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I have read somewhere (unfortunately I cannot locate the exact text anymore) that there exists an equality in distribution between the order statistics of any continuous distribution and the uniform distribution.

In other words, let $X_1,\ldots,X_n$ be identically distributed random variables, where $X_{1:n},\ldots,X_{n:n}$ are their order statistics so that $X_{1:n}<\ldots<X_{n:n}$. Then, there exists an equality in distribution

\begin{equation*} \frac{X_{i:n}}{X_{n:n}}\overset{d}{=} Y_{i:n-1}, \quad \forall 1\leq i\leq n-1 \end{equation*} where $Y_i\sim \text{Uniform}(0,1)$. In other words, the ratio $\frac{X_{i:n}}{X_{n:n}}$ has supposedly the same distribution as the $i\text{th}$ order statistic of the $n-1$ independent and uniformly distributed random variates $U_1,\ldots,U_{n-1}$. Is this true?

Can we then say that $\frac{X_{1:n}}{X_{n:n}},\ldots,\frac{X_{n-1:n}}{X_{n:n}}$ are i.i.d. and distributed as $\sim \text{Uniform}(0,1)$?

$\textbf{Edit:}$

I managed to track down the text I was thinking of. It is in the book "A First Course in Order Statistics" by Barry C. Arnold , N. Balakrishnan and H. N. Nagaraja. Specifically, I am thinking of equations (2.4.1) and (2.4.2) in Chapter 2, pp. 21-22.

Please have a look on Google Books (I do not want to copy-paste the whole thing due to copyright): 2.4. Some Properties of Order Statistics

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  • $\begingroup$ That seems wrong -- the ratio takes values outside $[0,1]$, the denominator could be concentrated around, zero, ... -- are you sure you don't mean values of the cumulative distribution function? Something along these lines might be true for those (but then again, those values are uniformly distributed over $[0,1]$, so that wouldn't be terribly interesting...) $\endgroup$ – joriki Jun 23 '18 at 16:53
  • $\begingroup$ Not sure which ratio you mean takes values outside $[0,1]$? Yes, with distribution I mean the CDF. $\endgroup$ – index Jun 23 '18 at 17:48
  • $\begingroup$ I mean the ratio $\frac{X_{i:n}}{X_{n:n}}$. It could take any value, no? $\endgroup$ – joriki Jun 23 '18 at 22:06
  • $\begingroup$ @joriki By definition, the $(X_{i:n})_{1\leq i\leq n}$ are sorted in non-decreasing order. So all the ratios are (assuming the original $X_i$'s take value in $[0,\infty)$ between $0$ and $1$. $\endgroup$ – Clement C. Jun 24 '18 at 0:55
  • $\begingroup$ @ClementC.: But there's no assumption about positivity in the question. Also, even with that assumption, there would be easy counterexample -- for instance a uniform distribution on $[0,1]\cup[9,10]$, where about half of the ratios would be $\gt\frac9{10}$. $\endgroup$ – joriki Jun 24 '18 at 6:40

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