Expected value of $X$ given $X > Y$

Question

I'm doing some research and I'm trying to compute a closed form for $ \mathbb{E}[ X \mid X > Y] $ where $X$, $Y$ are independent normal (but not identical) random variables. Is this known?

Answer 1

Hint: you can directly do the double integrals in the district $x>y$ for $xf(x)$, not so hard, I remember.

Answer 2

yes, $\mathbb{E}(X|X > y)$ has a closed form as the expectation of a truncated normal. However, integrating that expression times the pdf of $Y$, is difficult. The normal r.v. are arbitrary. Does anyone have a satisfactory answer or is there no known closed form?

Answer 3

Explicitly, we have for $X \sim \operatorname{Normal}(\mu_x, \sigma_x^2)$ and $Y \sim \operatorname{Normal}(\mu_y, \sigma_y^2)$, $$\operatorname{E}[X \mid X > Y] = \int_{y=-\infty}^\infty \int_{x=y}^\infty x f_{X,Y}(x,y) \, dx \, dy$$ where $$f_{X,Y}(x,y) = \frac{1}{2\pi \sigma_x \sigma_y} \exp \biggl(-\frac{(x-\mu_x)^2}{2\sigma_x^2} - \frac{(y-\mu_y)^2}{2\sigma_y^2} \biggr)$$ is a bivariate normal density with zero correlation (since $X$, $Y$ are independent). This integral does not, for general parameters, have an elementary closed form.

Answer 4

One can reduce the problem to the computation of $E(U\mid U\gt aV-b)$ where $(U,V)$ are i.i.d. standard normal. Let $\varphi$ denote the standard normal PDF, then $u\varphi(u)=-\varphi'(u)$ hence $$E(U;U\gt aV-b)=\int_\mathbb R\varphi(v)\mathrm dv\int_{av-b}^\infty u\varphi(u)\mathrm du=\int_\mathbb R\varphi(v)\varphi(av-b)\mathrm dv.$$ Computing the product $\varphi(v)\varphi(av-b)$ and using the change of variable $w=cv$ with $$c=\sqrt{1+a^2},$$ one gets $$E(U;U\gt aV-b)=\varphi(b/c)/c.$$ On the other hand, $U-aV$ is centered normal with variance $c^2$ hence $$P(U\gt aV-b)=\Phi(b/c),$$ where $\Phi$ denotes the standard normal CDF. Thus, $$E(U\mid U\gt aV-b)=\frac{\varphi(b/c)}{c\Phi(b/c)}.$$ For general independent normal random variables $X$ and $Y$, note that $$E(X\mid X\gt Y)=\mu_X+\sigma_XE(U\mid U\gt aV-b),$$ with $$a=\sigma_Y/\sigma_X,\qquad b=(\mu_X-\mu_Y)/\sigma_X,$$ that is, $$ E(X\mid X\gt Y)=\mu_X+\tau\sigma_X\frac{\varphi(\tau\nu)}{\Phi\left(\tau\nu\right)},\qquad\nu=\mu_X-\mu_Y,\qquad\tau=\frac{\sigma_X}{\sqrt{\sigma_X^2+\sigma_Y^2}}.$$