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}
finding $P(M=m)$ in a sequence of independent random variables
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probability
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0I 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. – 2012-04-24
1 Answers
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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 $\mathrm P(M\geqslant2)=\mathrm P(X_0\gt X_1)=\ldots$