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On a straight stick of length $1$, choose $n-1$ independent uniformly random points. Cut the stick at those points, yielding $n$ fragments.

Let $\mathbb{E}_n$ be the expectation of the product of lengths of the fragments.

Simulations suggest that $\mathbb{E}_n=\prod\limits_{k=n}^{2n-1}\frac1k$. For example, $\mathbb{E}_5=\frac{1}{5\cdot6\cdot7\cdot8\cdot9}$.

This is an elegant result, so I suspect there is an intuitive explanation, but it eludes me.

Is there an intuitive explanation for $\mathbb{E}_n=\prod\limits_{k=n}^{2n-1}\frac1k$ ?

My thoughts:
I have only been able to prove the result for $n=2$ and $n=3$, and my proofs are not intuitive.

$\color{red}{n=2}$:
Hold the stick horizontally.
Let the distance of the random point from the left end be $x$. $$\mathbb{E}_2=\int_0^1x(1-x)\mathrm dx=\frac16=\prod\limits_{k=2}^{2(2)-1}\frac1k$$

$\color{red}{n=3}$:
Hold the stick horizontally.
Let the distance of the 1st chosen point from the left end be $x$.
Let the distance of the 2nd chosen point from the left end be $y$.
(We make no assumption about which of $x$ and $y$ is greater.)

For a fixed $x$ value, the expectation of the product of fragment lengths is $$P(y<x)\times\int_0^xy(x-y)(1-x)\mathrm dy+P(y>x)\times\int_x^1x(y-x)(1-y)\mathrm dy$$

where $P(y<x)=x$ and $P(y>x)=1-x$.

Then we integrate this expectation from $x=0$ to $x=1$:

$$\mathbb{E}_3=\int_0^1\left(x\int_0^xy(x-y)(1-x)\mathrm dy+(1-x)\int_x^1x(y-x)(1-y)\mathrm dy\right)\mathrm dx=\frac{1}{60}=\prod\limits_{k=3}^{2(3)-1}\frac1k$$

Context:
I recently answered a question involving products of lengths of random line segments. That made me think of this question.

Dan
  • 22,158

2 Answers2

10

You first pick $n-1$ random points which splits the interval $[0,1]$ into $n$ segments $U_1,\ldots,U_n$ of lengths $L_i=|U_i|$. You wish to compute the expected value $\mu=E[L_1\cdots L_n]$. We can assume the segments $U_i$ are sorted in increasing order, although that doesn't really matter.

Next, similarly pick $n$ points $X_1,\ldots,X_n$ randomly in $[0,1]$. The probability that $X_i\in U_i$ is $L_i$, and since the $X_i$ are picked independently, the probability that $X_i\in U_i$ for all $i$ equals $\mu=E[L_1\cdots L_n]$.

The above is the same as first picking $2n-1$ points randomly, and then selecting at random the $n$ points $X_1,\ldots,X_n$ from these. The remaining $n-1$ points split the interval into segments $U_i$, and the probability that $X_i\in U_i$ for all $i$ is the same as the points $X_1,\ldots,X_n$ being every second point out of the $2n-1$ in increasing order starting at the lowest. The probability of this, ie of $X_i$ being $n$ specific points out of the $2n-1$ picked in increasing order, is $$ \mu=\frac{1}{\binom{2n-1}{n}\cdot n!}=\frac{1}{(2n-1)\cdots n}. $$

Einar Rødland
  • 9,707
7

Here is a proof of the claim. (Unfortunately, it is not that intuitive.)

Let $X_{(1)} < X_{(2)} < \ldots < X_{(n-1)}$ denote the $n-1$ uniformly random points, sorted in ascending order. The joint density of $(X_{(1)}, \ldots, X_{(n-1)})$ is well-known:

$$ f_{X_{(1)}, \ldots, X_{(n-1)}}(y_1, \ldots, y_{n-1}) = (n-1)! $$

whenever $0 < y_1 < \ldots < y_{n-1} < 1$. Then the expected product is

\begin{align*} \mathbb{E}_n &= (n-1)! \int_{0<y_1<\ldots<y_{n-1}<1} y_1(y_2-y_1)\cdots(y_{n-1}-y_{n-2})(1-y_{n-1}) \, \mathrm{d}y_1 \cdots \mathrm{d}y_{n-1} \end{align*}

By repeatedly applying the formula $\int_{0}^{a} \frac{y^k}{k!} (a - y) \, \mathrm{d}y = \frac{a^{k+2}}{(k+2)!} $, we get

\begin{align*} \mathbb{E}_n &= (n-1)! \int_{0<y_2<\ldots<y_{n-1}<1} \frac{y_2^3}{3!}(y_3-y_2)\cdots(1-y_{n-1}) \, \mathrm{d}y_2 \cdots \mathrm{d}y_{n-1} \\ &= (n-1)! \int_{0<y_3<\ldots<y_{n-1}<1} \frac{y_3^5}{5!}(y_4-y_3)\cdots(1-y_{n-1}) \, \mathrm{d}y_3 \cdots \mathrm{d}y_{n-1} \\ &\quad \vdots \\ &= (n-1)! \int_{0<y_{n-1}<1} \frac{y_{n-1}^{2n-3}}{(2n-3)!}(1-y_{n-1}) \, \mathrm{d}y_{n-1} \\ &= (n-1)! \cdot \frac{1}{(2n-1)!}, \end{align*}

proving the desired equality.

Sangchul Lee
  • 167,468