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I am trying to show that $E_\hat{F}(\hat{se}^2_B) = \hat{se}_\infty^2$. $B$ is the number of bootstrap replications. $\hat{se}_\infty^2$ is the ideal bootstrap variance. $\hat{se}_B=\{\sum_{b=1}^B[\hat{\theta}^*(b)-\hat{\theta}^*(\cdot)]^2/(B-1)\}^{\frac{1}{2}}$ where $\hat{\theta}^*(\cdot)=\sum_{b=1}^B \hat{\theta}^*(b)/B$.

The goal of this is to go on to show that $\hat{se}_B$ always has a greater standard deviation than $\hat{se}_\infty$.

The question comes from problem 6.2 of Efron and Tibshirani's An Introduction to the Bootstrap (page 57). I am working through this book on my own to prepare for some research work and for personal edification.

I know this shouldn't be a difficult thing to show, but something isn't clicking with me. This is an expectation over an empirical distribution $\hat{F}$ so it seems that I should simply take $\frac{1}{B}\sum_{b=1}^B \hat{se}_B^2$. However, I realize I'm not even sure what $\hat{se}_\infty^2$ should look like outside of simple cases like the sample mean. I'm missing something fundamental here.

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For fixed $\hat{F}$, the $\hat{\theta}^*(b)$ for different $b$ are $B$ IID random variables. Moreover, $\hat{\theta}^*(\cdot)$ is the mean of these random variables, $\hat{se}_B^2$ is the sample variance of these random variables, and $\hat{se}_{\infty}^2$ is the true variance of one of these random variables (with respect to $\hat{F}$). Thus, the statement $E_{\hat{F}}(\hat{se}_B^2) = \hat{se}_{\infty}^2$ follows from the fact that for any random variable, the expectation of the sample variance is the true variance.

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