Let $U_1, U_2, U_3 \sim \text{Unif}(0, 1)$. What is the CDF of $M = \max(U_1, U_2, U_3)$?

Question

I have the following problem:

Let $U_1, U_2, U_3 \sim \text{Unif}(0, 1)$, and let $L = \min(U_1, U_2, U_3)$ and $M = \max(U_1, U_2, U_3)$.

Find the marginal CDF and marginal PDF of $M$.

The solution says the following:

The event $M \le m$ is the same as the event that all three $U_1,U_2,U_3$ are at most $m$, so the CDF of $M$ is $F_M (m) = m^3$.

How is it that the CDF of $M$ is $m^3$ rather than $m$? This multiplies the max of each of $U_1, U_2, U_3$, rather than just taking the max of the function $M$ itself, so I'm unsure about the reasoning behind this.

I would greatly appreciate it if people could please take the time to clarify this.

Answer 1

$P(M \le m) = P(U_1 \le m, U_2 \le m, U_3 \le m)$ since your answer is $m^3$ are assume that they are not only identically distributed but also are independent, then: $P(U_1 \le m, U_2 \le m, U_3 \le m)=P(U_1 \le m)^3 = m^3$. Because if the maximum is $\le$ then something than all of them are $\le$ then that thing.

Answer 2

$m^{3}$ is correct. $P(M \leq m)=m^{3}$ for $0 <m<1$, $0$ for $m \leq 0$ and $1$ for $m \geq 1$.