Lagmental Vicfred

Uniform Order Statistics Have Beta Densities by Vicfred

The kth smallest of n independent uniform samples follows a beta law. I will separate the object being defined from the consequence being claimed.

Set-up

Continuous-time processes are described by finite-dimensional distributions plus path regularity. A Poisson process \(N_t\) has independent increments, while Brownian motion \(B_t\) has Gaussian increments.

$$ U_{(1)}\le\cdots\le U_{(n)},\qquad U_i\stackrel{\mathrm{iid}}{\sim}\operatorname{Unif}(0,1) $$

I read the first line as input and the second as output. The symbols \(\forall\) and \(\exists\) are not interchangeable, and neither may be upgraded silently to \(\Longleftrightarrow\).

$$ U_{(k)}\sim\operatorname{Beta}(k,n+1-k) $$

The calculation

This is the algebraic core of the note. Once this line is correct, the surrounding interpretation has something solid to refer to.

$$ f_{U_{(k)}}(x)=\frac{n!}{(k-1)!(n-k)!}x^{k-1}(1-x)^{n-k}\mathbf1_{(0,1)}(x) $$

What survives abstraction

The aligned summary deliberately puts the datum and conclusion on different rows. Mathematically, this is the distinction between specifying an object and proving a property of it.

$$ \begin{aligned} \mathsf{D}\;&:\quad U_{(1)}\le\cdots\le U_{(n)},\qquad U_i\stackrel{\mathrm{iid}}{\sim}\operatorname{Unif}(0,1),\\[5pt] \mathsf{C}\;&:\quad U_{(k)}\sim\operatorname{Beta}(k,n+1-k). \end{aligned} $$

The boundary

Matching means and variances does not identify a distribution. Independence, increment laws, and sample-path properties are separate ingredients.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] U_{(k)}\sim\operatorname{Beta}(k,n+1-k) \end{gathered}} $$

The notation is dense, but it is doing honest work: every delimiter records scope and every index records dependence. Removing one should require a mathematical reason.

This article was posted on Mon 03 August 2020. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.