Some Optimal Conditions for the ASCLT

Let \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$X_{1},X_{2},\ldots $$\end{document}X1,X2,… be independent random variables with \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${E}X_{k}=0$$\end{document}EXk=0 and \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\sigma _{k}^{\,2}:={E}X_{k}^2<\infty $$\end{document}σk2:=EXk2<∞ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(k\ge 1)$$\end{document}(k≥1). Set \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$S_k=X_1+\cdots +X_k$$\end{document}Sk=X1+⋯+Xk and assume that \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$s_{k}^{\,2}:={E}S_k^2\rightarrow \infty $$\end{document}sk2:=ESk2→∞. We prove that under the Kolmogorov condition 1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} |X_n|\le L_n, \quad L_n =o(s_n/(\log \log s_n)^{1/2}) \end{aligned}$$\end{document}|Xn|≤Ln,Ln=o(sn/(loglogsn)1/2)we have 2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \frac{1}{\log s_{n}^{\,2}}\sum _{k=1}^{n}\frac{\sigma _{k+1}^{\,2}}{s_{k}^{\,2}}f\left( \frac{S_{k}}{s_{k}}\right) \rightarrow \frac{1}{\sqrt{2\pi }}\int _{\mathbb {R}}f(x)e^{-x^2/2}\,\textrm{d}x\quad \mathrm {a.s.}\end{aligned}$$\end{document}1logsn2∑k=1nσk+12sk2fSksk→12π∫Rf(x)e-x2/2dxa.s.for any almost everywhere continuous function \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f: {\mathbb R} \rightarrow {\mathbb R}$$\end{document}f:R→R satisfying \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$|f(x)|\le e^{\gamma x^2}$$\end{document}|f(x)|≤eγx2, \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\gamma <1/2$$\end{document}γ<1/2. We also show that replacing the o in (1) by O, relation (2) becomes generally false. Finally, in the case when (1) is not assumed, we give an optimal condition for (2) in terms of the remainder term in the Wiener approximation of the partial sum process \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\{S_n, \, n\ge 1\}$$\end{document}{Sn,n≥1} by a Wiener process.


Introduction
Let X 1 , X 2 , . . .be i.i.d.random variables with E X 1 = 0, E X 2  1 = 1 and put S n = X 1 + • • • + X n .In its simplest form, the almost sure central limit theorem states that for any x ∈ R where I denotes indicator function and (x) is the standard normal distribution function.Relation (3) was proved by Brosamler [7] and Schatte [17] under slightly stronger moment assumptions and by Lacey and Philipp [11] under finite variances.
In past decades, a wide theory of a.s.limit theorems of the type (3) has been developed and several extensions and improvements in (3) were proved.In particular, it turned out that under mild moment assumptions (see (5)), any weak limit theorem for partial sums S n of independent random variables X 1 , X 2 , . . .has an almost sure version where the weights  (Ibragimov and Lifshits [9], Lifshits [12]).The last paper also proves the optimality of the moment assumption assumed in [6,12], showing the surprising role of loglog moments in ASCLT theory.
A generalization of (3) in another direction was given by Schatte [18], who proved that if X 1 , X 2 , . . .are i.i.d.random variables with E X 1 = 0, E X 2 1 = 1, E|X 1 | 3 < ∞ and f : R → R is a function almost everywhere continuous with respect to the Lebesgue measure satisfying then letting S n = n k=1 X k we have Berkes, Csáki and Horváth [5] showed that γ < 1/4 in (8) can be replaced by γ < 1/2 and showed also that the assumption E|X 1 | 3 < ∞ can be dropped.Ibragimov and Lifshits [10] proved that the finiteness of the integral in (9) plus the assumption that f (x)e −H x 2 is nonincreasing for some H > 0 is sufficient in (9); on the other hand, the finiteness of the integral itself is not, even if f is continuous and P(X 1 = 1) = P(X 1 = −1) = 1/2.The purpose of the present paper is to study the problem for independent, not necessarily identically distributed random variables with finite variances and to give an optimal condition for the ASCLT.More precisely, we will prove the following theorem.
Then, we have for any almost everywhere continuous function f : R → R satisfying If (10) only holds with O instead of o, then (11) is generally false.
In terms of the bounds for the random variables X n , Theorem 1 provides an optimal condition for the ASCLT (11).Note that condition (10) (see, e.g., [13], p. 272) which, in view of the results of Marcinkiewicz and Zygmund [16] and Weiss [20], is also optimal.This indicates a strong connection between the LIL and ASCLT, a connection which will become very clear by the proof of Theorem 1.As we will see, however, the LIL (13) itself does not imply (11) and neither does the strong approximation where W is a Wiener process defined on the same probability space as the sequence (X n ) and {S(t), t ≥ 0} is the partial sum process defined by (7).The approximation (14) was proved by Strassen [19] in case of i.i.d.sequences (X n ) with mean 0 and variance 1, and he showed that it implies a large class of refinements of the LIL.Major [14] proved that (14) holds also under the Kolmogorov condition (10).As we will see, in the case when (10) does not hold, the validity of ( 11) is closely connected with the order of Wiener approximability of the partial sums S n .We will namely prove the following results.
