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Trying to calculate E[(XY)^2] for X, Y ~ N(0, 1) and Cov(X, Y) = p.

oblixps

Member
May 20, 2012
38
Let X, Y be random variables distributed as N(0, 1) and let Cov(X, Y) = p. Calculate E[X2Y2].

I have that Cov(X, Y) = E[(X - 0)(Y - 0)] = E[XY] = p.

I have no idea how to continue on however. I can't think of a way to relate E[X2Y2] with E[XY]. I have considered the Var(XY) but that also involved E[X2Y2] and thus wouldn't help me in finding E[X2Y2].

Can someone offer a hint or two on how to proceed?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Let X, Y be random variables distributed as N(0, 1) and let Cov(X, Y) = p. Calculate E[X2Y2].

I have that Cov(X, Y) = E[(X - 0)(Y - 0)] = E[XY] = p.

I have no idea how to continue on however. I can't think of a way to relate E[X2Y2] with E[XY]. I have considered the Var(XY) but that also involved E[X2Y2] and thus wouldn't help me in finding E[X2Y2].

Can someone offer a hint or two on how to proceed?
In general a bivariate normal distribution is written as...

$\displaystyle f_{X,Y} (x,y) = \frac{1}{2\ \pi\ \sigma_{X}\ \sigma_{Y}\ \sqrt{1 - \rho^{2}}}\ e^{- \frac{z}{2\ (1-\rho^{2})}}\ (1)$

... where...

$\displaystyle z = \frac{(x - \mu_{X})^{2}}{\sigma_{X}^{2}} - 2\ \frac{\rho\ (x - \mu_{X})\ (y - \mu_{Y})}{\sigma_{X}\ \sigma{Y}} + \frac{(y - \mu_{Y})^{2}}{\sigma_{Y}^{2}}\ (2)$

$\displaystyle \rho= \text{cor}\ (X,Y) = \frac{V_{X,Y}}{\sigma_{X}\ \sigma_{Y}}\ (3)$

In Your case is $\mu_{X}=\mu_{Y}=0$, $\sigma_{X} = \sigma_{y} = 1$, $V_{X,Y}= \rho= p$ , so that it should be...

$\displaystyle E\ \{X^{2}\ Y^{2} \} = \frac{1}{2\ \pi\ \sqrt{1 - p^{2}}}\ \int_{- \infty}^{+ \infty} \int_{- \infty}^{+ \infty} x^{2}\ y^{2}\ e^{- \frac{x^{2} - 2\ p\ x\ y + y^{2}}{2\ (1-p^{2})}}\ d y\ d x\ (4) $

The solution of the integral (4) is not so easy and it is delayed in next posts...

Kind regards

$\chi$ $\sigma$