Continuous Joint Distribution: Find $E(X^2Y^2)$ Question: Suppose $X$ and $Y$ have a jointly continuous distribution with joint density:

$$f(x,y)= \frac 1 {2\pi \sqrt{1-p^2}} e^{ \frac {-(x^2 - 2pxy + y^2)}{2(1-p^2)}}$$

where $p$ is a constant for which $|p| &lt; 1$. Find $E(X^2Y^2)$.



I tried to find $\displaystyle E(X^2Y^2) =  \iint x^2y^2f(x, y) \,dx\,dy. $

where $f(x,y)$ is the joint density, but I'm not sure how to take the integral.

Any help would be appreciated!

Thanks!
According the expression of density function of $(X,Y)$, $(X,Y)$ is joint Gaussian distributed with $\mathbb{E}[X]=\mathbb{E}[Y]=0$, $\mathbb{E}[X^2]=\mathbb{E}[Y^2]=1$ and $\mathbb{E}[XY]=\rho$. Usually, if $(X_1,X_2,X_3,X_4)$ is multivariate
Gaussian distributed with $\mathbb{E}[X_i]=0, 1\le i\le 4$, then the following formula is convenient to calculate its forth moments:
$$ \mathbb{E}[X_1X_2X_3X_4]=\mathbb{E}[X_1X_2]\mathbb{E}[X_3X_4]+\mathbb{E}[X_1X_3]\mathbb{E}[X_2X_4]+\mathbb{E}[X_1X_4]\mathbb{E}[X_2X_3]. \tag{*}$$
In (*), let $X_1=X_2=X$ and $X_3=X_4=Y$, it is easy to get the follows:
$$ \mathbb{E}[X^2Y^2]=\mathbb{E}[X^2]\mathbb{E}[Y^2]+2(\mathbb{E}[XY])^2=1+2\rho^2.$$
\begin{align}
x = {} &amp; u + v \\
y = {} &amp; u -v \\[10pt]
dx\,dy = {} &amp; \left|\frac{\partial(x,y)}{\partial(u,v)}\right| \,du\,dv = 2\,du\,dv \\[10pt]
&amp; (x^2-2\rho xy + y^2) \\[4pt]
= {} &amp; (u+v)^2 - 2\rho(u+v)(u-v) + (u-v)^2 \\[4pt]
= {} &amp;  2(1-\rho)u^2 + 2(1+\rho) v^2
\end{align}
\begin{align}
&amp; \iint \exp \Big(-(x^2-2\rho xy + y^2)^2/(2(1-\rho^2))\Big) \,dx\,dy \\[10pt]
= {} &amp; \iint \exp \Big(-(2(1-\rho)u^2 + 2(1+\rho)v^2)/(2(1-\rho^2))\Big) (2\,du\,dv) \\[10pt]
= {} &amp; \iint \exp\left( \frac{-u^2}{1+\rho} \right) \exp\left( \frac{-v^2}{1-\rho} \right) (2\,du\,dv)
\end{align}
This can be written as a product of an integral with respect to $u$ and one with respect to $v$.

But you also have the factor $x^2y^2$. You need to write that in terms of $u$ and $v$ and then expand it and get a sum, so you then get a sum of integrals. ${}$