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Chapter 11: Problem 9

Prove that, if a 3 -by-3 transition matrix has the property that its column sums are 1 , then \((1 / 3,1 / 3,1 / 3)\) is a fixed probability vector. State a similar result for \(n\) -by- \(n\) transition matrices. Interpret these results for ergodic chains.

Short Answer

Expert verified
For a 3-by-3 matrix with column sums of 1, \\(\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)\\) is a fixed vector. Similarly, for an \(n\)-by-\(n\) matrix, \\(\left(\frac{1}{n}, \dots, \frac{1}{n}\right)\\) is fixed. In an ergodic chain, this represents the unique stationary distribution.

Step by step solution

01

Understanding the Matrix Properties

A 3-by-3 transition matrix with column sums equal to 1 means that the sum of each column is 1, which is a requirement for a stochastic matrix. In such matrices, every entry is non-negative and each column represents the transition probabilities from one state to another. Since sums are 1, each column can be interpreted as a probability distribution over states.
02

Identify the Fixed Probability Vector

A fixed probability vector for the matrix is a vector \(\mathbf{v} = (v_1, v_2, v_3)\) such that \(((1 / 3), (1 / 3), (1 / 3))\) satisfies the equation \[ M \cdot \mathbf{v} = \mathbf{v} \] where \(M\) is our transition matrix. This means multiplying \(M\) with \(\mathbf{v}\) should yield \(\mathbf{v}\) itself.
03

Calculate the Product

Let's substitute \(\mathbf{v}\) in the equation \[ M \cdot \begin{pmatrix} \frac{1}{3} \ \frac{1}{3} \ \frac{1}{3} \end{pmatrix} = \begin{pmatrix} \frac{1}{3} \ \frac{1}{3} \ \frac{1}{3} \end{pmatrix} \] Since the column sums of \(M\) are 1, each entry of the resulting vector \(Mi\) will also sum to 1/3 for each row. This assures us the vector \(\mathbf{v}\) is fixed under multiplication.
04

Extend to n-by-n Matrices

For any \(n\)-by-\(n\) transition matrix where each column sums to 1, a fixed probability vector is similarly, \(\mathbf{v} = \left( \frac{1}{n}, \frac{1}{n}, \dots, \frac{1}{n} \right)\). Here, the same logic applies - all entries in each row will add up uniformly due to the column sum being 1.
05

Interpret the Results for Ergodic Chains

In an ergodic chain, every state communicates with every other state. The stationary distribution is unique and the same principles apply here. Therefore, with column sums equal to 1, \(\mathbf{v} = \left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)\) (for a 3-state Markov chain) and \(\mathbf{v} = \left( \frac{1}{n}, \frac{1}{n}, \dots, \frac{1}{n} \right)\) for an \(n\)-state chain could represent the equilibrium or stationary distribution of an ergodic Markov chain.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fixed Probability Vector
In the context of a transition matrix, a fixed probability vector represents a state of equilibrium. Imagine you're working with a 3-by-3 transition matrix. This matrix has some interesting properties: its columns add up to 1, confirming it's a stochastic matrix. Now, suppose we identify
  • a fixed probability vector as \( \mathbf{v} = \left( \frac{1}{3}, \frac{1}{3}, \frac{1}{3} \right) \).
This vector is called 'fixed' because applying the matrix to it鈥攎ultiplying the matrix \( M \) by \( \mathbf{v} \)鈥攚ill result in the vector staying the same. Hence, it suggests a balance or equilibrium across the states, as probabilities don't change over time. This property shows stability and is vital for analyzing steady states in probabilistic systems.
Stochastic Matrix
A stochastic matrix is a type of matrix often used in probability and statistics. Each of its columns represents transition probabilities of moving from one state to another, and one important feature is that these columns must sum to 1. This ensures each column can be seen as a probability distribution.
  • For example, in our 3-by-3 matrix scenario, each column will have values adding up to 1.
  • This characteristic signifies that the system preserves total probability, just redistributing it across various states.
This concept is crucial because it means the system described by the matrix can evolve over time without losing total probability; questions about the long-term behavior of processes modeled by these matrices often revolve around such stability. A stochastic matrix doesn鈥檛 restrict its format; it can apply to any size, such as our extension to an \( n \)-by-\( n \) matrix.
Ergodic Markov Chain
An ergodic Markov chain has unique characteristics. All states within the chain are interconnected. Here, any state can be reached from any other state, a hallmark of 'communicating states.' This translates into a system where you can expect long-term probabilities to stabilize into a fixed pattern. Such chains typically possess a single, unique stationary distribution.
  • Importantly, for an ergodic chain with a transition matrix where each column sums to 1,
  • the fixed probability vector concept applies, implying that this vector also serves as the stationary distribution.
  • For our 3-state chain, this would mean every state has an equal likelihood of \( \frac{1}{3} \).
The beauty of ergodic chains lies in their robustness to initial conditions鈥攐ver time, the system 'forgets' its starting point and settles into a long-term behavior unaffected by where it began.

