Two independent random standard normal variable X and Y, find the probability that:
$P(X>2Y)$ and $P(X>2|Y|)$
I'm pretty sure the first one is $0.5$, because $X-2Y$ should also be a normal random variable with mean $0$, so there is $50%$ chance that it is positive.
However, for the second one, if I transform it into:
$P(-X/2<Y<X/2) = P[\Phi(X/2)-\Phi(-X/2)]$
$ =\int_{-\infty}^{+\infty}[\Phi(X/2)-\Phi(-X/2)] f(X) dX$
$ =\int_{0}^{1}[\Phi(X/2)-\Phi(-X/2)] d\Phi(X)$
Is it the correct way to go? I'm stuck here for the next step. Also I was wondering if there is some more intuitive approach to solve the second one. Thanks!
UPDATE: a quick Python simulation gives a result around 0.15
There is an intuitive way to solve this. The region where $X>2|Y|$ is an angular sector (draw it). Since the joint distribution of $X$ and $Y$ is rotationally symmetric (write the joint distribution in polar coordinates and see it does not depend on $\theta$), the probability $(X,Y)$ is in any angular sector is just the angle of that sector divided by $2\pi$.