# 7.4. Markov Chains

# Markov Chains

We will now study stochastic processes, experiments in which the outcomes of events depend on the previous outcomes. Such a process or experiment is called a **Markov Chain** or **Markov process**. The process was first studied by a Russian mathematician named Andrei A. Markov in the early 1900s.

A small town is served by two telephone companies, Mama Bell and Papa Bell. Due to their aggressive sales tactics, each month 40% of Mama Bell customers switch to Papa Bell, that is, the other 60% stay with Mama Bell. On the other hand, 30% of the Papa Bell customers switch to Mama Bell. The above information can be expressed in a matrix which lists the probabilities of going from one state into another state. This matrix is called a **transition matrix**.

The reader should observe that a transition matrix is always a square matrix because all possible states must have both rows and columns. All entries in a transition matrix are non-negative as they represent probabilities. Furthermore, since all possible outcomes are considered in the Markov process, the sum of the row entries is always 1.

Example 7.4.1

**Solution**

Example 7.4.2

**Solution**

Let *W* denote that Professor Symons walks and *B* denote that he rides his bicycle. We use the following tree diagram to compute the probabilities.

Monday |
Tuesday |
Wednesday |

The probability that Professor Symons walked on Wednesday given that he walked on Monday can be found from the tree diagram, as listed below.

*P*(Walked Wednesday | Walked Monday) = *P*(WWW) + *P*(WBW) = 1/4 + 1/8 = 3/8.

*P*(Bicycled Wednesday | Walked Monday) = *P*(WWB) + *P*(WBB) = 1/4 + 3/8 = 5/8.

*P*(Walked Wednesday | Bicycled Monday) = *P*(BWW) + *P*(BBW) = 1/8 + 3/16 = 5/16.

*P*(Bicycled Wednesday | Bicycled Monday) = *P*(BWB) + *P*(BBB) = 1/8 + 9/16 = 11/16.

We represent the results in the following matrix:

Alternately, this result can be obtained by squaring the original transition matrix.

We list both the original transition matrix *T* and *T ^{2}* as follows:

The reader should compare this result with the probabilities obtained from the tree diagram. Consider the following case, for example:

*P*(Walked Wednesday | Bicycled Monday) = *P*(BWW) + P(BBW) = 1/8 + 3/16 = 5/16.

It makes sense because to find the probability that Professor Symons will walk on Wednesday given that he bicycled on Monday, we sum the probabilities of all paths that begin with *B* and end in *W*. There are two such paths, and they are BWW and BBW.

Certain Markov chains, called **regular Markov chains**, tend to stabilize in the long run. It so happens that the transition matrix we have used in the the above examples is just such a Markov chain. The next example deals with the long term trend or steady-state situation for that matrix.

Example 7.4.3

**Solution**

*T*, it should stabilize.

When this happens, we say that the system is in steady-state or state of equilibrium. In this situation, all row vectors are equal. If the original matrix is an *n* by *n* matrix, we get *n* vectors that are all the same. We call this vector a **fixed probability vector** or the **equilibrium vector** *E*. In the above problem, the fixed probability vector *E* is . Furthermore, if the equilibrium vector *E* is multiplied by the original matrix *T*, the result is the equilibrium vector *E*. That is,

*ET* = *E*

or,

**Regular Markov Chains**

A Markov chain reaches a state of equilibrium if it is a **regular** Markov chain. A Markov chain is said to be a **regular Markov chain** if some power of it has only positive entries.

As we take higher powers of *T*, *T ^{n}*, as

*n*becomes large, approaches a state of equilibrium. The equilibrium distribution vector

*E*can be found by letting

*ET*=

*E*.

Example 7.4.4

If the initial market share for Mama Bell is 20% and for Papa Bell 80%, we’d like to know the long term market share for each company.

Let matrix *T* denote the transition matrix for this Markov chain, and *M* denote the matrix that represents the initial market share. Then *T* and *M* are as follows:

and

Since each month the town’s people switch according to the transition matrix *T*, after one month the distribution for each company is as follows:

After two months, the market share for each company is:

After three months the distribution is:

After four months the market share is:

After 30 months the market share is .

The market share after 30 months has stabilized to .

This means that:

Once the market share reaches an equilibrium state, it stays the same, that is, *ET* = *E*.

Can the equilibrium vector *E* be found without raising the transition matrix to large powers? The answer to this question provides us with a way to find the equilibrium vector *E*. The answer lies in the fact that *ET* = *E*. Since we have the matrix *T*, we can determine *E* from the statement *ET* = *E*. The example below illustrates this approach.

Example 7.4.5

*T*from the previous example, suppose , then

*ET*=

*E*gives us:

**Practice questions**

**1.** A survey of American car buyers indicates that if a person buys a Ford, there is a 60% chance that their next purchase will be a Ford, while owners of a GM will buy a GM again with a probability of 0.80. Express the buying habits of these consumers in a transition matrix.

**2.** The local police department conducts a campaign to reduce the rates of texting and driving in the community. The effects of the campaign are summarized in the transition matrix below:

If 35% of people in the community reported texting and driving before the campaign:

**a.** What is the percentage of people in the community that reported texting and driving after the campaign?

**b.** If the campaign were to be repeated multiple times, what is the long-range trend in terms of the lowest rate that texting and driving can be reduced to in this community?

Answers

**1. **

**2. a. ** 31%

** b.** 25%