Given i.i.d. random variables $X, Y \sim N(0, 1)$, find the conditional distribution of $X$ given that $X + Y > 0.$
First approach: Find the conditional CDF of $X$ given $X+Y >0$ :
$F_X(x | X+Y >0 ) = \mathbb{P}[X<x | X+Y > 0 ] = \mathbb{P}[X+Y >0 | X<x] \cdot \frac{\mathbb{P}[X<x]}{\mathbb{P}[X+Y >0 ]}$.
Notice that $X+Y \sim N(0,2)$ and so $\mathbb{P}[X+Y >0 ] = \frac{1}{2}$
However I cannot seem to calculate nicely $\mathbb{P}[X+Y >0 | X<x]$
Is there a better way to approach this problem? Can someone calculate $\mathbb{P}[X+Y >0 | X<x] $, I am going to differentiate it anyways so It can be as an integral.
Let $\Phi(x):=\mathbb P(X\le x)$ be the c.d.f. of the standard $\mathcal N(0,1)$ distribution, and recall that $\Phi(-x)=1-\Phi(x)$. A way to proceed could be (I leave it for you to justify the steps): \begin{align*} \mathbb P(X\le x\mid X+Y>0) &=\frac{\mathbb P(X\le x,X+Y>0)}{\mathbb P(X+Y>0)}\\[.4em] &=\frac{\displaystyle\int_{-x}^\infty\mathbb P(-y<X\le x)\,\mathrm d\Phi(y)}{\displaystyle\int_{-\infty}^\infty\mathbb P(-y<X)\,\mathrm d\Phi(y)}\\[.4em] &=\frac{\mathbb E[(\Phi(x)-\Phi(-Y))\mathbf 1_{\{Y\ge -x\}}]}{\mathbb E[\Phi(Y)]}\\[.4em] &=\frac{\Phi(x)^2–\mathbb E[\Phi(Y)\mathbf 1_{\{Y\le x\}}]}{\mathbb E[\Phi(Y)]}. \end{align*} To further simplify this one could use that $\Phi(Y)$ is uniformly distributed on $(0,1)$.