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I have been trying to find an expression for the covariance of two normally distributed variables $X$ and $Y$ if $cov(X,Y)=c$ then $cov(X,XY)=?$

I would greatly appreciate any help. Probably it must be something very simple that i am missing

Christoph Hanck
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vanbasten
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    Hint: First, you need to know something about the joint distribution of $X$ and $Y$. Try to find the conditional covariance of $X$ and $XY$ conditioned on $Y = y$. Then find the unconditional covariance of $X$ and $XY$ – Dilip Sarwate May 13 '18 at 21:17
  • Thank you. What I did might seem illogical but basically Cov(x,xy)=E(x^2*y)-E(x)E(xy) Then E(x^2)E(y)+cov(x^2,y)=var(x)E(y)+2E(x)cov[x,y]

    Using steins lemma for second term above

    And doing same thing for E(xy) I get express everything with two variables and no multiplication. But not sure if I am doing the right thing

     – vanbasten May 14 '18 at 21:58
  • I have no idea how you get to the second equation in your comment from the first one, and you still don't seem to understand my point about the joint distribution of $X$ and $Y$. For the case of independent $X$ and $Y$ (whether normal or not), I show in this answer that $\operatorname{cov}(X,XY)=\operatorname{var}(X)E[Y]$. For jointly normal but dependent $X$ and $Y$ , one can say more – Dilip Sarwate May 15 '18 at 15:49
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    If $X$ and $Y$ are jointly binormal, then the joint moment generating function of $(X,Y)$ is $$ M(s,t)=\exp(\mu_X s + \mu_Y t + \sigma_X^2 s^2/2 + \sigma_Y^2 t^2/2 + c st), $$ where $c=\operatorname{Cov}(X,Y)$. From this you can work out $$E(X^2 Y) = \frac{\partial^3}{\partial s^2\partial t}M(s,t)|_{s=0,t=0} $$ and $$ \operatorname{Cov}(X,XY)=E(X^2Y)-EXE(XY) $$ in terms of $\mu_X$, $\mu_Y$, $\sigma_X$, $\sigma_Y$ and $c$. – Jarle Tufto May 18 '18 at 20:26
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    But note, as also pointed out by @DilipSarwate, that the joint distribution is not known from the assumptions you are making. – Jarle Tufto May 18 '18 at 20:28

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