# Bootstrap for small sample

Does bootstrap method help for small sample? In my mind, bootstrap is a solution when you don't have belief in a normality assumption. If your data is random enough, it might be convincing to sample from your sample and get an empirical distribution for your statistic.

But if the size of your sample is small, even if you believe your sample is random, can you say bootstrap is the best way for small sample? Would there be a chance that your bootstrap result is biased?

• How small is your "small sample?" Apr 26, 2016 at 22:32

However, there are some situations where there is good reason to prefer bootstrap to usual asymptotic inference for small (or really any) samples. Some uses of bootstrap achieve what is called an asymptotic refinement. This refers to a situation where the bootstrap estimate of the distribution of your estimator converges faster than the usual asymptotic approximation. For example, if you have some asymptotically normal estimate, $\sqrt{n}(\hat{\theta} - \theta) \leadsto_d N(0,\sigma^2)$, then $$P( \sqrt{n}(\hat{\theta}-\theta)/\hat{\sigma} < z) = \Phi(z) + O(n^{-1}),$$ i.e. the size distortion of hypothesis tests decreases at rate $1/n$. Some versions of bootstrap can be shown to have size distortions that decrease at a rate faster than $1/n$. Specifically, using bootstrap to compute the distribution of a pivotal statistic (one whose asymptotic distribution is completely known, like the t-statistic) generally gives this faster convergence. In this example if for each bootstrap replicate of the data, you calculate $t_b = \sqrt{n}(\hat{\theta}_b - \hat{\theta})/\hat{\sigma}_b$, and let $\hat{F}$ be the CDF of the $t_b$, then
$$P( \sqrt{n}(\hat{\theta}-\theta)/\hat{\sigma} < z) = \hat{F}(z) + o(n^{-1})$$ Here we have a little $o$ where before there was a big $O$. This gives some hope that when bootstrapping pivotal statistics, the bootstrap might be more accurate for small samples than the usual asymptotic approximations.