Re: Probability of rth order statistic

Licas,

You have given the cdf of the kth order stat. To determine the prob
that X1 is ranked r, the probability should simply be

(n-1)C(k-1) * [F(x)]^(k-1) * [1-F(x)]^(n-k)

rather than the sum from k to n because I've to select (k-1) entries
from the remaining (n-1) to be lesser than X1 and the remaining MUST be
greater than X1. This seems to make the most sense to me.

Thanks,

"Luis A. Afonso" wrote:
> 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)

.

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