Suppose $X$ and $Y$ have a bivariate normal distribution with parameters ${\mu _X} = {\mu _Y} = 0$, ${\sigma _X}^2 = {\sigma _Y}^2 = 1$, and $\rho = {\rho _{X,Y}} \ne 0$. I'm asked to find the correlation coefficient between $XY$ and $X$.
So far I know $$\rho \left( {XY,Y} \right) = \frac{{{\text{Cov}}\left( {XY,X} \right)}}{{{\sigma _{XY}} \cdot {\sigma _X}}}$$
Now finding ${{\text{Cov}}\left( {XY,X} \right)}$ and ${\sigma _{XY}} = \sqrt {{\sigma _{XY}}^2}$ is where I'm stuck. $${\text{Cov}}\left( {XY,X} \right) = E\left[ {XY \cdot X} \right] – E\left[ {XY} \right] \cdot E\left[ X \right] = E\left[ {{X^2}Y} \right]$$
How do I continue and find $E\left[ {{X^2}Y} \right]$ and ${{\sigma _{XY}}^2}$? Do I need to integrate?
Best Answer
Another aprroach is to use the law of total expectation thus
$$\mathbb{E}[X^2Y]=\mathbb{E}[\mathbb{E}(X^2Y|Y)]=\mathbb{E}[y\mathbb{E}(X^2|Y)]$$
And the quantity
$$\mathbb{E}(X^2|Y)$$
is known because it is known how to factorize
$$f(x,y)=f(y)f(x|y)$$
into two gaussian densities
In your case you have
$$f_{XY}(x,y)=\frac{1}{2\pi\sqrt{1-\rho^2}}\text{exp}\left\{-\frac{1}{2(1-\rho^2)} \left[ x^2-2\rho xy+y^2 \right] \right\}=\dots=$$
$$=\frac{1}{\sqrt{2\pi}}e^{-y^2/2}\cdot\underbrace{ \frac{1}{\sqrt{2\pi}\sqrt{1-\rho^2}}e^{-(x-\rho y)^2/[2(1-\rho^2)]}}_{N(\rho y;1-\rho^2)}$$
thus
$$\mathbb{E}[X^2|Y]=\mathbb{V}[X|Y]+\mathbb{E}^2[X|Y]=1-\rho^2+\rho^2y^2$$
Concluding...
$$\mathbb{E}[X^2Y]=\mathbb{E}[Y(1-\rho^2)+\rho^2Y^3]=0$$