The distribution(or just expectation) of $\min\{U_{1}, U_{2}\}\times \min\{U_{1}, U_{3}\}$?

Question

I'm trying to find the expectation of $\min\{U_{1}, U_{2}\}\times \min\{U_{1}, U_{3}\}$, where $U_{1}, U_{2}, U_{3}$ are independent from $\mathrm{Unif}(0, 1)$, is there any idea?

Answer 1

I can start you off, use the Law of Total Expectation and this link : Expectation of Minimum of $n$ i.i.d. uniform random variables. (for the case of two iid uniforms)

\begin{align} E[\min\{U_1,U_2\}\min\{U_1,U_3\} ]=& \frac{1}{2}\Big(E[\min\{U_1,U_2\}\min\{U_1,U_3\} ~|~ U_1<U_2] \\ &+E[\min\{U_1,U_2\}\min\{U_1,U_3\} ~|~ U_2<U_1]\Big) \\ =& \frac{1}{2}\Big(E[U_1\min\{U_1,U_3\}] \\ &+E[U_2\min\{U_1,U_3\}~|~U_2<U_1]\Big). \end{align}

Now lets deal with $E[U_2\min\{U_1,U_3\}~|~U_2<U_1]$.

\begin{align} E[U_2\min\{U_1,U_3\}~|~U_2<U_1]=& E[U_2\min\{U_1,U_3\}~|~U_2<U_1,U_3<U_1] \\ &+E[U_2\min\{U_1,U_3\}~|~U_2<U_1,U_1<U_3] \\ =& E[U_2U_3]+E[U_2U_1~|~U_2<U_1] \end{align}

Now solve $E[U_2U_1~|~U_2<U_1]$ using a double integral.