correlation of bootstrap sample means

Question

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.

Answer 1

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:

$cov(\bar{x}_1^*,\bar{x}_2^*)=\frac{1}{n^2} (\sum_{i,j}{cov(x_i^{(1)},x_j^{(2)})})=\frac{1}{n}$

$var(\bar{x}^*_i)=\frac{1}{n^2}\left(n*var(x_1^{(1)})+n(n-1)*cov(x_1^{(1)},x_2^{(1)}) \right)\\ \quad \quad \ \ = \frac{2n-1}{n^2}$

$\rightarrow corr(\bar{x}_1^*,\bar{x}_2^*)=\frac{n}{2n-1}$