# Last step in an absorbing Markov chain

The preceding three posts are devoted to a problem involving absorbing Markov chains (finding the mean time to absorption and the probability of absorption). The links to these three posts are here, here and here. One method in solving this problem is to use the fundamental matrix. The same method can be used to solve a problem involving the last step before being absorbed. Here are two examples.

Example 1
Six coins are tossed all at once. The coins randomly fall heads or tails. For the coins that fall tails, you pick up and toss again. The process continues in such a way that after each toss, you pick up the coins that fall tails and toss again. The process stops until all coins show heads. Let $Y$ be the number of coins in the last toss. Determine the probability distribution of the random variable $Y$.

Note that $Y$ can range from 6 down to 1. If the first toss results in 6 heads, there is no need to toss again, meaning that $Y=6$, which is not very likely. The problem is to determine how likely it is for $Y$ to be a given value.

Example 2
A maze has 7 areas with two of the areas being food sources (area 5 and area 7)

A maze with two food sources (areas 5 and 7)

A mouse is placed into this maze. When the mouse is in an area with food, it stays there and does not leave. When the mouse is in an area with no food, it moves into the adjacent areas at random. In other words, if the area has $k$ exits, the mouse moves to one of these $k$ areas with probability $1/k$.

If the mouse is in one of the non-food areas, then the mouse will be absorbed into one of the food areas. We are interested in identifying the non-food area before reaching a food area. The areas adjacent to a food area would be 2, 4 and 6. These would be the entry points for food. When the mouse finds food, how likely is the entry point to be area 2, area 4 or area 6? More specifically, if the mouse is in area $i$ where $i=1,2,3,4,6$, determine the probability that the mouse transitions into a food area from area $k$ where $k=2,4,6$.

A common idea for these two examples and others like them is that we are interested in locating the transient state from which the process enters an absorbing state. We would like to find the probabilities of the various transient states being the entry point to absorption given an initial state. We discuss these two examples after demonstrating the method using a smaller example.

Illustrative Example

A Markov chain has 4 states (0, 1, 2 and 3). The process cycles among these states according to the following transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 \cr 0 & 1 & 0 & 0 & 0 \cr 1 & 0.3 & 0.4 & 0.2 & 0.1 \cr 2 & 0.1 & 0.2 & 0.3 & 0.4 \cr 3 & 0 & 0 & 0 & 1 \cr } \qquad$

There are two absorbing states (0 and 3) and two transient states (1 and 2). If the process starts at a transient state, the process will eventually be absorbed (be transitioned into an absorbing state). The question we want to answer is this. What is the transient state from which the process enters an absorbing state? In this example, the process can enter an absorbing state from state 1 or state 2. Given that the process is in state 1 initially, what is the probability that the process enters absorption from state 1? Likewise from state 2? The same question can be asked if the process starts at state 2 initially.

Let $\{ X_n,n=0,1,2,\cdots \}$ be the chain described by the above matrix. Let $T=\text{min}\{ n \ge 0: X_n=0 \text{ or } X_n=3 \}$. The quantity $T$ is the random time to absorption. The quantity $T-1$ is the time period before absorption. Thus $X_{T-1}$ is the last transient state before absorption. We calculate the following probabilities.

$P(X_{T-1}=1 \lvert X_0=1)$

$P(X_{T-1}=2 \lvert X_0=1)$

$P(X_{T-1}=1 \lvert X_0=2)$

$P(X_{T-1}=2 \lvert X_0=2)$

The first two probabilities form the conditional distribution of the last transient state given that the initial state is 1. The last two probabilities form the conditional distribution of the last transient state given that the initial state is 2. To find these probabilities, it will be helpful to perform calculations involving the fundamental matrix.

$Q=\left[\begin{array}{cc} 0.4 & 0.2 \\ 0.2 & 0.3 \\ \end{array}\right] \ \ \ \ \ \ \ \ I-Q=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}\right]- \left[\begin{array}{cc} 0.4 & 0.2 \\ 0.2 & 0.3 \\ \end{array}\right]= \left[\begin{array}{cc} 0.6 & -0.2 \\ - 0.2 & 0.7 \\ \end{array}\right]$

$W=(I-Q)^{-1}= \left[\begin{array}{cc} 0.6 & -0.2 \\ - 0.2 & 0.7 \\ \end{array}\right]^{-1}= \left[\begin{array}{cc} 35/19 & 10/19 \\ 10/19 & 30/19 \\ \end{array}\right]= \left[\begin{array}{cc} W_{11} & W_{12} \\ W_{21} & W_{22} \\ \end{array}\right]$

See here for more information on the calculation. The matrix $Q$ is the part of $\mathbf{P}$ consisting of the transient states. The matrix $I-Q$ is the difference between the identity matrix and $Q$. The matrix $W=(I-Q)^{-1}$, the inverse of $I-Q$, is the fundamental matrix for the absorbing Markov chain.

The matrix $W=(I-Q)^{-1}$ gives the mean times spent in transient states. For example, if the process starts in state 1, it spend 35/19 = 1.842 periods in state 1 and 10/19 = 0.526 periods in state 2. In total, it spends on average 45/19 = 2.368 periods in transient states before being absorbed (if the initial state is 1). The following shows the calculation for the case of initial state being 1.

$\displaystyle P(X_{T-1}=1 \lvert X_0=1)=W_{11} \times P_{10}+W_{11} \times P_{13}=\frac{35}{19} (0.3+0.1)=\frac{14}{19}=0.7368$

$\displaystyle P(X_{T-1}=2 \lvert X_0=1)=W_{12} \times P_{20}+W_{12} \times P_{23}=\frac{10}{19} (0.1+0.4)=\frac{5}{19}=0.2632$

The calculation shows that if the initial state is 1, it is much more likely for the process to enter an absorbing state from state 1 than from state 2 (almost 74% chance). Before giving an indication why the calculation works, here’s the calculation for the initial state being 2.

$\displaystyle P(X_{T-1}=1 \lvert X_0=2)=W_{21} \times P_{10}+W_{21} \times P_{13}=\frac{10}{19} (0.3+0.1)=\frac{4}{19}=0.2105$

$\displaystyle P(X_{T-1}=2 \lvert X_0=2)=W_{22} \times P_{20}+W_{22} \times P_{23}=\frac{30}{19} (0.1+0.4)=\frac{15}{19}=0.7895$

If the process starts from state 2 instead, it is more likely (almost 79% chance) that the process enters an absorbing state from state 2.

Here’s the thought process for obtaining these probabilities. Consider $P(X_{T-1}=1 \lvert X_0=1)$. The starting transient state is 1 and the ending transient state is also 1. Then we look at the mean time in transient state $W_{11}$. From the ending transient state 1, the process can go into the absorbing state 0 or 3. Thus the desired probability is $W_{11} \times P_{10}+W_{11} \times P_{13}=W_{11} (P_{10}+P_{13})$.

If the starting state is 1 and the ending transient state is 2, we would take the sum of mean time $W_{12}$ times $P_{2k}$ for $k=0,3$. Thus $P(X_{T-1}=2 \lvert X_0=2)$ would be $W_{12} \times P_{20}+W_{12} \times P_{23}=W_{12} (P_{20}+P_{23})$.

In general $P(X_{T-1}=j \lvert X_0=i)$ is the sum of $W_{ij} \times P_{jk}$ over all absorbing states $k$ from which it is possible to go from $j$ to $k$. To get a sense of why this idea works, let’s look at the the probabilities of absorption calculated from the fundamental matrix $W$.

$Q \times R= \left[\begin{array}{cc} W_{11} & W_{12} \\ W_{21} & W_{22} \\ \end{array}\right] \times \left[\begin{array}{cc} P_{10} & P_{13} \\ P_{20} & P_{23} \\ \end{array}\right]= \left[\begin{array}{ccc} W_{11} \cdot P_{10}+W_{12} \cdot P_{20} & \text{ } & W_{11} \cdot P_{13}+W_{12} \cdot P_{23} \\ W_{21} \cdot P_{10}+W_{22} \cdot P_{20} & \text{ } & W_{21} \cdot P_{13}+W_{22} \cdot P_{23} \\ \end{array}\right]$

To make it easier to see, we rewrite the last matrix on the right as follows.

$Q \times R = \bordermatrix{ & 0 & \text{ } & 3 \cr 1 & W_{11} \cdot P_{10}+W_{12} \cdot P_{20} & \text{ } & W_{11} \cdot P_{13}+W_{12} \cdot P_{23} \cr 2 & W_{21} \cdot P_{10}+W_{22} \cdot P_{20} & \text{ } & W_{21} \cdot P_{13}+W_{22} \cdot P_{23} \cr } \qquad$

This is a 2 x 2 matrix indicating the probability of absorption based on the initial states (the rows) and the absorbing states (the columns). The first column gives the probabilities of absorption into state 0 and the second column gives the probabilities of absorption into state 3. Each of the four elements is a sum of two quantities. We can rearrange these 8 quantities to get the desired probabilities. For example, $P(X_{T-1}=1 \lvert X_0=1)$ would be the quantities corresponding to the initial transient state 1 and ending transient state 1, which would be $W_{11} \cdot P_{10}+W_{11} \cdot P_{13}$. The probability $P(X_{T-1}=1 \lvert X_0=2)$ would be the sum of the quantities corresponding to the initial transient state 2 and ending transient state 1, which would be $W_{21} \cdot P_{10}+W_{21} \cdot P_{13}$.

