Continuous Joint Distribution: Find $E(X^2Y^2)$

Question

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 $|p| < 1$. Find $E(X^2Y^2)$.

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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!

Answer 1

\begin{align} x = {} & u + v \\ y = {} & u -v \\[10pt] dx\,dy = {} & \left|\frac{\partial(x,y)}{\partial(u,v)}\right| \,du\,dv = 2\,du\,dv \\[10pt] & (x^2-2\rho xy + y^2) \\[4pt] = {} & (u+v)^2 - 2\rho(u+v)(u-v) + (u-v)^2 \\[4pt] = {} & 2(1-\rho)u^2 + 2(1+\rho) v^2 \end{align} \begin{align} & \iint \exp \Big(-(x^2-2\rho xy + y^2)^2/(2(1-\rho^2))\Big) \,dx\,dy \\[10pt] = {} & \iint \exp \Big(-(2(1-\rho)u^2 + 2(1+\rho)v^2)/(2(1-\rho^2))\Big) (2\,du\,dv) \\[10pt] = {} & \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. ${}$

Answer 2

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: $$ \mathbb{E}[X^2Y^2]=\mathbb{E}[X^2]\mathbb{E}[Y^2]+2(\mathbb{E}[XY])^2=1+2\rho^2.$$