The Brooks-Chacon Biting Lemma and the Baum-Katz Theorem Along Subsequences

1. Introduction

The following result (see [7]) quantifies the rate of convergence in the strong law of large numbers, through a complete convergent numerical Baum-Katz-type series along subsequences of norm bounded random variables, without any supplementary probabilistic hypothesis on their (in)dependence or on their distributions:

Theorem 0. On a complete probability space $(\Omega ,\mathcal{F},\mathbf{P})$ we consider a sequence of random variables ${\left(X_n\right)}_{n\ge 1}$ that is uniformly bounded in $L^p$ for some $0<p<2$, i.e.,

$$\underset{n\ge 1}{sup}\left|\right|X_n{\left|\right|}_p\le C\mathrm{for}\mathrm{some}C>0.$$

Then there exists a subsequence ${\left(Y_n\right)}_{n\ge 1}$ of ${\left(X_n\right)}_{n\ge 1}$ such that, for all $1<r\le p$, we have

$$\sum _{n=1}^{\infty }{n}^{p/r-2}\mathbf{P}\left(\left\{\omega \in \Omega :\left|\sum _{j=1}^n{Y}_j\left(\omega \right)\right|>\epsilon n^{1/r}\right\}\right)<\infty \mathrm{for}\mathrm{any}\epsilon >0.$$

(1)

Remarks. The complete convergence of the series in formula (1) implies that the subsequence ${\left(Y_n\right)}_{n\ge 1}$ satisfies the strong law of large numbers, i.e., $Y_n/n^{1/p}\to 0$$\mathbf{P}$-a.s. The parameter $r\in (1,2)$ is keeping Theorems 0 within the realm of laws of large numbers; indeed, if $p\ge r\ge 2$ then by the central limit theorem for subsequences, the series in formula (1) diverges for all $\epsilon >0$ even if ${\left(X_n\right)}_{n\ge 1}$ is an i.i.d. sequence with mean zero and finite variance. Also note that formula (1) trivially holds if $0<p<r<2$.

There are situations (see, e.g., the recent papers [4] and [5] and their references) when the sequence of random variables in question is not (uniformly) bounded in $L^p$ and yet satisfies a law of large numbers. In this case, Theorem 0 is not appropriate to quantify the rate of convergence in that law of large numbers. In addition, the examples in [6,8] and [3] show that Theorem 0 fails if one drops the $L^p$-uniform boundedness hypothesis, for any $0<p<2$. Inspired by the celebrated Komlós-Saks-type extension of the law of large numbers (cf. [4,5]), in Section 2 we shall prove a version of the Baum-Katz theorem under a special boundedness hypothesis, different from the $L^p$-boundedness condition required in Theorem 0. The idea is to construct a rich family of uniformly integrable subsequences of ${\left(X_n\right)}_{n\ge 1}$ as in [2], for which condition (1) holds; note that the hypotheses in [8] and [3] cannot produce Baum-Katz type theorems, as the families of subsequences therein are no longer uniformly integrable. Instead, we shall employ the methodology given by the celebrated Bitting Lemma of Brooks and Chacon (cf. [1]). A modification of this methodology is presented in Section 3; it will produce a second version of the Baum-Katz theorem under a Mazur-Orlicz-type hypothesis (cf. [4,5]).

2. Main Result

Theorem 1. Let $0<p<2$. On a complete probability space $(\Omega ,\mathcal{F},\mathbf{P})$ we consider a sequence of random variables ${\left(X_n\right)}_{n\ge 1}$ such that

$$\underset{n\ge 1}{lim\; sup}{\left|X_n\left(\omega \right)\right|}^p<\infty \mathrm{for}\mathrm{all}\omega \in \Omega .$$

Then there exists a subsequence ${\left(Y_n\right)}_{n\ge 1}$ of ${\left(X_n\right)}_{n\ge 1}$ such that Equation (1) holds for all $1<r\le p$. In particular, the subsequence ${\left(Y_n\right)}_{n\ge 1}$ satisfies the strong law of large numbers, i.e., $Y_n/n^{1/p}\to 0$$\mathbf{P}$-a.s.

Examples. (i) Note that the working hypothesis in Theorem 1 is equivalent to

$$\underset{n\ge 1}{sup}{\left|X_n\left(\omega \right)\right|}^p<\infty \mathrm{for}\mathrm{all}\omega \in \Omega .$$

In particular, one can see that Theorems 0 and 1 do not overlap and do not imply each other. Indeed, uniformly $L^p$-bounded sequences of functions can still have an infinite limsup, and vice-versa: there are sequences of functions, with finite limsup, that are not (uniformly) bounded in $L^p$ (see also [9]).