Theorem 2 Let X 1 , X 2 , . . .be independent, zero mean random variables and define σ 2 n and s 2 n as in Theorem 1. Assume that and there exists a Wiener process W such that Then, for any a.e.continuous function f : R → R satisfying (12) we have (11).On the other hand, replacing (16) with with any fixed function ψ(n) → ∞, (11) becomes generally false.
As the proof will show, the first half of Theorem 2 remains valid if f (x) = e max{x 2 /2−cx,0} , c > 0.
Theorem 3 Under the setup of Theorem 2, assume that (15) holds and there exists a Wiener process W such that S(t) = W (t) + o (t/ log log t) 1/2 a.s. as t → ∞. (18) Then (11) holds for any measurable function f : R → R such that the integral on the right side of ( 11) is finite and f (x) = e x 2 /2 g(x), where log g(x) is uniformly continuous on R.

Proof of the theorems
Proof of Theorem 1.By an argument in Schatte [18], it suffices to prove the theorem in the case f (x) = e γ x 2 , 0 < γ < 1/2.We first note that by the Kolmogorov exponential bounds (see, e.g., Loève [13], p. 266) the assumptions made on the sequence (X n ) in Theorem 1 imply that given any > 0, K > 0, we have for n ≥ n 0 ( , K ) For 1 ≤ k ≤ l, we put For a fixed δ > 0, we set and define the truncated variable Lemma 1 There exist a constant η > 0 and an integer k 0 such that Proof We follow [10].Define, similar to (21), 123 and put Clearly ξ l −ξ k,l = s k s l ξ k and thus using the mean value theorem and f Thus, the integral of |X | on the set On the other hand, |X | ≤ 2 f (M l ) by ( 25), and thus the integral of |X | on the set On the other hand, the independence of ξk and ξk,l implies that Now choosing δ sufficiently small, we have (2 + δ)γ < 1 and thus M l f (M l ) 2 ≤ (log s l ) 2−η for some constant η > 0 and l ≥ k ≥ k 0 .Together with (26), this yields (22).Now we prove (23).Let F + k denote the distribution function of |ξ k |.Integration by parts and using f (x) = e γ x 2 yields Applying (19) with K = 3, we get for a suitably small η ∈ (0, 1) and large enough k, provided , δ > 0 are chosen so small that (4γ as before, and thus (23) is proved.
Using Lemma 1, we can complete the proof of Theorem 1 by following the standard path of proving ASCLT's.Put Thus, we have for where the second relation follows from (23) and the Cauchy-Schwarz inequality.Let now Note that (10) , and thus, in Theorem 1 it suffices to prove relation (11) with the weight σ 2 k+1 /s 2 k replaced by d k in (29).Clearly Let 0 < δ < η/2.By the first relation of (28), the contribution of those terms on the right-hand side of (30) where for some δ > 0 On the other hand, since s k+1 s k → 1 we have that L = sup and thus by the second relation of (28) the contribution of terms on the right-hand side of (30) where for n ≥ n 0 by δ < η/2.Thus, using (31) and (32) we get for sufficiently large n , and thus, introducing the notation Since s +1 /s → 1, we can select a sequence (n ) such that s 2 n ∼ e 4/η , and then the Beppo Levi theorem implies that almost surely ∞ =1 T 2 n < ∞ and consequently Let F k denote the distribution function of ξ k = S k /s k and recall that we may fix and by (19) and its analogue for the lower tail we obtain that for large enough k which is o(1) if 2γ < 1 − .On the other hand, we have that and consequently, using integration by parts.By the CLT (note that the Kolmogorov condition (10) implies the Lindeberg condition), we have for fixed x when k → ∞, and further for |x| ≤ M k we have whose integral in (−∞, ∞) is finite when is small enough.Thus, the integral in (35) tends to zero by the dominated convergence theorem.The estimates in (37) also show that the first term in (35) tends to 0, and thus, (34) is proved.
We already proved (33) along the sequence (n ); thus, by (34) the relation is also valid along (n ) when → ∞.By the law of the iterated logarithm for (X n ) (valid by the Kolmogorov condition (10)), we have P ξk = ξ k i.o.= 0, and thus, along (n ).But since f > 0 and by s 2 n ∼ e 4 /η we have log s 2 n +1 / log s 2 n → 1, relation (38) holds along the whole sequence of integers, i.e., lim Thus, the first half of Theorem 1 is proved.
To prove the second half of Theorem 1, we use an example due to Weiss [20], showing the sharpness of the basic assumption in Kolmogorov's LIL.Let X 1 , X 2 , . . .be independent random variables with for a fixed α > 0. Then simple calculations show (see [20], p. 123) that, using the notations of Theorem 1, we have as k → ∞.Thus, (10) holds with o replaced by O. Also, in [20, Theorem 1] it is shown that if α is sufficiently large, then there is a δ > 0 such that almost surely for infinitely many n.Now let f (x) = exp(γ x 2 ) where 1 2+δ < γ < 1/2.If we take a fixed n which is large enough and satisfies (41), then using (39) and (40) we get that This completes the proof of Theorem 1.
