Markov Chains

Every year, 5% of the population of Town 1 moves to Town 2 while 2% of the population of Town 2 moves to Town 1. In 2000, Town 1 had 10,000 people and Town 2 had 15,000 people.

a) How many people were in each city in 2001?

• Town 1: $10000(0.95)+15000(0.2)=9800$
• Town 2: $10000(0.05)+15000(0.98)=15200$

b) As a sequence? (In thousands)

• $a_n=0.95a_{n-1}+0.02b_{n-1}{a}_0=10$
• $b_n=0.05a_{n-1}+0.98b_{n-1}{b}_0=15$

c) In vector form, ${\vec{y}}_n=\left[\begin{array}{c}{a}_n\\ b_n\end{array}\right]$

$$\begin{array}{rl}{\vec{y}}_{n+1}& =\left[\begin{array}{cc}.95& .02\\ .05& .98\end{array}\right]\left[\begin{array}{c}{a}_n\\ b_n\end{array}\right]=\left[\begin{array}{cc}.95& .02\\ .05& .98\end{array}\right]{\vec{y}}_n\\ A& =\left[\begin{array}{cc}.95& .02\\ .05& .98\end{array}\right]\\ \lambda & =\{0.93,1\}{\vec{v}}_{\dots }=\{\left[\begin{array}{c}-1\\ 1\end{array}\right],\left[\begin{array}{c}2\\ 5\end{array}\right]\}\text{Computation already performed.}\\ P& =\left[\begin{array}{cc}-1& 2\\ 1& 5\end{array}\right]\therefore P^{-1}=\left[\begin{array}{cc}\frac{5}{-7}& \frac{2}{7}\\ \frac{1}{7}& \frac{1}{7}\end{array}\right],D=\left[\begin{array}{cc}.93& 0\\ 0& 1\end{array}\right]\\ {\vec{y}}_n& =A^n\left[\begin{array}{c}10\\ 15\end{array}\right]\\ & =P{D}^n{P}^{-1}\left[\begin{array}{c}10\\ 15\end{array}\right]\\ & =\left[\begin{array}{cc}-1& 2\\ 1& 5\end{array}\right]{\left[\begin{array}{cc}.93& 0\\ 0& 1\end{array}\right]}^n\left[\begin{array}{cc}\frac{5}{-7}& \frac{2}{7}\\ \frac{1}{7}& \frac{1}{7}\end{array}\right]\left[\begin{array}{c}10\\ 15\end{array}\right]\\ & =\left[\begin{array}{cc}-1& 2\\ 1& 5\end{array}\right]\left[\begin{array}{cc}.93^n& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}\frac{5}{-7}& \frac{2}{7}\\ \frac{1}{7}& \frac{1}{7}\end{array}\right]\left[\begin{array}{c}10\\ 15\end{array}\right]\\ \left[\begin{array}{c}{a}_n\\ b_n\end{array}\right]& =\left[\begin{array}{c}\frac{5}{7}(10+4(0.93{)}^n)\\ \frac{5}{7}(25-4(0.93{)}^n)\end{array}\right]\text{It has been computed as.}\\ \therefore & \\ & \\ & a_n=\frac{5}{7}(10+4(0.93{)}^n),b_n=\frac{5}{7}(25-4(0.93{)}^n\end{array}$$

d) So what happens to the two towns?

$$\begin{array}{rl}\underset{n\to \infty }{\lim }{a}_n& =\underset{n\to \infty }{\lim }\frac{5}{7}(10+4(0.93{)}^n)\\ & =\frac{5}{7}(10)\\ & \\ \underset{n\to \infty }{\lim }{b}_n& =\underset{n\to \infty }{\lim }\frac{5}{7}(25-4(0.93{)}^n\\ & =\frac{5}{7}(25)\end{array}$$

---

5/11/24