How to Read a Transition Probability Matrix

The Long Run Behavior of Markov Chains

Marking A. Pinsky , Samuel Karlin , in An Introduction to Stochastic Modeling (Quaternary Edition), 2011

four.1.one Doubly Stochastic Matrices

A transition probability matrix is called doubly stochastic if the columns sum to one equally well equally the rows. Formally, P = ||P _ij || is doubly stochastic if

$P_{ij}\ge 0\text{and}\sum _k{P}_{i\mathrm{one\; thousand}}=\sum _k{P}_{kj}=1\text{for}\text{all}i,j$

Consider a doubly stochastic transition probability matrix on the Due north states 0, 1, …, N − 1. If the matrix is regular, then the unique limiting distribution is the uniform distribution π = (1/North, …, i/North). Because there is only i solution to π _j = ∑_{one thousand}π _k P _kj and σ _k π _chiliad = i when P is regular, we need but to check that π = (1/North, …, 1/Northward) is a solution where P is doubly stochastic in lodge to establish the claim. Past using the doubly stochastic characteristic ∑ _j P _jk = ane, we verify that

$\frac{1}{N}=\sum _j\frac{1}{\mathrm{Due\; north}}{P}_{j\mathrm{thousand}}=\frac{1}{N}.$

Equally an example, let Y_n be the sum of n independent rolls of a fair dice and consider the problem of determining with what probability Y_n is a multiple of 7 in the long run. Let X_n be the remainder when Y_n is divided by seven. Then, Ten_n is a Markov chain on u.s.a. 0, 1, …, 6 with transition probability matrix

The matrix is doubly stochastic, and it is regular (P² has only strictly positive entries), hence the limiting distribution is $\pi =\left(\frac{1}{7},\mathrm{\dots },\frac{\mathrm{ane}}{7}\right)$ . Furthermore, Y_northward is a multiple of 7 if and only if X_n = 0. Thus, the limiting probability that Y_n is a multiple of seven is $\frac{1}{7}$ .

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Discrete-Time Markov Chains

Oliver C. Ibe , in Markov Processes for Stochastic Modeling (Second Edition), 2013

4.12 Problems

4.1

Consider the following transition probability matrix:

$P=\left[\begin{array}{ccc}0.6& 0.2& 0.2\\ \mathrm{0.iii}& \mathrm{0.iv}& 0.3\\ 0.0& 0.3& 0.7\end{array}\right]$

a.

Give the state-transition diagram.

b.

Given that the process is currently in state one, what is the probability that it will be in state 2 at the cease of the third transition?

c.

Given that the procedure is currently in state 1, what is the probability that the first fourth dimension it enters state iii is the 4th transition?

iv.2

Consider the following social mobility problem. Studies point that people in a society tin can be classified as belonging to the upper form (state 1), center class (state ii), and lower class (state 3). Membership in any grade is inherited in the following probabilistic manner. Given that a person is raised in an upper-class family, he will take an upper-class family with probability 0.45, a middle-class family with probability 0.48, and a lower-class family with probability 0.07. Given that a person is raised in a heart-grade family, he volition have an upper-course family unit with probability 0.05, a middle-class family with probability 0.seventy, and a lower-class family with probability 0.25. Finally, given that a person is raised in a lower-grade family, he volition have an upper-form family with probability 0.01, a heart-class family with probability 0.50, and a lower-form family with probability 0.49. Determine the post-obit:

a.

The state-transition diagram of the process.

b.

The transition probability matrix of the process.

c.

The limiting-land probabilities. Interpret what they mean to the layperson.

iv.3

A taxi driver conducts his business in iii different towns 1, 2, and three. On any given day, when he is in town ane, the probability that the next rider he picks up is going to a identify in boondocks 1 is 0.3, the probability that the adjacent rider he picks up is going to boondocks ii is 0.two, and the probability that the adjacent passenger he picks up is going to boondocks 3 is 0.v. When he is in boondocks 2, the probability that the next passenger he picks up is going to town 1 is 0.one, the probability that the side by side rider he picks upward is going to boondocks two is 0.8, and the probability that the next passenger he picks up is going to town 3 is 0.1. When he is in boondocks iii, the probability that the next rider he picks up is going to boondocks 1 is 0.4, the probability that the next passenger he picks upward is going to boondocks 2 is 0.iv, and the probability that the next passenger he picks up is going to town iii is 0.2.

a.

Determine the state-transition diagram for the process.

b.

Give the transition probability matrix for the procedure.

c.

What are the limiting-land probabilities?

d.

Given that the taxi driver is currently in town 2 and is waiting to selection up his first customer for the twenty-four hours, what is the probability that the first time he picks up a passenger to boondocks 2 is when he picks up his 3rd passenger for the solar day?

eastward.

Given that he is currently in town 2, what is the probability that his tertiary passenger from now will be going to boondocks one?

four.4

The New England fall weather tin can be classified as sunny, cloudy, or rainy. A pupil conducted a detailed study of the conditions conditions and came up with the post-obit conclusion: Given that information technology is sunny on whatever given day, then on the following solar day it will exist sunny again with probability 0.five, cloudy with probability 0.three and rainy with probability 0.2. Given that it is cloudy on any given twenty-four hour period, and so on the following day it volition be sunny with probability 0.four, cloudy again with probability 0.3 and rainy with probability 0.3. Finally, given that it is rainy on any given day, so on the following solar day it will exist sunny with probability 0.two, cloudy with probability 0.5 and rainy again with probability 0.three.

a.

Requite the state-transition diagram of New England fall weather with the state "sunny" equally land i, the state "cloudy" as state two, and the state "rainy" as state iii.

b.

Using the same convention every bit in office (a), give the transition probability matrix of the New England autumn weather.

c.

Given that it is sunny today, what is the probability that it will be sunny iv days from at present?

d.

Decide the limiting-state probabilities of the weather.

iv.5

Consider the following transition probability matrix:

$P=\left[\begin{array}{ccc}\mathrm{0.iii}& 0.2& 0.5\\ \mathrm{0.i}& 0.8& 0.1\\ 0.4& \mathrm{0.four}& \mathrm{0.ii}\end{array}\right]$

a.

What is $P^n$ ?

b.

