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Dec 1 at 17:53 comment added Nate River @JamesMartin Ah, that’s a very nice independence argument.
Dec 1 at 16:37 comment added James Martin yes, that's what I meant :) $\,\,\,\,\,$
Dec 1 at 16:35 comment added mathworker21 @JamesMartin Thanks. And I think you mean "atomic" at the end.
Dec 1 at 16:30 comment added James Martin @mathworker21 You are welcome be as careful as you like! I was just giving you a brief indication of why it's true. I remember it as an exercise from the first course I took in probability, from Frank Kelly in 1992. Indeed, the non-atomic case is less elegant....
Dec 1 at 16:18 comment added mathworker21 @JamesMartin Ok, but you have to be careful. If running maximum was defined with strict inequality and the distribution is uniform over $\{1,2\}$, say, then the conditional probability that [$X_3$ is a running maximum] given that [$X_2$ is] is equal to $0$, but the unconditional probability is positive.
Dec 1 at 15:45 comment added James Martin @mathworker They are independent. The relative ordering of the first $n$ values (which is uniform on the $n!$ possibilities) carries no information about whether or not the $(n+1)$st is larger than all of them.
Dec 1 at 15:32 comment added mathworker21 @JamesMartin They are independent?
Dec 1 at 15:14 comment added James Martin In particular, the sequence $M_n$ is given by $M_n=B_1+\dots+B_n$ where the $B_i$ are independent random variables, $B_i\sim\text{Bernoulli}(1/i)$. (The events that $X_i$ is a running maximum are independent for different $i$.)
Dec 1 at 15:06 comment added Fedor Petrov The number of left-to-right maxima between $X_1,\dots,X_n$ is distributed as the number of cycles in the random permutation in $S_n$. This was studied a lot. See, for instance, arxiv.org/pdf/1903.04906 and references therein
Dec 1 at 13:36 history edited Iosif Pinelis CC BY-SA 4.0
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Dec 1 at 13:27 history edited Nate River CC BY-SA 4.0
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Dec 1 at 13:15 history edited Nate River CC BY-SA 4.0
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Dec 1 at 13:09 history asked Nate River CC BY-SA 4.0