Let $x$ and $y$ two i.i.d having an uniform distribution over $[0,1]$. Then what is the conditional expectation, $\mathbb{E}[x / y\ |\ x < y]$.
It seems to me, this should be:
$\int_{\{x < y\}}{x/y\ dP(\cdot\ |\ x <y)} = (1 /\mathbb{P}[x < y])\int_0^1\int_0^y{x/y\ dx dy}$. Is this correct? Where $\mathbb{P}[x < y]$ is the probability that $x < y$ which is 1/2.