On the Komlós–Révész SLLN for Ψ-Mixing Sequences

Abstract

The Komlós–Révész strong law of large numbers (SLLN) is proved for $\psi$-mixing sequences without a rate assumption.

1. Introduction and Result

Let ${\left\{X_k\right\}}_{k\in \mathbb{N}},$ $\mathbb{N}=\{1,2,3,\dots \},$ be a sequence of random variables defined on a probability space $(\mathsf{\Omega },\mathcal{F},P)$ and ${\parallel X\parallel }_p^p={E\left(\right|X|}^p)$. It was proved by Komlós and Révész in [1] that for centered, independent random variables $X_k$,

$${\displaystyle \frac{{\sum }_{k=1}^n{\parallel X_k\parallel }_2^{-2}{X}_k}{{\sum }_{k=1}^n{\parallel X_k\parallel }_2^{-2}}}{\to }_{n}0,\mathrm{a}.\mathrm{s}.$$

(1)

provided that ${\sum }_{k=1}^{\infty }{\parallel X_k\parallel }_2^{-2}$ diverges (see also [2], p. 75 and Ex. 13 on p. 137 in [3]). If $E\left(X_k\right)=m,$ $k\ge 1,$ then we have strong consistency:

$${\displaystyle \frac{{\sum }_{k=1}^n{\parallel X_k-m\parallel }_2^{-2}{X}_k}{{\sum }_{k=1}^n{\parallel X_k-m\parallel }_2^{-2}}}{\to }_{n}m,\mathrm{a}.\mathrm{s}.$$

and it was proved that this estimator has minimal variance. Nevertheless, as observed in the preface of [4], “For many phenomena in the real world, the observations are not independent…”, so the question on (1) is intriguing when there is a lack of independence. In [5], the Komlós–Révész theorem is investigated for pairwise independence by the method of subsequences. For martingale difference sequences in $L^p$, $p\in (1,2]$, by the Doob theorem (see Th. 2 on p. 246 in [3]), it is obtained in [6]. In [7], the case $p\in (0,1)$ and $p>2$ is also discussed for negatively dependent and mixing sequences. In the latter, some rate on mixing is assumed, so this paper aims to remove this restriction.

Let $\sigma$-fields $\mathcal{A},\mathcal{B}$ satisfy $\mathcal{A},\mathcal{B}\subset \mathcal{F}$. Recall the following strong measure of dependence:

$$\psi (\mathcal{A},\mathcal{B})=sup|{\displaystyle \frac{P(B\cap A)}{P\left(B\right)·P\left(A\right)}}-1|;P\left(B\right)·P\left(A\right)>0,A\in \mathcal{A},B\in \mathcal{B}.$$

It is well known that (see p. 124 in Vol. I, [4])

$${\psi }_m=\underset{J\ge 1}{sup}\psi ({\mathcal{F}}_1^J,{\mathcal{F}}_{J+m}^{\infty })=\underset{J\ge 1}{sup}sup{\displaystyle \frac{\Vert E\left(g\right|{\mathcal{F}}_1^J)-E\left(g\right){\Vert }_{\infty }}{{\Vert g\Vert }_1}},$$

where ${\mathcal{F}}_k^m$ denotes $\sigma$–field generated by $X_k,X_{k+1},\dots ,X_m$, $m\in \mathbb{Z}$ and the inner sup is taken over $g\in L_{\mathrm{real}}^1\left({\mathcal{F}}_{J+m}^{\infty }\right)$. We say that $\left\{X_k\right\}$ is $\psi$-mixing if $\underset{m\to \infty }{lim}{\psi }_m=0$. It is worth noticing that Kesten and O’Brien and Bradley gave examples of $\psi$–mixing sequences with an arbitrary rate of mixing (see Ch. 3 and Ch. 26 in [4]).

The following is the main result in this paper.

Theorem 1. 

Suppose $\left\{X_k\right\}$ is ψ–mixing. Let $\left\{a_k\right\}$ be a sequence satisfying $0<a_k^{p-1}E{\left|X_k\right|}^p\le C<\infty$ for $k\in \mathbb{N}$.

