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Suppose $X \sim N(\mu_X, \sigma_X^2)$ and $Y ~ \sim N(\mu_Y, \sigma_Y^2)$ are Gaussians with correlation $\rho$. Then this formula can be used to obtain the conditional density for $X | Y = y$.

Is there another nice formula for the conditional density $X | Y \geq y$ instead?

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  • $\begingroup$ Sorry in general no. The conditional CDF is $\displaystyle \Pr\{X \leq x|Y \geq y\} = \frac {\Pr\{X \leq x, Y \geq y\}} {\Pr\{Y \geq y\}}$. After differentiating with respect to $x$, we got the conditional density which will involve the integral of a bivariate Gaussian density. $\endgroup$ – BGM Mar 3 '17 at 3:53
  • $\begingroup$ But how do you know that this integral doesn't evaluate to something nicer, like it does for $Pr(X \leq x|Y=y)$? $\endgroup$ – user541020 Mar 3 '17 at 5:04
  • $\begingroup$ The integral of Gaussian density has no closed form except the integral limit is at the mean or infinities. I have no nice proof to you for this fact. So usually people can express the result in terms of some function like $\Phi$ which has no closed form available. $\endgroup$ – BGM Mar 3 '17 at 5:16

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