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The expectation over $x$ of a covariance between variables $A$ and $B$ (where the distribution of $B$ varies according to $x$) is equal to the covariance of $A$ with the expectation of $B$ over $x$:

$$ E_x[Cov(A,B)]=Cov(A,E_x[B]) $$

I wonder, then, whether it is also true that the variance of the covariance follows the same pattern? i.e.

$$ Var_x[Cov(A,B)]=Cov(A,Var_x[B]) $$

This seems like it might be true: can anyone confirm this?

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  • $\begingroup$ I'm not sure what you mean by $E_x[B]$. Does that mean the expectation of $B$ given that $A=x$? In other words, what is $x$ in this problem - is it a value of a variate, or perhaps a parameter of the $B$ distribution? $\endgroup$ Apr 28, 2017 at 15:30
  • $\begingroup$ Sorry, perhaps I wasn't very clear. By $E_x[B]$, I mean the expected value of $B$ across a series of different states, where each state is a different $x$. For instance, in the state of the world $x$, a factor $A$ (such as the global oil price, for instance), might have a particular covariance with the share value $B$ of a company. This covariance will differ depending on the state the world is in (e.g. whether Donald Trump is in power). Each state is a different $x$. $\endgroup$
    – Sprog
    Apr 28, 2017 at 15:34
  • $\begingroup$ This explanation makes it even less clear (and your use of a political event is a distraction that does nothing to elucidate the meaning). $\endgroup$
    – Ben
    Apr 17, 2018 at 2:28
  • $\begingroup$ Hi Ben, maybe it's clearer if the $x$ notation is just left out? Just write $E[Cov(A,B)]=Cov(A,E[B])$. Basically, I have in mind a situation in which the covariance may be strong or weak (because the distribution of $B$ can differ), so there are a range of possible values for the covariance. These possible values form a distribution with an expectation and a variance. $\endgroup$
    – Sprog
    Apr 17, 2018 at 11:55

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