$X，Y$ i.i.d. $N(0,\sigma^2)$. What's the probability of $P(X>|2Y|)$? We know that $X-2Y \sim N(0,5\sigma^2)$, but how to compute $P(X>|2Y|)$?
 A: $P[X>|2Y|] = P[X>0]*P[X^2 > 4Y^2 | X > 0] = 0.5*P[\frac{Y^2}{X^2}<0.25]$
Now, note that $X^2$ and $Y^2$ are both squares of standard normal variates which means they follow $\chi^2_{(1)}$ distribution. Now you can use the result that if U and V are independent and follow chi squared distribution with degrees of freedom $a$ and $b$ respectively then the ratio $\frac{U}{V}$ follows F-Distribution with parameters $a$ and $b$ respectively.
Now to find $P[\frac{Y^2}{X^2}<0.25]$ use the tables or if you're not allowed to do that just integrate the pdf of F-Distribution over the specified range. The integration should not be much time consuming since the degrees of freedom are (1,1) hence the pdf will reduce to a relatively simple function.
A: Here is a hint. Since $X$ and $Y$ are independent, their joint density is $f_{XY}(x,y)=f_X(x)f_Y(y)=\frac{1}{2\pi\sigma^2}\exp(-\frac{x^2+y^2}{2\sigma^2})$. Further, because of symmetry,
$$
\text{Prob}[X>|2Y|] = 2\text{Prob}[X>2Y,Y>0]
= 2 \int_0^{+\infty} \text{d}y \int_{2y}^{+\infty} \text{d}x f_{XY}(x,y)\,,
$$
which you can easily solve if you transform to polar coordinates.
Another way of looking at it is to think of $f_{XY}(x,y)$ as a perfectly round cake out of which you take a wedge with angle $2\text{arctan}(\frac{1}{2})$. The probability mass of that wedge should therefore be $\text{arctan}(\frac{1}{2})/\pi \approx 0.4636/3.1415=0.1476$. 