Summary

Before tackling the examples stated earlier, we summarize the ideas in the preceding section. Suppose that $\mathbf{P}$ is the transition probability matrix for a given absorbing Markov chain. For conceptual clarity, we can rewrite $\mathbf{P}$ in the following form.

$\mathbf{P}=\left[\begin{array}{cc} Q & R \\ \mathbf{0} & I \\ \end{array}\right]$

In this formulation, $Q$ consists of the one-step transition probabilities from transient states into transient states, $R$ consists of the one-step transition probabilities from transient states into absorbing states. Furthermore, all entries of $\mathbf{0}$ are zero and $I$ is the identity matrix of an appropriate size.

The next step is to obtain the fundamental matrix $W=(I-Q)^{-1}$, which is the inverse matrix of $I-Q$. The entries of $W$ are mean times spent in transient states. For example, the entry $W_{ij}$ is the mean time spent in state $j$ given that the initial state is $i$. As mentioned, the entries of the matrix $R$ are one-step probabilities from a transient state into an absorbing state. A typical entry of $R$ is $P_{ik}$ where $i$ is a transient state and $k$ is an absorbing state. As the above example demonstrates, the entries of $W$ and $R$ are used in finding the probability of the transient state from which the process enters an absorbing state.

Let $T=\text{min} \{ n \ge 0: X_n \text{ is an absorbing state} \}$, which is the random time to absorption. The goal is to find the probability of the form

$P(X_{T-1}=j \lvert X_0=i)$

where $i$ is the initial transient state and $j$ is the ending transient state (the last transient state before absorption). This probability is obtained by combining the appropriate entries of the matrices $W$ and $R$ in the following way:

$\displaystyle P(X_{T-1}=j \lvert X_0=i)=\sum \limits_{k} \biggl( W_{ij} \cdot P_{jk} \biggr)=W_{ij} \sum \limits_{k} P_{jk}$

where $k$ is over all relevant absorbing states, i.e. all absorbing states $k$ for which it is possible to transition from $j$ to $k$.

Examples

We now solve the two examples indicated at the beginning.

Example 1
Let $X_0=6$, i.e. the 0th toss involves 6 coins. Let $X_n$ be the number of coins in the $(n+1)$st toss. This is indeed a Markov chain since the next state depends only on the current state. The states decrease as more and more tosses are made. Thus the states of the chain are 6, 5, 4, 3, 2, 1 and 0 with state 0 being the absorbing state.

In each step in the chain, we observe the number of tails in tossing a number of coins. Thus each row in the transition probability matrix consists of the probabilities from a binomial distribution. Thus the following gives the transition probability matrix $\mathbf{P}$.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & 5 & 6 \cr 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \cr 1 & 1/2 & 1/2 & 0 & 0 & 0 & 0 & 0 \cr 2 & 1/4 & 2/4 & 1/4 & 0 & 0 & 0 & 0 \cr 3 & 1/8 & 3/8 & 3/8 & 1/8 & 0 & 0 & 0 \cr 4 & 1/16 & 4/16 & 6/16 & 4/16 & 1/16 & 0 & 0 \cr 5 & 1/32 & 5/32 & 10/32 & 10/32 & 5/32 & 1/32 & 0 \cr 6 & 1/64 & 6/64 & 15/64 & 20/64 & 15/64 & 6/64 & 1/64 \cr } \qquad$

The following shows the calculation to obtain the fundamental matrix $W$.

$Q= \left[\begin{array}{cccccc} 1/2 & 0 & 0 & 0 & 0 & 0 \\ 2/4 & 1/4 & 0 & 0 & 0 & 0 \\ 3/8 & 3/8 & 1/8 & 0 & 0 & 0 \\ 4/16 & 6/16 & 4/16 & 1/16 & 0 & 0 \\ 5/32 & 10/32 & 10/32 & 5/32 & 1/32 & 0 \\ 6/64 & 15/64 & 20/64 & 15/64 & 6/64 & 1/64 \end{array}\right] \ \ \ \ \ \ \ \ \ R = \bordermatrix{ & 0 \cr 1 & 1/2 \cr 2 & 1/4 \cr 3 & 1/8 \cr 4 & 1/16 \cr 5 & 1/32 \cr 6 & 1/64 \cr } \qquad$

$I-Q= \left[\begin{array}{cccccc} 1/2 & 0 & 0 & 0 & 0 & 0 \\ -2/4 & 3/4 & 0 & 0 & 0 & 0 \\ -3/8 & -3/8 & 7/8 & 0 & 0 & 0 \\ -4/16 & -6/16 & -4/16 & 15/16 & 0 & 0 \\ -5/32 & -10/32 & -10/32 & -5/32 & 31/32 & 0 \\ -6/64 & -15/64 & -20/64 & -15/64 & -6/64 & 63/64 \end{array}\right]$

$W=(I-Q)^{-1} = \bordermatrix{ & 1 & 2 & 3 & 4 & 5 & 6 \cr 1 & 2 & 0 & 0 & 0 & 0 & 0 \cr 2 & 4/3 & 4/3 & 0 & 0 & 0 & 0 \cr 3 & 10/7 & 4/7 & 8/7 & 0 & 0 & 0 \cr 4 & 152/105 & 72/105 & 32/105 & 112/105 & 0 & 0 \cr 5 & 942/651 & 472/651 & 272/651 & 112/651 & 672/651 & 0 \cr 6 & 2820/1953 & 1428/1953 & 928/1953 & 528/1953 & 192/1953 & 1984/1953 \cr } \qquad$

To solve the problem at hand, first define the time to absorption $T=\text{min} \{ n \ge 0: X_n=0 \}$. The following calculation gives the probabilities of the last step.

$\displaystyle P(Y=1)=P(X_{T-1}=1 \lvert X_0=6)=W_{61} \times P_{10}=\frac{2820}{1953} \times \frac{1}{2}=\frac{1410}{1953}=0.7220$

$\displaystyle P(Y=2)=P(X_{T-1}=2 \lvert X_0=6)=W_{62} \times P_{20}=\frac{1428}{1953} \times \frac{1}{4}=\frac{357}{1953}=0.1828$

$\displaystyle P(Y=3)=P(X_{T-1}=3 \lvert X_0=6)=W_{63} \times P_{30}=\frac{928}{1953} \times \frac{1}{8}=\frac{116}{1953}=0.0594$

$\displaystyle P(Y=4)=P(X_{T-1}=4 \lvert X_0=6)=W_{64} \times P_{40}=\frac{528}{1953} \times \frac{1}{16}=\frac{33}{1953}=0.0169$

\displaystyle \begin{aligned}P(Y=5)&=P(X_{T-1}=5 \lvert X_0=6)=W_{65} \times P_{50}=\frac{192}{1953} \times \frac{1}{32}=\frac{6}{1953}=0.003072 \end{aligned}

\displaystyle \begin{aligned}P(Y=6)&=P(X_{T-1}=6 \lvert X_0=6)=W_{66} \times P_{60}=\frac{1984}{1953} \times \frac{1}{64}=\frac{31}{1953}=0.01587 \end{aligned}

Starting a 6 coins, the reduction in coins in tosses is gradual. There is a 72% chance that the coins will be reduced to one coin in the last toss. The most likely way to end the game is from tossing the last single coin. However, the calculation does not indicate how long it will take to reach the last step of one coin.

Example 2
The diagram of the maze is repeated here.

A maze with two food sources (areas 5 and 7)

There are 7 states in this Markov chain, representing the areas in the maze. The following gives the transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 1 & 2 & 3 & 4 & 5 & 6 & 7 \cr 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \cr 2 & 0 & 0 & 1/2 & 0 & 0 & 0 & 1/2 \cr 3 & 0 & 1/3 & 0 & 1/3 & 0 & 1/3 & 0 \cr 4 & 0 & 0 & 1/2 & 0 & 1/2 & 0 & 0 \cr 5 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \cr 6 & 0 & 0 & 1/3 & 0 & 1/3 & 0 & 1/3 \cr 7 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$

The states 5 and 7 are the areas with food, thus are absorbing states. The next step is the calculation surrounding the fundamental matrix $W$.