(i) 

If $0<p<1$ and ${\sum }_{k=1}^{\infty }{a}_k$ converges, then

$${\displaystyle \frac{{\sum }_{k=1}^n{a}_k{X}_k}{{\sum }_{k=n}^{\infty }{a}_k}}{\to }_{n}0,\mathrm{a}.\mathrm{s}.$$

(2)

(ii) 

If $1<p\le 2$ and $E\left(X_k\right)=0,$ ${sup}_n{\left({\sum }_{k=1}^n{a}_k\right)}^{-1}{\sum }_{k=1}^n{a}_{k}E\left|X_k\right|<\infty$ and ${\sum }_{k=1}^{\infty }{a}_k$ diverges, then

$${\displaystyle \frac{{\sum }_{k=1}^n{a}_k{X}_k}{{\sum }_{k=1}^n{a}_k}}{\to }_{n}0,\mathrm{a}.\mathrm{s}.$$

(3)

(iii) 

If $p>2$ and $E\left(X_k\right)=0,$ ${sup}_n{\left({\sum }_{k=1}^n{a}_k\right)}^{-1}{\sum }_{k=1}^n{a}_{k}E\left|X_k\right|<\infty$ and ${\sum }_{k=1}^{\infty }{a}_k$ diverges and $0<a_{k}E{\left|X_k\right|}^2\le C<\infty$ for $k\in \mathbb{N}$, then (3) holds.

Remark 1. 

(a) It follows from the proof that in case $\left(i\right)$, ψ–mixing is not required.

(b) In fact, ${sup}_n{\left({\sum }_{k=1}^n{a}_k\right)}^{-1}{\sum }_{k=1}^n{a}_{k}E\left|X_k\right|<\infty$ if ${sup}_{k}E\left|X_k\right|<\infty$. The latter is also used in the case of pairwise independent random variables (see Theorem 2 in [5]) and can be omitted if ${\psi }_m=0$ for some $m\ge 1$.

(c) For $p=1$, Theorem 1 does not hold. Suppose that $\left\{X_k\right\}$ is a stochastic sequence $P(X_k=1)=1-P(X_k=0)=1/2$ constructed in the proof of Theorem 1 in [8]. Thus, if $\left\{g_n\right\}$ is a non-increasing sequence, $g_n\to 0$, then ${\psi }_n\asymp g_n$. Now, let $\left\{{\xi }_k\right\}$ be a sequence of independent random variables with $P({\xi }_k=k)={\displaystyle \frac{1}{2k}},P({\xi }_k=-k)={\displaystyle \frac{1}{2k}},P({\xi }_k=0)=1-{\displaystyle \frac{1}{k}}$ and independent of $X_k$. Set $Z_k={\xi }_{k}cos\left(X_k\pi \right)$, $k\ge 1$ so that $E\left(e^{it{Z}_k}\right)=E\left(e^{it{\xi }_k}\right)$. Thus, $E{Z}_k=0,E\left|Z_k\right|=1$. By Proposition 3.16 on p. 82 in Vol. I, [4] and by Theorem 6.2 on p. 193 in Vol. I, [4] we have that $\left\{Z_k\right\}$ is ψ-mixing (non-stationary) with rate $\asymp g_n$. Let $a_k\equiv 1.$ Now, (3) fails by the second Borel–Cantelli lemma (see [9] and p. 210 in [10]) since ${\sum }_{k\ge 1}P\left(\right|Z_k|\ge k)={\sum }_{k\ge 1}{\displaystyle \frac{1}{k}}=\infty$.

The following variant of the converse statement does not require strong mixing.

Proposition 1. 

Let $\left\{a_k\right\}$ be a sequence of positive reals. Suppose $\left\{X_k\right\}$ is an arbitrary dependent random sequence such that ${sup}_{k}E\left|X_k\right|=C<\infty$ and ${lim\; inf}_{n}E\left|{\sum }_{k=1}^n{a}_k{X}_k\right|=c>0$. If (3) holds, then ${\sum }_{k\ge 1}{a}_k$ diverges.

Set $q=|{\displaystyle \frac{p}{p-1}}|$. Theorem 1 applied with $a_k^{-1}=\left(E\right|X_k{{|}^p)}^{{\displaystyle \frac{1}{p-1}}}$ and, in case $\left(iii\right)$ with $a_k^{-1}=E|X_k{|}^2\vee (E|X_k{{|}^p)}^{{\displaystyle \frac{1}{p-1}}}$, yields

Corollary 1. 

