Elisabeth Esparza

Answered

2022-07-19

Non-geometric way to calculate expected value of breaks?
"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?

Answer & Explanation

Damarion Pierce

Expert

2022-07-20Added 11 answers

Step 1
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 $\begin{array}{rl}x& \le y\\ x& \le y-x\\ x& \le 1-y\end{array}$
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 ${\int }_{0}^{1/3}{\int }_{2x}^{1-x}x\phantom{\rule{thinmathspace}{0ex}}dy\phantom{\rule{thinmathspace}{0ex}}dx=\frac{1}{54},$ so the expected value of x is $\frac{\frac{1}{54}}{\frac{1}{6}}=\frac{1}{9}$.
Step 2
Next case. Suppose the interval [0,x] has middle length. This corresponds to two regions:
$\begin{array}{rl}x& \le y\\ y-x& \le x\\ x& \le 1-y\end{array}$
or $\begin{array}{rl}x& \le y\\ x& \le y-x\\ 1-y& \le x\end{array}$
Interestingly, these both have area 1/12 and for each of these we get $\iint x\phantom{\rule{thinmathspace}{0ex}}dA=\frac{5}{216}$. So the expected value of x in the middle case is $\frac{\frac{5}{216}}{\frac{1}{12}}=\frac{5}{18}$

Lillie Pittman

Expert

2022-07-21Added 4 answers

Step 1
For the ${k}^{th}$ smallest piece indicated by ${S}_{\left(k\right)}$, the expected length is given by
$\begin{array}{rl}\mathbb{E}\left(n\phantom{\rule{thinmathspace}{0ex}}{S}_{\left(k\right)}\right)& =\mathbb{E}\left({X}_{\left(k\right)}\right)\\ & =\sum _{i=0}^{k-1}\frac{1}{n-i}\end{array}$
which simplifies to the formula given in the book of 50 challenging problems.
Step 2
For the question asked, $n=3$ and calculate the sums for $k=1,2,3$, which we find to be
$\begin{array}{rl}\mathbb{E}\left({S}_{\left(1\right)}\right)& =\frac{1}{9}\\ \mathbb{E}\left({S}_{\left(2\right)}\right)& =\frac{5}{18}\\ \mathbb{E}\left({S}_{\left(3\right)}\right)& =\frac{11}{18}\end{array}$

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?