$Q = \bordermatrix{ & 1 & 2 & 3 & 4 & 6 \cr 1 & 0 & 1 & 0 & 0 & 0 \cr 2 & 0 & 0 & 1/2 & 0 & 0 \cr 3 & 0 & 1/3 & 0 & 1/3 & 1/3 \cr 4 & 0 & 0 & 1/2 & 0 & 0 \cr 6 & 0 & 0 & 1/3 & 0 & 0 \cr } \qquad \ \ \ \ \ \ \ \ \ R = \bordermatrix{ & 5 & 7 \cr 1 & 0 & 0\cr 2 & 0 & 1/2 \cr 3 & 0 & 0 \cr 4 & 1/2 & 0 \cr 6 & 1/3 & 1/3 \cr } \qquad$

$I-Q= \left[\begin{array}{ccccc} 1 & -1 & 0 & 0 & 0 \\ 0 & 1 & -1/2 & 0 & 0 \\ 0 & -1/3 & 1 & -1/3 & -1/3 \\ 0 & 0 & -1/2 & 1 & 0 \\ 0 & 0 & -1/3 & 0 & 1 \\ \end{array}\right]$

$W=(I-Q)^{-1} = \bordermatrix{ & 1 & 2 & 3 & 4 & 6 \cr 1 & 1 & 13/10 & 9/10 & 3/10 & 3/10 \cr 2 &0 & 13/10 & 9/10 & 3/10 & 3/10 \cr 3 & 0 & 3/5 & 9/5 & 3/5 & 3/5 \cr 4 & 0 & 3/10 & 9/10 & 13/10 & 3/10 \cr 6 & 0 & 1/5 & 3/5 & 1/5 & 6/5 \cr } \qquad$

The matrices $W$ and $R$ are all we need to finish the problem. Suppose that we place the mouse initially at either area 2, area 3 and area 6. We find the probabilities of the last transient state being 2, 4 and 6. Once again $T$ is the random time to absorption and is defined by $T=\text{min} \{ n \ge 0: X_n=5 \text{ or } X_n=7 \}$.

First, the starting state is 2.

$\displaystyle P(X_{T-1}=2 \lvert X_0=2]=W_{22} \cdot P_{27}=\frac{13}{10} \cdot \frac{1}{2}=\frac{6.5}{10}=0.65$

$\displaystyle P(X_{T-1}=4 \lvert X_0=2]=W_{24} \cdot P_{45}=\frac{3}{10} \cdot \frac{1}{2}=\frac{1.5}{10}=0.15$

$\displaystyle P(X_{T-1}=6 \lvert X_0=2]=W_{26} \cdot (P_{65}+P_{67})=\frac{3}{10} \cdot \frac{2}{3}=\frac{2}{10}=0.2$

Next, the starting state is 3.

$\displaystyle P(X_{T-1}=2 \lvert X_0=3]=W_{32} \cdot P_{27}=\frac{3}{5} \cdot \frac{1}{2}=\frac{1.5}{5}=0.3$

$\displaystyle P(X_{T-1}=4 \lvert X_0=3]=W_{34} \cdot P_{45}=\frac{3}{5} \cdot \frac{1}{2}=\frac{1.5}{5}=0.3$

$\displaystyle P(X_{T-1}=6 \lvert X_0=3]=W_{36} \cdot (P_{65}+P_{67})=\frac{3}{5} \cdot \frac{2}{3}=\frac{2}{5}=0.4$

Lastly, the starting state is 6.

$\displaystyle P(X_{T-1}=2 \lvert X_0=6]=W_{62} \cdot P_{27}=\frac{1}{5} \cdot \frac{1}{2}=\frac{1}{10}=0.1$

$\displaystyle P(X_{T-1}=4 \lvert X_0=6]=W_{64} \cdot P_{45}=\frac{1}{5} \cdot \frac{1}{2}=\frac{1}{10}=0.1$

$\displaystyle P(X_{T-1}=6 \lvert X_0=6]=W_{66} \cdot (P_{65}+P_{67})=\frac{6}{5} \cdot \frac{2}{3}=\frac{4}{5}=0.8$

The results agree with the intuition. If the mouse is placed in area 2 initially, it is most likely that the mouse enters food area from area 2 (65% chance). Because the mouse makes the moves at random, there is a small but not negligible chance that the entry to food is from area 4 or area 6.

The initial state of area 3 is more neutral. If the mouse is placed in area 3 at first, the entry to food is roughly equally likely to be from areas 2, 4 and 6, with area 6 having a slight edge. Note that area 6 has food areas on two sides.

If the mouse is placed in area 6 initially, the entry to food is mostly likely to be from area 6 since there are foods on two sides of area 6.

Practice Problems

Several practice problems to reinforce the concepts discussed here are available in a companion blog. See the problems at the end of this problem set.

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# Absorbing Markov chains

The preceding two posts are devoted to solving the problem of determining mean time to absorption and the problem of determining the probability of absorption (using first step analysis here and using fundamental matrix here). The Markov chains in these problems are called absorbing Markov chains. This post summarizes the properties of such chains.

What are Absorbing Markov Chains?

A state in a Markov chain is said to be an absorbing state if the process will never leave that state once it is entered. That is, if the state $i$ is an absorbing state, then $P_{ii}=1$. A Markov chain is said to be an absorbing Markov chain if it has at least one absorbing state and if any state in the chain, with a positive probability, can reach an absorbing state after a number of steps. The following transition probability matrix represents an absorbing Markov chain.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & 5 \cr 0 & 1 & 0 & 0 & 0 & 0 & 0 \cr 1 & 0.5 & 0 & 0.5 & 0 & 0 & 0 \cr 2 & 0 & 0.5 & 0 & 0.5 & 0 & 0 \cr 3 & 0 & 0 & 0.5 & 0 & 0.5 & 0 \cr 4 & 0 & 0 & 0 & 0.5 & 0 & 0.5 \cr 5 & 0 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$ …………………………………………………………. (1)

The above matrix can be interpreted as a small scaled random walk or a small scaled gambler’s ruin. It has two absorbing states, 0 and 5. Once the process reaches these states, it does not leave these states. The other states (1, 2, 3 and 4) can reach an absorbing state in a finite number of steps with a positive probability. For example, state 1 can reach state 0 in one step with probability 0.5. State 1 can reach state 5 in four steps with probability 1/16.

As we will see below, any non-absorbing state in an absorbing Markov chain will eventually reach an absorbing state will probability 1. Thus any non-absorbing state in an absorbing Markov chain is a transient state. That is, when starting in a non-absorbing state, the process will only spend finite amount of time in non-absorbing states.

In order to be an absorbing Markov chain, it is not sufficient for a Markov chain to have an absorbing state. It must also have the property that all non-absorbing states must eventually reach an absorbing state. The following is an example of a chain that is not an absorbing Markov chain.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & 5 \cr 0 & 0.5 & 0.3 & 0.2 & 0 & 0 & 0 \cr 1 & 0.4 & 0.3 & 0.3 & 0 & 0 & 0 \cr 2 & 0.3 & 0.4 & 0.3 & 0 & 0 & 0 \cr 3 & 0 & 0 & 0.3 & 0.3 & 0.4 & 0 \cr 4 & 0 & 0 & 0 & 0.4 & 0.3 & 0.3 \cr 5 & 0 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$ …………………………………………………………. (2)

Even though the above chain has one absorbing state (state 5), it is not an absorbing chain. Note that the states 0, 1 and 2 can never reach state 5.

Reaching Absorbing States with Certainty

Now we discuss several properties of such chains. The first one is that it does not matter what state the process is in initially, eventually the process reach an absorbing state, i.e. the process will be absorbed at some point in time. Suppose that the absorbing Markov chain has $t$ non-absorbing states and $r$ absorbing states. The transition probability matrix $\mathbf{P}$ can be written in the following format

$\mathbf{P}=\left[\begin{array}{rr} Q & R \\ \mathbf{0} & I \\ \end{array}\right]$ ……………………………………………………………………………………….. (3)

where $Q$ is a $t \times t$ matrix representing the transition probabilities from the non-absorbing states into the non-absorbing states and $R$ is a $t \times r$ matrix representing the transition probabilities from the non-absorbing states into the absorbing states. Furthermore, $\mathbf{0}$ is an $r \times t$ matrix consisting of zeros in its entries and $I$ is the $r \times r$ identity matrix. The following is the matrix in (1) arranged in the new format.

$\mathbf{P} = \bordermatrix{ & 1 & 2 & 3 & 4 & \text{ } & 0 & 5 \cr 1 & 0 & 0.5 & 0 & 0 & \text{ } & 0.5 & 0 \cr 2 & 0.5 & 0 & 0.5 & 0 & \text{ } & 0 & 0 \cr 3 & 0 & 0.5 & 0 & 0.5 & \text{ } & 0 & 0 \cr 4 & 0 & 0 & 0.5 & 0 & \text{ } & 0 & 0.5 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 0 & 0 & 0 & 0 & 0 & \text{ } & 1 & 0 \cr 5 & 0 & 0 & 0 & 0 & \text{ } & 0 & 1 \cr } \qquad$ …………………………………………………….. (4)

Decomposing $\mathbf{P}$ into the form in (3) is a useful technique. It makes certain calculations simple to perform. We will get to that in a minute. First we prove an important fact showing that the non-absorbing states in an absorbing chain are indeed transient.

