Another complicated problem on probabilitic conditioning.

Question

The real random variables $X$ and $Y$ are independent and have a uniform distribution $U([0,1])$. Find: $$\mathbb{E}\left[ (X-Y)^2|XY \right]$$

Solution: $$\mathbb{E}\left[ (X-Y)^2|XY \right]=\mathbb{E}\left[ X^2-2XY+Y^2|XY \right]=2\mathbb{E}[X^2|XY]-2XY$$ What to do next I don't know.

Answer 1

\begin{align} \int_0^1\mathrm dy\int_0^1\mathrm dx\,(x-y)^2\delta(xy-s) &= \int_0^1\mathrm dy\int_0^1\mathrm dx\,(x-y)^2\frac1y\delta\left(x-\frac sy\right)\\ &= \int_s^1\mathrm dy\,\left(\frac sy-y\right)^2\frac1y\\ &= \int_s^1\mathrm dy\,\left(\frac{s^2}{y^3}-2\frac sy+y\right)\\ &= \left[-\frac12\frac{s^2}{y^2}-2s\log y+\frac12y^2\right]_s^1\\ &= -\frac12s^2+\frac12+\frac12+2s\log s-\frac12s^2\\ &= 2s\log s-s^2+1 \end{align}

and \begin{align} \int_0^1\mathrm dy\int_0^1\mathrm dx\,\delta(xy-s) &= \int_0^1\mathrm dy\int_0^1\mathrm dx\,\frac1y\delta\left(x-\frac sy\right)\\ &= \int_s^1\mathrm dy\,\frac1y\\ &= -\log s\;. \end{align}

So

$$ \mathbb{E}\left[ (X-Y)^2\mid XY=s\right]=\frac{s^2-1}{\log s}-2s\;. $$

Here's a plot.