Markov chains and their applications

Question

I'm trying to solve a Markov chain question, I believe this a special kind of those chains, I've solved part a still I'm stuck in part 2.

A city is served by two newspapers, the Star and the Times. Each year the Star loses 30% of its subscribers to the Times and retains 70% of its subscribers. During the same time period, the Times loses 10% of its subscribers to the Star while retaining the other 90%.

a) Write down a Markov matrix that describes the transition of subscribers between the two newspapers each year.

b) Find the steady state vector for the matrix in (a), what percentage of the subscribers will subscribe to the Times? .How many years it takes to reach the steady state vector?

If anyone can kindly help.

Answer 1

The transition matrix is $$ A = \frac1{10} \begin{bmatrix} 7 & 3 \\ 1 & 9 \end{bmatrix} \ . $$ (The first row corresponds to the Star, the second row to the Times.)

If we define an initial distribution of readers, $(s_0,t_0)$, a row vector / a $1\times 2$-matrix, $s_0$ the percentage of the Star readers at time $0$, $t_0$ the corresponding percentage for the Times, then we are modeling $$ [s_n\ t_n]= [s_0\ t_0]A^n\ . $$ The characteristic polynomial of $A$ is $(x-1)(x-3/5)$ and some left eigenvectors are

• $(1,3)$ for the $0$, accepted, and
• $(1,-1)$ for the $3/5$, rejected.

Explicitly: $$ \begin{bmatrix} 1&3 \end{bmatrix} A =\frac 1{10} \begin{bmatrix} 1&3 \end{bmatrix} \begin{bmatrix} 7 & 3 \\ 1 & 9 \end{bmatrix} =\frac 1{10} \begin{bmatrix} 1\cdot 7+3\cdot 1 & 1\cdot 3+3\cdot 9 \end{bmatrix} = \begin{bmatrix} 1&3 \end{bmatrix} \ . $$ So we get stability with $[s_0\ t_0]=\frac 14[1\ 3]=[1/4\ 3/4]$.

How many years...?! Well, we must know the initial distribution, and the number of readers. Assuming we start with the proportion $1:3$, equilibrium, we remain so all the time. If there are only Star readers at the beginning, say some $40\; 000$, then after some twenty years we get the $1:3$ proportion. Sage code:

sage: A = 1/10 * matrix( QQ, 2, 2, [7, 3, 1, 9] )
sage: A.charpoly().factor()
(x - 1) * (x - 3/5)

sage: vector([40000., 0.]) * A^10
(10181.3985280000, 29818.6014720000)
sage: vector([40000., 0.]) * A^20
(10001.0968475320, 29998.9031524680)
sage: vector([40000., 0.]) * A^25
(10000.0852908641, 29999.9147091359)

(Well, it is hard to get some $1.968\dots$ readers in practice, so the computation should maybe multiply and truncate/round at each step...)