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Aug 9, 2011 at 21:23 comment added OctaviaQ (...cont) That gave me the exact probabilities for $X_{final}$ being each order stat. Then I used those weights to mix the 5th -8th order statistics for F(x), using a textbook formula to define the order distributions of F(x) (where the pdf of the 4kth order stat of n draws from cdf F(x) = $\frac{n!}{(k-1)!(n-k)!}F(x)^{k-1}(1-F(x))^{n-k}f(x)$
Aug 9, 2011 at 21:18 comment added OctaviaQ My brute force method: I figured that $X_{final}$ would be a mixture of predictable weights of order statistics of n*m draws from F(x). For example, for $n=4$ and $m=2$, we start with 8 independent draws from F(x), and $X_{final}$ > the 4th order statistic. To find the PR of it being each order stat 5-8, I wrote a computer script that wrote out every permutation of 1-8, and an algorithm that found $X_{final}$ for each permutation (using the order stats themselves as the comparisons) (cont...)
Aug 9, 2011 at 18:25 comment added Bogdan Lataianu yes, but there might be some interesting ideas
Aug 9, 2011 at 18:15 comment added OctaviaQ yeah, I figured it was now irrelevant, since you provided a much better solution. I'll see if I can find what I'd written.
Jul 31, 2011 at 8:54 comment added Bogdan Lataianu JandR you deleted a comment of yours in which you indicated an ad-hoc method using weights.
Jul 20, 2011 at 20:15 vote accept OctaviaQ
Jul 20, 2011 at 6:34 history tweeted twitter.com/#!/StackStats/status/93569328003293185
Jul 20, 2011 at 0:18 comment added Bogdan Lataianu may i ask where did you get this problem?
Jul 19, 2011 at 23:49 answer added Bogdan Lataianu timeline score: 4
Jul 19, 2011 at 19:16 history edited user88 CC BY-SA 3.0
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Jul 19, 2011 at 18:33 answer added highBandWidth timeline score: 1
Jul 19, 2011 at 18:14 history edited OctaviaQ
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Jul 19, 2011 at 18:05 history asked OctaviaQ CC BY-SA 3.0