Let $X\in \mathcal{N}(0,1)$, $Y\in \mathcal{N}(0,1)$, and are independent. Show that
\begin{align} \mathbb{E}[X\vert X>Y] = \frac{1}{\sqrt{\pi}}. \end{align}
I am not sure how to handle the inequality inside the conditional. So I did the following. Let $Z=X-Y$, then $Z\in\mathcal{N}(0,2)$ and now construct the bivariate Gaussian random variable $(X,Z)^{T}$ (is this allowed, does it follow that it is Gaussian?). Let us denote its mean vector $\mu$ and its covariance matrix $\mathbf{C}$. The covariance of $X$ and $Z$ is
\begin{align} \mathbb{E}[(X-\mu_{x})(Z-\mu_{Z})] &= \mathbb{E}[XZ] = \{\text{Double expectation}\} = \mathbb{E}[\mathbb{E}[XZ\vert X]] = \{\text{Take out what is known}\} \\ &= \mathbb{E}[X\;\mathbb{E}[Z\vert X]] = \mathbb{E}[X\;\mathbb{E}[X-Y\vert X]] = \mathbb{E}[X\;\{\mathbb{E}[X\vert X]-\mathbb{E}[Y\vert X]\}] \\ &= \{\text{Independence}\} = \mathbb{E}[X\;\{X-\mathbb{E}[Y]\}] = \mathbb{E}[X^{2}] - \mathbb{E}[X]\mathbb{E}[Y] = \mathbb{V}[X] - 0 = 1. \end{align}
Hence the covariance matrix becomes
\begin{align} \mathbf{C} = \begin{pmatrix}\mathbb{V}[X]&\mathbb{C}[X,Z] \\\mathbb{C}[X,Z] &\mathbb{V}[Z]\end{pmatrix} =\begin{pmatrix}1&1 \\1 &2\end{pmatrix}. \end{align}
The mean vector is just $\mu = \mathbf{0}$. Now I was thinking maybe I can use the conditional distribution of $X$ given $Z=z$, but that is an equality, not an inequality. So how do I do this?
The joint density of $(X,Y)$ is: $$f(x,y) = \frac{1}{2\pi} \mathrm e^{-\frac12(x^2+y^2)}$$
So you can calculate:
$$\begin{align} \mathbb{E}[X \mid X>Y] &= \frac{1}{\mathbb{P}(X>Y)} \iint_{x>y} \frac{1}{2\pi} x \mathrm e^{-\frac12(x^2+y^2)} \,\mathrm dx\,\mathrm dy \\[1ex]& = \frac{1}{1/2} \int_{y=-\infty}^{\infty} \int_{x=y}^{\infty} \frac{1}{2\pi} x \mathrm e^{-\frac12 x^2} \mathrm e^{-\frac12 y^2} \,\mathrm dx\,\mathrm dy \\[1ex] &= \frac{1}{\pi} \int_{-\infty}^{\infty} \mathrm e^{-\frac12 y^2}\int_{y}^{\infty} x\mathrm e^{-\frac12 x^2} \,\mathrm dx\,\mathrm dy \\[1ex]& = \frac{1}{\pi} \int_{-\infty}^{\infty} \mathrm e^{-y^2} \,\mathrm dy \\[1ex] &= \frac{1}{\sqrt{\pi}}\end{align}$$