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?
 A: A little late, but anyways... First, to simplify later calculations, rewrite the sample mean in terms of an expression containing the central moments under the empirical measure. Let $S_n = \frac{1}{n}\sum(X_i - \bar{X}_n) = 0$. Then
$$
\bar{X}_n = S_n +\bar{X}_n = \frac{1}{n}\sum (X_i - \bar{X}_n) + \bar{X}_n
$$
Now, Var$(\bar{X}_n^2) = E(\bar{X}_n^4) - (E\bar{X}_n^2)^2$. We'll tackle the first term. Note that $\bar{X}_n$ is the mean under the empirical measure, so we treat it as a constant when taking expectations.
$$
\begin{align}
E(\bar{X}_n^4) &= E(S_n + \bar{X}_n)^4 \\
&= E(S_n^4 + 4\bar{X}_nS_n^3 + 6\bar{X}_n^2S_n^2 + 4\bar{X}_n^3S_n + \bar{X}_n^4) \\
&= E(S_n^4) + 4\bar{X}_nE(S_n^3) + 6\bar{X}_n^2E(S_n^2) + \bar{X}_n^4
\end{align}
$$
where we used that $S_n = 0$ to drop the second-to-last term. In the following expansions, terms involving a product with $nS_n$ will not be written.
$$
\begin{align}
E(S_n^4) &= E\left(\frac{1}{n^4}\left[\sum(X_i - \bar{X}_n)^4  + \sum\sum(X_i - \bar{X}_n)^2(X_j - \bar{X}_n)^2\right]\right) \\
&= \frac{\hat{a}_4}{n^3} + \frac{3(n-1)\hat{a}_2^2}{n^3}\\
E(S_n^3) &= \left(\frac{1}{n^3}\sum(X_i - \bar{X}_n)^3\right) = \frac{\hat{a}_3}{n^2}\\
E(S_n^2) &= \left(\frac{1}{n^2}\sum(X_i - \bar{X}_n)^2\right) = \frac{\hat{a}_2}{n}
\end{align}
$$
These are straightforward sums of products with some combinatorics to count the number of terms. Doing similar calculations for the second term of the variance and putting it all together:
$$
Var(\bar{X}_n^2) = \frac{4\bar{X}_n^2\hat{a}_2}{n} + \frac{4\bar{X}_n\hat{a}_3}{n^2} + \frac{\hat{a}_4 + (2n - 3)\hat{a}_2^2}{n^3}
$$
A: I actually think that the book may in fact have the right answer. 
My suggested change to the process: 
$E(S_n^4) = E(\frac{1}{n^4}[\Sigma(X_i-\bar{X}_n)^4+\frac{1}{n^2}\frac{1}{n^4}\Sigma\Sigma(X_i-\bar{X}_n)^2(X_j-\bar{X_n})^2)$ actually simplifies to $\frac{\hat{\alpha_4}}{n^3} + \frac{\hat{\alpha}_2^2}{n^3}$ which subsequently gets canceled when you find that $E(\bar{X}_n^2)^2 = E((\bar{X}_n+S_n)^2)^2 = \bar{X}_n^4 + 2\bar{X}_n^2\frac{\hat{\alpha}_2}{n} + \frac{\hat{\alpha}_2^2}{n^2}$. 
Thus the final result is as the book gives:
$$v_{boot} = E(\bar{X}_n^4) - E(\bar{X}_n^2)^2 = \frac{4\bar{X}_n^2\hat{\alpha}_2}{n} + \frac{4\bar{X}_n\hat{\alpha}_3}{n^2} + \frac{\hat{\alpha}_4}{n^3}$$
Hope this clarifies the process. Please submit corrections or suggestions if somehow my reasoning was flawed.
A: I think AlexK is correct.
The term $\displaystyle \sum_{i\neq j}(X_i^*-\bar X_n)^2(X_j^*-\bar X_n)^2$ has $3n(n-1)$ parts in total (counting duplicates): From 4 factors you can chose 2 in $4\choose2$ ways and the remaining 2 in $2\choose 2$ ways so the coefficients of the distinct terms are 6.
From $n$ indices you can choose 2 indices in $n\choose 2$ ways. Hence there are $3n(n-1)$ terms of this form.
A: Although the idea behind the procedure proposed by AlexK and others is in the good direction, there are some key points that should be considered if you want to be strictly correct in this.

