# Independence between random vector and event

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?

• 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 $X$ – kimchi lover Nov 28 '18 at 22:40
• You also need independence between $U_k$ and$X_j$ for all indices. – herb steinberg Nov 28 '18 at 22:46

In short you need to establish whether: $${\forall u\in(0;1)~\forall x\in(0;1):\\\quad\mathsf P(\{U_1{>}U_2{>}U_3\})=\mathsf P(\{U_1{>}U_2{>}U_3\}\mid u{=}\max\{U_i\}_{i=1}^3,x{=}\min\{X_j\}_{j=1}^n)}$$
• $\min\{X_j)_{j=1}^n\}$ 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 $P(\{U_1{>}U_2{>}U_3\}\mid u{=}\max\{U_i\}_{i=1}^3) = P(\{U_1{>}U_2{>}U_3\})$ ? – Pedro Salgado Nov 29 '18 at 12:22
• 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 – Graham Kemp Nov 30 '18 at 0:23