Suppose $\left\{X_k\right\}$ is ψ-mixing.

(i) 

If $0<p<1$ and ${\sum }_{k=1}^{\infty }{\parallel X_k\parallel }_p^{-q}$ converges, then

$${\displaystyle \frac{{\sum }_{k=1}^n{\parallel X_k\parallel }_p^{-q}{X}_k}{{\sum }_{n=k}^{\infty }{\parallel X_k\parallel }_p^{-q}}}{\to }_{n}0,\mathrm{a}.\mathrm{s}.$$

(4)

(ii) 

If $1<p\le 2,$ ${sup}_n({\sum }_{k=1}^n\parallel X_k{\parallel }_p^{-q}{)}^{-1}{\sum }_{k=1}^n\parallel X_k{\parallel }_p^{-q}E\left|X_k\right|<\infty ,$ $E\left(X_k\right)=0$ and ${\sum }_{k=1}^{\infty }{\parallel X_k\parallel }_p^{-q}$ diverges, then

$${\displaystyle \frac{{\sum }_{k=1}^n{\parallel X_k\parallel }_p^{-q}{X}_k}{{\sum }_{k=1}^n{\parallel X_k\parallel }_p^{-q}}}{\to }_{n}0,\mathrm{a}.\mathrm{s}.$$

(5)

(iii) 

If $p\ge 2,$

$$\underset{n}{sup}({\displaystyle \sum _{k=1}^n}\parallel X_k{\parallel }_2^2\vee \parallel X_k{\parallel }_p^q{)}^{-1}{\displaystyle \sum _{k=1}^n}(\parallel X_k{\parallel }_2^2\vee \parallel X_k{\parallel }_p^q{)}^{-1}E\left|X_k\right|<\infty ,$$

$E\left(X_k\right)=0$ and ${\sum }_{k=1}^{\infty }(\parallel X_k{\parallel }_2^2\vee \parallel X_k{{\parallel }_p^q)}^{-1}$ diverges, then

$${\displaystyle \frac{{\sum }_{k=1}^n(\parallel X_k{\parallel }_2^2\vee \parallel X_k{{\parallel }_p^q)}^{-1}{X}_k}{{\sum }_{k=1}^n(\parallel X_k{\parallel }_2^2\vee \parallel X_k{{\parallel }_p^q)}^{-1}}}{\to }_{n}0,\mathrm{a}.\mathrm{s}.$$

(6)

We apply our result to the $\psi$-mixing Cramér model. Namely, we have sequence $\left\{{\eta }_k\right\}$ with $P({\eta }_k=1)=1-P({\eta }_k=0)=1/lnk,$ $k\ge 3,$ which is $\psi$-mixing. Let ${\zeta }_k={\eta }_{k}lnk$. It is easy to see that ${\sigma }_k^2=\mathrm{Var}\left({\zeta }_k\right)\sim lnk,$ $m_k=E\left({\zeta }_k\right)=1$ and ${\sum }_{k=3}^n{\displaystyle \frac{1}{{\sigma }_k^2}}\sim {\displaystyle \frac{n}{lnn}}$. Thus, for $\left\{{\zeta }_k\right\}$ in the case $p=2$, we obtain by (5) that SLLN for Cramér model independence can be replaced by $\psi$-mixing (see e.g., Lemma A.1 in [11]).

Lemma 1. 

$$\underset{n}{lim}{\displaystyle \frac{lnn}{n}}{\displaystyle \sum _{k=3}^n}{\eta }_k=1,\mathrm{a}.\mathrm{s}.$$

The paper is organized as follows. In the next section, there are some auxiliary results. These results are required in the proof of Theorem 1 in the last section.

2. Auxiliary Results

The following result for $p>1$ is in [12] and for $p=1$ see [9] (see also Theorem 2.20 on p. 40 in [13]).

Theorem 2. 