Obtain ${\varphi }_{\mathrm{xiii}}(\mathrm{five})$ , the mean occupancy time of state 3 upwards to five transitions given that the process started from state 1.

four.6

Consider the following transition probability matrix:

$P=\left[\begin{array}{ccc}0.5& 0.25& 0.25\\ 0.3& 0.3& \mathrm{0.iv}\\ 0.25& 0.5& 0.25\end{array}\right]$

a.

Calculate $p_{13}(3),p_{22}(\mathrm{two})$ , and $p_{32}(4)$ .

b.

Calculate $p_{32}(\infty )$ .

4.7

Consider the post-obit transition probability matrix:

$P=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0.75& 0& 0.25& 0\\ 0& 0.25& 0& 0.75\\ 0& 0& 0& \mathrm{i}\end{array}\right]$

a.

Put the matrix in the canonical form $P=\left[\begin{array}{cc}I& 0\\ R& Q\end{array}\right]$ .

b.

Calculate the expected assimilation times ${\mu }_2$ and ${\mu }_{\mathrm{iii}}$ .

4.8

Consider the following transition probability matrix:

$P=\left[\begin{array}{ccc}0.5& 0.25& 0.25\\ 0.3& \mathrm{0.three}& 0.4\\ 0.25& 0.5& 0.25\end{array}\right]$

a.

Calculate $f_{13}(4)$ the probability of first passage from state 1 to state 3 in iv transitions.

b.

Calculate the mean sojourn time in state two.

4.9

Permit $\left\{X_n\right\}$ be a Markov chain with the state infinite $\{\mathrm{one},2,3\}$ and transition probability matrix

$P=\left[\begin{array}{ccc}0& 0.4& \mathrm{0.half\; dozen}\\ 0.25& 0.75& 0\\ 0.4& 0& 0.6\end{array}\right]$

Let the initial distribution be $p(0)=[p_1(0),p_2(0),p_{\mathrm{three}}(0)]=[0.4,0.2,0.4]$ . Calculate the following probabilities:

a.

$P[X_1=2,X_3=\mathrm{two},X_{\mathrm{three}}=1|{\mathrm{Ten}}_0=\mathrm{one}]$

b.

$P[X_1=2,X_{\mathrm{three}}=2,{\mathrm{Ten}}_3=1]$

c.

$P[{\mathrm{Ten}}_1=2,X_4=2,X_6=2]$

four.10

On a given day Mark is cheerful, so-so, or glum. Given that he is cheerful on a given day, then he will be cheerful again the next solar day with probability 0.6, so-so with probability 0.2, and glum with probability 0.2. Given that he is and so-then on a given solar day, then he will be cheerful the adjacent day with probability 0.3, and then-so once again with probability 0.5, and glum with probability 0.2. Given that he is glum on a given day, then he will be so-so the next day with probability 0.5, and glum once again with probability 0.five. Permit country ane denote the cheerful state, state 2 denote the so-so state, and country iii denote the glum state. Let $X_n$ denote Marker's mood on the nth day, then $\{X_{\mathrm{northward}},n=0,1,2,\dots \}$ is a three-state Markov chain.

a.

Draw the state-transition diagram of the process.

b.

Give the state-transition probability matrix.

c.

Given that Mark was and so-so on Monday, what is the probability that he will exist cheerful on Midweek and Friday and glum on Sunday?

d.

On the long run, what proportion of time is Marking in each of his 3 moods?

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Markov Chains

Mark A. Pinsky , Samuel Karlin , in An Introduction to Stochastic Modeling (Quaternary Edition), 2011

Exercises

3.7.1

Consider the Markov chain whose transition probability matrix is given by

$\text{P}=\begin{array}{c}0\\ 1\\ 2\\ 3\end{array}\Vert \begin{array}{cccc}0& 1& 2& \mathrm{three}\\ \mathrm{ane}& 0& 0& 0\\ 0.1& 0.2& 0.5& \mathrm{0.two}\\ \mathrm{0.i}& 0.2& 0.6& 0.1\\ 0& 0& 0& \mathrm{one}\end{array}\Vert .$

The transition probability matrix respective to the nonabsorbing states is

$\text{Q}=\begin{array}{c}0\\ 1\end{array}\Vert \begin{array}{cc}1& 2\\ 0.2& \mathrm{0.v}\\ \mathrm{0.ii}& 0.6\end{array}\Vert .$

Summate the matrix changed to I − Q, and from this decide

(a)

the probability of absorption into country 0 starting from state 1;

(b)

the hateful time spent in each of states 1 and ii prior to absorption.

3.7.2

Consider the random walk Markov chain whose transition probability matrix is given by

$\text{P}=\begin{array}{c}0\\ \mathrm{i}\\ 2\\ 3\end{array}\Vert \begin{array}{cccc}0& 1& \mathrm{two}& 3\\ 1& 0& 0& 0\\ 0.3& 0& 0.7& 0\\ 0& 0.3& 0& 0.7\\ 0& 0& 0& 1\end{array}\Vert .$

The transition probability matrix corresponding to the nonabsorbing states is

$\text{Q}=\begin{array}{c}0\\ 1\end{array}\Vert \begin{array}{cc}1& 2\\ 0& 0.7\\ 0.3& 0\end{array}\Vert .$

Summate the matrix inverse to I − Q, and from this make up one's mind

(a)

the probability of absorption into land 0 starting from state one;

(b)

the mean fourth dimension spent in each of states i and 2 prior to assimilation.

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Markov Processes

Scott L. Miller , Donald Childers , in Probability and Random Processes (Second Edition), 2012

9.2 Calculating Transition and Country Probabilities in Markov Chains

The land transition probability matrix of a Markov concatenation gives the probabilities of transitioning from 1 state to some other in a unmarried time unit. It will be useful to extend this concept to longer fourth dimension intervals.

Definition 9.three: The n -stride transition probability for a Markoff chain is

(9.iv) $P_{i,j}^{\left(\mathrm{due\; north}\right)}=Pr\left(X_{k+1}=j|X_k=i\right).$

Also, define an n -step transition probability matrix P (north) whose elements are the n -stride transition probabilities in Equation (9.4).

Given the one-stride transition probabilities, information technology is straightforward to calculate higher order transition probabilities using the post-obit consequence.

