Timeline for Joint pdf of functions of order statistics
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4, 2013 at 21:58 | history | tweeted | twitter.com/#!/StackStats/status/408354620302700544 | ||
| Dec 4, 2013 at 21:51 | vote | accept | JohnK | ||
| Dec 4, 2013 at 17:23 | answer | added | whuber♦ | timeline score: 7 | |
| Dec 4, 2013 at 16:47 | history | edited | JohnK | CC BY-SA 3.0 |
edited body
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| Dec 4, 2013 at 16:36 | comment | added | wolfies | Second time this has come up - lol - and I learn not from my errors :) | |
| Dec 4, 2013 at 16:34 | comment | added | JohnK | @whuber It's the functions that I having trouble comprehending. And by functions I mean the difference between the cdf values$v_1,v_2$. Why is the unit interval split like that precisely? | |
| Dec 4, 2013 at 16:31 | comment | added | whuber♦ | And that's all the information you need: $F$ is the probability integral transform so indeed $F(Y_i)$ is distributed as the $i^\text{th}$ order statistic from a uniform distribution. For a non-rigorous (but nevertheless accurate) heuristic, think of this order statistic as dividing the interval $[0,1]$ into three bins of length $F(Y_i)$, $dY_i$ (an infinitesimal), and $1-F(Y_i)$, into which precisely $i-1$, $1$, and $n-i$ of the values fall (that's where the multinomial coefficients come from). | |
| Dec 4, 2013 at 16:28 | comment | added | JohnK | @wolfies This is precisely what I am suggesting. This is a general condition that holds and people who have seen order statistics, tolerance intervals and coverages before might be able to help me derive it. | |
| Dec 4, 2013 at 16:20 | comment | added | JohnK | @wolfies This is all the information I am given. | |
| Dec 4, 2013 at 16:10 | history | asked | JohnK | CC BY-SA 3.0 |