Independence of Minimum and Maximum of a set of random variables

Question

I have a little problem with an exercise:

Let $X_1, . . . , X_n$ independent, real-valued random variables with distribution functions $F_1, . . . , F_n$. Let the $X_i$ be uniform distributed in $[0,1]$

Let $M = max(X_1, . . . , X_n)$ and $m = min(X_1, . . . , X_n)$ and let $F_M(x) = \prod_{n}^{i=1}F_i(x)$ and $F_m(x) = 1 −\prod_{n}^{i=1}(1 − Fi(x))$

Are $m$ and $M$ independent?

I have alredy calculated $F_M, F_m$ and $f_M, f_m$.

I thought to prove whether they were independent or not by proving:

$f(m,M)=f(min(X_1,...,X_n),max(X_1,...,X_n))= f(min(X_1,...,X_n))*f(max(X_1,...,X_n))$

Is it right how I am proceeding? In case it would be right, I can not understand how to calculate $f(min(X_1,...,X_n),max(X_1,...,X_n))$.

Can anyone help me?

Answer 1

I assume you already computed $F_m$ and $F_M$, as mentioned in your question: for $x \in [0,1]$, we have respectively $F_M(x) = x^n$ and $F_m(x) = 1 - (1-x)^n$.

In particular, $\mathbb{P}(M \le \frac{1}{2}) = F_M\big(\frac{1}{2}\big) = \frac{1}{2^n}$, and $\mathbb{P}(m > \frac{1}{2}) = 1-F_m\big(\frac{1}{2}\big) = \frac{1}{2^n}$.

Ad absurdum, if $m$ and $M$ were independent, we would have $$\mathbb{P}\big(M \le \frac{1}{2} < m\big) = \mathbb{P}\big(M \le \frac{1}{2}\big)\mathbb{P}\big(m > \frac{1}{2}\big) = \frac{1}{4^n} > 0$$

Since we have almost surely $m \le M$, this probability is actually $0$, meaning that $m$ and $M$ are in fact not independent. This proof works for any integer $n \ge 1$.