Theorem nine.1 (Chapman–Kolmogorov Equation):

(9.5) $\begin{array}{cc}{P}_{i,j}^{\left(n\right)}=\sum _{\mathrm{thousand}}{p}_{i,k}^{\left(m\right)}{p}_{\mathrm{yard},j}^{\left(\mathrm{due\; north}-m\right)},& \text{for}\text{whatsoever}m=0,1,\mathrm{two},\dots ,n.\end{array}$

Proof: Outset, condition on the upshot that in the process of transitioning from state i to state j, the Markov chain passes through land k at some intermediate point in time. And so using the principle of total probability

(9.6) $Pr\left(X_{l+n}=j|X_l=i\right)=\sum _{\mathrm{grand}}Pr\left(X_{l+n}=j|X_l=k,{\mathrm{Ten}}_{l+m}=\mathrm{1000}\right)Pr\left(X_{l+m}=\mathrm{grand}|X_{\mathrm{grand}}=i\right).$

Using the Markov property, the expression reduces to the desired form:

(9.7) $Pr\left({\mathrm{10}}_{l+n}=j|X_l=i\right)=\sum _{k}Pr\left(X_{l+\mathrm{northward}}=j|X_{l+m}=k\right)Pr\left(X_{\mathrm{50}+m}=\mathrm{thou}|X_k=i\right).$

This effect can be written in a more compact course using transition probability matrices. Information technology is easily seen that the Chapman–Kolmogorov equations can be written in terms of the n -step transition probability matrices as

(ix.8) $P^{\left(\mathrm{north}\right)}=P^{\left(k\right)}{P}^{\left(n-m\right)}.$

So, starting with the fact that P ⁽¹⁾ = P, it follows that P (²) = p ⁽¹⁾ p ⁽ⁱ⁾ = p ⁱⁱ, and using induction, it is established that

(9.9) $P^{\left(n\right)}=P^{\mathrm{due\; north}}.$

Hence, nosotros can notice the northward -step transition probability matrix through matrix multiplication. If northward is large, it may be more convenient to compute P^northward via eigendecomposition. In many cases ^ane the matrix P can exist expanded as P = UΛU ^−one, where Λ is the diagonal matrix of eigenvalues and U is the matrix whose columns are the respective eigenvectors. Then,

(9.10) $P^{\mathrm{north}}=U{\Lambda }^{\mathrm{north}}{U}^{-\mathrm{i}}.$

Another quantity of interest is the probability distribution of the Markov chain at some fourth dimension instant k. If the initial probability distribution of the Markov chain is known, then the distribution at some later indicate in fourth dimension can hands be plant. Let π_j (chiliad) = Pr(10_thou =j) and π(k) be the row vector whose jth element is π_j (k). Then

(9.11) ${\pi }_j\left(\mathrm{chiliad}\right)=Pr\left(X_k=j\right)=\sum _{i}Pr\left(X_k=j|{\mathrm{Ten}}_0=i\right)Pr\left(X_0=i\right)=\sum _i{p}_{i,j}^{\left(k\right)}{\pi }_i\left(0\right),$

or in vector form,

(9.12) $\pi \left(\mathrm{one\; thousand}\right)=\pi \left(0\right)P^{\mathrm{thou}}.$

Example 9.vii (continuation of Example 9.2)

Think in Example 9.ii, the kid who purchased kid's meals at his favorite restaurant in order to collect a set up of 4 superhero action figures. Initially, earlier whatever meals are purchased, the child has no action figures and so the initial probability distribution is π(0) = (1,0,0,0,0). Repeated awarding of Equation (9.12) with the probability transition matrix given in Example 9.two results in

$\begin{array}{c}\pi \left(\mathrm{i}\right)=\left(0,1,0,0,0\right),\\ \pi \left(2\right)=\left(0,1/4,3/4,0,0\right),\\ \pi \left(3\right)=\left(0,1/\mathrm{xvi},9/16,\mathrm{three}/8,0\right),\\ \pi \left(4\right)=\left(0,1/64,21/64,\mathrm{nine}/\mathrm{xvi.3}/32\right),\\ \pi \left(5\right)=\left(0,1/256,45/256,75/128,15/64\right),\\ \pi \left(6\right)=\left(0,1/1024,93/1024,135/256,195/512\right),\end{array}$

and and then on. Information technology is to be expected that if the child buys enough meals, he will somewhen consummate the collection (i.e., go to country 4) with probability budgeted unity. This can easily be verified analytically by calculating the limiting course of P^k as k → ∞. Recall that for this instance, P is a triangular matrix and hence its eigenvalues are simply the diagonal entries. Hence, the diagonal matrix of eigenvalues is

$\Lambda =\left[\begin{array}{ccccc}0& 0& 0& 0& 0\\ 0& 1/4& 0& 0& 0\\ 0& 0& 1/\mathrm{two}& 0& 0\\ 0& 0& 0& \mathrm{iii}/\mathrm{iv}& 0\\ 0& 0& 0& 0& \mathrm{ane}\end{array}\right].$

It should exist articulate that $\underset{k\to \infty }{lim}{\Lambda }^k$ is a matrix with all zero entries except the one in the lower right corner, which is equal to 1. Using MATLAB (or another math package) to calculate the corresponding matrix of eigenvectors, it is institute that

$\underset{k\to \infty }{lim}{P}^k=U\left(\underset{k\to \infty }{lim}{\Lambda }^{\mathrm{grand}}\right)U^{-\mathrm{one}}=\left[\begin{array}{ccccc}0& 0& 0& 0& \mathrm{i}\\ 0& 0& 0& 0& 1\\ 0& 0& 0& 0& \mathrm{i}\\ 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 1\end{array}\right].$

Then, using the initial distribution of π(0) = (1,0,0,0,0), the land distribution equally X works out to exist

$\pi =\underset{k\to \infty }{lim}\pi \left(k\right)=\underset{k\to \infty }{lim}\pi \left(0\right)P^{\mathrm{one\; thousand}}=\left[\begin{array}{ccccc}0& 0& 0& 0& 1\end{array}\right].$

In Example 9.half-dozen, information technology was seen that as grand → ∞, the k-step transition probability matrix approached that of a matrix whose rows were all identical. In that case, the limiting product lim _{chiliad → ∞} π(0)P^k is the same regardless of the initial distribution π(0). Such a Markov chain is said to have a unique steady-land distribution, π. Information technology should exist emphasized that not all Markov bondage have a steady-land distribution. For example, the Poisson counting procedure of Example 9.i clearly does non, since whatsoever counting process is a monotonie non-decreasing function of time and, therefore, information technology is expected that the distribution should skew toward larger values equally time progresses.

