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I'm curious if my solution to this problem is correct:

$X_1,X_2,...,X_n$ are each independently, uniformly distributed over (0,1). Let $M = max (X_1, X_2,...,X_n)$. Show that the distribution function of $M, F_M(.)$ is given by:

$F_M(a)=a^n, \quad 0 \le a \le 1$

Solution?:

The probability that $M=a$ is given by the probability that $n-1$ of the $n$ variables are less than $a$, and that the remaining variable is $a$:

$f_M(a) \ = \ \binom{n}{n-1} F(a)^{n-1} f(a) \ = \ n F(a)^{n-1}$

Given that $f(a)=1$ for a uniform RV. Now for $F_M(a)$, we have the derivative of this so the distribution function will be its integral:

$F_M(a)=\int_{0}^{a} f_M(x)dx \ = \ \int_{0}^{a} nF(x)^{n-1} dx $

Since $F(x)$ is uniform on $(0,1)$:

$= \ n \int_{0}^{a} x^{n-1} dx=n(n)^{-1} x^n = x^n$

$F_M(a)=a^n$

Thoughts appreciated.

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3 Answers 3

up vote 1 down vote accepted

The probability that one of the values is exactly $a$ is $0$, so I think that the short argument below poses fewer difficulties. For minimum a similar argument will work. The idea you used can be adapted to computing other order statistics. I would be more comfortable using a limits argument based on the same insight.

The maximum of our $n$ random variables is $\le a$ if and only if they are all $\le a$. The probability of this is $a^n$.

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+1 related: math.stackexchange.com/questions/11871/… –  leonbloy Oct 17 '12 at 18:38

Actually, the prob that the max is less than $a$ is the same as the probability that all of them are less than $a$. Thus, $\mathrm{Pr}(X_{max}\leq a)=\mathrm{Pr}(X_{i}\leq a \forall i)$ = $\Pi_{i=1}^{n} \mathrm{Pr}(X_{i}\leq a)$ due to independence = $a^{n}$ due to the fact that they are all uniform (0,1)

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Thanks Ken, I see. Given $n-1$ smaller than $a$ the $n$th must be $a$ to ensure it is the maximum. Think I started backwards, I wonder is the above wrong or just unnecessary? –  conjectures Oct 17 '12 at 18:50
    
Most of your argument is okay, but you are making statements about continuous random variables which are not exactly right. The prob that the max $ = a$ is actually zero. That's why we work with statements like $X_{i} \leq a$ and so forth. –  Ken Dunn Oct 17 '12 at 18:54

Thanks for your answers.

Just on the use of $f(x)$ for a continuous distribution. I am aware of the argument why this is zero, but is it not correct to use the definition of a uniform rv, i.e. that $f(x)=1, 0 \le x \le 1$?

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That is correct. –  Ken Dunn Oct 17 '12 at 19:03

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