Conditional expectation of X given X is greater than Y

Question

I wonder if someone can help me with this question:

"Let $X,Y \sim \mathcal{N}(0,1)$ where X and Y are independent. Find $E[X | X>Y]$."

I know that $E[X|Y=y]= \int_{-\infty}^{\infty} x f_{X|Y}(x,y)dx$, but in this case one is conditioning on an inequality, and I don't know how to treat that. Thanks in advance!

Answer 1

You can calculate the appropriate conditional density directly

$$f_{X\mid X>Y}(x)=\frac{dF_{X\mid X>Y}(x)}{d x}=$$ $$=\lim_{\Delta x\to 0}\frac{F_{X\mid X>Y}(x+\Delta x)-F_{X\mid X>Y}(x)}{\Delta x}=\lim_{\Delta x\to 0}\frac{P(X<x+\Delta x\mid X>Y)-P(X<x\mid X>Y)}{\Delta x}=$$ $$=\lim_{\Delta x\to 0}\frac{P(x<X<x+\Delta x\mid X>Y)}{\Delta x}=$$ $$=\frac1{P(X>Y)}\lim_{\Delta x\to 0}\frac{P(x<X<x+\Delta x\cap X>Y)}{\Delta x}.$$

Now,

$$P(x<X<x+\Delta x\cap X>Y)=\int_{x}^{x+\Delta x}P(u>Y)\ f_X(u)\ du$$ $$=\int_{x}^{x+\Delta x}F_Y(u)\ f_X(u)\ du.$$

Then

$$\lim_{\Delta x\to 0}\frac{\int_{x}^{x+\Delta x}F_Y(u)\ f_X(u)\ du}{\Delta x}=$$ $$=\frac {d}{dx}\int_{-\infty}^x F_Y(u)\ f_X(u)\ du =f_X(x)F_Y(x).$$

That is, $$f_{X\mid X>Y}(x)=\frac{f_X(x)F_Y(x)}{P(X>Y)}=\frac{f_X(x)F_X(x)}{P(X>Y)}$$

because the two random variables has a common distribution.

Then, the conditional expectation is

$$E[X\mid X>Y]=2\int_{-\infty}^{\infty}xf_X(x)F_X(x)\ dx$$

because $P(X>Y)=\frac12$ .

So, $$E[X\mid X>Y]=2\int_{-\infty}^{\infty}xf_X(x)F_Y(x)\ dx=\frac1{\pi}\int_{-\infty}^{\infty}xe^{-\frac{x^2}2}\int_{-\infty}^xe^{-\frac{u^2}2} \ du\ dx.$$

Answer 2

Note that \begin{align*} E(X \mid X > Y) = \frac{E(X\mathbb{1}_{X>Y})}{P(X>Y)}= 2E(X\mathbb{1}_{X>Y}). \end{align*} Moreover, \begin{align*} E(X\mathbb{1}_{X>Y}) &= E\big(E(X\mathbb{1}_{X>Y} \mid X) \big)\\ &=E(X\Phi(X))\\ &=\int_{-\infty}^{\infty}x\Phi(x)\phi(x) dx, \end{align*} where $\Phi$ is the cumulative distribution of a standard normal and $\phi$ is the density.