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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.

Indranil Banerjee
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  • I think it is correct. You should look for $\mathbb E(U/V)$ where $(U,V)$ has the set ${(x,y)\in[0,1]^2\mid x<y}$ as support and has a constant PDF (wich will be $2=\frac{1}{\frac12}$) – drhab Jun 16 '15 at 15:11

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