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I've been thinking about problem that, conceptually is very simple. I have two random variables $F_{x}$ and $F_{y}$ both of which follow a Gaussian distribution of mean $0$ and variance $1$. Now, I wish to compute the expectation value, $\langle\sqrt{F_{x}^2 + F_{y}^2}\rangle$. It's been more than a while since I did any probability theory and am wondering if there is an easy way of doing this. I keep on running into nasty integrals, but I may be approaching the problem from the wrong direction.

Thanks,

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The integration is not bad. Change to polar coordinates. One integration by parts, and we are at an integral that is familiar. We don't even need to integrate by parts, since the integral with respect to $r$ can be found from the variance.

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Hi Andre, could you elaborate a bit more on how to find the probability density function for $\sqrt{F_{x}^2 + F_{y}^2}$? –  covertbob Oct 14 '12 at 23:15
    
One could find the pdf. But we don't need to. The expectation is the double integral of $\sqrt{x^2+y^2}$ times the joint density function, which is $\frac{1}{2\pi}e^{-(x^2+y^2)/2}$. Now change to polar coordinates. –  André Nicolas Oct 14 '12 at 23:40
    
Ah, thank you. The problem is trivial now. Just out of curiosity, how would you go about finding the pdf? –  covertbob Oct 14 '12 at 23:43
    
Let $W$ be our square root. We want the cdf of $W$, that is, $\Pr(\sqrt{X^2+Y^2}\le w)$. This is the double integral of the joint density over the circle centre the origin, radius $w$. To evaluate the double integral, change to polar coordinates. The $r \,dr\,d\theta$ gives us an integral that is easy by substitution $u=r^2/2$. So now we have the cdf, differentiate for the density. –  André Nicolas Oct 15 '12 at 0:24

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