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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?

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

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  • $\begingroup$ The expression $\mathbb P(X\mid X+Y>0)$ is meaningless. Are you trying to find the distribution of $X$ conditioned on $\{X+Y>0\}$? $\endgroup$
    – Math1000
    Mar 12, 2020 at 3:36
  • $\begingroup$ Yes, apologies for the confusion. $\endgroup$
    – Kreol
    Mar 12, 2020 at 12:20

1 Answer 1

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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}

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  • $\begingroup$ Hi Math1000, thank you so much for coming back to me ! Your answer covers the calculation of P(X) conditioned on {X+Y>0}, but does not answer my question about the joint probability notation. Could you please have a look ? Really appreciate your time. I also don`t quite understand how did you get line 2 ? $\endgroup$
    – Kreol
    Mar 12, 2020 at 12:22
  • $\begingroup$ $X+Y\sim\mathcal N(0,2)$ so $\mathbb P(X+Y>0)=\frac12$. $\endgroup$
    – Math1000
    Mar 12, 2020 at 23:16
  • $\begingroup$ I can't make sense of your other question because of the rampant abuse of notation. $\endgroup$
    – Math1000
    Mar 12, 2020 at 23:18
  • $\begingroup$ 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 $P(X\leq t|X+Y>0)$ $\endgroup$
    – Kreol
    Mar 13, 2020 at 1:18

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