# WLLN for bootstrap means of stationary ergodic processes?

Setup:$$\quad$$

Suppose that $$(X_n)$$ is a stationary ergodic process with $$E|X_1|<\infty$$.

Given $$X^{(n)}=(X_1, \dots, X_n)$$, select a standard Efron bootstrap subsample $$(X_{n,1}^*, \dots, X_{n,m(n)}^*)$$ by pulling $$m(n)$$ times with replacement from a uniform distribution $$U(\{X_1, \dots, X_n\})$$, i.e. $$X_{n,i}^* = X_{Z_{n,i}}, \quad Z_{n,i} \overset{\text{iid}}{\sim} U(\{1,\dots,n\}), \quad i=1,\dots,m(n),$$ where the $$Z_{n,i}$$ are independent of $$(X_n)$$ and form a triangular array with independent rows.

Let the bootstrap mean $$\mu_{m(n)}^*$$ be the sample mean of the bootstrap subsample, i.e. $$\mu_{m(n)}^* = \frac{1}{m(n)} \sum_{i=1}^{m(n)} X_{n,i}^*.$$

Question:

In the case that $$(X_n)$$ is a stationary ergodic process, are there any known results about when the following WLLN holds? $$\mu_{m(n)}^* \overset{P}{\longrightarrow} E[X]$$ as $$n \to \infty$$.

What I've found:

• In the case that the $$X_i$$ are i.i.d. and $$m(n) \to \infty$$, it is known that the WLLN above holds for any $$m(n) \to \infty$$ (e.g. see p.2848 of this 2003 survey by Csörgő and Rosalsky).

• Additionally, Einmal and Rosalsky later proved that $$\mu_{m(n)}^* - \frac{1}{n} \sum_{i=1}^n X_i \overset{P}{\longrightarrow} 0$$ holds for any $$(X_n)$$ (not necessarily independentent or identically distributed), provided $$m(n) \uparrow \infty$$ and $$\frac{X_n}{\sqrt{m(n)}} \overset{\text{a.s.}}{\longrightarrow} 0.$$ This, however, doesn't cover all stationary ergodic processes with $$E|X_1|<\infty$$.

• It seems as though the answer should be yes. I would suggest writing $X_n$ as $Y_n+Z_n$ where $Y_n$ is $X_n$ if $|X_n|\le m(n)^{1/3}$ and 0 otherwise; similarly $Z_n$ is $X_n$ if $|X_n|>m(n)^{1/3}$ and 0 otherwise. Then the Einmal and Rosalsky result applies to the Bootstrap averages of the $Y_n$, so all that remains is to check that the Bootstrap averages of the $Z_n$ approach 0 in probability. I believe that follows from Markov's inequality once you know that $\mathbb E Z_n\to 0$. Sep 25 at 19:57
• It works! Thank you :) Sep 25 at 23:00
• @Anthony Quas: If your comment answered the question you should post it as an answer, especially to keep this site from seeing this question as unanswered. Sep 27 at 8:21

It seems as though the answer should be yes. I would suggest writing $$X_n$$ as $$Y_n+Z_n$$ where $$Y_n$$ is $$X_n$$ if $$|X_n|\le m(n)^{1/3}$$ and 0 otherwise; similarly $$Z_n$$ is $$X_n$$ if $$|X_n|>m(n)^{1/3}$$ and 0 otherwise. Then the Einmal and Rosalsky result applies to the Bootstrap averages of the $$Y_n$$, so all that remains is to check that the Bootstrap averages of the $$Z_n$$ approach 0 in probability. I believe that follows from Markov's inequality once you know that $$\mathbb EZ_n\to 0$$.