(ii) The hypotheses in Theorem 1 are satisfied, e.g., by the working condition in the motivational papers [4] and [5], namely:

$$\underset{N\to \infty }{lim}N·\underset{n\ge 1}{sup}\mathbf{P}\left(\right|X_n|>N)=0.$$

This condition implies uniform boundedness in $L^0$ (tightness), i.e.,

$$\underset{N\to \infty }{lim}\underset{n\ge 1}{sup}\mathbf{P}\left(\right|X_n|>N)=0$$

and is implied by uniform integrability, i.e.,

$$\underset{N\to \infty }{lim}\underset{n\ge 1}{sup}\mathbf{E}(X_n·{\mathbf{1}}_{\left\{\right|X_n|>N\}})=0,$$

where $\mathbf{E}$ denotes the expectation with respect to $\mathbf{P}$. Also note that $L^p$-boundedness condition in Theorem 0 is stronger than the last three conditions, provided $p\in [1,2)$ (see, e.g., Example 4.2 in [4]).

Proof of Theorem 1. For any natural number $m\ge 1$, let us define the $\mathcal{F}$-measurable sets

$$A_m:=\left\{\omega \in \Omega :\underset{n\ge 1}{sup}{\left|X_n\left(\omega \right)\right|}^p\le m\right\}.$$

Assume that $r<p$ and fix $a>p/r-1$. As $\mathbf{P}\left(A_m\right)\to 1$ as $m\to \infty$ by hypothesis, we can choose an index $m_1\ge 1$ such that $\mathbf{P}\left(A_{m_1}\right)>1-2^{-a}$. By Fatou’s lemma we obtain:

$$\underset{n\ge 1}{sup}{\int }_{A_{m_1}}{\left|X_n\left(\omega \right)\right|}^{p}d\mathbf{P}\left(\omega \right)\le m_1.$$

(2)

We now apply the Bitting Lemma (cf. [1]) to the sequence ${\left(X_n\right)}_{n\ge 1}$ and the subset $A_{m_1}$ above, to obtain: a non-decreasing sequence ${\left(B_k^1\right)}_{k\ge 1}$ of subsets in $\mathcal{F}$ with $\mathbf{P}\left(B_k^1\right)\to 1$ as $k\to \infty$, and a subsequence ${\left(X_n^1\right)}_{n\ge 1}$ of ${\left(X_n\right)}_{n\ge 1}$ such that ${\left(X_n^1\right)}_{n\ge 1}$ is uniformly integrable on each of the subsets $A_{m_1}\cap B_k^1$, $k\ge 1$. This latter fact together with estimate (2) show that Theorem 0 applies to the sequence ${\left(X_n^1\right)}_{n\ge 1}$ and gives:

$$\sum _{n=1}^{\infty }{n}^{p/r-2}\mathbf{P}\left(\left\{\omega \in A_{m_1}\cap B_k^1:\left|\sum _{j=1}^n{X}_j^1\left(\omega \right)\right|>\epsilon n^{1/r}\right\}\right)<\infty \mathrm{for}\mathrm{any}\epsilon >0\mathrm{and}k\ge 1.$$

Next, we choose a natural number $m_2\ge 1$ such that $\mathbf{P}\left(A_{m_2}\right)>1-3^{-a}$, and another application of the Bitting Lemma, this time to ${\left(X_n^1\right)}_{n\ge 1}$, produces: a non-decreasing sequence ${\left(B_k^2\right)}_{k\ge 1}$ of subsets in $\mathcal{F}$ with $\mathbf{P}\left(B_k^2\right)\to 1$ as $k\to \infty$, and a subsequence ${\left(X_n^2\right)}_{n\ge 1}$ of ${\left(X_n^1\right)}_{n\ge 1}$, therefore a subsequence of ${\left(X_n\right)}_{n\ge 1}$ as well, such that ${\left(X_n^2\right)}_{n\ge 1}$ is uniformly integrable on each of the subsets $A_{m_2}\cap B_k^2$, $k\ge 1$, and

$$\sum _{n=1}^{\infty }{n}^{p/r-2}\mathbf{P}\left(\left\{\omega \in A_{m_2}\cap B_k^2:\left|\sum _{j=1}^n{X}_j^2\left(\omega \right)\right|>\epsilon n^{1/r}\right\}\right)<\infty \mathrm{for}\mathrm{any}\epsilon >0\mathrm{and}k\ge 1.$$