This concept of a steady-state distribution can exist viewed from the perspective of stationarity. Suppose at time k, the process has some distribution, π(k). The distribution at the side by side time instant is then π(k+ 1) = π(k)P. If π(k) = π(k+ i), then the procedure has reached a point where the distribution is stationary (independent of fourth dimension). This stationary distribution, π, must satisfy the relationship

(9.xiii) $\pi =\pi P.$

In other words, π (if it exists) is the left eigenvector of the transition probability matrix P, that corresponds to the eigenvalue λ = i. The next example shows that this eigenvector is not ever unique.

Example ix.eight (The Gambler's Ruin revisited)

Suppose a certain gambler has $five and plays confronting another player (the house). The gambler decides that he volition play until he either doubles his coin or loses it all. Suppose the house has designed this game of risk so that the gambler will win with probability p = 0.45 and the house will win with probability q = 0.55. Let 10_g exist the corporeality of money the gambler has afterwards playing the game k times. The transition probability matrix for this Markov chain is

$p=\left[\begin{array}{ccccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0.55& 0& 0.45& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0.55& 0& 0.45& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0.55& 0& 0.45& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0.55& 0& 0.45& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0.55& 0& 0.45& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0.55& 0& 0.45& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0.55& 0& 0.45& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0.55& 0& 0.45& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0.55& 0& 0.45\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\end{array}\right]$

This matrix has (2) repeated eigenvalues of λ = 1, and the corresponding eigenvectors are [10 0 0 0 0 0 0 0 0 0] and [00 0 0 0 0 0 0 0 0 l] Note that any linear combination of these will besides be an eigenvector. Therefore, any vector of the form [p00000000l–p] is a left eigenvector off and hence there is no unique stationary distribution for this Markov chain. For this case, the limiting form of the country distribution of the Markov chain depends on the initial distribution. The limiting form of P^k can easily be found to be

$\underset{k\to \infty }{lim}{p}^k\left[\begin{array}{ccccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0.9655& 0& 0& 0& 0& 0& 0& 0& 0& 0& \text{0}\text{.0345}\\ 0.9233& 0& 0& 0& 0& 0& 0& 0& 0& 0& \text{0}\text{.0767}\\ \text{0}\text{.8717}& 0& 0& 0& 0& 0& 0& 0& 0& 0& \text{0}\text{.1283}\\ \text{0}\text{.8087}& 0& 0& 0& 0& 0& 0& 0& 0& 0& \text{0}\text{.1913}\\ \text{0}\text{.7317}& 0& 0& 0& 0& 0& 0& 0& 0& 0& \text{0}\text{.2683}\\ \text{0}\text{.6376}& 0& 0& 0& 0& 0& 0& 0& 0& 0& \text{0}\text{.3624}\\ \text{0}\text{.5225}& 0& 0& 0& 0& 0& 0& 0& 0& 0& \text{0}\text{.4775}\\ \text{0}\text{.3819}& 0& 0& 0& 0& 0& 0& 0& 0& 0& \text{0}\text{.6181}\\ \text{0}\text{.2101}& 0& 0& 0& 0& 0& 0& 0& 0& 0& \text{0}\text{.7899}\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right]$

Using the initial distribution π(0) = [0 000010000 0] (that is, the gambler starts off in country 5), and then it is seen that the steady-country distribution is $\underset{k\to \infty }{lim}\pi \left(g\right)=\left[\begin{array}{ccccccccccc}0.7317& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0.2683\end{array}\right]$ . So, when the gambler starts with $five, he has most a 73% chance of losing all of his money and about a 27% chance of doubling his money.

As seen in Case 9.7, with some Markov chains, the limiting form of P ^k (as k → ∞) does not necessarily converge to a matrix whose rows are all identical. In that case, the limiting grade of the state distribution will depend on the starting distribution. In the example of the gambler's ruin problem, nosotros probably could take guessed this behavior. If the gambler had started with very piddling coin, nosotros would expect him to end up in the state of ruin with very high probability; whereas, if the gambler was very wealthy and the house had very piffling coin, we would expect a much greater chance of the gambler somewhen breaking the business firm. Accordingly, our intuition tells us that the probability distribution of the gamblers ultimate state should depend on the starting state.

In general, there are several different manners in which a Markov chain's country distribution can bear every bit k → ∞ In some cases, $\underset{\mathrm{thousand}\to \infty }{lim}\pi \left(k\right)$ does non be. Such would exist the example when the process tends to oscillate between two or more states. A second possibility, as in Case ix.7, is that $\underset{\mathrm{one\; thousand}\to \infty }{lim}\pi \left(\mathrm{yard}\right)$ does in fact converge to a fixed distribution, just the class of this limiting distribution depends on the starting distribution. The last case is when, $\underset{k\to \infty }{lim}\pi \left(\mathrm{yard}\right)=\pi$ That is the land distribution converges to some fixed distribution, π, and the grade of π is contained of the starting distribution. Here, the transition probability matrix, P, will have a single (not repeated) eigenvalue at λ = 1, and the corresponding eigenvector (properly normalized) volition exist the steady-land distribution, π. Furthermore, the limiting course of P^k will be ane whose rows are all identical and equal to the steady-state distribution, π. In the next department, we look at some conditions that must be satisfied for the Markov chain to accomplish a unique steady-state distribution.

Example 9.9

In this example, nosotros provide the MATLAB code to simulate the distribution of the queue length of the taxi stand described in Example ix.4. For this example, we take the number of arrivals per time unit, X, to be a Poisson random variable whose PMF is

$P_{\mathrm{10}}\left(k\right)=\frac{{\lambda }^{\mathrm{thou}}{\mathrm{eastward}}^{-\lambda }}{k!}.$

Recollect that in the taxi stand, 1 client is served per time unit (bold there is a least one client in the queue waiting to be served). The following code can be used to estimate and plot the PMF of the queue length. The average queue length was also calculated to be three.36 customers for an arrival rate of λ = 0.85 customers per fourth dimension unit. Effigy 9.3 shows a histogram of the PMF of the queue length for the same arrival charge per unit.

