how can we derive both the CDF of $M$ and the PDF of $M$

Question

Given $X_i \sim \operatorname{Uniform}[0, 1]$ for $i = 1, \dots, n$. What is the distribution of $M := \min(X_1, \dots, X_n)$?

I feel like I'm missing something and I've been stuck on this for two days

Answer 1

I guess the $X_i$'s are iid. Let $0 \leq m \leq 1$

$$F(m)=P(M \leq m)=1-P( M>m)=1-P(X_1>m,X_2>m,...,X_n>m)$$ because if the minimum of the $X_i$'s are greater than $m$, then all of them are greater than $m$, and reciprocally. Use the independence property and the fact that the $X_i's$ have the same distribution $$P(X_1>m,X_2>m,...,X_n>m)=\prod_{i=1}^nP(X_i>m)=P(X_1>m)^n$$

We have $P(X_1>m)=(1-m)$

Therefore, $$F(m)=P(M \leq m)=1-(1-m)^n$$ if $m>1$, then $F(m)=1$ if $m<0$, then $F(m)=0$

The cdf is then $$F'(m)=m(1-m)^{n-1}$$