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Let $U_1, U_2$ and $U_3$ be three independent uniformly $(0, 1)$ random variables.

Let $X_1,...,X_n$ be a sequence of independent uniformly $(0, 1)$ random variables.

Consider that $X_i$ and $U_j$ are independents, for all $i,j$.

Show that the event $\{U_1 > U_2 > U_3\}$ is independent of the $(U_{(3)},X_{(1)})$, where $U_{(3)} = \max\{U_1,U_2,U_3\}$ and $X_{(1)} = \min\{X_1,...,X_n\}$.

I have no idea to start. How'd be the definition of independence between events and random variables?

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    $\begingroup$ As for your last question: the rv. $X$ is independent of an event $A$ if all events $B$ defined in terms of $X$ are independent of $A$, such as $B=[X\lt t]$, or $B=[X\in S]$, or most generally, for all $B$ in the sigma field $\mathcal{F}(X)$ generated by $X$ $\endgroup$ – kimchi lover Nov 28 '18 at 22:40
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    $\begingroup$ You also need independence between $U_k$ and$X_j$ for all indices. $\endgroup$ – herb steinberg Nov 28 '18 at 22:46
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An event and a random variable are independent if for all supported values of the random variable, the conditional expectation of the event given the variable equals the margial probability of the event.

In short you need to establish whether: $${\forall u\in(0;1)~\forall x\in(0;1):\\\quad\mathsf P(\{U_1{>}U_2{>}U_3\})=\mathsf P(\{U_1{>}U_2{>}U_3\}\mid u{=}\max\{U_i\}_{i=1}^3,x{=}\min\{X_j\}_{j=1}^n)}$$

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  • $\begingroup$ $\min\{X_j)_{j=1}^n\}$ is independent of $\{U_1{>}U_2{>}U_3\}$. How can I prove that $\{U_1{>}U_2{>}U_3\}$ and $\max\{U_i\}_{i=1}^3$ are independent, or $P(\{U_1{>}U_2{>}U_3\}\mid u{=}\max\{U_i\}_{i=1}^3) = P(\{U_1{>}U_2{>}U_3\})$ ? $\endgroup$ – Pedro Salgado Nov 29 '18 at 12:22
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    $\begingroup$ Use symmetry. @PedroSalgato $\endgroup$ – Graham Kemp Nov 29 '18 at 22:37
  • $\begingroup$ how can I use? @graham-kemp $\endgroup$ – Pedro Salgado Nov 30 '18 at 0:05
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    $\begingroup$ Argue that that $\mathsf P(\{U_1>U_2>U_3\}\mid u{=}\max\{U_i\}_{i=1}^3)=\mathsf P(\{U_3>U_1>U_2\}\mid u{=}\max\{U_i\}_{i=1}^3)$ and so forth, by symmetry. @Pedro-Salgado $\endgroup$ – Graham Kemp Nov 30 '18 at 0:23

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