4

In "50 Challenging Problems in Probability", question #43 is the following:

"A bar is broken at random in two places. Find the average size of the smallest, of the middle-sized, and of the largest pieces."

The author gives what seems like a complicated geometric way of calculating the probabilities. He arrives at the solutions 1/9, 5/18, and 11/18. Is there a simpler, non-geometric way of calculating these probabilities?

Thomas Johnson
  • 339

2 Answers2

3

Here's a sketch. Pick points $0\le x\le y\le 1$. Now let's find the expected value of $x$ given that the interval $[0,x]$ is shortest. That is, we look at the region satisfying \begin{align*} x&\le y \\ x&\le y-x \\ x&\le 1-y \end{align*} You can draw your own pictures. We then get a triangle with $0\le x\le 1/3$, $2x\le y\le 1-x$. This triangle has area $1/6$ (by inspection) or by double-integration, and $$\int_0^{1/3}\int_{2x}^{1-x} x\,dy\,dx = \frac1{54},$$ so the expected value of $x$ is $\dfrac{\frac1{54}}{\frac16} = \dfrac19$.

Next case. Suppose the interval $[0,x]$ has middle length. This corresponds to two regions: \begin{align*} x&\le y \\ y-x&\le x \\ x&\le 1-y \end{align*} OR \begin{align*} x&\le y \\ x&\le y-x \\ 1-y&\le x \end{align*} Interestingly, these both have area $1/12$ and for each of these we get $\displaystyle\iint x\,dA = \dfrac5{216}$. So the expected value of $x$ in the middle case is $\dfrac{\frac5{216}}{\frac1{12}} = \dfrac5{18}$.

The last case follows by subtraction.

Ted Shifrin
  • 125,228
  • Can you give intuition behind dividing the expectation with the area? like why do we have to divide $\frac{1}{54}$ with $\frac{1}{6}$? or divide $\frac{5}{216}$ with $\frac{1}{12}$? – Sagar Chand Nov 26 '17 at 09:23
  • 1
    @SagarChand Expectation is average value; thinking of it differently, you have to make the net measure of your space be $1$. – Ted Shifrin Nov 26 '17 at 13:13
  • Thanks I see. I think this is because of the fact that pieces of length (y-x) and (1-y) can also be smallest. Dividing by this area ensures that we are taking all cases into account(as they are equally likely) – Sagar Chand Nov 26 '17 at 14:39
  • 1
    I guess I should be more explicit. We're doing the expectation given that the segment is shortest, so we must divide by the probability that that particular segment is shortest. – Ted Shifrin Nov 26 '17 at 14:59
  • Oh! I see now.

    Sorry for bothering you.

    Thanks again :)

     – Sagar Chand Nov 26 '17 at 15:14
  • What is the double integration over x for (the on that resulted in 1/54)? – 24n8 Apr 04 '20 at 01:03
  • @Iamanon Do the integration I did there, omitting the $x$. – Ted Shifrin Apr 04 '20 at 01:06
  • Hmm I get $\frac{1}{6}$, the area of the triangle enclosed by the 3 constraints. Adding the $x$ seems to make it an expected value integration. – 24n8 Apr 04 '20 at 01:13
  • Oh, I'm sorry. I misread your question. We are finding the expectation of $x$, the length of the shortest piece. – Ted Shifrin Apr 04 '20 at 01:17
  • Right. I just read again and saw that you mentioned that's the expectation. I'm having a hard time seeing why that resulted in the expectation. I think the double integration is what's causing the confusion for me because usually I see the expectation for a single variable without joint pdfs as just a single integral over a pdf, but here it appears the pdf is per unit area rather than per unit length. – 24n8 Apr 04 '20 at 01:21
  • How do you do the integral if we're doing the last case (without subtraction)? The area would be 1/6, x from 1/3 to 1, but I get $\frac{\frac{2}{27}}{\frac{1}{6}} = \frac{4}{9}$ integral is still the same range except outer integral is from 1/3 to 1, right? – Joe Apr 14 '23 at 18:13
  • @JoeVictor No. the area is $1/6$, but you have to break the double integral into two pieces. – Ted Shifrin Apr 14 '23 at 18:49
  • @TedShifrin Why? And break it around which point? I have a hunch it's 0.5 based on my drawing but I don't understand what's the mathematical reasoning – Joe Apr 15 '23 at 00:17
  • I'm not the best at it, but I figured it out. It's because the bounds of the y change from [2x, 1-x] to [x,1] after x=0.5. Thanks! – Joe Apr 15 '23 at 00:22
  • Your first limits are upside-down, but yes. – Ted Shifrin Apr 15 '23 at 00:24
2

The general solution has been discussed in a paper ``On the lengths of the pieces of a stick broken at random'', and some information is also there in Feller's book, vol. 2.

Also see here.

For the $k^{th}$ smallest piece indicated by $S_{(k)}$, the expected length is given by

\begin{align*} \mathbb{E}\left(n\, S_{(k)}\right) &= \mathbb{E}\left(X_{(k)}\right) \\ &= \sum_{i=0}^{k-1}\frac{1}{n-i} \end{align*} which simplifies to the formula given in the book of 50 challenging problems.

For the question asked, $n=3$ and calculate the sums for $k=1,2,3$, which we find to be \begin{align*} \mathbb{E}\left(S_{(1)}\right) &= \frac{1}{9}\\ \mathbb{E}\left(S_{(2)}\right) &= \frac{5}{18}\\ \mathbb{E}\left(S_{(3)}\right) &= \frac{11}{18} \end{align*}

gar
  • 4,988