1
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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
    Commented Mar 3, 2017 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
    Commented Mar 3, 2017 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
    Commented Mar 3, 2017 at 5:16

0

Reset to default

You must log in to answer this question.