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During solution of probability problem I came across following: Let $X_n$ - be a sequence of i.i.d (independent, identically distributed) random variables, uniformly distributed over $[0;1]$. Then is it true that random variables $m_n$, determined as $m_n=min\{X_1, ..., X_n\}$ are also independent or not?

If not, do you have any ideas of proving that such 𝑚𝑛 approaches 0 almost surely then? I was trying to proof this via fact mentioned above, and do not have anymore ideas..

Do you have any ideas about this one? This is very crucial for my solution, would appreciate any help!

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    $\begingroup$ $m_n \leq m_{n+1}$ so independence violates intuition. Technically, one can show that $m_n$'s are independent iff the random variables are constant w.p. $1$. $\endgroup$ Feb 5, 2021 at 0:02
  • $\begingroup$ @KaviRamaMurthy, and do you have any ideas of proving that such $m_n$ approaches 0 almost surely then? I was trying to proof this via fact mentioned above, and do not have anymore ideas.. $\endgroup$ Feb 5, 2021 at 0:05
  • $\begingroup$ For any $x>0$ we have $P(m_n >x)=P(X_1>x)^{n} \to 0$. So $m_n \to 0$ in probability. By monotonicity almost sure convergence also true. $\endgroup$ Feb 5, 2021 at 0:15
  • $\begingroup$ In my first comment above I meant to say $m_{n+1} \leq m_n$. $\endgroup$ Feb 5, 2021 at 0:16
  • $\begingroup$ @KaviRamaMurthy, may you please write a more detailed solution about almost sure convergence if that is possible. New to the topic and don't really get it on the go unfortunately. I've also want to notice that it is needed to prove almost sure convergence, in other words: $P(lim_{n \rightarrow \infty} m_n=0)=1$ $\endgroup$ Feb 5, 2021 at 0:19

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(This answer is in response to the comments by OP).

We have $0 \leq m_{n+1} \leq m_m \leq 1$. This implies that $m =\lim_{ n\to \infty} m_m$ exist almost surely. In particular, this gives convergence in probability. But $P(m_n >\epsilon) =(P(X_1>\epsilon)^{n} \to 0$ (since $x^{n} \to 0$ whenver $|x|<1$). It follows that $m_m \to 0$ in probability and $m_n \to m$ in probability. Hence, $m=0$ almost surely and $\lim_{ n\to \infty} m_n=0$ almost surely.

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