Suppose $\left\{X_k\right\}$ is ψ-mixing and $E\left(X_k\right)=0,$ $k\in \mathbb{Z}$. Let $\left\{b_n\right\},$ $b_0>0$ be an increasing to infinity sequence of real numbers such that for some $p\ge 1$,

(i) 

${\sum }_{k=1}^{\infty }{b}_k^{-2p}E\left(\right|X_k{|}^{2p})<\infty ;$

(ii) 

${\sum }_{k=1}^{\infty }{b}_k^{-2}{(b_k^2-b_{k-1}^2)}^{1-p}{\left(E\left(X_k^2\right)\right)}^p<\infty$;

(iii) 

${sup}_n{b}_n^{-1}{\sum }_{k=1}^{n}E\left|X_k\right|<\infty .$

Then, $b_n^{-1}{S}_n\to 0,$ $n\to \infty ,$ almost surely (a.s.).

The next result improves Lemma 2.1 slightly in [14].

Proposition 2. 

Suppose $\left\{X_k\right\}$ is ψ-mixing and $E\left(X_k\right)=0,$ $k\in \mathbb{Z}$. Let $b_n\in \mathbb{R},$ $b_0>0,$ $b_n\uparrow \infty ,$ and ${sup}_n{b}_n^{-1}{\displaystyle \sum _{k=1}^n}E\left|X_k\right|<\infty ,$ $p\in (1,2],$ ${\sum }_{k=1}^{\infty }{\displaystyle \frac{E|X_k{|}^p}{b_k^p}}<\infty ,$ then $b_n^{-1}{S}_n\to 0,$ $n\to \infty ,$ a.s.

Proof of Proposition 2. 

Set ${\overline{X}}_k=X_{k}I\left(\right|X_k|\le b_k)-E(X_{k}I\left(\right|X_k|\le b_k\left)\right).$ In view of Theorem 2, we can assume $1<p<2$.

$$\sum _{k\ge 1}P(X_k\ne X_{k}I\left(\right|X_k|\le b_k\left)\right)\le C\sum _{k\ge 1}{\displaystyle \frac{E|X_k{|}^p}{b_k^p}}<\infty .$$

Further,

$$\begin{array}{c}{\displaystyle b_n^{-1}|{\displaystyle \sum _{k=1}^n}E(X_{k}I\left(\right|X_k|\le b_k\left)\right)|\le b_n^{-1}{\displaystyle \sum _{k=1}^n}E\left(\right|X_k\left|I\right(|X_k|>b_k\left)\right)}\hfill \\ \le b_n^{-1}{\displaystyle \sum _{k=1}^n}{b}_k^{-p+1}E\left(\right|X_k{|}^{p}I\left(\right|X_k|>b_k\left)\right)\le b_n^{-1}{\displaystyle \sum _{k=1}^n}{b}_k^{-p+1}E{\left|X_k\right|}^p\to 0.\hfill \end{array}$$

So that

$$\sum _{k\ge 1}{\displaystyle \frac{E{\overline{X}}_k^2}{b_k^2}}\le \sum _{k\ge 1}{\displaystyle \frac{E(X_k^{2}I\left(\right|X_k|\le b_k\left)\right)}{b_k^2}}\le \sum _{k\ge 1}{\displaystyle \frac{E|X_k{|}^p}{b_k^p}}<\infty .$$

Now, apply Theorem 2 to $\left\{{\overline{X}}_k\right\}$ with $p=1$. □

By Lemma 3.6″ on p. 284 in [15] and Kronecker’s lemma (see e.g., [16], p. 236) we have the following SLLN for arbitrary dependent random variables (see also [17]).

Lemma 2. 

Suppose $\left\{X_k\right\}$ is a sequence of random variables, $p\in (0,1],$ $b_n\uparrow \infty$ and ${\sum }_{k=1}^{\infty }{\displaystyle \frac{E|X_k{|}^p}{b_k^p}}<\infty$. Then, $b_n^{-1}{S}_n\to 0,$ $n\to \infty ,$ a.s.

3. Proofs

Proof of Theorem 1. 

The key role in the proof of Theorem 1 is played by the Dini theorem (see e.g., [18], Theorem 4 on p. 127) and the Abel–Dini theorem (see e.g., [18], Theorem 1 on p. 125) (for a generalization, see Lemma 11 in [7]).