Thus, by induction, we constructed for each $i\ge 1$: an $\mathcal{F}$-measurable set $A_{m_i}$ with $\mathbf{P}\left(A_{m_i}\right)>1-{(i+1)}^{-a}$, a non-decreasing sequence ${\left(B_k^i\right)}_{k\ge 1}$ of subsets in $\mathcal{F}$ with $\mathbf{P}\left(B_k^i\right)\to 1$ as $k\to \infty$, and a subsequence ${\left(X_n^i\right)}_{n\ge 1}$ of ${\left(X_n^{i-1}\right)}_{n\ge 1}$ such that ${\left(X_n^i\right)}_{n\ge 1}$ is uniformly integrable on each of the subsets $A_{m_i}\cap B_k^i$, $k\ge 1$, and

$$\sum _{n=1}^{\infty }{n}^{p/r-2}\mathbf{P}\left(\left\{\omega \in A_{m_i}\cap B_k^i:\left|\sum _{j=1}^n{X}_j^i\left(\omega \right)\right|>\epsilon n^{1/r}\right\}\right)<\infty \mathrm{for}\mathrm{any}\epsilon >0\mathrm{and}k,i\ge 1.$$

(the convention is that ${\left(X_n^0\right)}_{n\ge 1}$ is precisely ${\left(X_n\right)}_{n\ge 1}$).

Now define $Y_n:=X_n^n$, $n\ge 1$; it follows that ${\left(Y_n\right)}_{n\ge 1}$ is a subsequence of ${\left(X_n\right)}_{n\ge 1}$ and, using a diagonal argument in the previous formula, we obtain that

$$\sum _{n=1}^{\infty }{n}^{p/r-2}\mathbf{P}\left(\left\{\omega \in A_{m_n}\cap B_k^n:\left|\sum _{j=1}^n{Y}_j\left(\omega \right)\right|>\epsilon n^{1/r}\right\}\right)<\infty \mathrm{for}\mathrm{any}\epsilon >0\mathrm{and}k\ge 1.$$

(3)

As $\mathbf{P}\left(B_k^n\right)\to 1$ as $k\to \infty$ for all $n\ge 1$, formula (3) and the dominated convergence theorem imply that

$$\sum _{n=1}^{\infty }{n}^{p/r-2}\mathbf{P}\left(\left\{\omega \in A_{m_n}:\left|\sum _{j=1}^n{Y}_j\left(\omega \right)\right|>\epsilon n^{1/r}\right\}\right)<\infty \mathrm{for}\mathrm{any}\epsilon >0.$$

(4)

Therefore, to prove that our series (1) converges for this particular subsequence ${\left(Y_n\right)}_{n\ge 1}$, it suffices to prove formula (4) with $A_{m_n}$ replaced by its set complement, i.e.,

$$\sum _{n=1}^{\infty }{n}^{p/r-2}\mathbf{P}\left(\left\{\omega \in \Omega \setminus A_{m_n}:\left|\sum _{j=1}^n{Y}_j\left(\omega \right)\right|>\epsilon n^{1/r}\right\}\right)<\infty \mathrm{for}\mathrm{any}\epsilon >0.$$

(5)

Indeed, as $\mathbf{P}\left(A_{m_n}\right)>1-{(n+1)}^{-a}>1-n^{-a}$ and $a>p/r-1>0$, the series in (5) is

$$\le \sum _{n=1}^{\infty }{n}^{p/r-2}\mathbf{P}\left(\left\{\omega \in \Omega \setminus A_{m_n}\right\}\right)\le \sum _{n=1}^{\infty }{n}^{p/r-2-a}<\infty ,$$

so the proof is achieved in the case $r<p$.

If $r=p$, then we modify the above methodology as follows: by induction we now choose $\mathcal{F}$-measurable sets $A_{m_i}$ with $\mathbf{P}\left(A_{m_i}\right)>i/(i+1)$ for all $i\ge 1$; as such, the Biting Lemma and the diagonal argument produce the subsequence ${\left(Y_n\right)}_{n\ge 1}$ and the following replacement of Equation (4):

$$\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n}}\mathbf{P}\left(\left\{\omega \in A_{m_n}:\left|\sum _{j=1}^n{Y}_j\left(\omega \right)\right|>\epsilon n^{1/r}\right\}\right)<\infty \mathrm{for}\mathrm{any}\epsilon >0.$$

To show that the series (1) converges for along the subsequence ${\left(Y_n\right)}_{n\ge 1}$, it suffices to prove the following replacement of Equation (5):

$$\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n}}\mathbf{P}\left(\left\{\omega \in \Omega \setminus A_{m_n}:\left|\sum _{j=1}^n{Y}_j\left(\omega \right)\right|>\epsilon n^{1/r}\right\}\right)<\infty \mathrm{for}\mathrm{any}\epsilon >0.$$

Indeed, by the new choice of $A_{m_n}$, the latter series is

$$\le \sum _{n=1}^{\infty }{\displaystyle \frac{1}{n}}\mathbf{P}\left(\left\{\omega \in \Omega \setminus A_{m_n}\right\}\right)\le \sum _{n=1}^{\infty }{\displaystyle \frac{1}{n(n+1)}}<\infty ,$$

and this achieves the proof in the case $r=p$.

