk , enabling use to be made of property (4.31). Alternatively, we will prove Lemma 4.15 (and later on Theorem 4.7 and Corollary 4.11) as it stands and assume also that (4.31) holds. We see that |Ek−1 exp(˙ıtYnk ) − 1| ≤ |Ek−1 [exp(˙ıtYnk ) − 1 − ı˙tYnk ]I(|Ynk | ≤ τ)+ |Ek−1 [exp(˙ıtYnk ) − 1]I(|Ynk | > τ)| + |tEk−1Ynk I(|Ynk | ≤ τ)| ≤ 1 2 I(|Ynk | ≤ τ) + 2Pk−1 (|Ynk | > τ) + ≤ t 2 Ek−1Ynk 2 +|tEk−1Ynk [I(|Ynk | ≤ τ) − I(|Xnk | ≤ τ)]| + +|tEk−1Ynk I(|Xnk | ≤ τ)|, where the first and the second terms are less than Y2 1 2 t (1 + τ2 )Ek−1 nk 2 2 1 +Ynk
and
2
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τ ), 2
2 Ynk 1 + τ2 , E k−1 2 τ2 1 +Ynk
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Page 84 of 94
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respectively. Moreover, we have (4.32)
|Ek−1Ynk I(|Xnk | ≤ τ)| = |ank − ank Pk−1 (|Xnk | ≤ τ)| = τ τ = |ank |Pk−1 (|Xnk | > τ) ≤ Pk−1 (|Ynk | ≥ ) 2 2
and (4.33)
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|Ek−1Ynk [I(|Ynk | ≤ τ) − I(|Xnk | ≤ τ)]| ≤ τ 3τ 3τ τ ≤ Ek−1 |Ynk |I( ≤ |Ynk | ≤ ) ≤ Pk−1 (|Ynk | ≥ ), 2 2 2 2
as (4.31) holds. Hence, |Ek−1 exp(˙ıtYnk ) − 1| ≤ 2 Ynk 1 2 1 + (τ/2)2 2 t (1 + τ ) + 2(1 + τ|t|) E k−1 2 2 (τ/2)2 1 +Ynk
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which gives (4.30). PROOF OF T HEOREM 4.7. We need to show that the assumptions of Theorem 4.7 imply (4.8). To this end it is sufficient to prove that (4.10) and (4.11) are satisfied. From Lemmas 4.12 and 4.15 it follows that (4.10) holds, and that Nn
Nn
∑ |Ek−1 exp(˙ıtYnk ) − 1|2 ≤ M2dnNn
∑ Ek−1
k=1
k=1
2 Ynk P −→ 0, 2 1 +Ynk
i.e. (4.28) is satisfied. Hence, and by Lemma 4.14, we get (4.11). Corollaries 4.8, 4.9 and 4.10 easily follow from Theorem 4.6 (see also [7], p. 93). In order to prove Corollary 4.11 we need the following auxiliary results. L EMMA 4.16.. (4.20) implies (4.19).
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PROOF. By Fn,k−1 -measurability of ank we obtain for every ε > 0 Nn
Nn
ε
Nn
ε
∑ Pk−1(|Ynk | ≥ ε) ≤ ∑ Pk−1(|Xnk | ≥ 2 ) + ∑ I(|ank | ≥ 2 ).
k=1
But
k=1
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k=1
! ε ε −→ 0 P ∑ I(|ank | ≥ ) ≥ ε = P max |ank | ≥ 2 2 1≤k≤Nn k=1 Nn
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as (4.29) holds. Hence, we get the desired implication. L EMMA 4.17.. (4.15) and (4.20) imply that
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Nn
Ynk
P
∑ Ek−1 1 +Y 2 −→ 0.
