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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 $E|X_1|<\infty$.

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    $\begingroup$ 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 $\mathbb E Z_n\to 0$. $\endgroup$
    – Anthony Quas
    Commented Sep 25, 2022 at 19:57
  • $\begingroup$ It works! Thank you :) $\endgroup$
    – zxmkn
    Commented Sep 25, 2022 at 23:00
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    $\begingroup$ @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. $\endgroup$
    – Peter O.
    Commented Sep 27, 2022 at 8:21

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Answered in comments above

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 $\mathbb EZ_n\to 0$.

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