Theorem 1
In an absorbing Markov chain, the probability that the process will reach an absorbing state is 1. In other words, the probability of being in a non-absorbing state approaches zero as $n$ goes to infinity, i.e. $Q^n \rightarrow \mathbf{0}$ as $n \rightarrow \infty$.

Here’s a proof. For each non-absorbing state $k$, let $C_k$ be the minimum number of steps to transition from state $k$ to an absorbing state. For example, in the matrix (1), the minimum number of steps to go from state 1 to an absorbing state (state 0) is 1 with probability 0.5. In the same example, the minimum number of steps to go from state 3 to an absorbing state (state 5) is 2 with probability 0.25.

For each non-absorbing state $k$, let $A_k$ be the probability that, starting from state $k$, the process will take more than $C_k$ steps to reach an absorbing state. The probability $A_k$ must be less than 1. If $A_k=1$, the process for certain will not reach an absorbing state in $C_k$ steps, contradicting the property of $C_k$. Thus $A_k<1$.

Let $C$ be the maximum of all $C_k$ and let $A$ be the maximum of all $A_k$. Taking maximum is possible here since the Markov chain has a finite state space. The probability of the process taking more than $C$ steps to reach an absorbing state is less than or equal to $A$. This follows from the property of $C_k$ and $A_k$.

Furthermore, the probability of the process taking more than $2 \times C$ steps to reach an absorbing state is less than or equal to $A^2$. This follows from the fact that the Markov chain is memoryless. After taking $C$ steps and if the process still not in an absorbing state, the probability of the process taking additional $C$ or more steps to reach an absorbing state is less than or equal to $A$. Thus multiply the two probabilities gives $A^2$. By an inductive reasoning, the probability of the process taking more than $n \times C$ steps to reach an absorbing state is less than or equal to $A^n$. Since $A<1$, the numbers $A^n$ approach zero as $n$ goes to infinity.

It follows that as the integer $n$ increases without bound, the probability that the process will take more than $n$ steps to reach an absorbing state approaches zero. Turning it around, the probability that the process will reach an absorbing state in $n$ or fewer steps approaches 1. As a result, for any non-absorbing states $i$ and $j$, the transition probabilities $P_{ij}^n$ approaches zero as $n$ goes to infinity. Thus $Q^n \rightarrow \mathbf{0}$ as $n \rightarrow \infty$. $\square$

The above theorem shows that in an absorbing Markov chain, the process will eventually land in an absorbing state (it cannot cycle among the non-absorbing states indefinitely). Thus in an absorbing Markov chain, the time spent in the non-absorbing states is finite. As a result, non-absorbing states in an absorbing Markov chain are called transient states. Since the time spent in transient states is finite, one of the problems of interest is the mean time spent in transient states. Of course, the answer depends on the initial state.

More Properties of Absorbing Markov Chains

Another important property is the fundamental matrix that can be computed from the transition probability matrix of an absorbing chain. Given the transition probability matrix $\mathbf{P}$, decompose $\mathbf{P}$ according to (3). Recall that $Q$ consists of the transition probabilities from the transient states to the transient states. Derive $I-Q$ where $I$ is the identity matrix of the same dimension as $Q$. The following three theorems capture more important properties of absorbing Markov chains.

Theorem 2
In an absorbing Markov chain with transition probability matrix $\mathbf{P}$, the matrix $I-Q$ has an inverse $W$, i.e. $W=(I-Q)^{-1}$. The matrix $W$ has the following properties:

• $W=I+Q+Q^2+Q^3+\cdots$
• An entry $W_{ij}$ in the matrix $W$ is the mean time spent in the transient state $j$ given that the process starts at the transient state $i$. More specifically, $W_{ij}=E[\text{number of times state } j \text{ is visited} \lvert X_0=i]$.

For an absorbing Markov chain with transition probability matrix $\mathbf{P}$, the matrix $W=(I-Q)^{-1}$ is called the fundamental matrix of the absorbing Markov chain. A proof of Theorem 2 is sketched out in the preceding post. So we will not repeat it here.

Once the fundamental matrix is obtained by taking the inverse of $I-Q$, we can use it to solve two problems, stated in the following two theorems. The ideas of a proof are also sketched out in the preceding post.

Theorem 3 – mean time to absorption
In an absorbing Markov chain with transition probability matrix $\mathbf{P}$, consider the fundamental matrix $W=(I-Q)^{-1}$. The following calculation is of interest.

• Given that the process starts in the transient state $i$, consider the row of the matrix $W$ that corresponds to state $i$. The sum of all entries of $W$ on that row is the mean time spent in transient states given that the process start in state $i$. The sum is the mean time to absorption.
Theorem 4 – probabilities of absorption
In an absorbing Markov chain with transition probability matrix $\mathbf{P}$, consider the fundamental matrix $W=(I-Q)^{-1}$. The following calculation is of interest.

• Consider the matrix $R$ in the decomposition of $\mathbf{P}$. The matrix $W \times R$, the product of $W$ and $R$, gives the probabilities of absorption.
• More specifically, the row in $W$ corresponding to transient state $i$ multiplying the column in $R$ corresponding to the absorbing state $k$ results in the probability of being absorbed into state $k$ given that the process start in state $i$.
• Even more specifically, assuming that the transient states are $0,1,2,\cdots,t-1$, the following gives the probability of absorption into state $k$ given that the process starts in state $i$.
$\displaystyle \sum \limits_{j=0}^{t-1} \biggl( E[\text{number of times state } j \text{ is visited} \lvert X_0=i] \times P_{jk} \biggr)$

Examples

Examples are available in the preceding posts (here and here). We work three examples.

Example 1
An urn initially contains 7 blue balls and 5 red balls. One ball is selected at random one at a time from the urn. If the selected ball is not red, the ball is put back into the urn. If the selected ball is red, it is removed from the urn. The process continues until all red balls are removed. Determine the expected number of selections in order to have all red balls removed.

Let $X_n$ be the number of red balls in the urn after the $n$th selection. The states are 0, 1, 2, 3, 4 and 5. The initial state is $X_0=5$. The goal is to reach the absorbing state of 0 (all red balls are removed). The following is the transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & 5 \cr 0 & 1 & 0 & 0 & 0 & 0 & 0 \cr 1 & 1/8 & 7/8 & 0 & 0 & 0 & 0 \cr 2 & 0 & 2/9 & 7/9 & 0 & 0 & 0 \cr 3 & 0 & 0 & 3/10 & 7/10 & 0 & 0 \cr 4 & 0 & 0 & 0 & 4/11 & 7/11 & 0 \cr 5 & 0 & 0 & 0 & 0 & 5/12 & 7/12 \cr } \qquad$

There is only one absorbing state, the state 0. The states 1, 2, 3, 4 and 5 are transient states. Identify the matrix $Q$, which is the transition probabilities from the transient states into the transient states. Derive the matrix $I-Q$ where $I$ is a 5 x 5 identity matrix. The following is the inverse of $I-Q$.

$(I-Q)^{-1} = \bordermatrix{ & 1 & 2 & 3 & 4 & 5 \cr 1 & 8 & 0 & 0 & 0 & 0 \cr 2 & 8 & 9/2 & 0 & 0 & 0 \cr 3 & 8 & 9/2 & 10/3 & 0 & 0 \cr 4 & 8 & 9/2 & 10/3 & 11/4 & 0 \cr 5 & 8 & 9/2 & 10/3 & 11/4 & 12/5 \cr } \qquad$

The 5th row of $(I-Q)^{-1}$ gives the mean time spent in the transient states. The sum of the entries on the 5th row is 1259/60 = 20.9833. Thus it takes about 21 steps on average to remove all red balls from the urn.

Example 2
An urn initially contains 7 blue balls and 5 red balls. One ball is selected at random one at a time from the urn. If the selected ball is not red, the ball is put back into the urn. If the selected ball is red, it is removed from the urn and then a blue ball is put into the urn. The process continues until all red balls are removed. Determine the expected number of selections in order to have all red balls removed. Compare the expected value with the expected value in Example 1. Give an explanation for the difference.

As in Example 1, let $X_n$ be the number of red balls in the urn after the $n$th selection. The states are 0, 1, 2, 3, 4 and 5. The initial state is $X_0=5$. The goal is to reach the absorbing state of 0 (all red balls are removed). The following is the transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & 5 \cr 0 & 1 & 0 & 0 & 0 & 0 & 0 \cr 1 & 1/12 & 11/12 & 0 & 0 & 0 & 0 \cr 2 & 0 & 2/12 & 10/12 & 0 & 0 & 0 \cr 3 & 0 & 0 & 3/12 & 9/12 & 0 & 0 \cr 4 & 0 & 0 & 0 & 4/12 & 8/12 & 0 \cr 5 & 0 & 0 & 0 & 0 & 5/12 & 7/12 \cr } \qquad$

One big difference from Example 1 is that each red ball selected is replaced by a blue ball. Thus the urn always has 12 balls whereas in Example 1 the number of balls is dwindling. As before, identify $Q$ and derive $I-Q$. The following is the inverse of $I-Q$.

