So I encountered the following problem in an app with probability puzzles:
Given independent $X, Y \sim U(0, 1)$, what is the correlation of $X$ and $Y$ given that $X + Y \geq 1$? Simulating some data showed me that it is $-0.5$. However, I'd like to know how to derive this from the distributions. I tried to derive conditional distributions $f_{X|Y \geq 1-X}$ and $f_{Y|X \geq 1-Y}$, but I somehow keep ending up with identical expressions for both conditional distributions and also with correlations of $0$.
How do I solve a problem like this?
If $X+Y\geq 1$, the two random variables cannot be truly independent. Their joint pdf reads $$ f_{X,Y}(x,y)=2\Theta(x+y-1)\ , $$ with the Heaviside theta $\Theta(z)=1$ if $z\geq 0$ and $=0$ otherwise.
Computing the covariance of the two $$ \mathrm{Cov}(X,Y)=\langle XY\rangle-\langle X\rangle\langle Y\rangle $$ requires $$ \langle XY\rangle=\iint_0^1 dx\ dy\ 2xy\Theta(x+y-1) =5/12 $$ and $$ \langle X\rangle=\langle Y\rangle=2/3\ , $$ from which $$ \mathrm{Cov}(X,Y)=\frac{5}{12}-(2/3)^2=-1/36\ . $$ To compute the correlation coefficient, we need the standard deviation of $X$ and $Y$, computed as $$ \sigma_X=\sigma_Y=\sqrt{\langle X^2\rangle-(\langle X\rangle)^2}=\sqrt{1/2-4/9}=\frac{1}{3\sqrt{2}}\ , $$ from which the correlation coefficient $$ \rho=\frac{\mathrm{Cov}(X,Y)}{\sigma_X\sigma_Y}=-\frac{1}{36}\times 18=-1/2. $$