Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $P(M=m)$

share|improve this question
    
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

1 Answer 1

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 $\mathrm P(M\geqslant2)=\mathrm P(X_0\gt X_1)=\ldots$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.