# Which distribution has its maximum uniformly distributed?

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.

• It is worth noting that the law of any such $X$ must be unique, so "constructing a distribution" will give you the (only) answer.
– πr8
Sep 3 '20 at 7:10
• @πr8 can you elaborate why such a law has to be unique? Sep 3 '20 at 15:34
• math.stackexchange.com/questions/1397427/… Sep 3 '20 at 18:11
• if a maximum $Y$ is a scalar, how can it have a distribution? Sep 4 '20 at 1:00
• @develarist it's not a scalar. It's a sequence defined by $n$ and $X$. Sep 4 '20 at 11:50

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) $$F_{X_{(n)}}(x)=[F_X(x)]^n$$, so for a standard uniform you need $$F_X(x)=x^{1/n}$$ for $$0 (and $$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.