Effigy ix.3. Histogram of the queue length for the taxi stand up of Example 9.4 assuming a Poisson arrival procedure with an average arrival charge per unit of 0.85 arrivals per time unit.

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Reconstructing the Phylogeny

Grady Weyenberg , Ruriko Yoshida , in Algebraic and Detached Mathematical Methods for Modern Biological science, 2015

12.2.1.2 Jukes-Cantor 1969

The simplest model of DNA evolution is the JC (or JC69) model. In addition to the Markov process assumptions, it likewise assumes that there is but a single transition rate that governs all types of commutation [38]. This supposition implies a transition matrix of the form,

$Q=\left(\begin{array}{llll}\hfill -\mathrm{one}\hfill & \hfill 1/\mathrm{iii}\hfill & \hfill 1/3\hfill & \hfill \mathrm{ane}/\mathrm{iii}\hfill \\ \hfill 1/\mathrm{iii}\hfill & \hfill -\mathrm{ane}\hfill & \hfill \mathrm{i}/3\hfill & \hfill \mathrm{ane}/3\hfill \\ \hfill 1/3\hfill & \hfill 1/\mathrm{iii}\hfill & \hfill -1\hfill & \hfill 1/\mathrm{iii}\hfill \\ \hfill 1/\mathrm{iii}\hfill & \hfill \mathrm{i}/\mathrm{three}\hfill & \hfill 1/3\hfill & \hfill -1\hfill \end{array}\right).$

We index the rows and columns of matrices for nucleotide models in the following order: adenine, guanine, cytosine, and thymine/uracil. In this example information technology does not matter, but in subsequent models this becomes important.

Below, nosotros implement a function that calculates the transition probability matrix function P(d) and employ information technology to gauge the stationary distribution for the JC model. The characteristic timescale of the system (i.eastward., the parameter of the time t in the continuous fourth dimension Markov chain) is 1, and the probability matrix has converged quite well at a distance d = 100. (Beware: Attempting to evaluate the function at Inf, the R symbol representing infinity, appears to cause R to enter an infinite loop.)

library(expm)

DNA.alphabet <-c("a", "m", "c", "t")

Q <-matrix(1/three, nrow = 4, ncol = iv)# create 4x4 matrix and fill with ane/iii

diag(Q) <- -1 # set diagonal to -1

colnames(Q) <-rownames(Q) <- DNA.alphabet

P <-function(d)expm(d * Q) # Implement P(d)

P(100) # Characteristic timescale   is 1, and so 100 is ''close'' to infinity.

##   a   m   c   t

## a 0.25 0.25 0.25 0.25

## g 0.25 0.25 0.25 0.25

## c 0.25 0.25 0.25 0.25

## t 0.25 0.25 0.25 0.25

Thus, we observe that the stationary distribution for the JC model is compatible on all possible character states, π = (ane/4,1/4,1/four,1/iv). Under this model, because the Markov concatenation is time reversible, we can remember that the initial probability of this model is π = (1/four,1/4,1/4,ane/4).

Another interesting and useful thing happens if a Markov process has a stationary distribution π, and the following relationship with the charge per unit matrix Q is true.

(12.one) $\begin{array}{l}\hfill {\pi }_i{Q}_{ij}={\pi }_j{Q}_{ji},\forall i,j\end{array}$

This is known as the detailed balance equation, and when it holds the process is reversible. If a process is reversible, it means that once information technology has converged to the equilibrium distribution, the "arrow of time" disappears: there is no way to make up one's mind if a character'due south procedure Ten(d) is indexed in the proper direction, or if the time index has been reversed.

Reversibility turns out to exist a desirable belongings if we want to study molecular evolution. Consider a most-recent common ancestor, with two daughter lineages. If the processes describing sequence development are reversible, then we exercise not need to consider the two lineages separately. The reversibility ways that we tin care for the daughters as endpoints of one long Markov chain that goes "up" one lineage to the MRCA, and back "down" the other lineage. If the model was non reversible, then this would not be a valid simplification. We would need to model each lineage discretely, in the correct orientation from ancestor to descendant. It turns out that JC, along with all the other models we will hash out, is reversible.

Do 12.4

Show that the JC rate matrix satisfies the detailed balance condition.

Recall that P _ij (d) represents the probability of ending in country j when starting in land i, if the distance between the sequences is d. Considering the model is reversible, nosotros can use this matrix to observe the likelihood of observing a particular pair of characters in a sequence alignment, assuming the sequences are separated by altitude d. If the sites in the alignment are independent, then the likelihood of the unabridged alignment is the product of the private site likelihoods. That is, if the character pair i,j occurs in the alignment n _ij times, so the likelihood of the entire alignment is given by

$\mathrm{50}(d)=\prod _{\forall i,j}{P}_{ij}{(d)}^{{\mathrm{due\; north}}_{ij}}.$

For a variety of reasons, both theoretical and relating to numerical stability, it is more common to work with the log-likelihood

(12.two) $\begin{array}{l}\hfill l(d)=log\mathrm{50}(d)=\sum _{\forall i,j}{n}_{ij}log{P}_{ij}(d)\end{array}$

Readers familiar with statistical methods will virtually likely conceptualize the side by side step: we are now in a position to employ a sequence alignment to estimate the evolutionary altitude separating the two sequences. We will do this past attempting to observe the distance d which maximizes the likelihood, given the observed alignment data.

First, we implement a function that counts the occurrences of the various substitutions betwixt two sequences, and another that calculates the log-likelihood, given d and the substitution counts.

nMatrix <-function(alignment, i, j) {

  alignment <- alignment[c(i, j), ]

  x <-utilise(alignment, 2:1,function(y)as.DNAbin(DNA.alphabet) %in%

  y)

dimnames(10) <-list(Deoxyribonucleic acid.alphabet,Zip,labels(alignment))

tcrossprod(x[, , i], x[, , 2])

}

JC.lnL <-function(d, nm)sum(nm *log(P(d)))

Applying the nMatrix function to the hemoglobin information, the commutation count matrix for the HBA1 and HBA2 sequences is obtained.

