Expectation of $\min(X,Y)$ conditional on $\max(X,Y)$

Question

Let $X_1,X_2$ be two independent uniformly distributed random variables on $[0,1]$. What is the expectation of $\min(X_1,X_2)$ if at least one of them is known to exceed $0.5$, i.e $\max(X_1,X_2)>0.5.$ Apparently the answer is $11/24$ but I have been unable to verify this. Any hints/suggestions would be appreciated.

Answer 1

Letting $M_1 = \min(X_1, X_2)$ and $M_2 = \max(X_1, X_2)$, we have \begin{align*} P(M_1 < m_1, M_2 < m_2) &= P(M_2 < m_2) - P(M_1 > m_1, M_2 < m_2) \\ &=m_2^2 - (m_2 - m_1)^2 \\ &= 2m_1m_2 - m_1^2 \qquad \text{for } 0 < m_1 < m_2 < 1 \end{align*} Therefore the pdf is $$ f_{M_1, M_2}(m_1, m_2) = 2 \qquad \text{for } 0 < m_1 < m_2 < 1 $$ So $$ E[M_1\mid M_2 > 0.5] = \frac{\int_{0.5}^1 \int_{0}^{m_2} 2m_1 \, dm_1 \, dm_2}{\int_{0.5}^1 \int_0^{m_2} 2 \, dm_1 \, dm_2} = \frac{7}{18} $$

Answer 2

A graphic approach:

Considering that we can call U=max and V=min, the conditional law $V|U>\frac{1}{2}$ is easily derived by the ratio of the two areas, one of the which is the purple area and the other depends on the values of v

Thus the Conditional CDF is the following

$$F_{V|U>\frac{1}{2}}(v) = \begin{cases} 0, & \text{if $v<0$} \\ \frac{4}{3}v, & \text{if $0 \leq v <\frac{1}{2} $} \\ 1-\frac{4}{3}(1-v)^2, & \text{if $\frac{1}{2} \leq v <1$} \\ 1, & \text{if $v \geq 1$} \end{cases}$$

Leading to the same Expectation's result of $\frac{7}{18}$