Re: Probability of rth order statistic

On 3 Dec 2005 19:53:49 -0800, "Patrik" <hosanagar@xxxxxxxxx> wrote:

> I have N variables X1, X2,..., Xn which are all iid random variables. I
> want to figure out the probability that X1 is ranked "r" in ascending
> order among the N items.

Isn't it the same for each X_i?
Isn't it simply 1/N for any X_i?

Or am I missing something?

Here's the rest of the post.
> The CDF of the distribution for these X
> variables is F. I can think of 2 ways and I can't figure out which one
> (if any) is correct:
>
> 1. The number of ways in which I can choose (r-1) variables to place
> above X1 (i.e, having values less than X1) are (N-1) choose (r-1),
> i.e., (N-1)C(r-1). Having chosen these variables, I have to set the
> corresponding probabilities for values less than and greater than X1:
>
> Pr(rank=r) = (N-1)C(r-1) * {F(X1)^(r-1)} * {(1-F(X1))^(n-r)
>
> 2. Someone else suggested I use the pdf of the rth order statistic
> which is given by:
>
> N!/{(N-r)!(r-1)!} * {F(X1)^(r-1)} * {(1-F(X1))^(n-r) * f(b)
>
> Are either of these correct? I am inclined to use the first one because
> the second one is really Prob (rth order stat = X1) which is different.
> Maybe, I should use Prob(r-1 order stat < X1) * Prob(r+1 order stat >
> X1) but do I even need to do that (can formula suffice). I'd appreciate
> any input.


--
Rich Ulrich, wpilib@xxxxxxxx
http://www.pitt.edu/~wpilib/index.html
.

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