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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.

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  • $\begingroup$ 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 $\endgroup$
    – BCLC
    Mar 18, 2021 at 11:18
  • $\begingroup$ @JohnSmithKyon If $U$ ~ $Unif(0,1)$ then $P(U \le m) = m$ for $m \in (0,1)$. $\endgroup$
    – user13
    Mar 18, 2021 at 12:50
  • $\begingroup$ @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.) ? $\endgroup$
    – BCLC
    Mar 21, 2021 at 8:30
  • $\begingroup$ @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 $3 \le 10, 5 \le 10, 9 \le 10$. $\endgroup$
    – user13
    Mar 21, 2021 at 11:12
  • $\begingroup$ @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? $\endgroup$
    – BCLC
    Mar 21, 2021 at 23:42

2 Answers 2

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$P(M \le m) = P(U_1 \le m, U_2 \le m, U_3 \le m)$ 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.

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  • $\begingroup$ 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? $\endgroup$
    – BCLC
    Mar 18, 2021 at 11:16
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$m^{3}$ is correct. $P(M \leq m)=m^{3}$ for $0 <m<1$, $0$ for $m \leq 0$ and $1$ for $m \geq 1$.

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1
  • $\begingroup$ 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? $\endgroup$
    – BCLC
    Mar 18, 2021 at 11:17

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