(4.34)
k=1
nk
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PROOF. We note that for every 0 < ε < τ Ek−1Ynk I(|Ynk | ≤ τ) − Ek−1 Ynk = 2 1 +Ynk 3 Ynk I(|Ynk | ≤ τ) Ynk I(|Ynk | > τ) = Ek−1 + Ek−1 ≤ 2 2 1 +Ynk 1 +Ynk εEk−1
2 Ynk + (τ + 1)Pk−1 (|Ynk | ≥ ε). 2 1 +Ynk
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Hence, using (4.32) and (4.33), we get for every 0 < ε < 2τ Nn Ynk ≤ ∑ Ek−1 2 k=1 1 +Ynk
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Nn
2 Ynk τ 3τ Nn ≤ ε ∑ Ek−1 + 1+τ+ + ∑ Pk−1(|Ynk | ≥ ε), 2 2 2 k=1 1 +Ynk k=1
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which by (4.15) and (4.20) gives (4.34). PROOF OF C OROLLARY 4.11. Taking into account Corollary 4.10 and Lemmas 4.15 and 4.17 we only need to show that (4.15) is satisfied. First we note that for every k and all n 2 Ek−1Ynk I(|Xnk | ≤ τ) = Vark−1 Xnk I(|Xnk | ≤ τ) − a2nk Pk−1 (|Xnk | > τ), 2 − E2 Z . Since by (4.31) and (4.20) where Vark−1 Znk = Ek−1 Znk k−1 nk Nn
Nn
k=1
k=1
P
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∑ a2nk Pk−1(|Xnk | > τ) ≤ (τ/2)2 ∑ Pk−1(|Xnk | ≥ τ) −→ 0, then Go Back
P 2 Ek−1Ynk I(|Xnk | ≤ τ) −→ D,
(4.35) as (4.22) holds. But (cf (4.33))
τ 2 |Ek−1Ynk [I(|Ynk | ≤ τ) − I(|Xnk | ≤ τ)]| ≤ (3τ/2)2 Pk−1 |Ynk | ≥ . 2
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Hence, and by (4.20), we get Nn
(4.36)
P
∑ Ek−1Ynk2 I(|Ynk | ≤ τ) −→ D.
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k=1
Moreover, by (4.20), we have for every ε < τ Nn Nn 2 2 I(|Ynk | ≤ τ) − ∑ Ek−1Ynk I(|Ynk | ≤ ε) ≤ ∑ Ek−1Ynk k=1 k=1 Nn
P
≤ τ2 ∑ Pk−1 (|Ynk | ≥ ε) −→ 0, k=1
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and the same is true for every ε > τ; suffices to interchange ε and τ on the majorant side. Thus, using (4.20) and (4.36), it follows that Nn
(4.37)
P
∑ Ek−1Ynk2 I(|Ynk | ≤ ε) −→ D,
for all ε > 0. Contents
k=1
Since for every ε > 0 Nn 2 Ynk 1 Nn 2 E Y I(|Y | ≤ ε) ≤ E ∑ k−1 nk nk ∑ k−1 1 +Y 2 ≤ 1 + ε2 k=1 k=1 nk
≤
Nn
Nn
k=1
k=1
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∑ Ek−1Ynk2 I(|Ynk | ≤ ε) + ∑ Pk−1(|Ynk | ≥ ε),
then (4.20) and (4.37) imply (4.15).
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ACKNOWLEDGEMENTS . I am indebted to Professor Dominik Szynal for having introduced me to this subject and for his constant encouragement and useful advice. Also, I am grateful to Professor Zdzisław Rychlik for helpful discussion.
