Suppose that $X_i,$ $i= 1, 2 \dots$ are iid with finite positive variance,

then let $X_{i,n}^*$ $(i = 1, \dots, m)$ be a bootstrap sample of size m from $\lbrace X_1, \dots, X_n\rbrace$, then \begin{equation} \sqrt{m} \left( \frac{1}{m} \sum_{i = 1}^m X_{i,n}^* - \frac{1}{n} \sum _{i = 1}^n X_i\right) \longrightarrow N(0, \sigma^2) \text{ as } m \text{ and } n \rightarrow \infty \end{equation} we also have that: \begin{equation*} \sqrt{n} \left( \frac{1}{n} \sum_{i = 1}^n X_i - \mu \right)\longrightarrow N(0, \sigma^2) \text{ as } n \rightarrow \infty \end{equation*} by the central limit theorem. Then my question is do we also have that if $m = n$: \begin{equation*} \sqrt{n} \left( \frac{1}{n}\sum_{i = 1}^n X_{i,n}^* - \mu\right) \longrightarrow N(0, 2\sigma^2) \text{ as } n \longrightarrow \infty? \end{equation*}

Now, ideally I'd like to add the two sort of, however lack of independence and different m, n indices makes this difficult.

I tried simulating some data in R and I compared QQplots from which this seemed to hold, however possibly this is under certain conditions.

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    $\begingroup$ Your first line is not actually correct, as the $\sigma^2$ on the right of the $\rightarrow$ is a population number but, given a particular sample $X_1, \dots, X_n$, the variance of the sample - which is the relevant number here - need not equal $\sigma^2$. $\endgroup$ – jbowman Aug 9 '17 at 19:22
  • $\begingroup$ @jbowman Yeah you're right, I meant n to go to infinity here too, I've edited the question, thanks. $\endgroup$ – Sam Davenport Aug 9 '17 at 19:26
  • $\begingroup$ You could restrict the problem somewhat by making $m = \alpha n$ for some fixed $\alpha > 0$, well floor or ceiling to preserve integers, then you only have to have $n \to \infty$. But that might be more limiting than you want. $\endgroup$ – jbowman Aug 9 '17 at 19:43
  • $\begingroup$ Yeah that's true, in fact I'm most interested in the n = m case, or the a = 1 case. However, while I'd like to add the first two equations, because they're not independent, I don't know how to show that the third one holds under this assumption. $\endgroup$ – Sam Davenport Aug 9 '17 at 19:53
  • $\begingroup$ I admit I'm writing comments instead of doing the hard work to answer the question, but... it seems to me if you altered eq. 1 to have $s^2$ instead of $\sigma^2$ and got rid of the $n$ in $m, n \to \infty$, then use the fact that $s^2 \to \sigma^2$ as $n \to \infty$, you would get independence, because eq. 1 is true regardless of the actual value of $1/n \sum X_i$ (in some sense.) $\endgroup$ – jbowman Aug 9 '17 at 20:14

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