9
$\begingroup$

I have two normally distributed variables $X_1$ and $X_2$ with mean zero and covariance matrix $\Sigma$. I am interested in trying to calculate the value of $E[X_1^2 X_2^2]$ in terms of the entries of $\Sigma$.

I used the law of total probability to get $E[X_1^2 X_2^2] = E[X_1^2 E[X_2^2 | X_1]]$ but I'm not sure what the inner expectation reduces to. Is there another method here?

Thanks.

Edit: The variables are also multivariate normally distributed.

$\endgroup$
  • 5
    $\begingroup$ Do $X_1$ and $X_2$ enjoy a bivariate normal distribution also? (Just saying that $X_1$ and $X_2$ are normal with covariance matrix $\Sigma$ is not quite enough to conclude that the joint distribution is bivariate normal). $\endgroup$ – Dilip Sarwate Jan 9 '15 at 19:36
  • 1
    $\begingroup$ For the specific application I have in mind, $X_1$ and $X_2$ do have a bivariate normal distribution, by the multivariate central limit theorem. I forgot to mention this in my original post. $\endgroup$ – AGK Jan 9 '15 at 20:41
  • 1
    $\begingroup$ @AGK if you want to clarify your post, there is an "edit" button that lets you make changes. That's better for future readers who then don't have to look up key information in the comments beneath the question. $\endgroup$ – Silverfish Jan 9 '15 at 23:21
8
$\begingroup$

The expectation clearly is proportional to the product of the squared scale factors $\sigma_{11}\sigma_{22}$. The constant of proportionality is obtained by standardizing the variables, which reduces $\Sigma$ to the correlation matrix with correlation $\rho = \sigma_{12}/\sqrt{\sigma_{11}\sigma_{22}}$.

Assuming bivariate normality, then according to the analysis at https://stats.stackexchange.com/a/71303 we may change variables to

$$X_1 = X,\ X_2 = \rho X + \left(\sqrt{1-\rho^2}\right) Y$$

where $(X,Y)$ has a standard (uncorrelated) bivariate Normal distribution, and we need only compute

$$\mathbb{E}\left(X^2 (\rho X + \left(\sqrt{1-\rho^2}\right) Y)^2\right) = \mathbb{E}(\rho^2 X^4 + (1-\rho^2)X^2 Y^2 + c X^3 Y)$$

where the precise value of the constant $c$ does not matter. ($Y$ is the residual upon regressing $X_2$ against $X_1$.) Using the univariate expectations for the standard normal distribution

$$\mathbb{E}(X^4)=3,\ \mathbb{E}(X^2) = \mathbb{E}(Y^2)=1,\ \mathbb{E}Y=0$$

and noting that $X$ and $Y$ are independent yields

$$\mathbb{E}(\rho^2 X^4 + (1-\rho^2)X^2 Y^2 + c X^3 Y) = 3\rho^2 + (1-\rho^2) + 0 = 1 + 2\rho^2.$$

Multiplying this by $\sigma_{11}\sigma_{22}$ gives

$$\mathbb{E}(X_1^2 X_2^2) = \sigma_{11}\sigma_{22} + 2\sigma_{12}^2.$$


The same method applies to finding the expectation of any polynomial in $(X_1,X_2)$, because it becomes a polynomial in $(X, \rho X + \left(\sqrt{1-\rho^2}\right)Y)$ and that, when expanded, is a polynomial in the independent normally distributed variables $X$ and $Y$. From

$$\mathbb{E}(X^{2k}) = \mathbb{E}(Y^{2k}) = \frac{(2k)!}{k!2^k} = \pi^{-1/2} 2^k\Gamma\left(k+\frac{1}{2}\right)$$

for integral $k\ge 0$ (with all odd moments equal to zero by symmetry) we may derive

$$\mathbb{E}(X_1^{2p}X_2^{2q}) = (2q)!2^{-p-q}\sum_{i=0}^q \rho^{2i}(1-\rho^2)^{q-i}\frac{(2p+2i)!}{(2i)! (p+i)! (q-i)!}$$

(with all other expectations of monomials equal to zero). This is proportional to a hypergeometric function (almost by definition: the manipulations involved are not deep or instructive),

$$\frac{1}{\pi} 2^{p+q} \left(1-\rho ^2\right)^q \Gamma \left(p+\frac{1}{2}\right) \Gamma \left(q+\frac{1}{2}\right) \, _2F_1\left(p+\frac{1}{2},-q;\frac{1}{2};\frac{\rho ^2}{\rho ^2-1}\right).$$

The hypergeometric function times $\left(1-\rho ^2\right)^q$ is seen as a multiplicative correction for nonzero $\rho$.

$\endgroup$
  • 1
    $\begingroup$ Thanks for the detailed answer! I am also thinking about related questions with other polynomials, so this is a really helpful framework. That's a very clever transformation I hadn't seen before. Cool! $\endgroup$ – AGK Jan 9 '15 at 20:40
  • 3
    $\begingroup$ To help out your investigation, I have supplied the details for general polynomials. I was amused, when originally writing this answer, to realize that I learned this transformation from the elementary stats textbook by Friedman, Pisani, and Purves: we teach this to college freshmen! $\endgroup$ – whuber Jan 9 '15 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.