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Bibliography [1] Aerts M., Callaert H. 1989 The accuracy of the normal approximation for U-Statistics with a random summation index converging to a random variable. Acta Sci. Math. (Szeged) 53(3-4), 385–394. MR 91a:60056 [2] , The convergence rate of sequential fixed-width confidence intervals for regular functionals. Austral. J. Statist. 28 (1986), 97–106. MR 88a:62224 [3] , The exact approximation order in the central limit theorem for random U-Statistics. Sequential Anal. 5 (1986), 19–35. MR 87g:62031 [4] Ahmad I.A., Lin P.E., A Berry–Esseen type theorem. Utilitas Math. 11 (1977), 153–160. MR 55 #6525 [5] Aldous D.J., Weak convergence of randomly indexed sequences of random variables, Math. Proc. Camb. Philos. Soc. 83 (1978), 117-126. [6] Aldous D. J., Eagleson G. K., On mixing and stability of limit theorems. Ann. Probab. 6(2), 325-331. [7] Anscombe F.J. 1952 Large-sample theory of sequential estimation. Proc. Cambridge Philos. Soc. 98, 600-607. [8] 1953 Sequential estimation. J. Royal Stat. Soc., Sec. B, 15 (1953), 1-21. [9] Atakuziev D. On the convergence rate in a limit theorem for supercritical branching process and its application in limit theorems for sums of a random number of summands. (in Russian) Izv. Akad. Nauk UzSSR, Ser. fiz-mat nauk, No. 2 (1978), 3-8. [10] , Eine Abschätzung der Konvergenzgeschwindigkeit in einem integralen und einem lokalen Transportsatz. (in Russian) Doklady Akad. Nauk UzSSR 6 (1978), 3-5. [11] Azlarov T.A., Atakuziev D.A., Džamirzaev A.A, Shema summirovani sluqa$inyh veliqin s geometriqeski raspredelennym sluqa$inym indeksom. Sbornik: Predeljnye teoremy dl sluqa$ inyh processov, Taxkent, UzSSR (1977), 6-21. [12] Azlarov T.A., Džamirzaev A.A., Ob otnositeljno$i usto$iqivosti dl summ sluqa$inogo qisla sluqa$inyh veliqin, Izv. Akad. Nauk UzSSR, No. 2 (1972), 7-14. [13] , Ravnomernye ocenki v odno$i teoreme perenosa. Sbornik: Sluqa$inye processy i statistiqeskie vyvody, vypusk V, Taxkent, UzSSR (1975), 10-14. [14] Babu G.J., Ghosh M., A random functional central limit theorem for martingales. Acta Math. Acad. Sci. Hungar. 27 (1976), 301-306. [15] Barndorff-Nielsen O., On the limit distribution of the maximum of a random number of independent random variables. Acta Math. Acad. Sci. Hungar. 15 (1964), 399-403. [16] Bartma´nska B., Szynal D., On nonuniform estimates of the rate of convergence in the central limit theorem for functions of the average of independent random variables. Mathematical Statistics and Probability Theory, Proc. 6th Pannonian Symp. on Math. Statist., Bad Tatzmannsdorf, Austria, Sept.14–20, 1986. [17] Barsov S.S., On the accuracy of normal approximation of the distributions of random sums of random vectors. (in Russian) Teor. Verojatnost. Primenen. 30 (1985), 351-354. [18] Batirov H.B., Maneviˇc D.V., Obxie ocenoqnye teoremy dlja raspredelenija summ sluqa$inogo qisla sluqa$inyh slagaemyh. Matemat. Zametki 34 (1983), 145-152. [19] Bethmann, J. (1988). The Lindeberg-Feller theorem for sums of a random number of independent random variables in a triangular array. Theory Probab. Appl. 33, No. 2, 354–359. [20] Be´ska M., Kłopotowski A., Słomi´nski L., Limit theorems for random sums of dependent d-dimensional random vectors. Z. Wahrsch. Verw. Gebiete 61 (1982), 43-57. [21] Bikelis A., On estimates of the remainder term in the central limit theorem. Liet. Mat. Rink. 6 (1966), 323-346 [in Russian] [22] Billingsley P. Limit theorems for randomly selected partial sums. Ann. Math. Statist. 33(1962), 85-92. [23] Billingsley P., Convergence of Probability Measures. Willey: New York 1962. [24] Billingsley P., Probability and Measure. Wiley: New York 1979. [25] Blum J.R., Hanson D.L., Rosenblatt J.I., On the central limit theorem for the sum of a random number of independent random variables. Z. Wahrsch. Verw. Gebiete 1 (1963), 389-393. [26] Borodin A.N., Some limit theorems for the processes with random time. Teor. Verojatnost. Primenen. 24 (1979), 754-770. [27] Brown B.M., Eagleson G.K., Martingale convergence to infinitely divisible laws with finite variances. Trans. Amer. Math. Soc. 162 (1971), 449-453. [28] Butzer P.L., Schulz D., General random sum limit theorems for martingales with large O-rates. Z. Anal. und Anwend. 2 (1983), 97-109. [29] , The random martingale central limit theorem and weak law of large numbers with o-rates. Acta Sci. Math. 45 (1983), 81-94.
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