MHB Solve Autoregressive Model Problem

[Fernando Revilla]

I quote an unsolved problem from another forum (Algebra) posted on January 16th, 2013.

Got the following problem.

In a country you can live in three different citys, A, B and C, the population is constant.

Each year;

70% of the residents in city A stay, 20% move to city B and 10% move to city C
90% of the residents in city B stay, 5% move to city A and 5% move to city C
50% of the residents in city C stay, 45% move to city A and 5% move to city B

I am suppose to explain this as an autoregressive process.
Through some datamining i found that the process is an AR(3) process, with coefficents
2,1 -1.3725 0.2725

My question is, is it possible to solve this analytically, without Least squares trial and error?

I provide an algebraic approach to predict the behaviour in the future.

Denote P_{n}=(a_n,b_n,c_n)^t, where a_n,b_n,c_n are the poblations of A,B,C respectively in the year n. According to the hypothesis:

a_n=0.7a_{n-1}+0.05b_{n-1}+0.45c_{n-1}\\b_n=0.2a_{n-1}+0.9b_{n-1}+0.05c_{n-1}\\c_n=0.1a_{n-1}+0.05b_{n-1}+0.5c_{n-1}

Equivalently

P_n=\begin{bmatrix}{0.7}&{0.05}&{0.45}\\{0.2}&{0.9}&{0.05}\\{0.1}&{0.05}&{0.5}\end{bmatrix}\;P_{n-1}=\dfrac{1}{20}\begin{bmatrix}{70}&{5}&{45}\\{20}&{90}&{5}\\{10}&{5}&{50}\end{bmatrix}\;P_{n-1} =MP_{n-1}

Then, P_n=MP_{n-1}=M^2P_{n-2}=\ldots=M^nP_0

As M is a Markov matrix, has the eigenvalue \lambda_1=1 and easily we can find the rest: \lambda_2=(11+\sqrt{3})/20 and \lambda_3=(11-\sqrt{3})/20. These eigenvalues are all simple, so M is diagonalizable in \mathbb{R}. If Q\in\mathbb{R}^{3\times 3} satisfies Q^{-1}AQ=D=\mbox{diag }(\lambda_1,\lambda_2,\lambda_3), then P_n=QD^nQ^{-1}P_0. Taking limits in both sides an considering that |\lambda_2|<1 and |\lambda_3|<1:

P_{\infty}:=\displaystyle\lim_{n \to \infty} P_n=Q\;(\displaystyle\lim_{n \to \infty}D^n)\;Q^{-1}P_0=Q\;\begin{bmatrix}{1}&{0}&{0}\\{0}&{0}&{0}\\{0}&{0}&{0}\end{bmatrix}\;Q^{-1}P_0

Not that for computing P_{\infty} we only need the first column (an eigenvalue v_1 associated to \lambda_1) of Q and the first row w_{1} of Q^{-1}. We get v_1=(19,42,8)^t and w_{1}=(1/69)(1,1,1). So,
$P_{\infty}=\begin{bmatrix}a_{\infty} \\ b_{\infty}\\ c_{\infty}\end{bmatrix}=$ $\dfrac{1}{69}\begin{bmatrix}{19}&{*}&{*} \\ {42}&{*}&{*} \\ {8}&{*}&{*}\end{bmatrix}\;\begin{bmatrix}{1}&{0}&{0} \\ {0}&{0}&{0} \\ {0}&{0}&{0}\end{bmatrix}\;$ $\begin{bmatrix}{1}&{1}&{1}\\{*}&{*}&{*}\\{*}&{*}&{*}\end{bmatrix}\begin{bmatrix}a_{0}\\b_{0}\\c_{0}\end{bmatrix}=$ $\dfrac{1}{69}\begin{bmatrix}{19(a_0+b_0+c_0)}\\{42(a_0+b_0+c_0)}\\{8(a_0+b_0+c_0)}\end{bmatrix}$

which represents the tendency of the poblations of A,B and C as n\to \infty.

 

[I like Serena]

Hmm, since it is given that the population remains constant, doesn't it suffice that:

$P_1 = P_0$​

That is,

$MP_0 = P_0$​

So the population ratios correspond to the eigenvector belonging to eigenvalue 1?

 

[Fernando Revilla]

Hmm, since it is given that the population remains constant, doesn't it suffice that: $P_1 = P_0$

Remains constant means $a_n+b_n+c_n=a_{n-1}+b_{n-1}+c_{n-1}$ for all $n\geq 1$, different from $(a_n,b_n,c_n)=(a_{n-1}b_{n-1},c_{n-1})$.