Let $X_1, \dots, X_n \sim_{iid} \chi^2_{m}$ be a random sample from a chi-squared distribution with $m$ degrees of freedom (d.f.). I was wondering if there's any known result for the order statistics

$$\max_{1 \le i \le n} X_i, \min_{1 \le i \le n} X_i $$

respectively as a function of $m$ and also $n$?

And finally, is there any known result for the ratio of this two order statistics:

$$ \frac{\max_{1 \le i \le n} X_i}{\min_{1 \le i \le n} X_i }?$$

P.S. I'm primarily interested how these three quantities above behave w.r.t. increasing d.f. $m$ and w.r.t. increasing sample size $n$.

In this regard, I've looked into this question on stats.SE, but couldn't make it helpful.

Any references would also be appreciated.


1 Answer 1


A book that may contain answers to all the questions you have is 'Order Statistics' by David and Nagaraja.

For you first question, the easiest way is to rely on the fact that order statistics can be written as the inverse cdf of the corresponding order statistic of a uniform sample, i.e. $$X_{(1)}, \ldots, X_{(n)} = (F^{-1}(U_{(1)}),\ldots, F^{-1}(U_{(n)})),$$

where the equality sign is meaning equality in distribution. Also, the joint distribution for two order statistics is well known and an interesting exercise.

  • $\begingroup$ Hi, thanks for your answer. I agree that joint distribution of the min and max can be calculated, and hence their ratio too. But the thing is, the variables being $\chi_m^2,$ the ratio distribution is mathematically very complicated. The real question is how they depend as a function of the d.f. $m?$ I'll check the book, but scouring the internet didn't produce a useful answer to this. $\endgroup$ Aug 7, 2020 at 16:14
  • 1
    $\begingroup$ Ok, I see the edit to you question that stresses the points that you want. I do not really know what you want to do with it, but this reference (arxiv.org/pdf/1207.7209.pdf) could be helpful. Further, as your data is i.i.d. the book 'extremes and related properties of random sequences and processes' may also help you. $\endgroup$ Aug 7, 2020 at 16:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.