Given two independent normal random variables $X$ and $Y$, what is $P(X\leq x\mid X>Y)$? As the title says, I'm looking for the distribution of $X$ given that $X>Y$.
 A: Assume that $X$ and $Y$ are independent $\mathrm{N}(0,1)$ rv's. Denote by $\phi$ and $\Phi$ the corresponding pdf and df, respectively. By definition,
$$
  P(X\leq x\mid X>Y) = \frac{P(X\leq x,X>Y)}{P(X>Y)} \ \, .
$$
Since $X$ and $Y$ are iid, by symmetry we have $P(X>Y)=P(X<Y)$. But we know that $P(X>Y)+P(X<Y)=1$ (the probability of a tie is zero). Hence, $P(X>Y)=1/2$. 
Now,
$$
\begin{align*}
P(X\leq x,X>Y) &= \mathrm{E}[P(X\leq x,X>Y\mid Y)] \quad \text{(total expectation)} \\
  &= \int_{-\infty}^\infty P(X\leq x,X>Y\mid Y=y)\,\phi(y)\,dy \\
  &= \int_{-\infty}^\infty P(X\leq x,X>y\mid Y=y)\,\phi(y)\,dy \quad \text{(use what you know)} \\
  &= \int_{-\infty}^\infty P(y<X\leq x)\,\phi(y)\,dy \quad \text{(independence)} \\
  &= \int_{-\infty}^\infty (\Phi(x)-\Phi(y)) I_{(-\infty,x]}(y)\,\phi(y)\,dy \\
  &= \Phi^2(x) - \int_{-\infty}^x \Phi(y)\,\phi(y)\,dy \, . \\
\end{align*}
$$
Therefore,
$$
  P(X\leq x\mid X>Y) = 2\left(\Phi^2(x) - \int_{-\infty}^x \Phi(y)\,\phi(y)\,dy\right) \, .
$$
Integrating by parts, we have
$$
  \int_{-\infty}^x \Phi(y)\,\phi(y)\,dy = \left[\Phi(y)\Phi(y)\right]_{-\infty}^x - \int_{-\infty}^x \Phi(y)\,\phi(y)\,dy \, ,
$$
yielding
$$
  \int_{-\infty}^x \Phi(y)\,\phi(y)\,dy = \frac{\Phi^2(x)}{2} \, ,
$$
and finally
$$
  P(X\leq x\mid X>Y) = \Phi^2(x) \, .
$$
