$\mathbb{E}[x|x>y]$ where both $x$ and $y$ are standard normally distributed variables. I am interested in finding $\mathbb{E}[x|x>y]$ given that $x \sim \mathcal{N}(0,1)$ and $y\sim \mathcal{N}(0,1)$. Furthermore I know that these variables are not correlated with each other. I know the result is $\frac{1}{\sqrt{\pi}}$ but I am not sure how to get there. What I have done so far is the following:
$\int_{-\infty}^{\infty}\frac{1}{1-F_{x}(y)}\int_{y}^{\infty}x f_x(x)dx f_y(y)dy$ however somehow this result does not give me $\frac{1}{\sqrt{\pi}}$, please help :).
 A: Hint: Note that $\mathbb{E}[x | x > y] = \mathbb{E}[\max\{x,y\}]$.  Try to take it from here, if you can.  If you want to see the rest, hover below:

 Let $\phi$ be the p.d.f. of a standard normal and $\Phi$ be the c.d.f. Then the c.d.f. of $\max\{x,y\}$ is $\Phi(x)^2$, implying the p.d.f. is its derivative, $2\phi(x)\Phi(x)$.  This implies that the mean is 2\begin{align}\int_{-\infty}^\infty x\phi(x) \Phi(x) \,dx = 2\int_{-\infty}^\infty \phi(x)^2 \,dx \end{align} by integration by parts (with $u = \Phi(x)$ and $dv = x\phi(x) \,dx$).  Recalling $\phi(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2 / 2}$ and integrating gives the answer.


Edit:  A solution adapting your work so far:
Note that $$\mathbb{E}[x | x > y] = \frac{\mathbb{E}[x \cdot \mathbf{1}_{x > y}]}{\mathbb{P}[x > y]}$$ where $\mathbb{1}_{x > y}$ is in the indicator random variable for the event $x > y$.  By symmetry, the denominator is $1/2$, so we have \begin{align}\mathbb{E}[x | x> y] = 2 \mathbb{E}[x \cdot \mathbf{1}_{x > y}] &= 2 \int_{-\infty}^\infty \int_{y}^\infty x \phi(x) \phi(y) dx dy \\&= 2 \int_{-\infty}^\infty x\phi(x)\int_{-\infty}^x \phi(y) dy dx \\
&= 2 \int_{-\infty}^\infty x\phi(x) \Phi(x) dx.\end{align}
Basically the $1/(1 - F_x(y))$ shouldn't be there; it should be one over a probability outside of the integrals entirely.
