correlation of bootstrap sample means Given a sample $\{x_1,\dots,x_n\}$, $z_1$ and $z_2$ are two bootstrap realizations of sample means, that is,
$$z_1 = \frac{1}{n}\sum\{x\in\text{bootstrap sample 1}\}$$
$$z_2 = \frac{1}{n}\sum\{x\in\text{bootstrap sample 2}\}$$
, how to compute the $corr(z_1, z_2)$?
UPDATE
To make sure I understand the problem correctly, here is what I'm trying to solve:

Here is what I tried, as @probabilityislogic suggested,
$$z = \frac{1}{n}\sum(x_ik_i) ;k_1+k_2+\cdots+k_n = n$$
, since this is a bootstrap sample, so $x_i$ are all constant and $k_i$ are random variables that
$$(k_1,k_2,\cdots,k_n) \sim Mult(k_1k_2\cdots k_n|n, p_1,p_2,\cdots,p_n)$$
. I tried to compute the mean and variance like this,
\begin{align*}
E(z) &= \sum_k\begin{pmatrix}n\\ k\end{pmatrix}p^kz\\
&= \sum_k\begin{pmatrix}n\\ k\end{pmatrix}p^k(\frac{1}{n}\sum_{i=1}^n x_ik_i)\\
&= \frac{1}{n}\left[\sum_k\begin{pmatrix}n\\ k\end{pmatrix}p^k(x_1k_1+\cdots+x_nk_n)\right]\\
&= \frac{1}{n}[np_1x_1+\cdots+np_nx_n]\\
&= p_1x_1+\cdots+p_nx_n
\end{align*}
, and since each $k_i$ could be any number in $[0,n]$, then $p_i = \frac{1}{n+1}$, right? If so, 
$$E(z) = \frac{n}{n+1}\bar x$$
.
To compute covariance,
\begin{align*}
Cov(z_1, z_2) = \sum_{k_{z1}, k_{z2}}\left[\text{Pr}(k_{z1},k_{z2})(z_1 - E(z))(z_2 - E(z))\right]
\end{align*}
, where $\text{Pr}(k_{z1}, k_{z2})$ is the joint probability of $k$ for $z_1,z_2$. Am I going the right way here for covariance of $z_1$ and $z_2$?
I can't seem to figure this out here, seems too complicated to me.
 A: I used to think the answer should be 0, but I see what I did wrong before. The problem is not looking for the conditioned correlation, so without loss of generality, we can look at the case where $\mu=0,\sigma^2=1$:
Let $x_i^{(k)}$ be sample $i$ from the $k$th bootstrap.
Then $\bar{x}_k^* = \frac{1}{n} \sum_{i}{x_i^{(k)}}$.
Key values we need to compute are:
$E\ x_i^{(k)}=E(E[x_i^{(k)}|\mathbf{x}])=E \bar{x}=0$
$var(x_i^{(k)})=E(var(x_i^{(k)}|\mathbf{x}))+var(E[x_i^{(k)}|\mathbf{x}])=\frac{n-1}{n}+\frac{1}{n}=1$
$cov(x_i^{(1)},x_j^{(2)})=E(x_i^{(1)}-0)(x_j^{(2)}-0)=E(E[x_i^{(1)}x_j^{(2)}|\mathbf{x}])=E\bar{x}^2=\frac{1}{n}$
(same is true for $(x_i^{(k)},x_j^{(k)}) \, i\neq j$) 
With above, calculation for the following should be easy:
$cov(\bar{x}_1^*,\bar{x}_2^*)=\frac{1}{n^2} (\sum_{i,j}{cov(x_i^{(1)},x_j^{(2)})})=\frac{1}{n}$
$var(\bar{x}^*_i)=\frac{1}{n^2}\left(n*var(x_1^{(1)})+n(n-1)*cov(x_1^{(1)},x_2^{(1)}) \right)\\
\quad \quad \ \ = \frac{2n-1}{n^2}$
$\rightarrow  corr(\bar{x}_1^*,\bar{x}_2^*)=\frac{n}{2n-1}$
