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I have three independent uniforms variables, $U, V, W$ that take their values in $[0,1]$. I also have $X = \max(U, W)$ and $Y = \max(V,W)$.

I need to compute the conditional expectation of $W$ knowing (the $\sigma$-algebra generated by) $X$ and $Y$. I have proven that it can be written as $f(X,Y)$ with $f$ measurable.

Is there any way to compute this without too much calculations? I have already computed $E(W \mid X)$ but I can't find any simple way to use this. Do I have to make the full calculation?

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up vote 1 down vote accepted

Here is a hands-on approach. One looks for the mean value of $W$ conditionally on $(X,Y)=(x,y)$, with $x$ and $y$ in $[0,1]$.

  • If $x=y$, then almost surely $U\lt W$ and $V\lt W$, in which case $W=x$.

  • If $x\lt y$, either $U\lt W\lt V$ (case 1), or $W\lt U\lt V$ (case 2).

    • In case 1, $(X,Y)=(W,V)$ hence $W=x$.
    • In case 2, $(X,Y)=(U,V)$ and $W$ is conditioned to be in the interval $[0,U]$ hence the conditional mean of $W$ is $\frac12U=\frac12x$.
    • Cases 1 and 2 are equiprobable by exchangeability of $(U,V,W)$ hence when $x\lt y$, the conditional mean of $W$ is $\frac12(x+\frac12x)=\frac34x$.
  • If $y\lt x$, by symmetry the conditional mean of $W$ is $\frac34y$.

  • Hence, when $x\ne y$, the conditional mean of $W$ is $\frac34\min\{x,y\}$.

Finally, $$ \mathbb E(W\mid X,Y)=\tfrac14\min\{X,Y\}\cdot(3+\mathbf 1_{X=Y}). $$

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