If $X \sim N(0, \sigma^2)$ and $Y \sim N(0, \sigma^2)$ are independent, how can we find the expectation $$E \left(\frac{X }{\sqrt{X^2+Y^2} }\right)\,?$$
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2For completeness sake, the PDF of $X/\sqrt{X^2+Y^2}$ is $1/(\pi\sqrt{1-x^2})$ for $-1<x<1$. – COOLSerdash Aug 18 '19 at 06:53
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From basic geometry it's just the cos of a random angle (reducing calculation of the pdf to a previously solved problem), not that you need it to get the expectation. The pdf is indeed as COOLSerdash indicates. – Glen_b Aug 19 '19 at 04:03
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You don't need to make use of $S$ and/or $T$. The conditional expectation $E\left[\frac{X}{\sqrt{X^2+Y^2}}\bigg\vert Y\right]$ is $0$ because PDF of $X$ is an even function and the expression we want to take the expectation is an odd function, which makes the overall expression in the integral an odd function, so the integration will be $0$. A detail: we can use this symmetry argument when the integral expression converges, and it is converging. Finally, use the law of iterated expectation to show that the overall expectation is $0$, i.e. $$E\left[\frac{X}{\sqrt{X^2+Y^2}}\right]=E\left[E\left[\frac{X}{\sqrt{X^2+Y^2}}\bigg\vert Y\right]\right]=E[0]=0$$
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