Problem source

A stick is broken at a point, chosen at random, along its length. Find the probability that the ratio, $R$, of the length of the shorter piece to the length of the longer piece is less than $r$. Find the probability density  function for $R$, and calculate the  mean and variance of $R$.

Solution source

Let $X \sim U[0, \tfrac12]$ be the shorter piece, so $R = \frac{X}{1-X}$, and

\begin{align*}
&& \mathbb{P}(R \leq r) &= \mathbb{P}(\tfrac{X}{1-X} \leq r) \\
&&&= \mathbb{P}(X \leq r - rX) \\
&&&= \mathbb{P}((1+r)X \leq r) \\
&&&= \mathbb{P}(X \leq \tfrac{r}{1+r} ) \\
&&&= \begin{cases} 
0 & r < 0 \\
\frac{2r}{1+r} & 0 \leq r \leq 1 \\
1 & r > 1
\end{cases} \\
\\
&& f_R(r) &= \begin{cases} \frac{2}{(1+r)^2} & 0 \leq r \leq 1 \\
0 & \text{otherwise} \end{cases}
\end{align*}

Let $Y \sim U[\tfrac12, 1]$ be the longer piece, then $R = \frac{1-Y}{Y} = Y^{-1} - 1$ and

\begin{align*}
\E[R] &= \int_{\frac12}^1 (y^{-1}-1) 2 \d y \\
&= 2\left [\ln y - y \right]_{\frac12}^1 \\
&= -2 + 2\ln2 +2\frac12 \\
&= 2\ln2 -1 \\
\\
\E[R^2] &= \int_{\frac12}^1 (y^{-1}-1)^2 2 \d y\\
&= 2\left [-y^{-1} -2\ln y + 1 \right]_{\frac12}^1 \\
&= 2 \left ( 2 - 2\ln 2+\frac12\right) \\
&= 3-4\ln 2 \\
\var[R] &= 3 - 4 \ln 2 -(2\ln 2-1)^2 \\
&= 2 - 4(\ln 2)^2
\end{align*}