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Question: For $U_1 , \dots, U_n$ i.i.d. $U \sim \mathrm{unif}[0,1]$, we want to find the asymptotic distribution of $Z_n = n(1-U_{(n)})$ where $U_{(n)} = \max(U_1 , ... , U_n)$

I found this: Asymptotic distribution of uniform order statistics But find it is not detailed enough as to the steps to take. Could I have some more detail?

EDIT: Found this giving one key intermediate step. Thank you for solutions!

https://math.stackexchange.com/questions/313390/probability-density-of-the-maximum-of-samples-from-a-uniform-distribution

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  • $\begingroup$ Include the self study tag. $\endgroup$ Mar 13 '18 at 4:51
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    $\begingroup$ This is a well known result. There is a lot of research on extremes. In my Stanford thesis back in the late 1970s I worked on results for dependent sequences which included an example of a uniform first order autoregressive process. A lot of the theory appears in the book by Leadbetter, Lindgren and Rootzen published by Springer in 1983.. $\endgroup$ Mar 13 '18 at 6:03
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    $\begingroup$ I should also note that asymptotic results are often used to approximate the exact distributions which don't always have a nice closed form (considered the normal distribution for example. You can note that this uniform distribution example is one where the exact distribution is given in closed form for every value of n. $\endgroup$ Mar 13 '18 at 6:22
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For $z \in [0,n]$ \begin{align*} P(Z_n \le z) &= 1-P\left(U_{(n)} \le 1 - \frac{z}{n}\right) \\ &= 1 - \prod_{i=1}^nF_{U_i}\left( 1 - \frac{z}{n}\right) \\ &= 1 - \left(1 - \frac{z}{n}\right)^n \\ &\to 1- e^{-z} \end{align*} as $n \to \infty$.

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    $\begingroup$ Quick correction $z\in (0,n) $ - and the limit is exponential distribution with mean parameter $1$ $\endgroup$ Mar 13 '18 at 5:45
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    $\begingroup$ Thanks for the support @probabilityislogic (pun intended) $\endgroup$
    – Taylor
    Mar 13 '18 at 13:07
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For each i P[$U_i$ < x]=x for all x in [0, 1] and P[U$(n)$ < x] = $x^n$ since the maximum is less than x iff every $U_i$ is < x (here is where iid is required). So P(n[1 - $U(n)$] < x) = P(1-$U(n) < x/n$) =

P[$U(n)$ > (1 - $x/n)]$ = 1-$(1-x/n)^n$. Then take the limit as n approaches infinity.

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  • $\begingroup$ @Taylor went ahead and took the limit for you. $\endgroup$ Mar 13 '18 at 5:06
  • $\begingroup$ We were both working on this at the same time. I took longer to finish it correctly. $\endgroup$ Mar 13 '18 at 5:09
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    $\begingroup$ hate when that happens (+1) $\endgroup$
    – Taylor
    Mar 13 '18 at 15:56
  • $\begingroup$ Thank you! Sorry, just accepted the first one :( Both are equally good (+1) $\endgroup$ Mar 13 '18 at 17:01
  • $\begingroup$ That is not a problem. Taylor's answer deserves a check mark. I gave him an up vote. $\endgroup$ Mar 13 '18 at 17:41

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