A stick is broken into two pieces, at a uniformly random chosen break point. Find the CDF.

Question

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 $[1/2,1]$?

Answer 1

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: $$\{L<t\} = \{U<t, 1-U<t\}.$$

Answer 2

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

Answer 3

Suppose the length of the stick is $1$ unit and it lies on the interval $[0,1]$. A break point can be arbitrarily chosen on the interval $(a,b) \subset [0,1]$ with probability $\frac{b-a}{1}$.

If the longer longer length is denoted by $L$, then $$F(l) = P\left(\frac{1}{2} < L <l\right) = 2 \cdot \frac{l-\frac{1}{2}}{1} = 2l-1$$

We multiply by $2$ because the longer side is either on the left side or on the right.