What is $P(X+Y>0 \mid X>0)$ given that $X,Y$ two different normal? X follows $N(0,\sigma_x^2)$ and Y follows $N(0,\sigma_y^2)$, and $X$ and $Y$ are independent. What is $P(X+Y>0 \mid X>0)$? If $\sigma_x^2=\sigma_y^2$ we can use symmetry and easily get answer 3/4, but what if $\sigma_x^2 \neq \sigma_y^2$?
 A: Let $X:=-\sigma_XN$ and $Y:=\sigma_YN'$, so $N,N'$ are independent $\mathcal N(0,1)$ variables. Let also $a:=\frac{\sigma_X}{\sigma_Y}>0$. Then
$$X+Y>0\iff N'>aN.$$
Hence $$\Bbb P(X+Y>0\mid X>0)=\frac{\Bbb P(N'>aN,N<0)}{\Bbb P(N<0)}=1-2\,\Bbb P(N'<aN,N<0).\tag{$\star$}$$
Integrating out $N'$, we are thus reduced to studying the map
$$f(a):=\Bbb P(N'<aN,N<0)=\mathbb E\!\left[\Phi(aN)\,\mathbf 1_{\left\{N<0\right\}}\right]\!,\qquad a>0,$$
where $\Phi$ is the c.d.f. of $\mathcal N(0,1)$, so $\Phi'(x)=\frac1{\sqrt{2\pi}}\,\mathrm e^{-\frac{x^2}2}$ for all $x\in\Bbb R$.

*

*We have $f(1)=\Bbb E\!\left[\Phi(N)\mathbf 1_{\left\{\Phi(N)<\frac12\right\}}\right]=\frac18$, since $\Phi(N)$ is uniformly distributed on $(0,1)$.

*An application of the dominated convergence theorem shows that $f$ is differentiable on $(0,\infty)$, with
$$f'(a)=\frac 1{\sqrt{2\pi}}\,\Bbb E\!\left[N\mathrm e^{-\frac12a^2N^2}\mathbf 1_{\{N<0\}}\right]=\frac1{2\pi}\left[-\frac1{2(1+a^2)}\mathrm e^{-\frac12(1+a^2)x^2}\right]_{x\to-\infty}^{x=0}=-\frac1{4(1+a^2)\pi}.$$
From 1. and 2., we deduce that
$$f(a)=f(1)+\int_1^af'(t)\,\mathrm dt=f(1)-\frac1{4\pi}\Bigl[\arctan t\Bigr]_{t=1}^{t=a}=\frac3{16}-\frac1{4\pi}\arctan a.$$
Recalling $(\star)$ and the value of $a$, we conclude that
$$\mathbb P(X+Y>0\mid X>0)=\frac58+\frac1{2\pi}\,\arctan\frac{\sigma_X}{\sigma_Y}.$$
