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If you draw n realizations from U(0,1), then will the sorting of those values (on average) be separated by $\frac{1}{n+1}$ between two consecutive values. E.g., if n=5, then the lowest one is realized at 1/6, the second lowest one at 2/6 etc.

Furthermore, could I take the above values (1/6, 2/6..., 5/6) set them equal to the c.d.f. of any other distribution, then solve for x (five values in total), and it will yield the order statistics for n=5 values for that the new distribution?

I'm only using intuition here, so my gut feeling might be wrong. The stuff I've read online is too technical for me.

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To my mind, the easiest way to think about this is to imagine the interval closed to a circle and to consider the point where it's broken up as an $(n+1)$-th random variable. That reveals the symmetry of the situation; it's now apparent that the intervals between any two consecutive numbers, including the case where one of them is the boundary at $0$ or $1$, are on the same footing and thus by symmetry must have mean $1/(n+1)$.

And yes, you can transfer this to another distribution by equating these values to its cumulative distribution function. I'm not sure why you keep mentioning $1/6$, $2/6$, though; you wouldn't be equating those means to the cumulative distribution function but the actual realizations from $U(0,1)$.

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    thx. what i meant to say: Take the mean of the ith order statistic. Taking$5$realizations from the exp (one time), and then ordering them and then then setting them equal to the c.d.f. isn't clean IMO. I'm interested in the long term behavior of the ith order statistic from the exp: So, taking$5$samples from U(0,1), ordering them, taking 5 samples again, ordering them again (repeating this process an infinite amount of time). Then, computing the average value of the highest number (of all intinite samples of the max), the average value of the second highest number etc.2012-10-26