# Is the expectation of $xy/(x^2 + (x+y)^2)$ for i.i.d. standard normal $x, y$ finite?

Let $$x, y$$ be i.i.d. standard normal random variables. Is the following expectation bounded?

$$\mathbb{E}\Big[ \frac{xy}{x^2 + (x+y)^2}\Big]$$

I used Wolfram alpha and simulation to compute the above expectation and in both cases, I got the value -0.2. I'm not sure if I can trust this result because ratio distributions are often heavy-tailed.

Another thing that I've observed is that although the ratio $$x/y$$ has Cauchy distribution whose expectation is undefined, Wolfram Alpha computes the expectation $$\mathbb{E}[x/y]$$ as zero.

Not just the expectation is bounded; there isn’t even a singularity in the function itself. Numerator and denominator are both homogeneous functions of degree $$2$$, so in polar coordinates the value depends only on the angular coordinate and not on the radial coordinate. With $$x=r\cos\phi$$ and $$y=r\sin\phi$$, we have
$$\frac{xy}{x^2+(x+y)^2}=\frac{\cos\phi\sin\phi}{\cos^2\phi+(\cos\phi+\sin\phi)^2}=\frac{\sin2\phi}{3+\cos2\phi+2\sin2\phi}\;.$$
The denominator is strictly positive for all angles. In fact, we can integrate over $$\phi$$ to compute the expectation; the result is $$-\frac15$$, in agreement with what you found. This applies to any rotationally symmetric joint distribution of $$x,y$$, not just to the standard normal one, since the function is independent of the radial coordinate.
• Is it possible to extend this analysis to the case where we have more than 2 i.i.d. standard normal random variables? For example for n = 3, is the expectation of $\frac{xy + yz + xz}{x^2 + (x+y)^2 + (x+y+z)^2}$ bounded? – KRL Jan 13 at 0:26
• @KRL: Certainly. Again, numerator and denominator are both homogeneous functions of degree $2$ and thus don't depend on the radial coordinate, and the denominator is positive in all directions, so there's no singularity at the origin. You can use spherical coordinates $x=r\sin\theta\cos\phi$, $y=r\sin\theta\sin\phi$ and $z=r\cos\theta$, but I can't get Wolfram|Alpha to evaluate the resulting complicated two-dimensional integral. – joriki Jan 13 at 1:09
• Is the reverse of this argument also true? For example, can we say the expectation of $\frac{(x+y)^2}{(x^2+(x+y)^2)^2}$ is unbounded because the numerator is proportional to $r^2$ and the denominator is proportional to $r^4$? @joriki – KRL Jan 14 at 21:05