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]
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: $$\begin{aligned}\mathsf{Cov}\end{aligned} \left(\bar{x},\bar{y}\right)=\sigma_{x}\sigma_{y}\rho_{x,y}$$