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JennySton
04-18-2010, 11:07 AM
Hi all!

I have a question about bootstrapping.

Let the variance estimate based on B bootstrap samples be \widehat{SE}^2_B = \frac{1}{B-1}\sum_{b=1}^B (\hat{\theta}^*(b) - \hat{\theta}^*(\cdot))^2 , where B is the number of bootstrap samples, \hat{\theta}^*(b) is the estimate based on the bth bootstrap sample and \hat{\theta}^*(\cdot) = \frac{1}{B}\sum_{b=1}^B \hat{\theta}^*(b). Also let the ideal bootstrap variance be \widehat{SE}^2_\infty = SE_{\hat{F}}(\hat{\theta}^*)^2.

I have to show that
1) E_{\hat{F}}(\widehat{SE}^2_B)= \widehat{SE}^2_\infty
which means that the bootstrap expected value of the variance estimate that is based on B bootstrap samples is the same as the ideal bootstrap variance. (Note, that the expected value is w.r.t. \hat{F}).

2) E_{F}(\widehat{SE}^2_B)= E_F(\widehat{SE}^2_\infty)
which means that variance estimate that is based on B bootstrap samples and the ideal bootstrap variance have same expected value with respect to the true CDF F.

3) Var_{F}(\widehat{SE}^2_B)=> Var_F(\widehat{SE}^2_\infty)
which means that variance estimate which is based on B bootstrap samples has larger variance than the the ideal bootstrap variance with respect to the true CDF F.

I was able to solve the first problem using the fact that the \hat{\theta}^*(b) are iid and E(\hat{\theta}^*(b)) = E(\hat{\theta}^*) \forall b . Then with some algebra I could show the result. However, I am not quite sure how to solve the other two problems. I was first thinking I could use the result from 1) but I think that does not work because in 1) we take the expected value with respect to \hat{F} but in 2) and 3) with respect to F.