# calculating the expectation

Let $Y$ and $X$ be independent centered normal random variables with variances $\frac{\sigma^2}{1-\theta^2}$ and 1 respectively. How can I compute

$$E\left[\frac{YX}{Y^2 + (\theta Y+X)^2}\right]$$

where $\theta$ is a constant?

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What are their expectation? – Davide Giraudo Feb 29 '12 at 22:00
@DavideGiraudo: I forgot to add their expectation. Thanks – neticin Feb 29 '12 at 22:06

Let $Z = Y/\sigma_Y = \sqrt{1-\theta^2} Y/\sigma$ (I'm assuming here that $|\theta| < 1$) to express this as $X Z/(a X^2 + b X Z + c Z^2)$ for some (rather messy) constants $a,b,c$ where $X$ and $Z$ are independent standard normal and $a x^2 + b x y + c y^2$ is a positive definite quadratic form. We can write $X = R \cos(\phi)$ and $Z = R \sin(\phi)$ where $R$ and $\phi$ are independent, and $\phi$ is uniform on $[0,2\pi]$. Then your expected value is $$\frac{1}{2 \pi} \int_0^{2 \pi} \frac{\cos(\phi)\sin(\phi)}{a\ \cos^2(\phi) + b\ \cos(\phi)\sin(\phi) + c\ \sin^2(\phi)}\ d\phi$$ which can be calculated using residues.