A hard problem on conditional expectation

Question

$U_{1}, U_{2}$ obeys a bivariate normal distribution with zero mean, variance $1$ and covariance $\rho$. How can we get $ E(U_{1}\mid U_{1}>U_{2}) $? The hint is that we can use the fact that $U_{1}\mid U_{2} $ obeys $N(\rho U_{2},(1-\rho^2))$ and $P(U_{1}>U_{2})=1/2$. And for ease of notation, can use $Z$ obeys standard normal distribution. The answer is $2\rho E[Z \Phi(- ((1-\rho)^{1/2} /(1+\rho)^{1/2})Z )] +2(1-\rho)^{1/2}E[Z\Phi(((1+\rho)^{1/2} /(1-\rho)^{1/2})Z)] $. But how can we get this?

Answer 1

Hint Let $X=U_1$ $Y=U_2$ Then $$P(X|X>Y) = \frac{P(Y<X|X)\, P(X)}{P(X>Y)}=\frac{\Phi \left( \frac{X-\rho X}{\sqrt{1-\rho^2} }\right) \phi(X)}{1/2} $$

where $\phi(\cdot)$ is the standard normal density, and $\Phi(\cdot)$ the distribution function.

The expectation of $X|X>Y$ is $\int X f(X) dX$ where $f(X)$, the pdf, is given above. That is

$$E(X|X>Y) = 2 \int X \, \Phi \left( \frac{X-\rho X}{\sqrt{1-\rho^2} }\right) \phi(X) dX$$

which, given that $\phi(X)$ is itself a pdf (that of a standard gaussian, say $Z$), (think of $\Phi$ as an arbitrary function) is

$$ 2 \, E \left[ Z \Phi \left(\frac{Z-\rho Z}{\sqrt{1-\rho^2} }\right) \right] = 2 \, E \left[ Z \, \Phi\left( \sqrt{\frac{1-\rho }{1+\rho}\, } Z \right)\right] $$