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.
Best Answer
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)