A plot of the JC log-likelihood for the HBA1 and HBA2 alignment is shown in Figure 12.5.

Figure 12.5. Log-likelihood part for the JC model and the sequences HBA1 and HBA2.

N12 <-nMatrix(hemo.aligned, "HBA1", "HBA2")

### substitution pair count for HBA1 & HBA2

N12

##   a   g   c   t

## a 95   2   one   0

## thousand   2 150   0   3

## c   2   3 207   4

## t   0   ii   half dozen 98

We tin now employ numerical optimization to find the maximum likelihood (ML) guess for the distance separating the two sequences.

optimize(JC.lnL, 0:1, nm = N12, maximum = TRUE)

## $maximum

## [1] 0.04478

##

## $objective

## [one] -130.3

We estimate that each character site in the alignment has experienced about 0.045 substitutions, total, along both lineages leading from the mutual ancestral hemoglobin gene to the modern day HBA1 and HBA2.

In fact, it is fairly like shooting fish in a barrel to obtain a closed-form solution for the JC trouble. This solution is presented in many other books (east.g., [4]), and we volition not hash out it at length here, except to note the final equation for calculating the JC distance between two sequences:

(12.3) $\begin{array}{l}\hfill d=-\frac{3}{4}ln\left(1-\frac{4}{\mathrm{three}}p\right).\end{array}$

Hither, p is the proportion of sites in the alignment that have unlike characters.

The dist.dna function from the ape packet computes the JC distance using the airtight-form solution and agrees with our numerical solution.

dist.dna(hemo.aligned[c("HBA1", "HBA2"), ], model = "JC")

Do 12.five

The matrix exponential of the JC rate matrix is given by the formula

$P_{ij}(d)=\left\{\begin{array}{ll}0.25+0.75{\text{e}}^{-4d/3}& \text{if}i=j,\\ 0.25-0.25{\text{e}}^{-4d/\mathrm{three}}& \text{if}i\ne j.\end{array}\right.$

Show that Eq. (12.3) solves the likelihood optimization trouble for the JC model.

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Special Random Processes

Oliver C. Ibe , in Fundamentals of Applied Probability and Random Processes (Second Edition), 2014

Section 12.7 Discrete-Time Markov Bondage

12.36

Determine the missing elements denoted past x in the post-obit transition probability matrix:

$\mathit{P}=\left[\begin{array}{cccc}\hfill \mathit{x}\hfill & \hfill \mathrm{ane}/3\hfill & \hfill 1/3\hfill & \hfill \mathrm{i}/\mathrm{iii}\hfill \\ \hfill 1/10\hfill & \hfill \mathit{x}\hfill & \hfill 1/5\hfill & \hfill 2/\mathrm{five}\hfill \\ \hfill \mathit{x}\hfill & \hfill \mathit{x}\hfill & \hfill \mathit{x}\hfill & \hfill 1\hfill \\ \hfill 3/\mathrm{five}\hfill & \hfill \mathrm{ii}/\mathrm{five}\hfill & \hfill \mathit{10}\hfill & \hfill \mathit{x}\hfill \end{array}\right]$

12.37

Draw the country-transition diagram for the Markov chain with the post-obit transition probability matrix.

$\mathit{P}=\left[\begin{array}{cccc}\hfill 1/2\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1/2\hfill \\ \hfill 1/\mathrm{two}\hfill & \hfill 1/2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill \mathrm{ane}/4\hfill & \hfill 0\hfill & \hfill \mathrm{i}/2\hfill & \hfill \mathrm{i}/4\hfill \\ \hfill 0\hfill & \hfill \mathrm{one}/\mathrm{ii}\hfill & \hfill \mathrm{ane}/4\hfill & \hfill \mathrm{i}/4\hfill \end{array}\right]$

12.38

Consider a Markov chain with the state-transition diagram shown in Figure 12.21.

a.

Give the transition probability matrix.

b.

Identify the recurrent states.

c.

Place the transient states.

Effigy 12.21. Figure for Problem 12.38

12.39

Consider the Markov chain with the land-transition diagram shown in Figure 12.22.

a.

Listing the transient states, the recurrent states, and the periodic states.

b.

Identify the members of each chain of recurrent states.

c.

Requite the transition probability matrix of the process.

d.

Given that the process starts in state i, either determine the numerical value of the probability that the process is in state eight later an infinitely big number of transitions or explain why this quantity does not exist.

Figure 12.22. Figure for Problem 12.39

12.40

Consider the three-state Markov chain shown in Figure 12.23:

a.

Place the transient states, the recurrent states, the periodic states, and the members of each chain of recurrent states.

b.

Either make up one's mind the limiting-state probabilities or explicate why they do not exist.

c.

Given that the process is currently in land one, determine P[A], the probability that information technology will be in state 3 at least once during the next two transitions.

Effigy 12.23. Effigy for Problem 12.40

12.41

Consider the Markov chain shown in Figure 12.24:

a.

Which states are transient?

b.

Which states are periodic?

c.

Does state 3 have a limiting-state probability? If so, determine this probability.

d.

Assuming that the process begins in state four, determine the z-transform of the PMF of K, where K is the number of trials up to and including the trial in which the process enters state 2 for the second time.

Figure 12.24. Figure for Problem 12.41

12.42

Find the limiting-country probabilities associated with the post-obit transition probability matrix:

$\mathit{P}=\left[\begin{array}{ccc}\hfill 0.4\hfill & \hfill \mathrm{0.three}\hfill & \hfill 0.3\hfill \\ \hfill \mathrm{0.three}\hfill & \hfill 0.4\hfill & \hfill 0.3\hfill \\ \hfill 0.3\hfill & \hfill 0.3\hfill & \hfill 0.4\hfill \end{array}\right]$

12.43

Consider the following transition probability matrix:

$\mathit{P}=\left[\begin{array}{ccc}\hfill 0.6\hfill & \hfill 0.2\hfill & \hfill 0.2\hfill \\ \hfill \mathrm{0.three}\hfill & \hfill 0.4\hfill & \hfill \mathrm{0.iii}\hfill \\ \hfill 0.0\hfill & \hfill \mathrm{0.three}\hfill & \hfill 0.7\hfill \end{array}\right]$

a.

Give the state-transition diagram.

b.

Given that the process is currently in country 1, what is the probability that it will be in state 2 at the end of the third transition?

c.

