A square has sides $ l=\frac14$. The probability to pick a point at random on perimeter is equal to the length of the side. Let X= the x coordinate of the point picked. Find the CDF.
This question has been solve before, yet I'm having difficulty understanding the solution given.
$P(X=0)= \frac14$
This I understand but:
For $0\lt x \le 0.25:$
$P(X=0)+P(0\lt x \le 0.25)= \frac14 + 2x$
What is the reasoning behind the 2x?
I understand the probability of picking a point at x=0.125 is half.
On
If $0\leq x<\frac14$ then: $$F(x)=\Pr(X\leq x)=\Pr(X<0)+\Pr(X=0)+\Pr(0<X\leq x)=0+\frac14+2x$$
Here $\frac14$ is the measure of $\{\langle 0,t\rangle\mid 0<t\leq \frac14\}$.
Here $2x$ is the sum of the measures of $\{\langle s,0\rangle\mid 0<s\leq x\}$ and $\{\langle s,\frac14\rangle\mid 0<s\leq x\}$.
Denoting the $3$ mentioned sets by $A,B,C$ we have: $$X\leq x\iff \langle X,Y\rangle \in A\cup B\cup C$$
Also note that the $A,B,C$ are disjoint.
Clearly, $\mathbb{P}[X=0] = \mathbb{P}[X = 1/4] = 1/4$ since this means you picked a point on the corresponding vertical edge.
So, $\mathbb{P}[X \le 0] = 1/4$ and $\mathbb{P}[X < 1/4] = 0.75$, and you need linear growth in the middle because of the uniformity of the pick.