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I know different forms of this question has been asked before on this website, but I don't know how to apply the answers there to this specific form.

Anyway, let $X_1, ..., X_n \sim F$ be iid, with finite expectation $\mu$ (finite variance is not mentioned). Consider the sample mean $\bar{X}_n$ as an estimator of $\mu$.

We know that we can approximate $P(\sqrt{n} (\bar{X}_n - \mu) \leq x) := G_n(x)$ by $P(\sqrt{n}(\bar{X}^*_n - \bar{X}_n) \leq x | X_1, ..., X_n) := G^*_n(x)$, where $\bar{X}^*_n$ is the mean of the bootstrap sample based on the sample $X_1, .., X_n$ by using the Central Limit Theorem.

I know that we have $\lim_{n\rightarrow \infty} \sup_x |G^*_n(x) - G_n(x)| = 0$, so the bootstrap works for the sample mean.

The question is if we need the centering: can we approximate $P(\sqrt{n} \bar{X}_n \leq x) := H_n(x)$ by $P(\sqrt{n}\bar{X}^*_n \leq x | X_1, ..., X_n) := H^*_n(x)$, i.e. do we have $\lim_{n\rightarrow \infty} \sup_x |H^*_n(x) - H_n(x)| = 0$?

The problem I'm having is that the Central Limit Theorem can't be nicely applied to $H_n(x)$ (and $H^*_n(x)$), since for $n$ big we have that $H_n(x)$ is approximately $\mathcal{N}(\sqrt{n}\mu, \sigma^2)$-distributed if I'm correct. Thus the expectation depends on $n$.

It seems too simple (and incorrect) to just say $\lim_{n \rightarrow \infty} H_n(x) = 0$ and $\lim_{n \rightarrow \infty} H^*_n(x) = 0$ for all $x \in \mathbb{R}$, thus $$\lim_{n\rightarrow \infty} \sup_x |H^*_n(x) - H_n(x)| = 0.$$

So my question: can the centering be skipped? If so, why?

Thanks in advance!

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The reason bootstrap works is because it tries to approximate the sampling distribution of $\sqrt{n}(\bar{X}_n-\mu)$. And this is achieved by approximating the limiting (normal) distribution itself. This is best understood in case you see the Edgeworth expansion for $G^*_n(x)$ and $G_n(x)$. The leading terms are both $\Phi(x/\sigma)$, where $\Phi()$ is the standard normal cdf and $\sigma^2 = Var(X)$.

So, removing the centering does not make practical sense as because the limiting distribution becomes an improper distribution. Obviously it works in the sense you have written above. Obviously different centerings can be applied, but that is a different issue. I hope this helps.

Thanks!

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