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

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.

$$P(M \le m) = P(U_1 \le m, U_2 \le m, U_3 \le m)$$ 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.
$$m^{3}$$ is correct. $$P(M \leq m)=m^{3}$$ for $$0 , $$0$$ for $$m \leq 0$$ and $$1$$ for $$m \geq 1$$.