# Order Statistics: How to calculate expected value of a function involving first and second order statistics

I am currently stuck with a challenging problem. I have n values drawn i.i.d. from a distribution F(x). Let $$v_1$$ be the nth order statistic (highest value) and let $$v_2$$ be the n-1 order statistic (second highest value). Then, what is the expected value of a function of them:

1. What is the expected value of $$\frac{v_1 - v_2}{v_1}$$ : $$E[\frac{v_1 - v_2}{v_1}]$$
2. What is the the expected value of $$\frac{v_1 - v_2}{v_1} * \frac{v_2}{2}$$ $$\frac{v_1 - v_2}{v_1} * \frac{v_2}{2}$$ I would highly appreciate it if you could give me some direction or explain how to calculate it.
• You have to compute the integrals (or sums, depending on the type of distribution). If you have a particular $F$ in mind, we could potentially go much further towards an answer.
– whuber
Apr 29, 2021 at 13:48
• @StephanKolassa unfortunatelly, I made a spelling error and it is not possible to cancel $v_2$ Apr 29, 2021 at 13:52
• @whuber I wanted to solve the issue in general, but what if we assume that $F(x) = x, \; x\in [0,1]$, so a uniform distribution Apr 29, 2021 at 13:54
• By “second order statistic” do you mean the second-lowest of $n$ iid observations from $X$? If so the usual notation is $X_{(2)}$. Or if you mean the second-highest, then $X_{(n-1)}$. Apr 29, 2021 at 14:01
• @MattF. My apologies, I was referring to the highest and second-highest value, so $X_{n}, X_{n-1}$ Apr 29, 2021 at 14:08

## 1 Answer

Let $$u$$ be the second-highest order statistic, and $$v$$ the highest order statistic.

Then we can write the joint pdf for $$u$$ and $$v$$ as the $$i=n-1$$, $$j=n$$ case of the theorem 5.4.6 here, $$n(n-1)F(u)^{n-2}f(u)f(v)$$

So we calculate the expectations by integrating this expression times $$(v-u)/v$$ or $$(u/2)(v-u)/v$$, integrating over the region with $$u.

For the uniform distribution these integrals give $$1/n$$ and $$(n-1)/2(n+1)^2$$.

• +1. But it may be worth noting this doesn't work for discrete distributions, where the possibility of ties makes the formulas more complicated.
– whuber
Apr 29, 2021 at 14:51
• @whuber & Matt F. thank you very much! Apr 29, 2021 at 15:01
• Here's the most recent archive.org link for the pdf, which is a set of lecture notes on order statistics by Prof. Hung Chen, National Taiwan University. This may be useful if the URL goes dead (link rot is a problem on our site, sadly!) Apr 29, 2021 at 23:31
• @Silverfish, if the link goes dead, Wikipedia also has the formula, with some explanation: en.wikipedia.org/wiki/… Apr 29, 2021 at 23:48
• When you say "the uniform distribution", I think you mean one with $0$ as the lower end of the support Apr 30, 2021 at 0:37