# Compute $\mathbb{E} \left(\min(X, Y) | \max(X, Y) \right)$ for $(X,Y)$ i.i.d. uniform on $(0,1)$

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?

• How did you get $\mathcal F$? – Kavi Rama Murthy Dec 3 '18 at 8:35
• @KaviRamaMurthy It is the $\sigma$-algebra made of $\max(X, Y)$. It's just the new symbol for $\sigma(\max(X,Y))$ – Hendrra Dec 3 '18 at 8:37
• My question is why it is $\{\emptyset, [0,1]\}$. This is not true. – Kavi Rama Murthy Dec 3 '18 at 8:38
• 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 $\frac 1 4$. – Kavi Rama Murthy Dec 3 '18 at 8:44
• 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 $\mathbb E(\min(X,Y)\mid\max(X,Y))=\frac12\max(X,Y)$. – drhab Dec 3 '18 at 9:23

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.

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