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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.
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    $\begingroup$ 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. $\endgroup$
    – whuber
    Apr 29, 2021 at 13:48
  • $\begingroup$ @StephanKolassa unfortunatelly, I made a spelling error and it is not possible to cancel $v_2$ $\endgroup$
    – Snowrabbit
    Apr 29, 2021 at 13:52
  • $\begingroup$ @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 $\endgroup$
    – Snowrabbit
    Apr 29, 2021 at 13:54
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    $\begingroup$ 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 $X_{(n-1)}$. $\endgroup$
    – Matt F.
    Apr 29, 2021 at 14:01
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    $\begingroup$ @MattF. My apologies, I was referring to the highest and second-highest value, so $X_{n}, X_{n-1}$ $\endgroup$
    – Snowrabbit
    Apr 29, 2021 at 14:08

1 Answer 1

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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 $u<v$.

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

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    $\begingroup$ +1. But it may be worth noting this doesn't work for discrete distributions, where the possibility of ties makes the formulas more complicated. $\endgroup$
    – whuber
    Apr 29, 2021 at 14:51
  • $\begingroup$ @whuber & Matt F. thank you very much! $\endgroup$
    – Snowrabbit
    Apr 29, 2021 at 15:01
  • $\begingroup$ 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!) $\endgroup$
    – Silverfish
    Apr 29, 2021 at 23:31
  • $\begingroup$ @Silverfish, if the link goes dead, Wikipedia also has the formula, with some explanation: en.wikipedia.org/wiki/… $\endgroup$
    – Matt F.
    Apr 29, 2021 at 23:48
  • $\begingroup$ When you say "the uniform distribution", I think you mean one with $0$ as the lower end of the support $\endgroup$
    – Henry
    Apr 30, 2021 at 0:37

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