$(I-Q)^{-1} = \bordermatrix{ & 1 & 2 & 3 & 4 & 5 \cr 1 & 12 & 0 & 0 & 0 & 0 \cr 2 & 12 & 6 & 0 & 0 & 0 \cr 3 & 12 & 6 & 4 & 0 & 0 \cr 4 & 12 & 6 & 4 & 3 & 0 \cr 5 & 12 & 6 & 4 & 3 & 12/5 \cr } \qquad$

Since the initial state is 5, we focus on the 5th row of $(I-Q)^{-1}$. The sum of that row is 137/5 = 27.4. Thus it takes on average 27.4 selections to remove all red balls. Note that the process takes longer than in Example 1. By replacing each selected red ball with a blue ball, it becomes harder to remove red balls especially when the process is in the lower states.

Example 3
An urn initially contains 4 blue balls and 2 red balls. One ball is selected at random one at a time from the urn. Each selected ball is replaced by a ball of the opposite color. The process continues until all balls in the urn are of the same color.

• Determine the probability that the process ends with the urn containing only red balls.
• Determine the expected number of selections in order for the urn to consist of balls of the same color.

Let $X_n$ be the number of red balls in the urn after the $n$th selection. There are 7 states in this chain – 0, 1, 2, 3, 4, 5 and 6. The initial state is $X_0=2$. The process will eventually reach either state 0 or state 6. We wish to find the probability of being absorbed into state 6 first. States 0 and 6 are the absorbing states. States 1, 2, 3, 4, and 5 are transient states. The following is the transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & 5 & 6 \cr 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \cr 1 & 1/6 & 0 & 5/6 & 0 & 0 & 0 & 0 \cr 2 & 0 & 2/6 & 0 & 4/6 & 0 & 0 & 0 \cr 3 & 0 & 0 & 3/6 & 0 & 3/6 & 0 & 0 \cr 4 & 0 & 0 & 0 & 4/6 & 0 & 2/6 & 0 \cr 5 & 0 & 0 & 0 & 0 & 5/6 & 0 & 1/6 \cr 6 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$

As before, identify $Q$ and derive $I-Q$. The following is the inverse of $I-Q$.

$W=(I-Q)^{-1} = \bordermatrix{ & 1 & 2 & 3 & 4 & 5 \cr 1 & 48/13 & 105/13 & 10 & 90/13 & 30/13 \cr 2 & 42/13 & 126/13 & 12 & 108/13 & 36/13 \cr 3 & 3 & 9 & 13 & 9 & 3 \cr 4 & 36/13 & 108/13 & 12 & 126/13 & 42/13 \cr 5 & 30/13 & 90/13 & 10 & 105/13 & 48/13 \cr } \qquad$

To answer the probability question, multiply the matrices $W$ and $R$. The following is the result.

$W \times R=\left[\begin{array}{rrrrr} 48/13 & 105/13 & 10 & 90/13 & 30/13 \\ 42/13 & 126/13 & 12 & 108/13 & 36/13 \\ 3 & 9 & 13 & 9 & 3 \\ 36/13 & 108/13 & 12 & 126/13 & 42/13 \\ 30/13 & 90/13 & 10 & 105/13 & 48/13 \end{array}\right] \times \left[\begin{array}{rr} 1/6 & 0 \\ 0 & 0 \\ 0 & 0 \\ 0 & 0 \\ 0 & 1/6 \end{array}\right]=\left[\begin{array}{rr} 8/13 & 5/13 \\ 7/13 & 6/13 \\ 1/2 & 1/2 \\ 6/13 & 7/13 \\ 5/13 & 8/13 \end{array}\right]$

With the initial state being 2, the probability of being absorbed into state 6 (all red balls) is 6/13 = 0.4615. The expected time to absorption (all balls of the same color) is the sum of the second row of $(I-Q)^{-1}$, which is 36. Thus on average it will take 36 selections of balls for the urn to consist of balls of the same color.

Practice Problems

Practice problems to reinforce the concepts discussed here are available in a companion blog. There are two problem sets. The first one is here and the second one is here.

Dan Ma math

Daniel Ma mathematics

Dan Ma Markov chains

Daniel Ma Markov chains

$\copyright$ 2018 – Dan Ma

# First step analysis and fundamental matrix

A great number of problems involving Markov chains can be evaluated by a technique called first step analysis. The general idea of the method is to break down the possibilities resulting from the first step (first transition) in the Markov chain. Then use the law of total probability and Markov property to derive a set of relationship among the unknown variables. This technique is demonstrated in this previous post in several examples. In this post, the technique is further discussed. An alternative approach based on the fundamental matrix is also emphasized.

This previous post shows how to use first step analysis to solve two problems: to determine the probability of absorption from transient states and to determine the expected time spent in transient states. This post shows that first step analysis leads to an effective approach of using fundamental matrix for solving the same two problems. We illustrate using an example.

Illustrative Example

We start with Example 3 discussed in the previous post. It involves the following transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 \cr 0 & 1 & 0 & 0 & 0 & 0 \cr 1 & 0.4 & 0.3 & 0.1 & 0.1 & 0.1 \cr 2 & 0.1 & 0.1 & 0.5 & 0.2 & 0.1 \cr 3 & 0.1 & 0.2 & 0.1 & 0.5 & 0.1 \cr 4 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$

The Markov chain represented by $\mathbf{P}$ cycles through 5 states. Whenever the process reaches state 0 or state 4, it stays there and not move. These two states are called absorbing states. The other states (1, 2 and 3) are called transient states because the process stays in each of these states a finite amount of time. When the process starts at one of these states, it will eventually transition (be absorbed) into one of the absorbing states (0 or 4). Thus the matrix $\mathbf{P}$ is called an absorbing Markov chain.

A natural question: how likely that the process is absorbed into state 0 versus state 4? Of course the answer to this question depends on the starting state.

Another question: if starting at state 1, 2 or 3, how many steps on average it will take for the process to be absorbed? In other words, what is the mean time to absorption? Of course, the answer will depend on the starting state.

Let $T$ the time it takes the process to be absorbed into 0 or 4. More specifically let $T$ be defined by $T=\{ n \ge 0: X_n=0 \text{ or } X_n=4 \}$. If the starting state is $i$, let $U_i$ be the probability of the process being absorbed into state 0 and let $V_i$ be the expected time to absorption (into state 0 or state 4). The quantities $U_i$ and $V_i$ can also be defined as follows:

$U_i=P[X_T=0 \lvert X_0=i] \ \ \ \ i=1,2,3$

$V_i=E[T \lvert X_0=i] \ \ \ \ \ \ \ \ \ \ i=1,2,3$

The quantities $U_i$ are the probabilities of absorption into state 0 and the quantities $V_i$ are the expected times to absorption (from each transient state). Applying the first step analysis produces the following two sets of equations.

\displaystyle \begin{aligned}&U_1=0.4+0.3 \times U_1+0.1 \times U_2+0.1 \times U_3 \\&U_2=0.1+0.1 \times U_1+0.5 \times U_2+0.2 \times U_3 \\&U_3=0.1+0.2 \times U_1+0.1 \times U_2+0.5 \times U_3 \end{aligned}

\displaystyle \begin{aligned}&V_1=1+0.3 \times V_1+0.1 \times V_2+0.1 \times V_3 \\&V_2=1+0.1 \times V_1+0.5 \times V_2+0.2 \times V_3 \\&V_3=1+0.2 \times V_1+0.1 \times V_2+0.5 \times V_3 \end{aligned}

We rearrange the equations by putting the variables $U_i$ or $V_i$ on one side and the constant terms on the other side.

$\displaystyle \begin{array}{rcr} \displaystyle 0.7 \times U_1-0.1 \times U_2-0.1 \times U_3 & = & 0.4 \\ \displaystyle -0.1 \times U_1+0.5 \times U_2-0.2 \times U_3 & = & 0.1 \\ \displaystyle -0.2 \times U_1-0.1 \times U_2+0.5 \times U_3 & = & 0.1 \end{array} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1a)$

$\displaystyle \begin{array}{rcr} \displaystyle 0.7 \times V_1-0.1 \times V_2-0.1 \times V_3 & = & 1 \\ \displaystyle -0.1 \times V_1+0.5 \times V_2-0.2 \times V_3 & = & 1 \\ \displaystyle -0.2 \times V_1-0.1 \times V_2+0.5 \times V_3 & = & 1 \end{array} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2a)$

The probabilities of absorption into state 0 are found by solving the system of equations (1a). The mean times to absorption are found by solving the system of equations (2a). It will be informative to write (1) and (2) in matrix notations.

$\left[\begin{array}{rrr} 0.7 & -0.1 & -0.1 \\ -0.1 & 0.5 & -0.2 \\ -0.2 & -0.1 & 0.5 \end{array}\right] \left[\begin{array}{c} U_1 \\ U_2 \\ U_3 \end{array}\right] = \left[\begin{array}{c} 0.4 \\ 0.1 \\ 0.1 \end{array}\right] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1b)$

$\left[\begin{array}{rrr} 0.7 & -0.1 & -0.1 \\ -0.1 & 0.5 & -0.2 \\ -0.2 & -0.1 & 0.5 \end{array}\right] \left[\begin{array}{c} V_1 \\ V_2 \\ V_3 \end{array}\right] = \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2b)$

Note that the same matrix appears on the left side of both (1b) and (2b). The quantities $U_i$ and $V_i$ can be found by multiplying both sides by the following inverse matrix.

