I am trying to solve the problem indicated in the title. I found a similar one in this link, What is $P(X+Y>0 \mid X>0)$ given that $X,Y$ two different normal?
My problem differs from the linked one as $X$ and $Y$ are correlated in my problem. But I un unclear about from which step should the solution to my problem differs from the linked one. I guess $\rho$ should show up in the last step. Could someone help? Thanks!
Hint:
(1) the joint density function of $X,Y$ is given as $$f_{X,Y}(x,y)= \frac{1}{2\pi\sigma^2\sqrt{(1-\rho^2)}}\exp{\left(-\frac{1}{2\sigma^2(1-\rho^2)}(x^2-2\rho xy+y^2)\right)}$$
(2) by the definition of conditional probability, we have $$P(X+Y>0|X<0)=\frac{P(X+Y>0,X<0)}{P(X<0)}$$ The denominator is $\frac12$ by symmetry, the numerator can be evaluated as a double integral on the set $\left\{(x,y)\in \mathbb R^2:x+y>0,x<0\right\}$.
Without loss of generality we take $\sigma^2=1.$ Write $Y=\rho X+\sqrt{1-\rho^2}Z$ where $Z\sim N(0,1)$ is independent of $X$. Denoting $\cos \theta=\sqrt{\frac{1+\rho}{2}}$ and $\sin \theta=\sqrt{\frac{1-\rho}{2}}$ with $0<\theta<\pi/2$ we can write $$\Pr(X+Y>0|X<0)=\Pr(X\cos \theta+Z\sin \theta>0|X<0)=\frac{\theta}{\pi}.$$
