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I have a fairly simple question that I'm currently unable to solve.

I have to calculate:

$\ Cov(Y,Y+e^X)$

To get here I've had to calculate:

$\ P_Y(y)$

$\ P_X(x)$

$\ Var(X)$

$\ Var(Y)$

Bonus info: X and Y are independent!

Having solved the above my idea is to calculate $\ Cov(Y,Y+e^X)$, so:

$\ Cov(Y,Y+e^X)= $$\ Cov(Y,Y)+Cov(Y,e^X)$ $\ =Var(Y)+Cov(Y,e^X)$

Since I already have $Var(Y)$ calculated, what I need is:

$Cov(Y,e^X)$

My idea is to use the rule:

$Cov(aX,Y)=a·Cov(X,Y)$

This would be smart considering that X,Y are independent.

But this is where uncertainty kicks in. The "problem" is that I'm unsure how I'm allowed to rewrite this since all examples I can find is either a constant that is multiplied or additive. Now X is lifted as an exponent.

The issue is that if I just blindly use the rule I get the correct result, but to me it's just as much about the method as it is about the answer.

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    $\begingroup$ If $X$ and $Y$ are independent then $e^X$ and $Y$ are independent. $\endgroup$ Dec 27 '17 at 15:35
  • $\begingroup$ Thank you for the reply! Much appreciated. $\endgroup$
    – Fauré
    Dec 27 '17 at 16:47
  • $\begingroup$ Could you post it as an answer so that I can mark the question as answered? $\endgroup$
    – Fauré
    Jan 2 '18 at 13:09
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If $X$ and $Y$ are independent random variables, then $f(X)$ and $g(Y)$ are also independent for any nice enough (Borel measurable) functions $f$ and $g$. In particular, $\text{Cov}(f(X),g(Y))=0$. In your example $$\text{Cov}(Y,Y+e^X)=\text{Cov}(Y,Y)+\text{Cov}(Y,e^X)=\text{Var}(Y)+0$$ etc.

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