Suppose $U_1, U_2 \sim \mathcal{U}[0,1]$ and both random variables are independent. I want to compute $E[U_1 \mid U_1 < U_2]$. I think the correct ansewr is
$$ E[U_{1}\mid U_{1}<U_{2}]=E[U_{1}\mid U_{1}=\min\{U_{1},U_{2}\}]=E[\min\{U_{1},U_{2}\}]=\int_{0}^{1}2(1-x)xdx=\frac{1}{3}. $$
But, I was wondering why the following reasoning leads to incorrect solution. If we fix $U_2 =u$, then $E[U_1\mid U_1 <u]=u/2$. Each $u$ happens equally likely, i.e., $U_2\sim \mathcal{U}[0,1]$, so $E[U_1\mid U_1 < U_2] = E[U_2]/2=1/4$. Or, more formally, $$ E[U_{1}\mid U_{1}<U_{2}]=\int_{0}^{1}\int_{0}^{u}\frac{x}{u}dxdu=\int_{0}^{1}\frac{u}{2}du=\frac{1}{4}. $$
Where did I make a mistake?
This is the mistake.
$U_2$ corresponds to $y$ when you pick a random point $(x,y)$ in the $x$-$y$ plane in the triangular region bound by $x\geq0$, $x\leq y$, and $y\leq 1$. It is easy to see that the distribution of $y$ is not $\mathcal{U}[0,1]$.