Density of Gaussian Random variable conditioned on sum I am struggling with this simple problem.
I have two Gaussian independent random variables $X \sim \mathcal{N}(0,\sigma_x^2,), Y \sim \mathcal{N}(0,\sigma_y^2,)$. I have to find the density of $X$ given that $X + Y > 0$.
I know that $X, X+Y$ shall be jointly normal distributed and I also know the forms of conditional distribution of $X | X+Y=z$  
https://en.wikipedia.org/wiki/Multivariate_normal_distribution
https://stats.stackexchange.com/questions/17463/signal-extraction-problem-conditional-expectation-of-one-item-in-sum-of-indepen 
but somehow I am confused because the condition that I have is that $X+Y > 0$ and not of the form $X+Y = z$.
I feel that some integration shall have to be performed but I am not sure how. Any pointers shall be very helpful. It will also help if I can get the conditional mean, variance if not the entire density function.
Thanks
 A: $Z=X+Y$ is normally distributed with mean $0$ and variance $\sigma_x^2+\sigma_y^2$
Conditioned on $Z=z$, you would have $X$ being normally distributed with mean $z\frac{ \sigma_x^2}{\sigma_x^2+\sigma_y^2}$ and variance ${ \frac{\sigma_x^2 \sigma_y^2}{\sigma_x^2 + \sigma_y^2}}$
Conditioned on $Z>0$, $Z$ would have a half-normal distribution with mean $\sqrt{(\sigma_x^2+\sigma_y^2)\frac2\pi}$ and variance $(\sigma_x^2+\sigma_y^2)\left(1-\frac2\pi\right)$, so $X$ would have a mean of $\frac{ \sigma_x^2}{\sqrt{\sigma_x^2+\sigma_y^2}}\sqrt{\frac2\pi}$  and a variance of $\sigma_x^2      { \frac{\sigma_x^2 \left(1-\frac2\pi\right)+\sigma_y^2}{\sigma_x^2 + \sigma_y^2}}$.  
But the conditional distribution of $X$ given $X+Y \gt 0$ would not have a standard distribution: it would be right-skewed though could take negative values.  For example with $\sigma_x^2=400$ and $\sigma_y^2=100$ it might look something like this, with the mean highlighted:

A: Since you are now conditional on an event with non-zero probability $\displaystyle \frac {1} {2}$, you should start with the conditional CDF:
$$ \begin{align} 
F_{X|X + Y > 0}(x) &= \Pr\{X \leq x|X + Y > 0\} \\
&= \frac {\Pr\{X \leq x, X + Y > 0\}} {\Pr\{X + Y > 0\}} \\
&= 2\int_{-\infty}^x \Pr\{Y > -u\}f_X(u)du\\
&= 2\int_{-\infty}^x F_Y(u)f_X(u)du
\end{align}$$
By differentiating with respect to $x$, we obtain the conditional density:
$$ f_{X|X + Y > 0}(x) 
= \frac {\partial} {\partial x}F_{X|X + Y > 0}(x)
= 2F_Y(x)f_X(x) 
= \frac {2} {\sigma_X}\Phi\left(\frac {x} {\sigma_Y}\right)
\phi\left(\frac {x} {\sigma_X}\right)$$
where $\Phi, \phi$ are the CDF and pdf of standard normal respectively. So this essentially is just a skew normal distribution.
https://en.wikipedia.org/wiki/Skew_normal_distribution
Comparing the parametrization with wiki's one, we have 
$$ \omega = \sigma_X, \xi = 0, \alpha = \frac {\sigma_X} {\sigma_Y}$$ 
so the conditional mean is given by
$$ \xi + \omega\frac {\alpha} {\sqrt{1 + \alpha^2}} \sqrt{\frac {2} {\pi}}
= \sqrt{\frac {2} {\pi}} \frac {\sigma_X^2} {\sqrt{\sigma_Y^2 + \sigma_X^2}}$$
Now put $\sigma_X^2 = 400, \sigma_Y^2 = 100$, we have
$$ E[X|X + Y >0] = \sqrt{\frac {2} {\pi}} \frac {400} {\sqrt{100 + 400}} = 40\sqrt{\frac {2} {5\pi}} \approx 14.273 $$
so it should agree with the numerical result from Henry.