Given that the process is currently in state 1, what is the probability that the first time information technology enters state 3 is the quaternary transition?

12.44

Consider the following social mobility problem. Studies indicate that people in a society can exist classified as belonging to the upper class (state 1), middle class (land ii), and lower grade (state 3). Membership in any form is inherited in the post-obit probabilistic manner. Given that a person is raised in an upper-course family, he will take an upper-form family with probability 0.45, a centre-class family with probability 0.48, and a lower-class family unit with probability 0.07. Given that a person is raised in a middle-class family, he will take an upper-class family unit with probability 0.05, a middle-form family with probability 0.70, and a lower-grade family with probability 0.25. Finally, given that a person is raised in a lower-class family, he volition have an upper-class family with probability 0.01, a eye-grade family with probability 0.50, and a lower-class family with probability 0.49. Determine the following:

a.

The land-transition diagram of the procedure

b.

The transition probability matrix of the process

c.

The limiting-land probabilities. Translate what they mean to the layperson.

12.45

A taxi driver conducts his business in three different towns 1, two, and 3. On any given day, when he is in boondocks 1, the probability that the next passenger he picks upwards is going to a identify in boondocks 1 is 0.iii, the probability that the side by side rider he picks upwards is going to town 2 is 0.2, and the probability that the next passenger he picks up is going to town 3 is 0.5. When he is in town 2, the probability that the next rider he picks up is going to town 1 is 0.ane, the probability that the side by side passenger he picks up is going to boondocks ii is 0.8, and the probability that the next rider he picks up is going to boondocks 3 is 0.1. When he is in town 3, the probability that the side by side passenger he picks up is going to town 1 is 0.4, the probability that the next passenger he picks up is going to town 2 is 0.4, and the probability that the next passenger he picks up is going to boondocks 3 is 0.2.

a.

Determine the country-transition diagram for the process.

b.

Give the transition probability matrix for the process.

c.

What are the limiting-state probabilities?

d.

Given that the taxi commuter is currently in town two and is waiting to pick up his first customer for the mean solar day, what is the probability that the commencement time he picks upwardly a passenger to town 2 is when he picks up his third passenger for the twenty-four hour period?

e.

Given that he is currently in town 2, what is the probability that his third passenger from now will be going to town one?

12.46

New England fall weather can exist classified every bit sunny, cloudy, or rainy. A student conducted a detailed report of the atmospheric condition design and came upward with the following conclusion: Given that information technology is sunny on whatsoever given day, then on the following day it will be sunny again with probability 0.5, cloudy with probability 0.3 and rainy with probability 0.two. Given that it is cloudy on any given mean solar day, then on the post-obit day it volition be sunny with probability 0.4, cloudy again with probability 0.3 and rainy with probability 0.3. Finally, given that information technology is rainy on any given twenty-four hour period, then on the following solar day information technology will be sunny with probability 0.2, cloudy with probability 0.5 and rainy again with probability 0.3.

a.

Give the country-transition diagram of New England autumn weather with the state "sunny" as state 1, the country "cloudy" as state two, and the state "rainy" as state 3.

b.

Using the same convention as in function (a), give the transition probability matrix of New England autumn atmospheric condition.

c.

Given that it is sunny today, what is the probability that it volition be sunny four days from now?

d.

Determine the limiting-state probabilities of the weather.

12.47

A educatee went to a gambling casino with $3. He wins $1 at each round with a probability p and loses $1 with a probability one   − p. Being a very cautious player, the student has decided to stop playing when he doubles his original $3 (i.east., when he has a total of $six) or when he loses all his coin.

a.

Give the land-transition diagram of the procedure.

b.

What is the probability that he stops after beingness ruined (i.due east., he lost all his money)?

c.

What is the probability that he stops after he has doubled his original amount?

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Markov Chains

Sheldon Grand. Ross , in Introduction to Probability Models (Twelfth Edition), 2019

4.iv.1 Limiting Probabilities

In Example four.8 we considered a two-country Markoff chain with transition probability matrix

$\mathbf{P}=\left(\begin{array}{cc}\hfill 0.7\hfill & \hfill \mathrm{0.iii}\hfill \\ \hfill 0.4\hfill & \hfill \mathrm{0.six}\hfill \end{array}\right)$

and showed that

${\mathbf{P}}^{(4)}=\left(\begin{array}{cc}\hfill 0.5749\hfill & \hfill 0.4251\hfill \\ \hfill 0.5668\hfill & \hfill 0.4332\hfill \end{array}\right)$

From this information technology follows that ${\mathbf{P}}^{(8)}={\mathbf{P}}^{(4)}\cdot {\mathbf{P}}^{(4)}$ is given (to three significant places) by

${\mathbf{P}}^{(8)}=\left(\begin{array}{cc}\hfill 0.571\hfill & \hfill 0.429\hfill \\ \hfill 0.571\hfill & \hfill 0.429\hfill \end{array}\right)$

Note that the matrix ${\mathbf{P}}^{(8)}$ is almost identical to the matrix ${\mathbf{P}}^{(4)},$ and that each of the rows of ${\mathbf{P}}^{(\mathrm{viii})}$ has almost identical values. Indeed, it seems that $P_{ij}^{\mathrm{north}}$ is converging to some value as $\mathrm{northward}\to \infty$ , with this value not depending on i. Moreover, in Example iv.22 we showed that the long-run proportions for this chain are ${\pi }_0=\mathrm{iv}/\mathrm{vii}\approx .571,{\pi }_1=\mathrm{three}/7\approx .429$ , thus making it announced that these long-run proportions may also exist limiting probabilities. Although this is indeed the case for the preceding chain, information technology is not always truthful that the long-run proportions are as well limiting probabilities. To see why not, consider a two-country Markov chain having

$P_{0,1}=P_{1,0}=1$

Because this Markov concatenation continually alternates between states 0 and 1, the long-run proportions of fourth dimension information technology spends in these states are

