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Let $\bar{X}_n$ denote the sample mean of n iid random variables. Let $\bar{X}^*_n$ be the bootstrap sample mean. Does $\left|\mathbb{P}\left(n^{1/2}\bar{X}_n\leq x\right)-\mathbb{P}\left(n^{1/2}\bar{X}^*_n\leq x\right)\right|$ converge to $0$ in probability?

I am asking this question because I know from [Bickel, P.J. and D.A. Freedman (1981), Some asymptotic theory for the bootstrap] that the bootstrap can be used to approximate the distribution of the sample mean in the sense that $$\left|\mathbb{P}\left(n^{1/2}\left(\bar{X}_n-\mu\right)\leq x\right)-\mathbb{P}\left(n^{1/2}\left(\bar{X}^*_n-\bar{X}_n\right)\leq x\right)\right|\overset{\mathbb{P}}{\rightarrow}0$$ and I was wondering whether the centering is needed. Maybe to be able to use the Central Limit Theorem?

Thanks a lot!

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What is $\mathbb{P}\left(n^{1/2}\bar{X}_n\leq x\right)$ actually? Is it asymptotically normal? And what can I say about $\mathbb{P}\left(n^{1/2}\bar{X}^*_n\leq x\right)$? – kelu Oct 14 '12 at 14:21
Keep in mind that the mean needs to exist for the distribution and so do higher order moments. – Michael Chernick Oct 15 '12 at 15:54
Yes, sorry, let's assume $\mu$ is finite, forgot that. But does that tell me anything more about my actual question? – kelu Oct 15 '12 at 20:20
Cross-posted at…. – whuber Oct 29 '12 at 11:42

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