What is $P(X > 0 \mid X + Y > 0)$ given that $X, Y$ are i.i.d standard normal?

Question

Suppose $X, Y$ are i.i.d and $X,Y\sim N(0,1)$. What is $P(X > 0 \mid X + Y > 0)?$

What I got so far is listed below:

$\begin{align*} P(X > 0 \mid X + Y > 0) &= \frac{P(X+Y>0\mid X>0)\cdot P(X > 0)}{P(X+Y > 0)}\\ &= P(X+Y > 0\mid X > 0)\\&=P(Y > -X\mid X>0) \\&=\int_{0}^{\infty}\int_{y=-x}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}dydx\end{align*}$

However, I don't know how to calculate this integral. Is there anything that I did wrong in the process? And also is there any easier way to approach this kind of problem?

Answer 1

The answer is obviously $3/4$ without needing any calculation.

The reasoning is that the joint density is radially symmetric, so that the following probabilities are all equal: $$\Pr[0 < Y < X] = \Pr[0 < X < Y] = \Pr[X < Y < 0] = \Pr[Y < X < 0]\\ = \Pr[0 < Y < -X] = \Pr[0 < -X < Y] = \Pr[-X < Y < 0] = \Pr[Y < -X < -0].$$ In exactly $3$ of these cases, $X > 0$ and $X+Y > 0$; in half of these cases $X+Y > 0$. So the conditional probability is $3/4$.