This question is relevant to bivariate normal distribution. I figured out most of the steps but am currently stuck on finding the closed form of a double integral.

Let $x$ and $y$ be of a bivariate normal distribution. Note that $x$ and $y$ are not independent, we denote $\rho$ to be the correlation between the two.

I would like to find the closed form of (approximation of) $\int_{-\infty}^\infty \int_a^b xy f(x,y) d_x d_y$

Here $f(x,y)$=$\frac{1}{2 \pi \sigma_x \sigma_y \sqrt(1-\rho^2)}exp(-\frac{1}{2(1-\rho^2)}[\frac{(x-\mu_x)^2}{\sigma_x^2}+\frac{(y-\mu_y)^2}{\sigma_y^2}-\frac{2 \rho (x-\mu_x) (y-\mu_y)}{\sigma_x \sigma_y}])$

That is, I would like some help in simplifying

$\int_{-\infty}^\infty \int_a^b xy \frac{1}{2 \pi \sigma_x \sigma_y \sqrt(1-\rho^2)}exp(-\frac{1}{2(1-\rho^2)}[\frac{(x-\mu_x)^2}{\sigma_x^2}+\frac{(y-\mu_y)^2}{\sigma_y^2}-\frac{2 \rho (x-\mu_x) (y-\mu_y)}{\sigma_x \sigma_y}])d_xd_y$

  • $\begingroup$ Have you tried directly asking WolframAlpha to evaluate? If it can be done by hand, you might be able to re-express the problem as letting $X \sim N(\mu_x,\sigma_x^2),Y=\mu_y+\rho (X-\mu_x)/\sigma_x+ \sqrt{1-\rho^2} \sigma_y Z$ where $Z$ is indep. of $X$ and distributed $Z\sim N(0,1)$. Or something like that, making sure that the marginals of X and Y line up and the correlation is $\rho$ $\endgroup$ – Ray Oct 6 '16 at 1:26
  • 1
    $\begingroup$ Integrate $y$ first. Then it's a standard Normal integral for $x$. You can write down the answer simply by looking at it! $\endgroup$ – whuber Oct 6 '16 at 13:06

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