Question A stick of length 1 is broken at a uniformly random point, yielding two pieces. Let X and Y be the lengths of the shorter and longer pieces, respectively, and let R = X/Y be the ratio of the lengths of X and Y. Find the CDF and PDF of R.
The solution given in the textbook was as follows:
Let $U \sim \text{Uniform}(0,1)$.
\begin{align*} P(R \le r) &= P(X(1+r) \le r) \\ &= P(\text{min}(U,1-U) \le r/(1+r))\\ &= 1 - P(U > r/(1+r), U < 1/(1+r)) \\ &= 1 - \frac{1}{1+r} + \frac{r}{1+r}. \end{align*}
So $F_R(r) = 2r(1+r)^{-1}$ for $0 < r < 1$. And $f_R(r) = 2(1+r)^{-2} $.
I understood how the PDF and CDF were calculated, but I did not understand the steps that were taken from the first step to the third step. Can someone give the steps that happen in between, basically expand the steps in between?
It is easy to see that $R \le r \iff \frac X Y \le r$. Recall as well that $Y = 1-X$ so the previous inequality is equivalent to $$ X(1+r) \le r. $$ This explains the first equality.
Note that $X = \min \{ U, 1-U \}$, yielding the second inequality. To see this, recall that $U$ is the uniformly random point that the stick is broken at, and $X$ is the length of the shorter piece. The length of the stick is $1$, so the lengths of the two pieces are $U$ and $1-U$.
To see the third step, note that $ \Pr ( \min \{U, 1-U \} \le r / (1+r) ) = 1 - \Pr ( \min \{U, 1-U \} > r / (1+r) ) $. We also have that $ \min \{ U, 1-U \} > r/(1+r) $ if and only if $ U > r/(1+r)$ and $1-U > r/(1+r)$.