# Bootstrap variance of squared sample mean

The following is question 8 of chapter 8 in Wasserman's All of Statistics:

Let $T_n = \overline{X}_n^2$, $\mu = \mathbb{E}(X_1)$, $\alpha_k = \int|x - \mu|^kdF(x)$, and $\hat{\alpha}_k = n^{-1}\sum_{i=1}^n|X_i - \overline{X}_n|^k$.

Show that $$v_{\mathrm{boot}} = \frac{4\overline{X}_n^2\hat{\alpha}_2}{n} + \frac{4\overline{X}_n\hat{\alpha}_3}{n^2} + \frac{\hat{\alpha}_4}{n^3} \>.$$

He previously defines $v_{\mathrm{boot}} = \frac{1}{B}\sum_{b=1}^B(T_{n,b}^* - \frac{1}{B}\sum_{r=1}^BT_{n,r}^*)^2$, where $T_{n,i}^*$ is the desired statistic from the $i$th bootstrap replication of the sample $X_1,...,X_n$.

It seems that the question as stated does not make sense: how can there be a formula for the bootstrap variance if the quantity requires simulation? Perhaps he meant to ask for the variance of the sampling distribution, but I get $\frac{\sigma^2}{n}$ for that. Any hints on how to intepret or solve this?

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Hi Alex, welcome to the site. Here is what I think the question is asking: Find $v_{\mathrm{boot}}$ which is the variance under the (empirical) measure $\hat F_n$. The "previously defined" version is simply the Monte Carlo estimate of $v_{\mathrm{boot}}$ rather than the quantity itself. Even so, I think what you'll find is that there are at least two other typos lurking: (1) The definition of $\hat \alpha_k$ should probably not include the modulus and (2) I believe the last term on the right-hand size is not quite correct. :) –  cardinal Apr 8 '12 at 19:20
(When I looked on that page of the book, I saw what I think was at least one other typo in the question before that as well. Also, this same problem, with the same [conjectured] errors appears to be reproduced in Wasserman's All of Nonparametric Statistics as well, on page 39. Once you've completed the exercise, you might consider sending a note to the author so that he can add it to the errata.) –  cardinal Apr 8 '12 at 19:23
Thank you, @cardinal. But for $v_{boot}$ I then obtain $V_{\hat{F}_n}(\overline{X}_n^2) = \mathbb{E}_{\hat{F}_n}(\overline{X}_n^4) - \mathbb{E}_{\hat{F}_n}(\overline{X}_n^2)^2 = 0$! (I reason that $\mathbb{E}_{\hat{F}_n}(\overline{X}_n^2) = n^{-2}\sum\sum\mathbb{E}_{\hat{F}_n}(X_i^*)\mathbb{E}_{\hat{F}_n}(X_j^*) = \overline{X}_n^2$ for samples from $\hat{F}_n$. Same thinking for $\mathbb{E}_{\hat{F}_n}(\overline{X}_n^4)$. I cannot figure out my error. For the previous question, do you mean that the author doesn't distinguish between the random variable $\hat{\theta}$ and its observed value? –  AlexK Apr 8 '12 at 22:24
@whuber and cardinal, thank you each for the explanations, I understand much better now. I believe I've worked out the question (the last term is indeed different) and will notify the author once I type it up. Is it accepted to post the solution here as well? Finally, how can I credit you for the help? –  AlexK Apr 10 '12 at 18:39
Hi AlexK. Yes, please do post your solution as an answer! I was hoping this would be the outcome. I do not want to speak for @whuber too much (though I doubt he'll mind in this instance), but do not be concerned with "crediting" us. I, for one, am happy to see you've arrived at a positive result and have benefitted from the site. I hope you'll continue to frequent it and participate. Cheers. –  cardinal Apr 15 '12 at 2:27
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