Gaussian estimate conditioned to stay positive Given a centered gaussian variable $X$ with variance $\sigma^2$ it is very easy to compute the conditional expectation
$$\mathbb {E}[X|X <0]=-\sqrt {2/\pi}\sigma. $$
Now I want to generalize this result for a 2-dimensional centered gaussian vector $(X,Y)$ with a fixed covariance covariance Matrix such that $\mathbb {E}[X^2]=\mathbb {E}[Y^2]=\sigma>0$ and $\mathbb {E}[XY]=c>0$. Can I compute or give an estimate of
$$\mathbb {E}[X|X <0,Y<0]? $$
Below you will find a very nice solution of this problem. Still open the problem to generalize the idea for a $n-$dimensional centered gaussian vector, that is, how to compute or estimate
$$\mathbb {E}[X_1|X_1<0,X_2<0,\dots,X_n<0]?$$
 A: Suppose $f_{X,Y}(x,y; \sigma,c)$ is the bivariate normal pdf with mean $(0,0)$ and $c = \frac{\rho}{\sigma^2}$ and $f_{X}(x;\sigma)$ and $f_{Y}(y;\sigma)$ are the corresponding marginal normal pdf's with mean $0$.
$$\begin{align}\color{red}{E(X,X<0|Y<0)} &= \int\limits_{-\infty}^{0}xf_{X|Y}(x|y<0)dx
\\\\ &= \int\limits_{-\infty}^{0}x \frac{f_{X,Y}(x,y<0)}{f_{Y}(y<0)}dx
\\\\ &= \int\limits_{-\infty}^{0} x \frac{\int\limits_{-\infty}^{0}f_{X,Y}(x,y)dy}{\int\limits_{-\infty}^{0}f_{Y}(y)dy}dx 
\\\\ &= \int\limits_{-\infty}^{0}x\frac{\int\limits_{-\infty}^{0}f_{X,Y}(x,y)dy}{0.5}dx
\\\\ &= \int\limits_{-\infty}^{0}2xf_{X}(x)\Phi(\frac{-\rho x}{\sqrt{1-\rho^2}})dx\end{align}$$
where $\Phi$ is the normal cdf with parameters, mean $ = 0$ and variance $=\sigma^2$.
Now, based on this answer, let the RHS be $h(\rho)$. Then using Leibniz rule, we get,
$$\begin{align}\frac{\partial h(\rho)}{\partial \rho} &= \int\limits_{-\infty}^{0}2xf_{X}(x)f_{X}(\frac{-\rho x}{\sqrt{1-\rho^2}})\left(\frac{-x}{(1-\rho^2)^{3/2}}\right)dx
\\\\ &= \frac{-2}{2\pi\sigma^2(1-\rho)^{3/2}} \int\limits_{-\infty}^{0} x^2\exp\left(\frac{-x^2}{2\sigma^2(1-\rho^2)}\right)dx
\\\\ &= \frac{-\sigma}{\sqrt{2\pi}}\end{align}$$ 
So, $h(\rho) = \frac{-\sigma\rho}{\sqrt{2\pi}} + K$, where $h(0) = \frac{-\sigma}{\sqrt{2\pi}} \implies K = \frac{-\sigma}{\sqrt{2\pi}}$.
$$E(X,X<0|Y<0) = h(\rho) = \frac{-\sigma(\rho+1)}{\sqrt{2\pi}}$$
$$E(X,X<0|Y<0) = E(X\mathbb{I}(X<0)|Y<0) = \frac{E(X\mathbb{I}(X<0)\mathbb{I}(Y<0))}{P(Y<0)}$$
$$E(X|X<0,Y<0) = \frac{E(X\mathbb{I}(X<0)\mathbb{I}(Y<0))}{P(X<0,Y<0)} = \frac{E(X,X<0|Y<0)P(Y<0)}{P(X<0,Y<0)}$$
$$\implies E(X|X<0,Y<0) = \frac{h(\rho)\frac{1}{2}}{\Phi_{X,Y}(0,0)}$$
where $\Phi_{X,Y}$ is the bivariate normal cdf with mean $(0,0)$, variance $\sigma_X^2 = \sigma_Y^2 = \sigma^2$ and covariance $c = \rho \sigma^2$.
Have a look at this answer to get the value of $\Phi_{X,Y}(0,0)$.
$$\Phi_{X,Y}(0,0) = \frac{1}{2\pi}\arcsin\rho + \frac{1}{4}$$
Finally,
$$E(X|X<0,Y<0) = \frac{\frac{-\sigma(\rho+1)}{\sqrt{2\pi}}}{\frac{1}{\pi}\arcsin\rho + \frac{1}{2}}$$
