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.

• i think this is the spirit of OP's question is like: so what event, if not $P(M \le m)$, for $m \in (0,1)$, would give a probability of $m$? i think we'd get an $m$ if we had something like... $P(U_i \le U_j \le U_k)=1$ for any distinct $i,j,k \in \{1,2,3\}$ but this would violate the independence assumption. i asked this in comments below
– BCLC
Mar 18, 2021 at 11:18
• @JohnSmithKyon If $U$ ~ $Unif(0,1)$ then $P(U \le m) = m$ for $m \in (0,1)$. Mar 18, 2021 at 12:50
• @user13 also $P(U_1 \le U_2 \le U_3) = 1$ implies $P(M \ge m) = m$ because $\{M \ge m\} = \{U_3 \ge m\}$ ($\mathbb P$-a.s.) because $M=U_3$ ($\mathbb P$-a.s.) ?
– BCLC
Mar 21, 2021 at 8:30
• @JohnSmithKyon I think you misunderstand the meaning of indexes. The fact that we denote the random variables by $U_1,U_2,U_3$ does not imply that $U_1 \le U_2$ or $U_1 \le U_3$. They can be denoted by $X, Y, Z$. The only thing we know about them is that they are i.i.d. uniformly distributed random variables. Hence, your assertion that $P(U_1 \le U_2 \le U_3) = 1$ (can read it as $P(X \le Y \le Z) = 1$) is false. I will try to explain the logic of the answer. Suppose we have three numbers $3,5,9$ then saying that the $max\{3,5,9\} \le 10$ is the same as saying $3 \le 10, 5 \le 10, 9 \le 10$. Mar 21, 2021 at 11:12
• @user13 I never said $U_1 \le U_2 \le U_3$. I said '$U_1 \le U_2 \le U_3$ implies (...)'. Do you know what I mean? (1,2,3) is arbitrary. I still think $U_3 \le U_1 \le U_2$ a.s. implies $P(M \le m) = m$. Am I wrong?
– BCLC
Mar 21, 2021 at 23:42

$$P(M \le m) = P(U_1 \le m, U_2 \le m, U_3 \le m)$$ 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.
• so what event, if not $P(M \le m)$, for $m \in (0,1)$, would give a probability of $m$? i think this is the spirit of OP's question. i think we'd get an $m$ if we had something like... $P(U_i \le U_j \le U_k)=1$ for any distinct $i,j,k \in \{1,2,3\}$ but this would violate the independence assumption?
$$m^{3}$$ is correct. $$P(M \leq m)=m^{3}$$ for $$0 , $$0$$ for $$m \leq 0$$ and $$1$$ for $$m \geq 1$$.
• so what event, if not $P(M \le m)$, for $m \in (0,1)$, would give a probability of $m$? i think this is the spirit of OP's question. i think we'd get an $m$ if we had something like... $P(U_i \le U_j \le U_k)=1$ for any distinct $i,j,k \in \{1,2,3\}$ but this would violate the independence assumption?