GS 2006/Math - Collins

Lab 5
Markov Chains


Preliminaries

This is based on Section 11.6 in the text. You should read both this lab and the material in 11.6.

Basics

The purpose is to study the random transitions of system between various states and to try to make predictions of future states. We start with k states which we number from 1 to k. The main assumption is that if at any time we are in state i then the probability that we will be in state j at the next time is given by a constant probability pji. For example, suppose we are trying to model a disease which has 3 states: 1 = well, 2 = sick, 3 = very sick. Then we might have the following transition probabilities:
From
Well Sick Very Sick
0.6 0.2 0 Well
0.4 0.7 0.1 Sick To
0 0.1 0.9 Very Sick
Note that each column sums to 1 as there is always a transition from one state to another. Now let x(n)i be the probability that at time n you are in state i. Then using conditional probabilities we have:

x(n+1)i = pi1 x(n)1 + pi2 x(n)2 + pi3 x(n)3 + pi4 x(n)4 + ... + pik x(n)k.

In matrix form, if we let x(n) be the k by 1 matrix of values x(n)1, x(n)2, ... , x(n)k, (called the nth state vector) we have simply

x(n+1) = P x(n),

where P is the matrix of all the transition probabilities. From the example above, suppose that initially you are well, so your initial state vector is x(0) = (1,0,0)T. Then using the equation above, we have

x(1) = P x(0) = (0.6, 0.4, 0)T
x(2) = P x(1) = (0.44, 0.52, 0.04)T
x(3) = P x(2) = (0.368,0.544, 0.088 )T

Notice that the sum of the entries is 1 in each one.
Thus, if each transition takes 1 week, then the results say that after 3 weeks you have probability 0.368 that you are still well, probability 0.544 that you are sick and probability 0.088 that you are very sick.

A natural question is what is the state vector in the long run? i.e. if we let n go to infinity, what do we get? One way to find out is to keep multiplying by P until we get a stationary solution, i.e. the solution doesn't change. The other way is to be smart and realize that if we find such a solution z then since it doesn't change, we must have z = Pz. Thus we can find this steady-state distribution by solving (I-P)z = 0 for a vector z where the sum of the entries is 1.

For our example that means we solve:
0.4 z1 - 0.22 = 0
-0.5 z1 + 0.3 z2 - 0.1 z3 = 0
-0.1 z2 + 0.1 z3 = 0
This has the general solution: z1 = 0.5 t, z2 = t and z3 = t. Thus since we want the sum to be 1, we take t = 2/5, so that z = (1/5, 2/5, 2/5)T and this is the steady-state vector for this problem.

MATLAB

To do all this in MATLAB you first enter the matrix P. If you want to see what happens over time from a starting distribution, enter the distribution as x (a column matrix). Then for each transition you just type

x = P*x

If you want to find the stead-state vector, use rref to solve (I-P)z = 0 via

rref(eye(k)-P)

where k is the number of states. Then from the rref solution, you'll get a general solution. You'll then have to find the particular solution so that the sum of the terms is 1. For example, from the P above we have

P = [0.6 0.2 0; 0.4 0.7 0.1; 0 0.1 0.9];
rref(eye(3)-P)
1 0 -0.5
0 1 -1.0
0 0  0.0
Thus the general solution is z1 = 0.5t, z2 = 1.0t and z3 = t. The sum of these terms is 2.5t, so to get a sum of 1, we choose t = 0.4, thus z = (0.2, 0.4 0.4)T.

Exercises

1. From 11.6 #2ab
2. From 11.6 #8
3. From 11.6 #T2ab (from the Technology Exercises)


Mail: ccollins@math.utk.edu