# 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.

$$\operatorname{var}(\bar{\bf{x}})=\frac{1}{n}\Sigma$$ 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: $$\operatorname{Corr}(\bar{x}_1, \bar{x}_2) = \frac{\operatorname{Cov}(\bar{x}_1, \bar{x}_2)}{\sqrt{\operatorname{Var}(\bar{x}_1)}\sqrt{\operatorname{Var}(\bar{x}_2)}} = \frac{\alpha/n}{(1/n)^{1/2}\cdot (1/n)^{1/2}} = \alpha$$
• @stefgehrig The distribution of the correlation coefficient can be highly skewed for small $m$. For example, the mean of the distribution for $m=5$ and $\alpha = 0.5$ is about $0.452$, I think. Jun 25, 2021 at 17:44