$\left[\begin{array}{rrr} 0.7 & -0.1 & -0.1 \\ -0.1 & 0.5 & -0.2 \\ -0.2 & -0.1 & 0.5 \end{array}\right]^{-1}=\left[\begin{array}{rrr} 230/141 & 60/141 & 70/141 \\ 90/141 & 330/141 & 150/141 \\ 110/141 & 90/141 & 340/141 \end{array}\right]$

Multiplying (1b) and (2b) by the inverse matrix produces the following.

$\left[\begin{array}{c} U_1 \\ U_2 \\ U_3 \end{array}\right] = \left[\begin{array}{rrr} 230/141 & 60/141 & 70/141 \\ 90/141 & 330/141 & 150/141 \\ 110/141 & 90/141 & 340/141 \end{array}\right] \times \left[\begin{array}{c} 0.4 \\ 0.1 \\ 0.1 \end{array}\right] =\left[\begin{array}{c} 35/47 \\ 28/47 \\ 29/47 \end{array}\right]=\left[\begin{array}{c} 0.745 \\ 0.596 \\ 0.617 \end{array}\right] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1c)$

$\left[\begin{array}{c} V_1 \\ V_2 \\ V_3 \end{array}\right] = \left[\begin{array}{rrr} 230/141 & 60/141 & 70/141 \\ 90/141 & 330/141 & 150/141 \\ 110/141 & 90/141 & 340/141 \end{array}\right] \times \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] =\left[\begin{array}{c} 360/141 \\ 570/141 \\ 540/141 \end{array}\right]=\left[\begin{array}{c} 2.55 \\ 4.04 \\ 3.83 \end{array}\right] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2c)$

The systems of equations (1a) and (2a) can certainly be solved by using other methods (e.g. Gaussian elimination). The inverse matrix approach is clean and can be implemented using software. More importantly, using the inverse matrix leads to a general description of the algorithm based on fundamental matrix.

How is the matrix in (1b) and (2b) obtained? Recall that the equations in (1a) and (2a) are obtained by rearranging the equations from the first step analysis. The first step analysis is of course based on the original transition probability matrix $\mathbf{P}$. The matrix in question is obtained from the transition probability matrix $\mathbf{P}$. The next two sections show how.

Before stating the algorithm, we continue to use the above example as illustration. The following is a rewrite of the transition probability matrix $\mathbf{P}$. Note that the transient states are grouped together.

$\mathbf{P} = \bordermatrix{ & 1 & 2 & 3 & \text{ } & 0 & 4\cr 1 & 0.3 & 0.1 & 0.1 & \text{ } & 0.4 & 0.1 \cr 2 & 0.1 & 0.5 & 0.2 & \text{ } & 0.1 & 0.1 \cr 3 & 0.2 & 0.1 & 0.5 & \text{ } & 0.1 & 0.1 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 0 & 0 & 0 & 0 & \text{ } & 1 & 0 \cr 4 & 0 & 0 & 0 & \text{ } & 0 & 1 \cr } \qquad=\left[\begin{array}{rr} Q & R \\ \mathbf{0} & I \\ \end{array}\right]$

The matrix $Q$ is the 3 x 3 matrix of transition probabilities for the 3 transient states 1, 2 and 3. The matrix $Q$ describes the transition probabilities between transient states. The matrix $R$ is a 3 x 2 matrix that shows the transition probabilities from transient to absorbing states. The matrix $\mathbf{0}$ is a 2 x 3 matrix consisting of 0’s. The matrix $I$ is an identity matrix describing the absorbing states (in this case 2 x 2).

The key matrix is the matrix $I-Q$ where $I$ is a 3 x 3 identity matrix.

$I-Q=\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]-\left[\begin{array}{rrr} 0.3 & 0.1 & 0.1 \\ 0.1 & 0.5 & 0.2 \\ 0.2 & 0.1 & 0.5 \end{array}\right]=\left[\begin{array}{rrr} 0.7 & -0.1 & -0.1 \\ -0.1 & 0.5 & -0.2 \\ -0.2 & -0.1 & 0.5 \end{array}\right]$

The matrix $W=(I-Q)^{-1}$, the inverse of $I-Q$, is called the fundamental matrix for the absorbing chain in question. As shown above, the fundamental matrix is used in finding the probabilities of absorption and the mean times to absorption.

The Algorithm

Recall that an absorbing Markov chain has at least one absorbing state such that every non-absorbing state will eventually transition into an absorbing state (i.e. will eventually be absorbed). The non-absorbing states in such a chain are called transient states.

Consider an absorbing Markov chain with transition probability matrix $\mathbf{P}$. Suppose that the chain has $N+1$ states with transient states $0,1,2,\cdots,y-1$ and absorbing states $y,y+1,\cdots,N$. Such labeling is for convenience. As the above example shows, the rearranging of the states is not necessary as long as one can keep the transient states apart from the absorbing states.

Write the matrix $\mathbf{P}$ in the following form

$\mathbf{P}=\left[\begin{array}{rr} Q & R \\ \mathbf{0} & I \\ \end{array}\right]$

where $Q$ consists of $P_{ij}$ where $0 \le i,j \le y-1$ and $R$ consists of $P_{ij}$ where $0 \le i \le y-1$ and $y \le j \le N$. In other words, the matrix $Q$ contains the one-step transition probabilities between transient states and $R$ contains the one-step transition probabilities from transient states into absorbing states. The matrix $\mathbf{0}$ is a $(N+1-y) \times y$ matrix consisting of 0’s. The matrix $I$ is an $(N+1-y) \times (N+1-y)$ identity matrix describing the absorbing states.

Consider the matrix $I-Q$ where $I$ is the $y \times y$ identity matrix. Then $W=(I-Q)^{-1}$, the inverse of $I-Q$, is the fundamental matrix for the absorbing Markov chain.

Here is one important property of the matrix $W=[ W_{ij} ]$.

An element $W_{ij}$ of $W$ represents the total time the chain spends in the transient state $j$ if the starting state is the transient state $i$.

This leads to the following observation.

The sum of all entries in a row of $W$ is the total time the chain is expected to spend in all the transient states if the starting state is the transient state corresponding to that row.

Here’s the property concerning probability of absorption.

The matrix $U=W \times R$, the product of the matrix $W$ and the matrix $R$, gives the probabilities of absorption. For example, the element $U_{ij}$ of $U$ is the probability of being absorbed into the state $j$ if the starting state is $i$.

To make the mathematical properties even more precise, consider $T=\{ n \ge 0: y \le X_n \le N \}$. The variable $T$ is the random number of steps the process takes to reach an absorbing state. Define the following quantities.

$U_{ij}=P[T=j \lvert X_0=i]$

$V_{i}=E[T \lvert X_0=i]$

where $i$ is any transient state ($0 \le i \le y$) and $j$ is any absorbing state ($y \le j \le N$). Here’s how to tie $U_{ij}$ and $V_i$ to the above stated facts.

The quantity $V_i$ is simply the sum of all entries of the row corresponding to the transient state $i$ in the matrix $W$. Equivalently let $\mathbf{V}$ be the column vector with entries $V_i$. Then $\mathbf{V}=W \times \mathbf{1}$ where $\mathbf{1}$ is the column vector consisting of all 1’s.

The quantities $U_{ij}$ are simply the entries in the matrix $U=W \times R$.

To further illustrate the technique, the next section has examples. It is critical that calculator or software be available for computing inverse matrix. Here is a useful online matrix calculator.

Examples

Example 1
This is another calculation example to illustrate the technique. Use the following transition probability matrix to compute the quantities $U_{ij}=P[T=j \lvert X_0=i]$ (the probabilities of being absorbed into states 0 and 4) and $V_{i}=E[T \lvert X_0=i]$ (the expected time spent in transient states 1,2 and 3 given the starting state is $i$). Of course $T$ is the random time to absorption defined by $T=\{ n \ge 0: X_n=0 \text{ or } X_n=4 \}$.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 \cr 0 & 1 & 0 & 0 & 0 & 0 \cr 1 & 0.3 & 0.45 & 0.1 & 0.1 & 0.05 \cr 2 & 0.05 & 0.1 & 0.2 & 0.1 & 0.55 \cr 3 & 0.1 & 0.1 & 0.1 & 0.1 & 0.6 \cr 4 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$

The following shows the matrices that are used in calculation.

