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Let $X, Y$ be independent random variables with uniform distribution on the interval $[0, 1]$. My task is to find $$\mathbb{E} \big(\min(X, Y) | \max(X, Y) \big).$$

I think it can be done in the following way $$\mathbb{E} \big(\min(X, Y) | \max(X, Y) \big) = \mathbb{E}\big(\min(X, Y)|\sigma(\max(X,Y) \big) = \mathbb{E}\big(\min(X, Y)|\mathcal{F} \big).$$ Of course $\mathcal{F} = \{ \emptyset, [0,1] \}$ thus $\min(X, Y)$ is $\mathcal{F}$-measurable. That leads us to $$\mathbb{E}\big(\min(X, Y)|\mathcal{F} \big) = \min(X, Y).$$ Is my reasoning correct? That won't stand for other distributions will it?

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  • $\begingroup$ How did you get $\mathcal F$? $\endgroup$ – Kavi Rama Murthy Dec 3 '18 at 8:35
  • $\begingroup$ @KaviRamaMurthy It is the $\sigma$-algebra made of $\max(X, Y)$. It's just the new symbol for $\sigma(\max(X,Y))$ $\endgroup$ – Hendrra Dec 3 '18 at 8:37
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    $\begingroup$ My question is why it is $\{\emptyset, [0,1]\}$. This is not true. $\endgroup$ – Kavi Rama Murthy Dec 3 '18 at 8:38
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    $\begingroup$ A set like $\{\max (X,Y) < \frac 1 2\} =\{X< \frac 1 2 \text {and} Y< \frac 1 2 \}$ belongs to $\mathcal F$. This set is neither empty nor $[0,1]$ because its probability is $\frac 1 4$. $\endgroup$ – Kavi Rama Murthy Dec 3 '18 at 8:44
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    $\begingroup$ In a less formal way you can state that $\min(X,Y)$ has uniform distritution on $[0,z]$ under the condition that $\max(X,Y)=z$. The corresponding expectation of that distribution is $\frac12z$, leading to the conclusion that $\mathbb E(\min(X,Y)\mid\max(X,Y))=\frac12\max(X,Y)$. $\endgroup$ – drhab Dec 3 '18 at 9:23
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drhab has already commented how to informally derive the solution. I will formally prove that this is in fact the correct solution.

To this end, let $Z = \max(X,Y)$. We want to show $$ E[\min(X,Y)|\sigma(Z)] = \frac{Z}{2} \quad \text{a.s.} $$

The right hand side is obviously $\sigma(Z)$-measurable. So all we need to show is $$ E[\min(X,Y)\mathbb{1}_F] = E\left[\frac{Z}{2}\mathbb{1}_F\right] $$ for all $F \in \sigma(Z)$.

We will show this property for $F = \{Z \leq a \}$ with $a \in \mathbb{R}$. This is enough because these sets form a $\pi$-system (meaning it's $\cap$-stable) generating $\sigma(Z)$ and we can use a dynkin system argument to get the property for all $F \in \sigma(Z)$. Let me know if I should go into more detail here.

Because $X$ and $Y$ are iid $U([0,1])$, $Z$ has the Lebesgue density $f_Z(z) = 2z\mathbb{1}_{[0,1]}(z)$. So we get for the right hand side $$ E\left[\frac{Z}{2}\mathbb{1}_F\right] = \frac12 \int_{-\infty}^azdP^Z(z) = \int_{-\infty}^a z^2 \mathbb{1}_{[0,1]}(z)dz\\ = \begin{cases} 0 \quad &a < 0\\ \frac13 a^3 \quad &a \in [0,1]\\ \frac13 \quad &a > 1 \end{cases}. $$

For the left hand side, we get, using the symmetry of $X$ and $Y$ and $F = \{Z \leq a \} = \{X \leq a \} \cap \{Y \leq a \}$, $$ E[\min(X,Y)\mathbb{1}_F] = 2E[\min(X,Y)\mathbb{1}_F\mathbb{1}_{\{X \leq Y\}}] = 2E[X \mathbb{1}_{\{X \leq a\}}\mathbb{1}_{\{Y \leq a\}}\mathbb{1}_{\{X \leq Y\}}]\\ = 2\int_0^1\int_0^1 x \mathbb{1}_{\{x \leq a\}} \mathbb{1}_{\{y \leq a\}}\mathbb{1}_{\{x \leq y\}} dy dx\\ = 2 \int_0^1 x \mathbb{1}_{\{x \leq a\}} \int_x^1 \mathbb{1}_{\{y \leq a\}} dy dx. $$ Now you can distinguish the same $3$ cases as above and notice that you arrive at the same values.

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    $\begingroup$ (+1) The phrase "a $\cap$-stable generator of $\sigma(Z)$" could be replaced by the more standard "a $\pi$-system generating $\sigma(Z)$. $\endgroup$ – Did Dec 3 '18 at 10:51

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