There are $2$ random variables $X\sim N(0,1)$ and $Y\sim N(0,1)$. These $2$ random variables are correlated, which can be given as $Corr(x,y) = \rho$

How do I compute an expectation of the below form?

$E[f(X) | Y > H]$ ?

$f(X)$ - It's a function of $X$.

$Y > H$ is a condition for computing the expectation. $H$ is a constant

How should I go about this?

If the condition did not exist I could have written this in the Integration Form. But with the condition how do I do that?

  • $\begingroup$ Well I am looking at how to compute the Integration to come to a closed form solution if there is one. Or at least how do I integrate to find the expectation? $\endgroup$ – Debdipta Majumdar Aug 15 '17 at 14:24
  • $\begingroup$ If $(X,Y)$ is itself normal, then $Y=\rho X+\sigma Z$ where $(X,Z)$ is standard normal and $\rho^2+\sigma^2=1$ with $\sigma>0$ hence $$E(f(X);Y>h)=E(f(X);Z>(h-\rho X)/\sigma)=E(f(X)(1-\Phi((h-\rho X)/\sigma))$$ and $$E(f(X)\mid Y>h)=\frac{E(f(X)(1-\Phi((h-\rho X)/\sigma)))}{1-\Phi(h)}=\frac{E(f(X)\Phi((\rho X-h)/\sigma))}{\Phi(-h)}$$ $\endgroup$ – Did Aug 15 '17 at 14:31
  • $\begingroup$ @MANMAID As every standard normal couple, yes. $\endgroup$ – Did Aug 15 '17 at 14:41
  • $\begingroup$ @Did - Is it possible to come up with a closed form solution of this if f(x) allows? Looks like the Numerator is going to be a double integral $\endgroup$ – Debdipta Majumdar Aug 15 '17 at 14:48
  • $\begingroup$ No, the numerator is the simple integral $$\int_\mathbb Rf(x)\Phi((\rho x-h)/\sigma)\varphi(x)dx$$ where $\varphi$ and $\Phi$ are the standard normal PDF and CDF, respectively. $\endgroup$ – Did Aug 15 '17 at 14:54

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