*

*You cannot arbitrarily decide over which statistic can or can't be considered to be "treated it as a constant when taking expectations" (As it happens with $\overline{X_n}$ respect to $S_n$).

*In the process you are also taking an expectation over an statistic that you know is defined as zero and still expect get something from it (in this case $S_n$, $S_n^2$,...).

*In general, the process of getting an estimator consist in first getting an expression for the statistic, as for example $\mathbb{V}(\overline{X_n})=\sigma^2/n$, and then transform it into an estimator, $\rightarrow \mathbb{\hat{V}}(\overline{X_n})=\hat{\sigma}^2/n$.

You can obtain an expression of this kind (with the desired form requested from the exercise) using
$$\mathbb{V}(\overline{X_n}^2)=\mathbb{E}\left(\overline{X_n} - \mu + \mu\right)^4-\mathbb{E}^2\left(\overline{X_n} - \mu + \mu\right)^2
$$
From this point, the procedure is quite similar to the proposed by AlexK, which requires taking $\mu$ instead of $\overline{X_n}$ and $\overline{X_n} - \mu$ instead of $S_n$. In this case $\mu$ is allowed to be taken out from expectation. Then
\begin{align} 
\mathbb{E}(\overline{X_n}^4) =& \mathbb{E}\left(\overline{X_n} - \mu + \mu\right)^4\\
=& \mathbb{E}\left(\overline{X_n} - \mu\right)^4 + 4 \mu \mathbb{E}\left(\overline{X_n} - \mu\right)^3 + 6\mu^2 \mathbb{E}\left(\overline{X_n} - \mu\right)^2 + 4\mu^3 \underbrace{\mathbb{E}\left(\overline{X_n} - \mu\right)}_{=0} + \mu^4
\end{align}
and
\begin{align} 
\mathbb{E}^2(\overline{X_n}^2) =& \mathbb{E}^2\left(\overline{X_n} - \mu + \mu\right)^2\\
=& \mathbb{E}^2\left(\overline{X_n} - \mu\right)^2+ 2\mu^2 \mathbb{E}\left(\overline{X_n} - \mu\right)^2 + \mu^4
\end{align}
where
$$
\mathbb{E}\left(\overline{X_n} - \mu\right)^2 = \frac{1}{n} \alpha_2
$$
$$
\mathbb{E}\left(\overline{X_n} - \mu\right)^3 = \frac{1}{n^2} \alpha_3
$$
$$
\mathbb{E}\left(\overline{X_n} - \mu\right)^4 = \frac{1}{n^3} \left[\alpha_4 + 3 (n-1) \alpha_2^2\right]
$$
All this gives as result
$$
\mathbb{V}(\overline{X_n}^2)= \frac{1}{n^3} \left[\alpha_4 + \left(2n-3\right) \alpha_2^2\right] + 4 \mu \frac{1}{n^2} \alpha_3+ 4\mu^2 \frac{1}{n} \alpha_2
$$
which then can be used as ansatz for the estimator
$$
\mathbb{\hat{V}}(\overline{X_n}^2)= \frac{1}{n^3} \left[\hat{\alpha}_4 + \left(2n-3\right) \hat{\alpha_2}^2\right] + 4 \overline{X_n} \frac{1}{n^2} \hat{\alpha_3}+ 4\overline{X_n}^2 \frac{1}{n} \hat{\alpha_2}
$$
Notice that in this solution I skipped the summation and combinatorial calculations, as it has been already discussed here. Also notice that I used the definition $\mathbb{E}\left(X_i - \mu\right)^k = \alpha_k$.