$Q = \bordermatrix{ & 1 & 2 & 3 \cr 1 & 0.45 & 0.1 & 0.1 \cr 2 & 0.1 & 0.2 & 0.1 \cr 3 & 0.1 & 0.1 & 0.1 \cr } \qquad \ \ \ \ R = \bordermatrix{ & 0 & 4 \cr 1 & 0.3 & 0.05 \cr 2 & 0.05 & 0.55 \cr 3 & 0.1 & 0.6 \cr } \qquad$

$I-Q=\left[\begin{array}{rrr} 0.55 & -0.1 & -0.1 \\ -0.1 & 0.8 & -0.1 \\ -0.1 & -0.1 & 0.9 \end{array}\right]$

$W=(I-Q)^{-1}=\left[\begin{array}{rrr} 1420/743 & 200/743 & 180/743 \\ 200/743 & 970/743 & 130/743 \\ 180/743 & 130/743 & 860/743 \end{array}\right]$

Here’s the matrix calculation to obtain the desired quantities.

$\left[\begin{array}{r} V_1 \\ V_2 \\ V_3 \end{array}\right]=\left[\begin{array}{rrr} 1420/743 & 200/743 & 180/743 \\ 200/743 & 970/743 & 130/743 \\ 180/743 & 130/743 & 860/743 \end{array}\right] \times \left[\begin{array}{r} 1 \\ 1 \\ 1 \end{array}\right]=\left[\begin{array}{r} 2.4226 \\ 1.7497 \\ 1.5747 \end{array}\right]$

$U=W \times R=\left[\begin{array}{rr} 908/1486 & 578/1486 \\ 243/1486 & 1243/1486 \\ 293/1486 & 1193/1486 \end{array}\right]=\left[\begin{array}{rr} 0.6110 & 0.3890 \\ 0.1635 & 0.8365 \\ 0.1972 & 0.8028 \end{array}\right]=\left[\begin{array}{rr} U_{1,0} & U_{1,4} \\ U_{2,0} & U_{2,4} \\ U_{2,0} & U_{3,4} \end{array}\right]$

Example 2
A fair coin is tossed repeatedly until the appearance of 4 consecutive heads. Determine the mean number of tosses required.

Consider the 5-state Markov chain with states 0, H, HH, HHH and HHHH. The state 0 means that the last toss is not a head. These 5 states are labeled 0, 1, 2, 3, 4 and 5, respectively.

$\mathbf{P} = \bordermatrix{ & 0 & \text{1=H} & \text{2=HH} & \text{3=HHH} & \text{4=HHHH} \cr 0 & 0.5 & 0.5 & 0 & 0 & 0 \cr 1 & 0.5 & 0 & 0.5 & 0 & 0 \cr 2 & 0.5 & 0 & 0 & 0.5 & 0 \cr 3 & 0.5 & 0 & 0 & 0 & 0.5 \cr 4 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$

Note that when a toss results in a tail, the process bounces back to the state 0. Otherwise, it advances to the next state (e.g. HH to HHH). The transient states are 0, 1, 2 and 3. State 4 is the absorbing state. The natural question is: on average how many tosses does it take to reach state 4 (to get 4 consecutive heads)? It is informative to obtain the matrix $I-Q$ and to find its inverse.

$I-Q=\left[\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]-\left[\begin{array}{rrrr} 0.5 & 0.5 & 0 & 0 \\ 0.5 & 0 & 0.5 & 0 \\ 0.5 & 0 & 0 & 0.5 \\ 0.5 & 0 & 0 & 0 \end{array}\right]=\left[\begin{array}{rrrr} 0.5 & -0.5 & 0 & 0 \\ -0.5 & 1 & -0.5 & 0 \\ -0.5 & 0 & 1 & -0.5 \\ -0.5 & 0 & 0 & 1 \end{array}\right]$

$(I-Q)^{-1} = \bordermatrix{ & 0 & \text{1=H} & \text{2=HH} & \text{3=HHH} \cr 0 & 16 & 8 & 4 & 2 \cr 1 & 14 & 8 & 4 & 2 \cr 2 & 12 & 6 & 4 & 2 \cr 3 & 8 & 4 & 2 & 2 \cr } \qquad$

The sum of the first row of $(I-Q)^{-1}$ is 30. It takes on average 30 tosses to get 4 consecutive heads if starting from scratch. After getting the first head, the process spends on average 28 tosses (the sum of the second row) before getting 4 consecutive heads. Even with three consecutive heads obtained, the chain still spends on average 16 tosses before reaching the state HHHH! On the surface, this may seem unbelievable. Remember that whenever a toss results in a tail, the process reverts back to 0 and the process essentially starts over again.

Example 3 (Occupancy Problem)
This example revisits the occupancy problem, which is discussed here. The occupancy problem is a classic problem in probability. The setting of the problem is that $k$ balls are randomly distributed into $N$ cells (or boxes or other containers) one at a time. After all the $k$ balls are distributed into the cells, we examine how many of the cells are occupied with balls. In this particular example, the number of balls is not fixed. Instead, balls are thrown into the cells one at a time until all cells are occupied. On average how many balls have to be thrown to achieve this?

More specifically, let $n=6$ (cells). On average, how many balls have to be thrown until all 6 cells are occupied? Let $X_n$ be the number of occupied cells after $n$ balls are thrown. The following is the transition probability matrix describing this Markov chain.

$\displaystyle \mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & 5 & 6 \cr 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 1 & 0 & \frac{1}{6} & \frac{5}{6} & 0 & 0 & 0 & 0 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 2 & 0 & 0 & \frac{2}{6} & \frac{4}{6} & 0 & 0 & 0 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 3 & 0 & 0 & 0 & \frac{3}{6} & \frac{3}{6} & 0 & 0 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 4 & 0 & 0 & 0 & 0 & \frac{4}{6} & \frac{2}{6} & 0 \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 5 & 0 & 0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6} \cr \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \cr 6 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$

For a more detailed discussion on this matrix, see the previous post on the occupancy problem. The transient states here are 0, 1, 2, 3, 4 and 5. Extract the matrix $Q$ with these 5 transient states. Derive the matrix $I-Q$. The following is the inverse of $I-Q$.

$(I-Q)^{-1} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & 5 \cr 0 & 1 & 6/5 & 3/2 & 2 & 3 & 6 \cr 1 & 0 & 6/5 & 3/2 & 2 & 3 & 6 \cr 2 & 0 & 0 & 3/2 & 2 & 3 & 6 \cr 3 & 0 & 0 & 0 & 2 & 3 & 6 \cr 4 & 0 & 0 & 0 & 0 & 3 & 6 \cr 5 & 0 & 0 & 0 & 0 & 0 & 6 \cr } \qquad$

The sum of the first row (the row for state 0) is 14.7. If starting with all 6 cells empty, it will take on average the throwing of 14.7 balls to have all 6 cells occupied. If there is already one ball in the 6 cells, then it will take on average 13.7 balls to fill the 6 cells. The decrease in the number of throws is not linear. When there are already 4 occupied cells, it will take on average 9 throws to fill the cells. When there is only one empty cell, it will take on average 6 more throws to fill the cells.

Because the number of cells is 6, the problem can be interpreted as rolling a die. The question can be restated: in rolling a fair die, how many rolls does it take on average to have each side of the die appeared at least one?

There is yet another interpretation of the problem of throwing balls into cells until all cells are occupied. This is precisely the coupon collector problem. The 6 cells in the example would be like 6 different coupon types (e.g. promotional prizes that come with the purchase of a product). When such a product is purchased, a prize (a coupon) is given at random. How many units of the product do you need to buy in order to have collected each of the coupon types at least once? For a fuller discussion on the coupon collector problem, see this blog post in a companion blog.

Example 4 (Gambler’s Ruin)
Consider a small scaled gambler’s ruin problem. Say, $N=8$ and $p=0.48$. That is, the gambler makes a series of one-unit bets against the house such that the probability of winning a bet is 0.48 and such that the gambler stops whenever his total capital is 0 units (the gambler is in ruin) or 8 units.

Let $X_n$ be the gambler’s capital (the number of units) after the $n$th bet. The following is the transition probability matrix describing this Markov chain.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \cr 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \cr 1 & 0.52 & 0 & 0.48 & 0 & 0 & 0 & 0 & 0 & 0 \cr 2 & 0 & 0.52 & 0 & 0.48 & 0 & 0 & 0 & 0 & 0 \cr 3 & 0 & 0 & 0.52 & 0 & 0.48 & 0 & 0 & 0 & 0 \cr 4 & 0 & 0 & 0 & 0.52 & 0 & 0.48 & 0 & 0 & 0 \cr 5 & 0 & 0 & 0 & 0 & 0.52 & 0 & 0.48 & 0 & 0 \cr 6 & 0 & 0 & 0 & 0 & 0 & 0.52 & 0 & 0.48 & 0 \cr 7 & 0 & 0 & 0 & 0 & 0 & 0 & 0.52 & 0 & 0.48 \cr 8 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \cr } \qquad$

There are two absorbing states in this chain, state 0 and state 8 (ruin or winning all units). The transient states are 1, 2, 3, 4, 5, 6 and 7. As discussed in this post, $Q$ is the matrix consisting of all transition probabilities $P_{ij}$ from the transient states into the transient states. Then derive the matrix $I-Q$ and compute its inverse.

