# Bootstrap approximation sample mean: skip the centering?

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?

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 $\sigma^2 = Var(X)$.