Q. 9.10 - A First Course in Probability

Question

A certain person goes for a run each morning. When he leaves his house for his run, he is equally likely to go out either the front or the back door, and similarly, when he returns, he is equally likely to go to either the front or the back door. The runner owns 5 pairs of running shoes, which he takes off after the run at whichever door he happens to be. If there are no shoes at the door from which he leaves to go running, he runs barefooted. We are interested in determining the proportion of time that he runs barefooted. (a) Set this problem up as a Markov chain. Give the states and the transition probabilities. (b) Determine the proportion of days that he runs barefooted. 

Step-by-Step Solution

Answer

(a) States are given below and transition matrix is$P=\left[\begin{array}{cccc}\frac{5}{6}& \frac{1}{6}& 0& 0\\ \frac{5}{6}& \frac{1}{6}& 0& 0\\ 0& 0& \frac{5}{6}& \frac{1}{6}\\ 0& 0& \frac{5}{6}& \frac{1}{6}\end{array}\right]$

(b)The proportion of days that he runs barefooted is $\frac{1}{6}$.

Part (a) Step 1: Given Information

We need to write Markov chain and give transition probabilities.

Part (a) Step 2: Simplify

Markov chain $X_n:n\ge 0$ that has four states

$0$- running with shoes and left from the front door

$1-$running barefooted and left from the front door

$2-$running with shoes and left from the back door

$3-$running barefooted and left from the back door

Now calculating the transition matrix. Let us consider that he has been running with shoes and that he left from the front door. When he comes home, he has $5/6$ of chances to leave with shoes from the front door and of chances to run barefooted from the front door next time. Use similar logic to obtain that the transition matrix is

$P=\left[\begin{array}{cccc}\frac{5}{6}& \frac{1}{6}& 0& 0\\ \frac{5}{6}& \frac{1}{6}& 0& 0\\ 0& 0& \frac{5}{6}& \frac{1}{6}\\ 0& 0& \frac{5}{6}& \frac{1}{6}\end{array}\right]$

Part (b) Step 1: Given Information

We have to find the proportion of days that he runs barefooted.

Part (b) Step 2: Simplify

Let's find stationary distribution. We are going to solve the system  which implies this set of equalities

$$\begin{gathered} {\mathrm{\pi }}_0=\frac{5}{6}{\mathrm{\pi }}_0+\frac{5}{6}{\mathrm{\pi }}_1 \\ {\mathrm{\pi }}_1=\frac{1}{6}{\mathrm{\pi }}_0+\frac{1}{6}{\mathrm{\pi }}_1 \\ {\mathrm{\pi }}_2= +\frac{5}{6}{\mathrm{\pi }}_2+\frac{5}{6}{\mathrm{\pi }}_3 \\ {\mathrm{\pi }}_3=+\frac{1}{6}{\mathrm{\pi }}_2+\frac{1}{6}{\mathrm{\pi }}_3 \end{gathered}$$

Now solving the equation with ${\sum }_i{\pi }_i=1$

${\mathrm{\pi }}_1+{\mathrm{\pi }}_3=\frac{1}{6}$

This is the probability that he leaves barefooted.