Proof of Theorem 3. Let, for the purposes of this proof, [t] denote the function which equals By s n+1 /s n → 1 we have [t]/t → 1 as t → ∞ and ( 18) and the LIL for the Wiener process imply |S(t)| = O((t log log t) 1/2 ) a.s.Thus, by (15) and the mean value theorem we get for By our assumption, f (x) = e x 2 /2 g(x), where log g(x) is uniformly continuous on R. Thus, for n → ∞ we have, uniformly for Thus, for the proof of the first half of Theorem 2 it remains to show Fix 0 < ε < 1.By ( 16), there exists an a.s.finite random variable T = T (ε) such that Let Clearly, for t ≥ T we have As f ≥ 1 everywhere, it follows that for t ≥ T and by γ < 1/2 it follows that Hence, applying the ASCLT (47) for f , it follows from (49) that both the liminf and limsup of for every x, and thus, (50), (51) and the dominated convergence theorem imply Thus, (48) is proved and the proof of the first half of Theorem 2 is completed.
As we noted after Theorem 2, the first half of the result remains valid for f (x) = e max{x 2 /2−cx,0} , c > 0. Relation (43) is easily modified in this case, and instead of 0 < ε < 1, we can use 0 < ε < A and then e γ (|x|+1) 2 in (50), (51) can be replaced by For the proof of the second half of Theorem 2, we need a lemma.
Lemma 2 Let ψ : R + → R + be a function with lim x→∞ ψ(x) = ∞.Then, there exists a sequence (X n ) of independent r.v.'s with mean zero and finite variances such that setting for some Wiener process W , but for f (x) = |x| p , p > 2 and any δ > 0 we have Proof We use a construction similar to that used in Lifshits [12].Since there exists a nondecreasing slowly varying function and thus since n j+1 ψ(n j+1 ) −(α−2) n j ψ(n j ) −(α−2) → 2 by the slow variation of ψ.Thus, Also, for k ≥ k 0 we have Since S * n is a sum of independent, symmetric random variables, we have P(S * n k−1 ≥ 0) ≥ 1/2 and thus Moreover, noting that for n k ≤ n < n k+1 the partial sum S * n does not change, (56) and the latter bound implies that for sufficiently large n Let now W be a Wiener process independent of {Y k , k ≥ 1} and set S n = W (n) + S * n .Clearly, {S n , n ≥ 1} is the partial sum sequence of a sequence (X n ) of independent random variables with mean zero and finite variances and (58)-(60) show that (52)-(54) hold.To prove (55) with f (x) = x p , let us observe that by (54) and ψ(x) = o(log log x) 1/2 , the set of integration in (55) contains for n ≥ n 0 the set E n = {0 ≤ W (n) ≤ √ n}, whose probability converges to (1) − (0) > 1/4 as n → ∞.Thus, by (60) and the independence of W (n) and S * n the probability of the set exceeds 1  16 ψ(n) −α for n ≥ n 0 and on E * n we have Thus, proving (55).This completes the proof of Lemma 2.
To prove the second half of Theorem 2, we will follow the argument in the proof of Theorem 1.Let δ > 0, ψ(n) = o(log log n) 1/2 and let (X n ) be the sequence provided by Lemma 2. Let f (x) = e γ x 2 (0 ≤ γ < 1/2), put s k = √ k and define M k , ξ k , ξ k,l , ξk , ξk,l as in the proof of Theorem 1.We claim that Lemma 1 remains valid in the present case.The proof of (22) requires only trivial changes, since the argument there does not use the Kolmogorov condition (10).To prove (23), we set and estimate the integral of f ( ξk ) 2 separately on B k and C k .We first note that B n is identical with the set of integration in (55) and by relation (54) on the set B k we have for sufficiently large k and consequently, we have on B k by (54) and (61), Thus, we have to estimate the integral of exp(2γ W (k) 2 /k) on the set B k where, according to the previous relations we have (61), i.e. we get, introducing the standard normal random variable 2+δ) .
Next we estimate the integral of f ( ξ 2 k ) on the set C k .Let us observe that in view of (54) and ψ(n) = o(log log n) 1/2 , on the set C k we have for sufficiently small δ and sufficiently large k, .
On the other hand, on C k we have 2+δ) .
Thus, for sufficiently small δ we get, using γ < 1/2, Funding Open access funding provided by ELKH Alfréd Rényi Institute of Mathematics.
d n are determined by the norming sequence (b n ) and D n = n k=1 d k .For (b n ) growing with polynomial speed we have d n = 1/n (see Berkes and Dehling [6]); assuming only b n ↑ ∞, (4) holds with d k = (b k+1 − b k )/b k