Find correlation given conditional probability of joint normal I came across this interesting question in an interview:
Given $X$ and $Y$, these two independent standard normal. We have the following probability of $P(X>0| X+Y>0) = 0.75$. One can get this easily by draw a 2d plane and find out the required area.
Now, if $X$ and $Y$ are joint normal with correlation $\rho$, and we are given $P(X>0| X+Y>0) = 0.8$, what is the value of $\rho$? 
We can find this by writing out the pdf of the joint normal $(X, Y)$ and compute the required probability and solve for $\rho$. I want to know if there is a more intuitive way other than the cumbersome double integral? 
I am thinking about making some transformation of $X$ and $Y$, but I don't have much clue how. 
 A: $\newcommand{\PM}{\mathbb{P}}$I don't know if the elaboration I provide is an intuitive way of getting the answer. I think it is clear from your question that we have $X\sim N(0,1)$ and $Y\sim N(0,1)$ in the second case as well. We have  (by definition):
\begin{align}
\PM(X>0 |X+Y>0) = \frac{\PM(X>0, X+Y>0)}{\PM(X+Y>0)}
\end{align}
The numerator can be rewritten using the Law of Total Probability:
\begin{align}
\PM(X>0 , X+Y>0) = \PM(X>0,X+Y>0,Y>0) + \PM(X>0,X+Y>0,Y<0)
\end{align}
And that can be simplified further:
\begin{align}
\PM(X>0,X+Y>0,Y>0)  = \PM(X>0,Y>0)
\end{align}
and on the other hand:
\begin{align}
 \PM(X>0,X+Y>0,Y<0) &= \PM(X+Y>0, Y<0) \\
&=\PM(Y<0|X+Y>0)\PM(X+Y>0)\\
&= (1-\PM(Y>0|X+Y>0))\PM(X+Y>0)\\
&=(1-\PM(X>0|X+Y>0))\PM(X+Y>0)\\
\end{align}
Note that we know without calculations that $\PM(X+Y>0)=\frac{1}{2}$ (why?). So putting everything together:
\begin{align}
\PM(X>0 |X+Y>0)  &= \frac{\PM(X>0,Y>0)+(1-\PM(X>0|X+Y>0))\PM(X+Y>0)}{\PM(X+Y>0)}\\
&= 2\PM(X>0,Y>0) + 1-\PM(X>0|X+Y>0)
\end{align}
Finally we get:
\begin{align}
\PM(X>0 |X+Y>0) = \frac{1}{2}+\PM(X>0,Y>0)
\end{align}
Now we are happy, because there is a known  closed form of $\PM(X>0,Y>0)$ in the case of $X,Y\sim N(0,1)$, namely:
\begin{align}
\PM(X>0,Y>0) = \frac{1}{4}+\frac{\arcsin(\rho)}{2\pi}
\end{align}
So we need to solve:
\begin{align}
\frac{3}{4}+\frac{\arcsin(\rho)}{2\pi} = \frac{4}{5}
\end{align}
We can solve this and finally get the result, namely: 
\begin{align}
\rho = \frac{\sqrt[]{5}-1}{4}
\end{align}
Just playing with the probability rules got us very far. (And using the result for $\PM(X>0,Y>0)$ of course)
A: For approaches "less cumbersome than the double gaussian integral", start from the fact that if the 2D standard normal distribution is invariant by the rotations, thus, for every $(X_0,Y_0)$ standard normal and every angular sector $S$ of angle $\vartheta$, $$P((X_0,Y_0)\in S)=\vartheta/(2\pi)$$
If $(X,Y)$ is standard normal, this yields $P(X>0\mid X+Y>0)$, using $(X_0,Y_0)=^d(X,Y)$, as follows.


*

*The event $\{X+Y>0\}$ corresponds to an angular sector of angle $\pi$.

*The event $\{X>0,X+Y>0\}$ corresponds to an angular sector of angle $3\pi/4$.


Hence, $$P(X>0\mid X+Y>0)=(3\pi/4)/\pi=3/4$$
Likewise, if $(X,Y)$ is centered normal with unit variances and correlation $\varrho$, then, considering $\sigma=\sqrt{1-\varrho^2}$, one gets $$(X,Y)=^d(X_0,\varrho X_0+\sigma Y_0)$$ The event $\{X+Y>0\}$ again corresponds to an angular sector of angle $\pi$. On the other hand, $$\{X>0,X+Y>0\}=\{X_0>0,X_0+\varrho X_0+\sigma Y_0>0\}=\{X_0>0,X_0+\tau Y_0>0\}$$ with $\tau=\sigma/(1+\varrho)$, which corresponds to an angular sector of angle $\vartheta$ in $(\frac12\pi,\pi)$ with $$\tan\vartheta=-\tau=-\sqrt{\frac{1-\varrho}{1+\varrho}}$$
The double angle formula $$\cos(2\vartheta)=\frac{1-\tan^2\vartheta}{1+\tan^2\vartheta}$$ then readily yields 

$$\varrho=\cos(2\vartheta)=\cos(2\pi P(X>0\mid X+Y>0))$$

For example, if $P(X>0\mid X+Y>0)=0.8$, one gets $$\varrho=\cos(2\pi/5)=(\sqrt5-1)/4$$
