$\def\Cov{\operatorname{Cov}}\def\Corr{\operatorname{Corr}}\def\Var{\operatorname{Var}}$ Suppose that $X$ and $Y$ have a joint normal distribution with $E(X)=E(Y)=0, \Var(X)=\Var(Y)=1, \Corr(X,Y)=\rho$, how can I find $\Corr(XY,Y)$ and $\Corr(X^{2},Y^{2})$

I have no idea how to compute them, does anyone could help me? Thanks.

  • $\begingroup$ Do you know that $\operatorname{corr}(XY,Y)=\dfrac{\operatorname{cov}(XY,Y)}{\sqrt{\operatorname{var}(XY)\operatorname{var}(Y)}}$? If not, then you must have run into difficulty after you wrote that down and thought about how to evaluate it. But you haven't told us that, and you should have. If you don't know that that's how correlation is defined, then the way out of your difficulty begins with learning that. $\endgroup$ – Michael Hardy Dec 12 '14 at 1:16
  • $\begingroup$ Sorry I forgot to mention I know the formula, but I cannot find $Cov(XY,Y)$ $\endgroup$ – user133140 Dec 12 '14 at 1:24
  • $\begingroup$ More generally, for $Cov(X Y, Y)$, see also: stats.stackexchange.com/questions/116535/… $\endgroup$ – wolfies Dec 12 '14 at 11:29

$\def\cov{\operatorname{cov}}\def\corr{\operatorname{corr}}\def\var{\operatorname{var}}\def\E{\operatorname{E}}$ $$ \cov(XY,Y) = \E(XY^2)-\E(XY)\E(Y) = \E(XY^2) = \E(\E(XY^2\mid Y)) = \E(Y^2\E(X\mid Y)). $$ So we need to know that $\E(X\mid Y) = \rho Y$. Then you have $\rho\E(Y^3) = 0$.

With the bivariate normal distribution you have $$ \E\left(\frac{X-\mu_X}{\sigma_X} \mid Y\right) = \rho \left( \frac{Y-\mu_Y}{\sigma_Y} \right). $$ Is that something you've seen before?

  • $\begingroup$ Why $XY^{2}$ becomes $XY$ after using the law of total expectations? $\endgroup$ – user133140 Dec 12 '14 at 1:50
  • $\begingroup$ What I did is keeping $XY^{2}$ remaining $\endgroup$ – user133140 Dec 12 '14 at 1:53
  • $\begingroup$ Sorry --- that was a typo. $\endgroup$ – Michael Hardy Dec 12 '14 at 1:53
  • $\begingroup$ Alright, now I know what mistake I have made, but why $E(Y^{3})=0$? I have no idea how to compute it, thanks again $\endgroup$ – user133140 Dec 12 '14 at 2:01
  • $\begingroup$ The distribution of $Y^3$ is identical to that of $-Y^3$; hence its expectation is $0$, provided only that $\operatorname{E}(|Y^3|)<\infty$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 12 '14 at 2:03

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