how to calculate $E(X+Y|X<x,Y<y)$

Question

consider $X\sim U(0,1), Y\sim U(0,1-x))$ for $E(X+Y|X<x,Y<y)$ I'd consider $E(X|X<x,Y<y)+E(Y|X<x,Y<y)$ $f_{x,y} = f_{y|x}f_x = 1/(1-x)$

$E(X|X<x,Y<y) = E(X|X<x) = \int_0^x xdx = x^2/2$

$E(Y|X<x,Y<y) = \int_0^{y}\int_0^{x}y/(1-x)dxdy = \int_0^{x}1/(1-x)dx\int_0^{y}ydy = \frac{-\ln(1-x)y^2}{2}$

Why is this logic wrong?

Answer 1

Don't use $x,y$ as the bound variables, when you have them in the conditions. Integrate with bound variables of $s,t$ to avoid confusion.

$\begin{align}\mathsf E(X+Y\mid X<x,Y<y) &= \mathsf E((X+Y)\mathbf 1_{X<x,Y<y})/\mathsf P(X<x,Y<y)\\ &=\dfrac{\int_{0}^{x}\int_0^{\min\{y,1-s\}}(s+t)/(1-s)\,\mathrm d t\,\mathrm d s}{\int_{0}^{x}\int_0^{\min\{y,1-s\}}1/(1-s)\,\mathrm d t\,\mathrm d s}\end{align}$