# Intuition behind m-out-of-n bootstrap

I am trying to get some intuition on why m-out-of-n bootstrap works but haven't been able to find good explanation. I would really appreciate any input on this.

I think I do understand what bootstrap is about -- estimating how $$\sqrt{n}(T_n(X_1,...,X_n)-T(X;F))$$ behaves using $$\sqrt{n}(T_n(X_1^*,...,X_n^*)-T(X;\hat{F_n}))$$. ($$X_1,...,X_n$$ drawn from $$F$$, the true CDF. And $$X_1^*,...,X_n^*$$ drawn from $$\hat{F_n}$$, the ECDF). From my understanding, when $$T$$ is a smooth function, bootstrap works fine. Sometimes when T is non-smooth (such as extreme order statistics, or $$|\mu|$$), m-out-of-n bootstrap can "smooth" things out and works.

My main question is:

1. Why does m-out-of-n bootstrap "smooth" things out?

I have two more things that I want to make sure I am understanding correctly.

1. Since only $$m$$ samples are drawn, how can the behavior (variability, etc.) of $$T_m(X_1^*,...,X_m^*)$$ resemble that of a sample statistics using $$n$$ observations ($$T_n(X_1,...,X_n)$$). Or is it only known that asymptotically they are the same?

2. When using m-out-of-n bootstrap method to find CI, do we need to scale the variance of $$\sqrt{m}(T_m(X_1^*,...,X_m^*;\hat{F_n})-T(X;\hat{F_n}))$$ by $$\frac{n}{m}$$ since we're drawing a smaller sample size from $$\hat{F_n}$$?

Hope my questions are clear.

I would argue that it's not so much that the $$m$$ of $$n$$ bootstrap does smoothing as that it makes smoothing unnecessary.
There are two components to the $$m$$ of $$n$$ bootstrap. The first is sampling just $$m$$ observations; the second is knowing the convergence rate.
A big part of the advantage of the subsampling is being able to handle the correct rate. If a statistic is $$\sqrt{n}$$-consistent and based on iid observations, the ordinary bootstrap pretty much has to work (Chapter 3.6 of van der Vaart & Wellner does this)
So, if you are looking to bootstrap the maximum, you need to know that it converges faster than $$\sqrt{n}$$ when you have a hard maximum. For example, with $$U[0,\theta]$$ you have $$n(X_{(n}-\theta)=O_p(1)$$. That means you need to scale the variance by $$m^2/n^2$$, not $$m/n$$.
• Thank you so much! This really helps. So far I've only studied statics converging at $\sqrt{n}$ rate. Do you happen know any book or classic example that talks about statics converging at a different rate? I really want to look more into this. Thanks again. – RevealedPreference Jul 9 at 0:59