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suppose $\{X_r,r\geq0\}$ be a sequence of independent random variables with same distribution. if $M=min\{n\geq 1;X_0\geq X_1\geq\ldots \geq X_{n-1}<X_n\}$, how can i find $P(M=m)$

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I think this is an exercise in Ross, but all orders $X_{\pi(0)} \ge .... \ge X_{\pi(n)}$ are equally likely and you can get it from that. You probably want continuous distributions, I think it is a big mess otherwise. – mike Apr 24 '12 at 17:08
up vote 1 down vote accepted

Hint: For continuous distributions, for every $m\geqslant1$, $[M\geqslant m]=[X_0\gt X_1\gt\cdots\gt X_{m-1}]$ up to null events, and, due to symmetry reasons, the probability of the RHS is straightforward.

For example, for $m=2$, $[X_0\gt X_1]$ and $[X_1\gt X_0]$ have the same probability (why?) and they make a partition of the whole probability space (true?), hence $\mathrm P(M\geqslant2)=\mathrm P(X_0\gt X_1)=\ldots$

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