**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](https://www.emis.de/journals/HOA/IJMMS/Volume2003_45/825942.pdf) by Csörgő and Rosalsky).
* Additionally, Einmal and Rosalsky later [proved](https://www.tandfonline.com/doi/full/10.1081/SAP-200064490?scroll=top&needAccess=false) 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 $E|X\_1|<\infty$.



WLLN for bootstrap means of stationary ergodic processes?
**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 $\mathbb EZ\_n\to 0$.


