Re: Probability of rth order statistic

The unmistakable impression that I draw from the Richard Ulrich´s feedback (as questions) to the problem posted by *Patrik* is that he is *virgin* about the classic *Statistics of Extremes*. The worse of all of this is that he could simply *think* in order that to avoid the *dunk* he posted (Isn´t it simply 1/N?).
Because I am a nice guy I will teach him (the following is a translation from a *elementary* Portuguese textbook on Statistics*).
***Let be a set of n random continuous variables (r.v.) i.i.d. (identically independently distributed) X1, X2, …,Xn with density f(x); therefore the Distribution Function is
____________F(x) = Integral (from -oo to x). f(t)dt
We will note X´k the kth r.v. - (kth ordinal statistics)- the value that one have k-1 values that are lesser than it. That is:
_______X´1<=X´2<=…<=X´n______________________
X´1 is the (first) minimum, X´2 the second, ..., X´n the nth minimum or maximum.
Because we put
________Fk(x) = prob.(X´k <= x) we have (Bernouilli Distribution):
________H is the probability that the number of values lesser or equal to x be greater or equal to k.
___H= sum (from k to n) nCj * [F(x)]^j * [1-F(x)]^(n-j)
____Therefore
____Fk(x)=[n! /((k-1)! (n-k)!]* F^(k-1)*(1-F)^(n-k)* f(x)
(for n sufficiently large a normal approach is possible).

That´s all (*folks*)

_________________licas (Luís A. Afonso)
.