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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 $E(X^2Y^2)$.

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!

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    Alternatively, find $(a,b,c)$ such that $(X,Y)=(aU,bU+cV)$ in distribution, where $(U,V)$ is i.i.d. standard normal (thus uses that $E(X)=E(Y)=0$), then note that $$E(X^2Y^2)=a^2(b^2E(U^4)+c^2E(U^2)E(V^2))=a^2(3b^2+c^2)$$ – 2017-01-16
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    There should be a linear transformation $\displaystyle \left. \begin{cases} u = ax+by \\ v = bx + cy \end{cases} \right\}$ that transforms $x^2-2\rho xy+y^2$ to $u^2+v^2.$ You'll notice the letter $b$ in two places, i.e. you can make the matrix symmetric. Then you'll have $e^{-(u^2+v^2)} = e^{-u^2} e^{-v^2}$ and then you can separate the integral with respect to $u$ from the integral with respect to $v$. (I used $\rho$ rather than $p$, as is more conventional.) – 2017-01-16
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    @Did Do you mind explaining why X transforms to aU, whereas Y transforms to bU + cV? (i.e. rather than X = aU+bV and Y=cU+dV?) – 2017-01-16
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    @MichaelHardy: Thanks! How do we know that we can make the matrix symmetric? – 2017-01-16
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    You may try to represent (X,Y) as (aU+bV,cU+dV) if you prefer. Of course, the representation as (aU,bU+cV) is equally valid and the computations it yields are simpler, but really, now, what is lacking from the picture is some genuine try from you to apply one of these suggestions, any of them, not the suggestions themselves. – 2017-01-16

2 Answers 2

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

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\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. ${}$