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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

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](https://stats.stackexchange.com/a/340131/6633) 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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