91Ó°ÊÓ

We need to write Markov chain and give transition probabilities.

Markov chain $X_n:nâ‰\yen 0$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

localid="1648119452522" $P=\left[\begin{array}{cccc}\frac{5}{6}& \frac{1}{6}& 0& 0\\ \frac{5}{6}& \frac{1}{6}& 0& 0\\ 0& 0& \frac{5}{6}& \frac{1}{6}\\ 0& 0& \frac{5}{6}& \frac{1}{6}\end{array}\right]$

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

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

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

Now solving the equation with ${∑}_i{\mathrm{π}}_i=1$

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

This is the probability that he leaves barefooted.