${\pi }_0={\pi }_{\mathrm{i}}=\mathrm{one}/2$

However,

$P_{0,0}^{\mathrm{north}}=\{\begin{array}{cc}1,\hfill & \text{if}\mathrm{northward}\text{is fifty-fifty}\hfill \\ 0,\hfill & \text{if}n\text{is odd}\hfill \end{array}$

and so $P_{0,0}^{\mathrm{north}}$ does non accept a limiting value as northward goes to infinity. In general, a concatenation that can merely render to a country in a multiple of $d>\mathrm{i}$ steps (where $d=\mathrm{two}$ in the preceding instance) is said to be periodic and does non have limiting probabilities. However, for an irreducible chain that is not periodic, and such chains are chosen aperiodic, the limiting probabilities will ever be and volition not depend on the initial state. Moreover, the limiting probability that the chain will be in land j will equal ${\pi }_j,$ the long-run proportion of time the chain is in state j. That the limiting probabilities, when they exist, will equal the long-run proportions tin can be seen by letting

${\alpha }_j=\underset{n\to \infty }{\lim }P({\mathrm{10}}_{\mathrm{north}}=j)$

and using that

$P({\mathrm{10}}_{\mathrm{northward}+\mathrm{ane}}=j)=\sum _{i=0}^{\infty }P(X_{n+1}=j|{\mathrm{Ten}}_n=i)P(X_n=i)=\sum _{i=0}^{\infty }{P}_{ij}P(X_{\mathrm{north}}=i)$

and

$1=\sum _{i=0}^{\infty }P(X_{\mathrm{due\; north}}=i)$

Letting $n\to \infty$ in the preceding two equations yields, upon assuming that nosotros can bring the limit within the summation, that

$\begin{array}{cc}\hfill {\alpha }_j=& \sum _{i=0}^{\infty }{\alpha }_i{P}_{ij}\hfill \\ \hfill 1=& \sum _{i=0}^{\infty }{\alpha }_i\hfill \end{array}$

Hence, $\{{\alpha }_j,j⩾0\}$ satisfies the equations for which $\{{\pi }_j,j⩾0\}$ is the unique solution, showing that ${\alpha }_j={\pi }_j,j⩾0.$

An irreducible, positive recurrent, aperiodic Markov chain is said to exist ergodic.

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Stochastic Processes

J. MEDHI , in Stochastic Models in Queueing Theory (Second Edition), 2003

one.6.two Equivalence of the 2 limiting forms

Let {Y_northward, due north ≥ 0} exist an irreducible and aperiodic chain with finite state infinite S and TPM P. Then from the ergodic theorem (Theorem 1.1) nosotros go that

$\begin{array}{cc}\underset{n\to \infty }{\lim }{p}_{ij}^{(n)}={\upsilon }_j,& i,j\in S\end{array}$

exists and is independent of i, and 5 = {ν ₁ , ν ₂ ,…} is the invariant distribution given past

(1.6.5) $\begin{array}{cc}\mathrm{Five}P=\mathrm{Five},& 5\mathrm{eastward}=V.\end{array}$

The Markov process {X(t) = Y _{Due north(t)} , t ≥ 0} is besides aperiodic and irreducible and has the same land space S. From the ergodic theorem (Theorem 1.4), nosotros go that

$\underset{t\to \infty }{\lim }{p}_{ij}(t)=u_j$

exists and is independent of i. Further, U = {u _one , u ₂ ,…} is a probability vector and U is given as the solution of

(1.6.6) $\begin{array}{c}UQ=\begin{array}{cc}0,& U\mathrm{eastward}=\mathrm{one},\end{array}\\ UQ=0\leftrightarrow U[\lambda (P-I)]=0\\ \leftrightarrow UP=U.\end{array}$

Thus, from (1.6.5) and (1.6.6), we get

$U\equiv \mathrm{Five}.$

In other words,

(1.6.vii) $\begin{array}{cc}\underset{t\to \infty }{\lim }{p}_{ij}(t)=\underset{n\to \infty }{\lim }{p}_{ij}^{(n)},& i,j\in S.\end{array}$

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Renewal Phenomena

Mark A. Pinsky , Samuel Karlin , in An Introduction to Stochastic Modeling (Fourth Edition), 2011

Exercises

7.2.1

Let {X_{due north} ; due north = 0, 1, …} be a 2-country Markov chain with the transition probability matrix

$P=\begin{array}{l}0\\ 1\end{array}\Vert \begin{array}{ll}0& 1\\ 1-a& a\\ b& 1-b\end{array}\Vert .$

State 0 represents an operating land of some system, while land 1 represents a repair land. We presume that the process begins in state X ₀ = 0, and so the successive returns to country 0 from the repair land course a renewal process. Make up one's mind the hateful duration of one of these renewal intervals.

7.2.2

A sure type component has two states: 0 = OFF and i = OPERATING. In state 0, the process remains at that place a random length of fourth dimension, which is exponentially distributed with parameter α, and and so moves to state i. The time in state one is exponentially distributed with parameter β, after which the process returns to country 0.

The organisation has ii of these components, A and B, with distinct parameters:

Component	Operating Failure Rate	Repair Charge per unit
A	βA	αA
B	βB	αB

In order for the organisation to operate, at least one of components A and B must exist operating (a parallel organisation). Presume that the component stochastic processes are independent of one another. Consider the successive instants that the system enters the failed state from an operating state. Use the memoryless holding of the exponential distribution to fence that these instants form a renewal process.

vii.2.3

Summate the mean number of renewals M (n) = Eastward [N (n)] for the renewal process having interoccurrence distribution

$p_1=0.4,p_2=\mathrm{0.ane},p_3=0.3,p_{\mathrm{iv}}=0.2$

for n = 1, 2, …, 10. Also summate u_northward = M (north) − M (n − 1).

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Markov Chains

Sheldon M. Ross , in Introduction to Probability Models (Tenth Edition), 2010

Proposition four.half-dozen

Consider an irreducible Markov chain with transition probabilities P_ij . If we can find positive numbers π _i , i ≥ 0, summing to ane, and a transition probability matrix Q = [Q_ij ] such that

(4.29) ${\pi }_i{\text{P}}_{ij}={\pi }_j{Q}_{ji}$

and so the Q_ij are the transition probabilities of the reversed concatenation and the π _i are the stationary probabilities both for the original and reversed chain.

The importance of the preceding proposition is that, by thinking backward, nosotros tin sometimes gauge at the nature of the reversed chain then use the prepare of Equations (iv.29) to obtain both the stationary probabilities and the Q_ij .

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