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Take $X_1,\ldots,X_n$ to be distinct observations, and let $X_1^*,\ldots,X_n^*$ be a bootstrapped sample (given a uniform resampling of cases with replacement) and let $\bar{X_n}^{*} = \frac{1}{n} \sum_{i=1}^n X_i^*$ (i.e. the mean of the bootstrapped sample). I would like to solve for the first moment of the distribution for $\bar{X_n}^{*}$ i.e. $E(\bar{X_n}^*)$. I can see that $$E(\bar{X_n}^* \mid X_1,\ldots,X_n) = E\left(\frac{1}{n}\sum_{i=1}^n X_i^* \mid X_1,\ldots,X_n\right)$$ $$= \frac{1}{n} \sum_{i=1}^n E(X_i^* \mid X_1,\ldots,X_n) = \frac{1}{n} \sum_{i=1}^n \left( \sum_{i=1}^n X_i P(X_i^* = X_i)\right) =\frac{1}{n} \sum_{i=1}^n \left( \sum_{i=1}^n \frac{X_i}{n} \right)$$ $$=\frac{1}{n} \sum_{i=1}^n \left( \bar{X}\right) =\frac{1}{n} n\bar{X} = \bar{X}.$$

However, I am having an issue solving for it unconditionally, i.e. $E(\bar{X_n}^*)$. Any recommendation on how to use the conditional distribution to solve for this?

Michael Hardy
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Elaine
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1 Answers1

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First, conditional on $X_1,\ldots,X_n$, $X_i^*$'s are IID with $$ E(X_i^*\mid X_1,\ldots,X_n)=\bar{X}\implies E\big(\bar{X}_n^*\mid X_1,\ldots,X_n\big)=\frac{1}{n}n\bar{X}=\bar{X}. $$ I find this simpler than your calculation. Next, iterated expectations give $$ E\big(\bar{X}_n^*\big)=E\big[E\big(\bar{X}_n^*\mid X_1,\ldots,X_n\big)\big]=E(\bar{X})=\frac{1}{n}\sum_{i=1}^nE(X_i). $$ This is as far as you can go unless you know something about the marginal distribution for each $X_i$.

yurnero
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