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Most popular questions from this chapter

A process moves on the integers \(1,2,3,4,\) and \(5 .\) It starts at 1 and, on each successive step, moves to an integer greater than its present position, moving with equal probability to each of the remaining larger integers. State five is an absorbing state. Find the expected number of steps to reach state five.

A study of the strengths of Ivy League football teams shows that if a school has a strong team one year it is equally likely to have a strong team or average team next year; if it has an average team, half the time it is average next year, and if it changes it is just as likely to become strong as weak; if it is weak it has \(2 / 3\) probability of remaining so and \(1 / 3\) of becoming average. (a) A school has a strong team. On the average, how long will it be before it has another strong team? (b) A school has a weak team; how long (on the average) must the alumni wait for a strong team?

In his book, Wahrscheinlichkeitsrechnung und Statistik, \(^{15}\) A. Engle proposes an algorithm for finding the fixed vector for an ergodic Markov chain when the transition probabilities are rational numbers. Here is his algorithm: For each state \(i\), let \(a_{i}\) be the least common multiple of the denominators of the non-zero entries in the \(i\) th row. Engle describes his algorithm in terms of moving chips around on the states-indeed, for small examples, he recommends implementing the algorithm this way. Start by putting \(a_{i}\) chips on state \(i\) for all \(i\). Then, at each state, redistribute the \(a_{i}\) chips, sending \(a_{i} p_{i j}\) to state \(j\). The number of chips at state \(i\) after this redistribution need not be a multiple of \(a_{i} .\) For each state \(i\), add just enough chips to bring the number of chips at state \(i\) up to a multiple of \(a_{i}\). Then redistribute the chips in the same manner. This process will eventually reach a point where the number of chips at each state, after the redistribution, is the same as before redistribution. At this point, we have found a fixed vector. Here is an example: $$ \mathbf{P}=\begin{array}{c} 1 & 2 & 3 \\ 1 \\ 2 \\ 3 \end{array}\left(\begin{array}{ccc} 1 / 2 & 1 / 4 & 1 / 4 \\ 1 / 2 & 0 & 1 / 2 \\ 1 / 2 & 1 / 4 & 1 / 4 \end{array}\right) $$ We start with \(\mathbf{a}=(4,2,4)\). The chips after successive redistributions are shown in Table 11.4 . We find that \(\mathbf{a}=(20,8,12)\) is a fixed vector. (a) Write a computer program to implement this algorithm. (b) Prove that the algorithm will stop. Hint: Let \(\mathbf{b}\) be a vector with integer components that is a fixed vector for \(\mathbf{P}\) and such that each coordinate of the starting vector \(\mathbf{a}\) is less than or equal to the corresponding component of b. Show that, in the iteration, the components of the vectors are always increasing, and always less than or equal to the corresponding component of \(\mathbf{b}\).

Assume that an experiment has \(m\) equally probable outcomes. Show that the expected number of independent trials before the first occurrence of \(k\) consecutive occurrences of one of these outcomes is \(\left(m^{k}-1\right) /(m-1) .\) Hint: Form an absorbing Markov chain with states \(1,2, \ldots, k\) with state \(i\) representing the length of the current run. The expected time until a run of \(k\) is 1 more than the expected time until absorption for the chain started in state \(1 .\) It has been found that, in the decimal expansion of pi, starting with the 24,658,601 st digit, there is a run of nine 7 's. What would your result say about the expected number of digits necessary to find such a run if the digits are produced randomly?

Consider the game of tennis when deuce is reached. If a player wins the next point, he has advantage. On the following point, he either wins the game or the game returns to deuce. Assume that for any point, player A has probability .6 of winning the point and player B has probability .4 of winning the point. (a) Set this up as a Markov chain with state 1: A wins; 2: B wins; 3: advantage \(\mathrm{A} ; 4:\) deuce; 5 : advantage \(\mathrm{B}\). (b) Find the absorption probabilities. (c) At deuce, find the expected duration of the game and the probability that \(\mathrm{B}\) will win.
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