$I-Q = \bordermatrix{ & 1 & 2 & 3 & 4 & 5 & 6 & 7 \cr 1 & 1 & -0.48 & 0 & 0 & 0 & 0 & 0 \cr 2 & -0.52 & 1 & -0.48 & 0 & 0 & 0 & 0 \cr 3 & 0 & -0.52 & 1 & -0.48 & 0 & 0 & 0 \cr 4 & 0 & 0 & -0.52 & 1 & -0.48 & 0 & 0 \cr 5 & 0 & 0 & 0 & -0.52 & 1 & -0.48 & 0 \cr 6 & 0 & 0 & 0 & 0 & -0.52 & 1 & -0.48 \cr 7 & 0 & 0 & 0 & 0 & 0 & -0.52 & 1 \cr } \qquad$

$W=(I-Q)^{-1} = \bordermatrix{ & 1 & 2 & 3 & 4 & 5 & 6 & 7 \cr 1 & 1.7444 & 1.4316 & 1.1429 & 0.8763 & 0.6303 & 0.4032 & 0.1935 \cr 2 & 1.5509 & 2.9825 & 2.3810 & 1.8257 & 1.3131 & 0.8399 & 0.4032 \cr 3 & 1.3413 & 2.5794 & 3.7223 & 2.8541 & 2.0528 & 1.3131 & 0.6303 \cr 4 & 1.1142 & 2.1426 & 3.0920 & 3.9683 & 2.8541 & 1.8257 & 0.8763 \cr 5 & 0.8681 & 1.6695 & 2.4092 & 3.0920 & 3.7223 & 2.3810 & 1.1429 \cr 6 & 0.6016 & 1.1569 & 1.6695 & 2.1426 & 2.5794 & 2.9825 & 1.4316 \cr 7 & 0.3128 & 0.6016 & 0.8681 & 0.1142 & 1.3413 & 1.5509 & 1.7444 \cr } \qquad$

The fundamental matrix $(I-Q)^{-1}$ gives a wealth of information about the prospect for the gambler in question. Suppose the gambler initially has 3 units in capital. The sum of the third row is 14.49, roughly 14.5. So on average the gambler makes 14.5 bets before reaching one of the two absorbing states (in ruin or owning all 8 units). With one bet = one period of time, the gambler spends on average 3.72 periods of time in state 3. Note that most of the time the gambler is in the lower capital states. The gambler that has initially 3 units only spends on average 0.6303 amount of time in state 7, an indication that the gambler does not have a great chance of winning all 8 units.

The matrix $W \times R$ gives the probabilities of absorption.

$R = \bordermatrix{ & 0 & 8 \cr 1 & 0.52 & 0 \cr 2 & 0 & 0 \cr 3 & 0 & 0 \cr 4 & 0 & 0 \cr 5 & 0 & 0 \cr 6 & 0 & 0 \cr 7 & 0 & 0.48 \cr } \qquad$

$W \times R = (I-Q)^{-1} \times R= \bordermatrix{ & 0 & 8 \cr 1 & 0.9071 & 0.0929 \cr 2 & 0.8065 & 0.1935 \cr 3 & 0.6975 & 0.3025 \cr 4 & 0.5794 & 0.4206 \cr 5 & 0.4514 & 0.5486 \cr 6 & 0.3128 & 0.6872 \cr 7 & 0.1627 & 0.8373 \cr } \qquad$

The first column of $W \times R$ gives the probabilities of ruin for the gambler. The smaller the initial capital, the greater the probability of ruin. With 3 units initially, the gambler has an almost 70% chance of ruin.

The fundamental matrix of an absorbing Markov chain is central to the technique described here. We now attempt to give an indication why the technique works.

Let $\mathbf{P}$ be the transition probability transition matrix for an absorbing Markov chain. Suppose that $\mathbf{P}$ is stated as follows

$\mathbf{P}=\left[\begin{array}{rr} Q & R \\ \mathbf{0} & I \\ \end{array}\right]$

such that $Q$ is the submatrix of $\mathbf{P}$ that consists of the one-step transition probabilities between transient states and $R$ is the submatrix of $\mathbf{P}$ that consists of the one-step transition probabilities from transient states into absorbing states. The fundamental matrix is $W=(I-Q)^{-1}$, which is the inverse of the matrix $I-Q$ where $I$ is the identity matrix of the same size as $Q$.

We sketch an answer to two questions. Why each entry $W_{ij}$ in the fundamental matrix $W$ is the mean time the process spent in state $j$ given that the process starts in state $i$? Why the matrix $W \times R$ gives the probabilities of absorption?

Note that the entries in $Q^n$, the $n$th power of the matrix $Q$, approach zero as $n$ becomes large. This is because $Q$ contains the one-step transition probabilities of the transient states (once in a transient state, the process has to leave the transient states at some point). We have the following equality.

$W=I+Q+Q^2+Q^3+\cdots+Q^n+\cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

To see the equality (3), fix transient states $i$ and $j$ with $i \ne j$. We show the following:

$W_{ii}=1+P_{ii}+P_{ii}^2+P_{ii}^3+\cdots+P_{ii}^n+\cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$

$W_{ij}=P_{ij}+P_{ij}^2+P_{ij}^3+\cdots+P_{ij}^n+\cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5)$

Suppose the process starts at state $i$. Define the indicator variables $Y_n$ and $Z_n$ for $n \ge 1$. At time $n$, if the process is in state $i$, let $Y_n=1$. Otherwise $Y_n=0$. Likewise, at time $n$, if the process is in state $j$, let $Z_n=1$. Otherwise $Z_n=0$. By definition,

$\displaystyle 1+Y_1+Y_2+Y_3+\cdots+Y_n+\cdots$

$\displaystyle Z_1+Z_2+Z_3+\cdots+Z_n+\cdots$

are the times spent in state $i$ and in state $j$, respectively. The 1 in the first random time accounts for the fact that the process is already in state $i$ initially. Because we are dealing with transient states, only a finite number of $Y_n$ and $Z_n$ can be 1’s. So these two random time variables are well defined. Each $Y_n$ is a Bernoulli random variable with $P[Y_n=1]=P_{ii}^n$. Likewise each $Z_n$ is a Bernoulli random variable with $P[Z_n=1]=P_{ij}^n$. The following gives (4) and (5).

\displaystyle \begin{aligned} W_{ii}&=1+E[Y_1]+E[Y_2]+\cdots+E[Y_n]+\cdots \\&=1+P_{ii}+P_{ii}^2+\cdots+P_{ii}^n+\cdots \end{aligned}

\displaystyle \begin{aligned} W_{ij}&=E[Z_1]+E[Z_2]+\cdots+E[Z_n]+\cdots \\&=P_{ij}+P_{ij}^2+\cdots+P_{ij}^n+\cdots \end{aligned}

Expressing equalities (4) and (5) in matrix form gives the equality (3). The following derivation shows that $W=(I-Q)^{-1}$.

\displaystyle \begin{aligned} W&=I+Q+Q^2+Q^3+\cdots+Q^n+\cdots \\&=I+Q (I+Q+Q^2+Q^3+\cdots+Q^n+\cdots) \\& = I+ Q W\end{aligned}

$W-Q W=I$

$W (I-Q)=I$

Thus the mean times spent in the transient states are obtained by taking the inverse of the matrix $I-Q$. Based on the derivation, the sum $I+Q+Q^2+Q^3+\cdots+Q^n$ for a sufficiently large $n$ would give a good approximation of the waiting time matrix $W$. Of course, computationally speaking, it is much easier to compute the inverse matrix of $I-Q$.

For the second question, it is instructive to examine the power of $\mathbf{P}$. By induction on the power $n$, $\mathbf{P}^n$ can be expressed as follows:

$\mathbf{P}^n=\left[\begin{array}{rr} Q^n & M_n \\ \mathbf{0} & I \\ \end{array}\right]=\left[\begin{array}{rr} Q^n & (I+Q+Q^2+Q^3+\cdots+Q^{n-1}) R \\ \mathbf{0} & I \\ \end{array}\right]$

The entries in the matrix $M_n$ are the $n$-step transition probabilities from a transient state to an absorbing state. For example, if the starting state is the transient state $i$ and $k$ is an absorbing state, then $P_{ik}^n$ is an entry of $M_n=(I+Q+Q^2+Q^3+\cdots+Q^{n-1}) R$. Starting at a transient state, the process will be absorbed at some point in time. So the matrix $M_n$ gives good approximation for probabilities of absorption when $n$ is large. Note that $M_n=(I+Q+Q^2+Q^3+\cdots+Q^{n-1}) R$ approaches $W \times R$ as $n$ goes to infinity.

The type of Markov chains discussed here are called absorbing Markov chains, the subject of next post.

Practice Problems

Practice problems to reinforce the concepts discussed here are available in a companion blog. There are two problem sets. The first one is here and the second one is here.

Dan Ma Markov chains

Daniel Ma Markov chains

$\copyright$ 2017 – Dan Ma