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?

"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 $\frac{\frac{1}{54}}{\frac{1}{6}}}={\displaystyle \frac{1}{9}$.

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 $\iint x\phantom{\rule{thinmathspace}{0ex}}dA={\displaystyle \frac{5}{216}}$. So the expected value of x in the middle case is $\frac{\frac{5}{216}}{\frac{1}{12}}}={\displaystyle \frac{5}{18}$

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 $\frac{\frac{1}{54}}{\frac{1}{6}}}={\displaystyle \frac{1}{9}$.

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 $\iint x\phantom{\rule{thinmathspace}{0ex}}dA={\displaystyle \frac{5}{216}}$. So the expected value of x in the middle case is $\frac{\frac{5}{216}}{\frac{1}{12}}}={\displaystyle \frac{5}{18}$

Lillie Pittman

Expert

2022-07-21Added 4 answers

Step 1

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

$\begin{array}{rl}\mathbb{E}\left(n\phantom{\rule{thinmathspace}{0ex}}{S}_{(k)}\right)& =\mathbb{E}\left({X}_{(k)}\right)\\ & =\sum _{i=0}^{k-1}\frac{1}{n-i}\end{array}$

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

$\begin{array}{rl}\mathbb{E}\left({S}_{(1)}\right)& =\frac{1}{9}\\ \mathbb{E}\left({S}_{(2)}\right)& =\frac{5}{18}\\ \mathbb{E}\left({S}_{(3)}\right)& =\frac{11}{18}\end{array}$

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

$\begin{array}{rl}\mathbb{E}\left(n\phantom{\rule{thinmathspace}{0ex}}{S}_{(k)}\right)& =\mathbb{E}\left({X}_{(k)}\right)\\ & =\sum _{i=0}^{k-1}\frac{1}{n-i}\end{array}$

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

$\begin{array}{rl}\mathbb{E}\left({S}_{(1)}\right)& =\frac{1}{9}\\ \mathbb{E}\left({S}_{(2)}\right)& =\frac{5}{18}\\ \mathbb{E}\left({S}_{(3)}\right)& =\frac{11}{18}\end{array}$

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