For $p\in (0,1)$ by the Dini theorem,

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{a_n^{p}E{\left|X_n\right|}^p}{{\left({\sum }_{k=n}^{\infty }{a}_k\right)}^p}}\le C{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{a_n}{{\left({\sum }_{k=n}^{\infty }{a}_k\right)}^p}}<\infty$$

since ${\sum }_{k=1}^{\infty }{a}_k<\infty$. Thus, (2) holds by Lemma 2.

Now, assume $p\in (1,2]$. By the Abel–Dini theorem,

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{a_n^{p}E{\left|X_n\right|}^p}{{\left({\sum }_{k=1}^n{a}_k\right)}^p}}\le C{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{a_n}{{\left({\sum }_{k=1}^n{a}_k\right)}^p}}<\infty$$

since ${\sum }_{k=1}^{\infty }{a}_k=\infty$. Set $b_n={\sum }_{k=1}^n{a}_k$. Since ${sup}_n{b}_n^{-1}{\sum }_{k=1}^n{a}_{k}E\left|X_k\right|<\infty ,$

$${\displaystyle \frac{{\sum }_{k=1}^n{a}_k{X}_k}{{\sum }_{k=1}^n{a}_k}}{\to }_{n}0,\mathrm{a}.\mathrm{s}.$$

holds by Proposition 2.

In the case that $p>2$, set $b_0=0.5b_1$. By Theorem 2 and the previous point, it is enough to prove that

$$S={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{{(b_n^2-b_{n-1}^2)}^{1-\frac{p}{2}}}{b_n^2}}{\displaystyle \frac{a_n^p{\left(E\left(X_n^2\right)\right)}^{\frac{p}{2}}}{b_n^p}}<\infty .$$

Let $\delta ={\displaystyle \frac{2ϵ}{p}},$ where $ϵ\in (0,1)$ is fixed. In view of $b_n\to \infty$, we have that there exist $C>0$ such that $a_{n}E\left(X_n^2\right)\le C{b}_n^{3-\delta },$ for each n. Therefore,

$$a_n^{p-1}E{\left(X_n^2\right)}^{{\displaystyle \frac{p}{2}}}\le C{b}_n^{{\displaystyle \frac{3p}{2}}-ϵ}{a}_n^{{\displaystyle \frac{p}{2}}-1}.$$

But ${(b_n^2-b_{n-1}^2)}^{1-{\displaystyle \frac{p}{2}}}={\left(a_n{b}_n(2-{\displaystyle \frac{a_n}{b_n}})\right)}^{1-{\displaystyle \frac{p}{2}}}$ so that by the Abel–Dini theorem,

$$S\le C{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{a_n{\left(a_n{b}_n\right)}^{1-\frac{p}{2}}}{b_n^{p+2}}}{b}_n^{{\displaystyle \frac{3p}{2}}-ϵ}{a}_n^{{\displaystyle \frac{p}{2}}-1}=C{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{a_n}{b_n^{1+ϵ}}}<\infty .$$

Since ${sup}_n{b}_n^{-1}{\sum }_{k=1}^n{a}_{k}E\left|X_k\right|<\infty ,$ Theorem 1 is proved. □

Proof of Proposition 1. 

Suppose ${\sum }_{k\ge 1}{a}_k<\infty$ and (3) holds. Choose $N_1$ such that ${\sum }_{k\ge n}{a}_k<{\displaystyle \frac{c}{3C}}$ for $n\ge N_1$. Next, choose $N_2\ge N_1$ such that $E|{\sum }_{k=1}^n{a}_k{X}_k|>c/2$ for $n\ge N_2$. Now, for a fixed $N\ge N_2$, we have

$${\displaystyle \sum _{k=N+1}^n}{a}_k{X}_k{\to }_n-{\displaystyle \sum _{k=1}^N}{a}_k{X}_k,\mathrm{a}.\mathrm{s}.$$

On the other hand, by Fatou’s lemma (cf. Theorem 5.1 on p. 218, [19]),

$$c/2<E|{\displaystyle \sum _{k=1}^N}{a}_k{X}_k|\le \underset{n}{lim\; inf}E|{\displaystyle \sum _{k=N+1}^n}{a}_k{X}_k|<c/3.$$

Thus, $c<0$. Contradiction. □