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I'm having trouble understanding how the CDF is found in the solution below:

We can assume the units are chosen so that the stick has length $1$. Let $L$ be the length of the longer piece, and let the break point be $U \sim Unif(0,1)$. For any $l \in [1/2,1]$, observe that $L<l$ is equivalent to $U<l,1-U<l$, which can be written as $1-l<U<l$. We can thus obtain $L$'s CDF as $$F_L(l) = P(L<l)=P(1-l<U<l)=2l-1$$

Can someone please explain why $L<l$ is equivalent to $U<l,1-U<l$? Isn't the break point $U$ in between $[1/2,1]$?

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    $\begingroup$ This question is asked quite a number of times. As you have specified the break point $U \sim \text{Unif}(0, 1)$, it's support will not suddenly change to $[1/2, 1]$. For a given break point with distance $U$ from one of the end point, the two broken piece has length $U$ and $1-U$ respectively. Since $L$ is the length of the longer piece, $L = \max\{U, 1-U\}$ and that's why the subsequent arguments follow. $\endgroup$
    – BGM
    May 27 '16 at 2:29
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    $\begingroup$ If you want the longer piece to have a length of 0.75, you can break the stick at 0.75 OR at (1 - 0.75) = 0.25 $\endgroup$
    – user137481
    May 27 '16 at 2:37
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By definition, $L$ is the max of $U$ and $1-U$. The max of two numbers is less than something if and only each of the numbers is less than that something. By this argument, we've shown: $$\{L<t\} = \{U<t, 1-U<t\}.$$

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$U$ is the break point which lies uniformly distributed on $(0;1)$.

$L$ is the length of the longer side of the break.   This is somewhere on $[\tfrac 1 2; 1)$ .

When the break point $U$ is less than $1/2$ the length of the longer stick is $1-U$, and other wise it is $U$.

So, for any $\tfrac 1 2\leq l\leq 1$, then the longer length being less than $l$ means that the break is both less that $l$ and greater than $1-l$ .

$$\mathsf P(L<l) ~=~ \mathsf P(1-l<U<l) \\ = ~~\,2l-1 \\ = ~ \frac{l-\tfrac 12}{1-\tfrac 12} $$

Which means $L\sim\mathcal {U}(\tfrac 12;1)$

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