3. A Variant of the Main Result

Proposition 2. Let $0<p<2$. On a complete probability space $(\Omega ,\mathcal{F},\mathbf{P})$ we consider a sequence of random variables ${\left(X_n\right)}_{n\ge 1}$ satisfying the following condition: for every subsequence ${\left({\tilde{X}}_n\right)}_{n\ge 1}$ of ${\left(X_n\right)}_{n\ge 1}$ and $n\ge 1$, there exists a convex combination $Z_n$ of $\left\{\right|{\tilde{X}}_n{|}^p,|{\tilde{X}}_{n+1}{|}^p,\dots \}$, such that ${sup}_{n\ge 1}{\left|Z_n\left(\omega \right)\right|}^p<\infty$ for all $\omega \in \Omega$. Then there exists a subsequence ${\left(Y_n\right)}_{n\ge 1}$ of ${\left(X_n\right)}_{n\ge 1}$ such that Equation (1) holds for all $1<r\le p$ and $Y_n/n^{1/p}\to 0$$\mathbf{P}$-a.s.

Examples. On $[0,1]$ endowed with the Lebesgue measure, the sequence $X_n\left(\omega \right)=n^2$ if $0<\omega <1/n$ and 0 otherwise, satisfies Theorem 2 because $X_n\to 0$ Lebesgue-a.s., yet it does not satisfy Theorem 1 with $p=1$ because it is not bounded in $L^1[0,1]$. As a matter of fact, both Theorems 1 and 2 may fail for unbounded sequences, e.g., $X_n\left(\omega \right)=n$.

Proof of Proposition 2. By hypothesis we can write

$$Z_n=\sum _{i\in I_n}{\lambda }_i^n{\left|{\tilde{X}}_{n+i}\right|}^p\mathrm{for}\mathrm{some}{\lambda }_i^n\ge 0\mathrm{with}\sum _{i\in I_n}{\lambda }_i^n=1,$$

where $I_n$ are finite subsets of $\{0,1,2\dots \}$. In addition, the sequence ${\left(Z_n\right)}_{n\ge 1}$ satisfies the condition ${sup}_{n\ge 1}{\left|Z_n\left(\omega \right)\right|}^p<\infty$ for all $\omega \in \Omega$. For any natural number $m\ge 1$, let us define the $\mathcal{F}$-measurable sets

$$A_m:=\left\{\omega \in \Omega :\underset{n\ge 1}{sup}{\left|Z_n\left(\omega \right)\right|}^p\le m\right\}.$$

We have $\mathbf{P}\left(A_m\right)\to 1$ as $m\to \infty$, so we can choose an index $m_1\ge 1$ such that $\mathbf{P}\left(A_{m_1}\right)>1-2^{-a}$ for some $a>p/r-1$ fixed, or $1/2$, according to $p>r$ or $p=r$, respectively. In both cases, by applying Fatou’s lemma we obtain:

$$\underset{n\ge 1}{sup}\sum _{i\in I_n}{\lambda }_i^n{\int }_{A_{m_1}}{\left|{\tilde{X}}_{n+i}\left(\omega \right)\right|}^{p}d\mathbf{P}\left(\omega \right)\le m_1.$$

Hence there is a subsequence ${\left({\overline{X}}_n\right)}_{n\ge 1}$ of ${\left({\tilde{X}}_n\right)}_{n\ge 1}$, therefore a subsequence of ${\left(X_n\right)}_{n\ge 1}$ as well, such that

$$\underset{n\ge 1}{sup}{\int }_{A_{m_1}}{\left|{\overline{X}}_n\left(\omega \right)\right|}^{p}d\mathbf{P}\left(\omega \right)\le m_1.$$

The remainder of the proof goes exactly as in the proof of Theorem 1, with Equation (2) replaced by the equation above, and applied to the subsequence ${\left({\overline{X}}_n\right)}_{n\ge 1}$.