# Joint pdf with condition

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 $$P(Z>0) = 1-F(Z\leq0):Z\in(-\infty,\infty)$$

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?

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}P(V,Z>0)dzdv=\int_{-\infty}^{\infty}\int_{0}^{\infty}P(V,Z)dzdv$$

• The expression $\mathbb P(X\mid X+Y>0)$ is meaningless. Are you trying to find the distribution of $X$ conditioned on $\{X+Y>0\}$? Mar 12, 2020 at 3:36
• Yes, apologies for the confusion. Mar 12, 2020 at 12:20

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}
• $X+Y\sim\mathcal N(0,2)$ so $\mathbb P(X+Y>0)=\frac12$. Mar 12, 2020 at 23:16
• Let $f_{XY}(x,y)$ be their joint pdf. Is this true ? $\int_{-\infty}^{\infty}\int_{0}^{\infty}f_{XY}(x,y)dxdy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f_{XY}(x>0,y)dxdy$ With this I am trying to understand the difference between $P(X,Y)$ and $P(X>0,Y)$. So for example if you ask me to prove your solution sums to 1, can I take an integral over $P(X\leq t|X+Y)$ with bounds for $X$ from ($-Y$;\infty), instead of $P(X\leq t|X+Y>0)$ Mar 13, 2020 at 1:18