There are a lot of questions here for the task of breaking a stick of length 1 into three parts such that they form a triangle. I came across a bit more complex version of it when the breaks are done consecutively. Namely, first we do the first break, then take the longer part and break it once again. I was trying to derive the CDF of second break, i.e. length of the left part after the second break.
This is what I have got so far. Let us denote $A$ as the length of the left part after the first break, and $B$ as the length of the left part after the second break. Thus, I have to find $F_B(x)=P(B<x)$.
I know that $A\sim U[0,1]$, thus
$$\begin{equation*} F_A(x)=\begin{cases} 0 \quad &\text{if} \, x \leq 0 \\ x \quad &\text{if} \, 0<x\leq1 \\ 1 \quad &\text{if} \, x>1 \\ \end{cases} \end{equation*}$$
It is also obvious that $B|A\sim U[0, \max(A,1-A)]$ and $$\begin{equation*} F_{B|A}(x|a)=\begin{cases} 0 \quad &\text{if} \, x \leq 0 \\ \frac{x}{\max(a,1-a)} \quad &\text{if} \, 0<x\leq\max(a,1-a) \\ 1 \quad &\text{if} \, x>\max(a,1-a) \\ \end{cases} \end{equation*}$$
Thus unconditioned CDF can be computed as $$F_B(x)=\int_{-\infty}^{+\infty}F_{B|A}(x|a)f_A(a)da=\int_0^1F_{B|A}(x|a)da$$
And here I am stuck since I cannot understand how this integral can be calculated.
Any help is much appreciated.
