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Covariances can be written in terms of a correlation coefficient $\rho$ and two standard deviations $\sigma$:

$Cov(a,b)=\rho_{ab}\sigma_a\sigma_b$ [Equation 1]

What happens when the covariance is between two population means?

Let $\bar{x}$ denote the average of the random variables $x_i$ (where $i$ is an index up to $n_x$, the number of $x$ variables), and let $\bar{y}$ denote the average of the variables $y_j$ (where $j$ is an index up to $n_y$, the number of $y$ variables). Let all the $x$ variables have an identical standard deviation. Call it $\sigma_x$. Although they have the same standard deviation, the $x$ values may be correlated or uncorrelated. Likewise, let all the $y$ variables have an identical standard deviation. Call it $\sigma_y$. Although all $y$ variables have the same standard deviation, the $y$ values may be correlated or uncorrelated.

I've read an article that seems to depend on the following being true:

$Cov(\bar{x},\bar{y})=\rho_{xy}\sigma_x\sigma_y$ [Equation 2]

where $\rho_{xy}$ is the correlation between a randomly-chosen pair of $x_i$ and $y_j$ variables.

How does Equation 2 follow from Equation 1?

I can see that the following is true, but not why this implies Equation 2:

$Cov(\bar{x},\bar{y})=\rho_{\bar{x}\bar{y}}\sigma_{\bar{x}}\sigma_{\bar{y}}$ [Equation 3]

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  • $\begingroup$ This question is confused. The question seeks the covariance between two population means ... but then is all about the covariance between sample means. $\endgroup$ – wolfies Oct 10 '17 at 10:40
  • $\begingroup$ Maybe it was phrased confusingly, but it is still about the covariance between two population means: $n_x$ and $n_y$ are the total numbers of each variable in the population, rather than a sample of them. $\endgroup$ – Sprog Oct 10 '17 at 12:55
  • $\begingroup$ Is the confusion because I used the notation $\bar{x}$ and $\bar{y}$ rather than $\mu_x$ and $\mu_y$? $\endgroup$ – Sprog Oct 10 '17 at 12:59
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Covariance is bilinear so that:

$$\begin{aligned}\begin{aligned}\mathsf{Cov}\end{aligned} \left(\bar{x},\bar{y}\right) & =\begin{aligned}\mathsf{Cov}\end{aligned} \left(\frac{1}{n_{x}}\sum_{i=1}^{n_{x}}x_{i},\frac{1}{n_{y}}\sum_{j=1}^{n_{y}}y_{j}\right)\\ & =\frac{1}{n_{x}}\frac{1}{n_{y}}\sum_{i=1}^{n_{x}}\sum_{j=1}^{n_{y}}\mathsf{Cov}\left(x_{i},y_{j}\right)\\ & =\frac{1}{n_{x}}\frac{1}{n_{y}}\sum_{i=1}^{n_{x}}\sum_{j=1}^{n_{y}}\sigma_{x}\sigma_{y}\rho_{x_{i},y_{j}}\\ & =\sigma_{x}\sigma_{y}\bar{\rho} \end{aligned} $$

where $\bar{\rho}=\frac{1}{n_{x}}\frac{1}{n_{y}}\sum_{i=1}^{n_{x}}\sum_{j=1}^{n_{y}}\rho_{x_{i},y_{j}}$.

If $\rho_{x_{i},y_{j}}$ does not depend from the indices $i,j$ then we can write $\rho_{x_{i},y_{j}}=\rho_{x,y}$ and find $\bar{\rho}=\rho_{x,y}$.

So in that case we end up with: $$\begin{aligned}\mathsf{Cov}\end{aligned} \left(\bar{x},\bar{y}\right)=\sigma_{x}\sigma_{y}\rho_{x,y}$$

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  • $\begingroup$ Hallelujah! That is very clear! $\endgroup$ – Sprog Oct 10 '17 at 9:12

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