Covariance across sample means? Say we have a multivariate normal distribution with non-zero covariance $\alpha$
$X \sim N
  \left(\mu = \left({\begin{array}{cc}
0 \\
1 \end{array} } \right), \Sigma = \left({\begin{array}{cc}
   1 & \alpha \\
   \alpha & 1 \\
  \end{array} } \right) \right)$
Now we take $m$ samples of same size $n$. Obviously the expected correlation of the two elements of $X$ in each sample of size $n$ is $\rho_1 = \rho_2 =  ... = \rho_m = \alpha/1^2 = \alpha$.
Now suppose we create new random variables by taking the average $\bar{X} = \left({\begin{array}{cc}
\bar{x_1} \\
\bar{x_2} \end{array} } \right)$ in each sample.
What is the expected correlation of the two elements of $\bar{X}$ among the $m$ data points we have now?
My intuitive response was that it is $\alpha$ as well, but after some thinking and simulations, I don't trust that intuition anymore. Approaching it formally, my idea would be to derive the formula for their covariance as
$\text{Cov}(\bar{x_1}, \bar{x_2}) = E(\bar{x_1} \bar{x_2}) - E(\bar{x_1}) \ E(\bar{x_2}) $
Once I know this covariance, I can calculate the correlation, since I know the variances of both $\bar{x}_1$ and $\bar{x}_2$ by the CLT, specifically here $\text{Var}(\bar{x}_1) = \text{Var} (\bar{x}_2) = 1/n$. However, I don't see any way to get to $E(\bar{x_1} \bar{x_2})$, which I need to calculate the covariance, which does not involve the covariance itself. For example, calculating the integral mentioned here requires a full specification of the joint PDF including covariance.
I am probably already off track though.

As in the multivariate setting, the variance-covariance matrix of a sample mean vector is:
$$
\operatorname{var}(\bar{\bf{x}})=\frac{1}{n}\Sigma
$$
where $\Sigma$ is the population variance-covariance matrix. Hence, the distribution of the mean vector is multivariate normal:
$$
\bar{\bf{x}}\sim \operatorname{N}\left(\bf{\mu}, \frac{1}{n}\Sigma\right)
$$
where $\bf{\mu}$ is the population mean vector.
In your case, $\Sigma = \left({\begin{array}{cc}
   1 & \alpha \\
   \alpha & 1 \\
  \end{array} } \right) $ and $\frac{1}{n}\Sigma = \left({\begin{array}{cc}
   1/n & \alpha/n \\
   \alpha/n & 1/n \\
  \end{array} } \right)$. So for the correlation we have:
$$
\operatorname{Corr}(\bar{x}_1, \bar{x}_2) = \frac{\operatorname{Cov}(\bar{x}_1, \bar{x}_2)}{\sqrt{\operatorname{Var}(\bar{x}_1)}\sqrt{\operatorname{Var}(\bar{x}_2)}} = \frac{\alpha/n}{(1/n)^{1/2}\cdot (1/n)^{1/2}} = \alpha
$$
