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Is this because when X and Y are independent, E(XY) = E(X)E(Y).

So Cov(X,Y)= E(XY)-E(X)E(Y)=0?

Or is there another way of solving this through using Law of Iterated Expectations?

  • 3
    $\begingroup$ Why doesn't the answer to your previous question address this one, too? After all, it's a generalization of this one with $b=0$. Is it really necessary to re-ask it? $\endgroup$
    – whuber
    Commented Oct 6, 2017 at 15:10

1 Answer 1


$Cov[X,Y] = E[XY] - E[X]E[Y]$

$E[XY] = E[E[XY|X]] = E[X * E[Y|X]]= E[X * E[Y]] = E[Y] * E[X] $ since $$ E[Y]$$ is a constant in the expression $$ E[X *E[Y]]$$.

Hope this helps.


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