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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?

Zhao Zhao
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  • I shouldn't forget $E[U_1~|~U_1<U_2]\neq E[U_1]$, I mean its clearer that $E[U_1~|~U_1<U_2,\ldots U_1<U_{100}]\neq E[U_1]$. – Daniel Adams Feb 21 '21 at 09:53

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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.

Daniel Adams
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    Thanks for your help!! But I don't quite understand the second equality you wrote... Should $\frac{1}{2}E[\min{U_1,U_3}]$ be $\frac{1}{2}E[U_1\min{U_1,U_3}]$? – Zhao Zhao Feb 18 '21 at 16:13
  • @ZhaoZhao that term is $E[U_2\min{U_1,U_3}]$ which because of independence is equal $E[U_2]E[\min{U_1,U_3}]$ – Daniel Adams Feb 19 '21 at 09:31
  • How can the conditional term $U_2<U_1$ be dropped in $E[U_2\min{U_1,U_3} ~|~ U_2<U_1]$? I know in $E[U_1\min{U_1,U_3} ~|~ U_1<U_2]$ the conditional term $U_1<U_2$ can be dropped because $U_2$ is no longer irrelevant. – Zhao Zhao Feb 19 '21 at 12:44
  • I think your right, how about now? – Daniel Adams Feb 19 '21 at 13:42
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    I think it's correct now, but I suppose an extra factor $\frac{1}{2}$ is needed, and I have added that in your answer. Thanks!! – Zhao Zhao Feb 19 '21 at 13:59
  • not sure where your getting any extra 1/2 from. But your welcome – Daniel Adams Feb 19 '21 at 14:08
  • The condition cannot be dropped like that. E.g. the conditional distribution of $U_1 \mid U_1 < U_2$ has a different distribution than the original $U_1$. The conditional distribution is like the minimum of the order statistics $U_{(1)}$ with sample size $2$. – BGM Feb 20 '21 at 13:51
  • @ZhaoZhao BGM is right, can you un-accept the answer? – Daniel Adams Feb 21 '21 at 09:51