Conditional normal distribution I hope to calculate $P(X+Y>0 \ |\  Y<0)$, where $X$, $Y$ are independent normal distribution with same mean ($\mu$) and variance ($\sigma$).
I tried to do it with direct integration. That is calculate $P(X+Y>0 \textrm{ and } Y<0)$ first. But failed after I eliminate the first integration sign.
Any other method to calculate this probability?
 A: I will take $\sigma = 1$ to avoid ambiguity as to whether the variance is $\sigma^2$ as most people write or $\sigma$ as the OP wants it, and assume for convenience that $\mu > 0$.


*

*The event $\{X + Y > 0\}$ occurs if the point $(X,Y)$ lies above the line $x+y = 0$, and has probability $\Phi(2\mu/\sqrt{2}) = \Phi(\sqrt{2}\mu)$ since $X + Y$ is normal with mean $2\mu$ and variance 2.

*The event $\{X > 0, Y > 0\} = \{X > 0\}\cap\{Y > 0\}$ occurs if the point $(X,Y)$ lies  in the first quadrant, and has probability $[\Phi(\mu)]^2$.

*Since $X$ and $Y$ are identically distributed, it follows that
$$P(\{X + Y > 0\}\cap\{Y < 0\}) = P(\{X + Y > 0\}\cap\{X < 0\}). $$ 

*Except for a set of zero probability, the event $\{X + Y > 0\}$ is the disjoint union of the events $\{X > 0, Y > 0\}, \{X + Y > 0\}\cap\{Y < 0\}$ and 
$\{X + Y > 0\}\cap\{X < 0\}$.  


Hence,
$$\begin{align*}
P\{X + Y > 0\} &= \Phi(\sqrt{2}\mu)\\
 &= P\{X > 0, Y > 0\} + P(\{X + Y > 0\}\cap\{Y < 0\}) +  P(\{X + Y > 0\}\cap\{X < 0\})\\
&= [\Phi(\mu)]^2 + 2P(\{X + Y > 0\}\cap\{Y < 0\})
\end{align*}
$$
and so $P(\{X + Y > 0\}\cap\{Y < 0\}) = \left.\left.\frac{1}{2}\right(\Phi(\sqrt{2}\mu) 
-  [\Phi(\mu)]^2\right)$, and since $P\{Y < 0\} = 1 -\Phi(\mu)$, we have that 
$$P\{X + Y > 0 \mid Y < 0\} = \frac{P(\{X + Y > 0\}\cap\{Y < 0\})}{P\{Y < 0\}}
= \frac{\Phi(\sqrt{2}\mu) 
-  [\Phi(\mu)]^2}{2(1 -\Phi(\mu))}$$
A: Writing $X$ and $Y$ as $X=\mu+\sigma X_0$ and $Y=\mu+\sigma Y_0$ where $X_0$ and $Y_0$ are i.i.d. standard gaussian random variables, one sees that the conditional probability is
$$
\mathrm P(X+Y>0\mid Y<0)=\frac{U(-\mu/\sigma)}{G(-\mu/\sigma)},
$$ 
where for every real $x$,
$$
G(x)=\int\limits_{-\infty}^xg(y)\mathrm dy,\qquad\qquad g(y)=\frac1{\sqrt{2\pi}}\mathrm e^{-y^2/2},
$$
and
$$
U(x)=\int\limits_{-\infty}^xg(y)\int\limits_{2x-y}^{+\infty}g(z)\mathrm dz\,\mathrm dy=G(x)-\int\limits_{-\infty}^xg(y)G(2x-y)\,\mathrm dy.
$$
Thus, $G$ is the usual CDF of a standard gaussian random variable. Some specific values are $U(-\infty)=0$, $U(0)=\frac18$ and $U(+\infty)=0$.
A: Since $\mathbb{P}(X+Y>0 \vert Y<0) = \frac{\mathbb{P}(X+Y>0 \land Y<0)}{\mathbb{P}(Y<0)}$, and the denominator being a normal CDF, the real problem is to compute $\mathbb{P}(X+Y>0 \land Y<0)$.
Note, that the pair $(U,V) = (X+Y,Y)$ has bivariate normal distribution with means $(2 \mu, \mu)$, variances $(2 \sigma^2, \sigma^2)$ and correlation $\rho = \frac{1}{\sqrt{2} \sigma^2} \mathbb{E}((X+Y-2\mu)(Y-\mu)) = \frac{1}{\sqrt{2} \sigma^2} \mathbb{E}((Y-\mu)^2) = \frac{1}{\sqrt{2}}$.
Thus we seek $\mathbb{P}(U > 0 \land V < 0) = \mathbb{P}(V < 0) - \mathbb{P}(U < 0 \land V < 0)$. The last part is the CDF of the bivariate normal, which has no closed form for generic $\mu$ and $\sigma$. So
$$
  \mathbb{P}(X+Y>0 \vert Y<0)  = 1 - \frac{\rm CDF_{U,V}(0,0)}{ \rm CDF_{V}(0)}
$$
Now using $\rm CDF$ of standard normal and standard bi-normal with correlation $\rho=\frac{1}{\sqrt{2}}$:
$$
  \mathbb{P}(X+Y>0 \vert Y<0)  = 1 - \frac{\rm CDF_{\mathcal{BN}(\rho)}(-\sqrt{2}\frac{\mu}{\sigma},-\frac{\mu}{\sigma})}{ \Phi(-\frac{\mu}{\sigma}) }
$$
The following plot shows $\mathbb{P}(X+Y>0 \vert Y<0)$ as a function of $\frac{\mu}{\sigma}$:

