Broken stick probability density

Question

A stick of length 1 is broken at an arbitrary point and the leftmost piece, say "m", is selected. Stick "m" is then broken at an arbitrary point and the leftmost piece "n" is selected. What is the probability density function for the length of "n"? What is the average length of "n"?

Answer 1

Let's use $M$ for the length of $m$, and $N$ for the length of $n$.

Given the length $M$, the conditional expected length of $N$ is clearly $E[N\mid M]=\frac{M}{2}$ so the overall expectation of this is clearly $E[N]=E\left[\frac{M}{2}\right]=\frac14$.

The density of $M$ is uniform on $[0,1]$, taking the value $1$. Given $M=x$, the density of $N$ is uniform, taking the value $\frac{1}{x}1$, so $\Pr[N\le y\mid M=x]=\frac{y}{x}$ for $0 \le y \le x$, and is $1$ when $x \le y$. So $$\Pr[N\le y] = \int_{x=y}^1 \frac{y}{x}\,dx +\int_{x=0}^y \,dx = y(1-\log_e y)$$ ant the density for $N$ is the derivative of this, namely $$-\log_e(y).$$

You can check the expected value, as $\displaystyle \int_0^1 -y \log_e(y) \,dy = \frac14$.

Answer 2

For $y \in (0,1)$

$P(n \le y)=P(n \le y|m>y).P(m>y)+P(n \le y|m\le y).P(m \le y)$

$P(n \le y|m\le y).P(m \le y)=1.P(m \le y)=P(m \le y)=y$

$P(n \le y|m>y).P(m>y)=\int_y^1P(n \le y|x>y)P(m=x)$

$=\int_y^1 \frac{y}{x}dx=-y \ln y$

$P(n \le y)=y(1-\ln y)$ $y \in (0,1)$