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Let's consider $Y_n$ the max of $n$ iid samples $X_i$ of the same distribution:

$Y_n = max(X_1, X_2, ..., X_n)$

Do we know some common distributions for $X$ such that $Y$ is uniformly distributed $U(a,b)$?

I guess we can always "construct a distribution" $X$ to enforce this condition for $Y$ but I was just wondering if a famous distribution satisfies this condition.

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    $\begingroup$ It is worth noting that the law of any such $X$ must be unique, so "constructing a distribution" will give you the (only) answer. $\endgroup$
    – πr8
    Sep 3, 2020 at 7:10
  • $\begingroup$ @πr8 can you elaborate why such a law has to be unique? $\endgroup$ Sep 3, 2020 at 15:34
  • $\begingroup$ math.stackexchange.com/questions/1397427/… $\endgroup$
    – shayaa
    Sep 3, 2020 at 18:11
  • $\begingroup$ if a maximum $Y$ is a scalar, how can it have a distribution? $\endgroup$
    – develarist
    Sep 4, 2020 at 1:00
  • $\begingroup$ @develarist it's not a scalar. It's a sequence defined by $n$ and $X$. $\endgroup$ Sep 4, 2020 at 11:50

2 Answers 2

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Let $F$ be the CDF of $X_i$. We know that the CDF of $Y$ is $$G(y) = P(Y\leq y)= P(\textrm{all } X_i\leq y)= \prod_i P(X_i\leq y) = F(y)^n$$

Now, it's no loss of generality to take $a=0$, $b=1$, since we can just shift and scale the distribution of $X$ to $[0,\,1]$ and then unshift and unscale the distribution of $Y$.

So what does $F$ have to be to get $G(y) =y$? We need $F(x)= x^{1/n}I_{[0,1]}$, so $f(x)=\frac{1}{n}x^{1/n-1}I_{[0,1]}$, which is a Beta(1/n,1) density.

Let's check

> r<-replicate(100000, max(rbeta(4,1/4,1)))
> hist(r)

enter image description here

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$F_{X_{(n)}}(x)=[F_X(x)]^n$, so for a standard uniform you need $F_X(x)=x^{1/n}$ for $0<x<1$ (and $0$ to the left and $1$ to the right of that interval), so $f_X(x)=\frac{1}{n}x^{\frac{1}{n}-1}$ on the unit interval and $0$ elsewhere.

It's a special case of the beta.

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