Say $X,Y$ are i.i.d and $X,Y \sim N(0,1)$. We need to find $P(X|X+Y>0)$. I set $Z=X+Y$ and $V=X$ and solved it with random variable transformation. Then to find conditional probability we need to calculate:
$P(V|Z>0) = \dfrac{P(V,Z)}{P(Z>0)}:Z>0$
Then denominator is $P(Z>0) = 1-F(Z\leq0):Z\in(-\infty,\infty)$
My question is about numerator:
Is $P(V,Z>0)$ equivalent to $P(V,Z):Z>0$ ? In other words, if I want to check that $P(V,Z>0)$ sums to 1, is it equivalent to checking that $P(V,Z)$ sums to one with a different integral range for $Z$? More precisely is the below correct?
$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}P(V,Z>0)dzdv=\int_{-\infty}^{\infty}\int_{0}^{\infty}P(V,Z)dzdv$
For $t\in\mathbb R$ we have \begin{align} \mathbb P(X\leqslant t\mid X+Y>0) &= \frac{\mathbb P(X\leqslant t,X+Y>0)}{\mathbb P(X+Y>0)}\\ &=2\cdot \mathbb P(X\leqslant t,X>-Y)\\ &= 2\cdot \mathbb P(X\leqslant t, X>Y)\\ &= 2\cdot \int_{-\infty}^t \int_{-\infty}^{t\wedge x}\frac1{2\pi} e^{-\frac12(x^2+y^2)}\ \mathsf dy\ \mathsf dx\\ &= \frac{1}{4} \left(\text{erf}\left(\frac{t}{\sqrt{2}}\right)+1\right)^2. \end{align}