<p>You are correct.</p>
<p>The bootstrap samples with replacement while subsampling samples without replacement.</p>
<p>Theoretically, when <span class="math-container">$m \ll n$</span>, it does not matter if you sample with or without replacement (if you sample with replacement and <span class="math-container">$m \ll n$</span> the probability of drawing the same observation twice can be made arbitrarily small as <span class="math-container">$\frac{m}{n} \to 0$</span>).</p>
<p>In practice, the two procedures usually provide very similar results. The key assumption that makes subsampling valid is that <span class="math-container">$m \to \infty$</span> and <span class="math-container">$\frac{m}{n} \to 0$</span> as <span class="math-container">$n \to \infty$</span>.
(To be precise if <span class="math-container">$\theta_n$</span> is your test statistic and <span class="math-container">$\theta$</span> is the true value and if <span class="math-container">$\tau_n (\theta_n - \theta)$</span> converges to a limit law, then <span class="math-container">$m \to \infty$</span>, <span class="math-container">$\frac{m}{n} \to 0$</span> and <span class="math-container">$\frac{\tau_m}{\tau_n} \to 0$</span> have to hold if <span class="math-container">$n \to \infty$</span>).</p>

CC BY-SA 4.0
2021-06-28T06:16:48.173Z
2021-06-28T07:00:44.643Z
2021-06-28T07:00:44.643Z
tdm
<p>My understanding is the distinction between subsampling and the m out of n bootstrap is that subsampling draws without replacement but the m out of n bootstrap does not.</p>
<p>If we are not in a situation with m = n, is there a reason to prefer the m out of n bootstrap? I am struggling to think of an intuitive reason why subsampling would not dominate, but I realize there may be some math justifying m out of n.</p>

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2021-06-28T16:15:10.930Z
You may want to look at jackknife (https://en.wikipedia.org/wiki/Jackknife_resampling) for similar intuitions. Sometimes, if you have a truly large sample you need to depend on m out of n sampling.
RegressForward
<econometrics><causal-inference>
Subsampling vs. m out of n bootstrap
Michael Gmeiner
# Subsampling vs. m out of n bootstrap

(Asked by: username_0 on 2021-06-27)

My understanding is the distinction between subsampling and the m out of n bootstrap is that subsampling draws without replacement but the m out of n bootstrap does not.

If we are not in a situation with m = n, is there a reason to prefer the m out of n bootstrap? I am struggling to think of an intuitive reason why subsampling would not dominate, but I realize there may be some math justifying m out of n.

**Tags:** econometrics, causal-inference

> **Comments:**
> **username_1**: You may want to look at jackknife (https://en.wikipedia.org/wiki/Jackknife_resampling) for similar intuitions. Sometimes, if you have a truly large sample you need to depend on m out of n sampling.  

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## 1 answers

### Answer by username_2 on 2021-06-28. Score: 6

You are correct.

The bootstrap samples with replacement while subsampling samples without replacement.

Theoretically, when $m \ll n$, it does not matter if you sample with or without replacement (if you sample with replacement and $m \ll n$ the probability of drawing the same observation twice can be made arbitrarily small as $\frac{m}{n} \to 0$).

In practice, the two procedures usually provide very similar results. The key assumption that makes subsampling valid is that $m \to \infty$ and $\frac{m}{n} \to 0$ as $n \to \infty$.
(To be precise if $\theta\_n$ is your test statistic and $\theta$ is the true value and if $\tau\_n (\theta\_n - \theta)$ converges to a limit law, then $m \to \infty$, $\frac{m}{n} \to 0$ and $\frac{\tau\_m}{\tau\_n} \to 0$ have to hold if $n \to \infty$).


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