# Chapman-Kolmogorov Equations

Stochastic processes and Markov chains are introduced in this previous post. Transition probabilities are an integral part of the theory of Markov chains. The post preceding this one is a beginning look at transition probabilities. This post shows how to calculate the $n$-step transition probabilities. The Chapman-Kolmogorov equations are also discussed and derived.

Introductory Example

Suppose $\left\{X_n: n=0,1,2,\cdots \right\}$ is a Markov chain with the transition probability matrix $\mathbf{P}$. The entries of the matrix are the one-step transition probabilities $P_{ij}$. The number $P_{ij}$ is the probability that the Markov chain will move to state $j$ at time $m+1$ given that it is in state $i$ at time $m$, independent of where the chain was prior to time $m$. Thus $P_{ij}$ can be expressed as the following conditional probability.

$(1) \ \ \ \ \ P_{ij}=P[X_{m+1}=j \lvert X_m=i]$

Thus the future state only depends on the period immediately preceding it. This is called the Markov property. Also note that $P_{ij}$ is independent of the time period $m$. Any Markov chain with this property is called a time-homogeneous Markov chain or stationary Markov chain.

The focus of this post is to show how to calculate the probability that the Markov chain will move to state $j$ at time $n+m$ given that it is in state $i$ at time $m$. This probability is denoted by $P_{ij}^n$. In other words, $P_{ij}^n$ is the probability that the chain will be in state $j$ after the Markov chain goes through $n$ more periods given that it is in state $i$ in the current period. The probability $P_{ij}^n$, as a conditional probability, can be notated as follows:

$(2) \ \ \ \ \ P_{ij}^n=P[X_{m+n}=j \lvert X_m=i]$

In (2), the $n$ in the $n$-step transition probabilities satisfies $n \ge 0$. Note that when $n=0$, $P_{ij}^0=1$ for $i=j$ and $P_{ij}^0=0$ for $i \ne j$. Including the case for $n=0$ will make the Chapman-Kolmogorov equations work better.

Before discussing the general method, we use examples to illustrate how to compute 2-step and 3-step transition probabilities. Consider a Markov chain with the following transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.6 & 0.2 & 0.2 \cr 1 & 0.3 & 0.5 & 0.2 \cr 2 & 0.4 & 0.3 & 0.3 \cr } \qquad$

Calculate the two-step transition probabilities $P_{02}^2$, $P_{12}^2$ and $P_{22}^2$. Then calculate the three-step transition probability $P_{02}^3$ using the two-step transition probabilities.

First, let’s handle $P_{02}^2$. We can condition on the first steps. To go from state 0 to state 2 in two steps, the chain must first go to an interim state and then from that state go to state 2.

\displaystyle \begin{aligned}P_{02}^2&=\sum \limits_{k=0}^2 P_{0k} \ P_{k2} \\&=P_{00} \ P_{02}+P_{01} \ P_{12}+P_{02} \ P_{22} \\&=0.6 \times 0.2+0.2 \times 0.2+0.2 \times 0.3 \\&=0.22 \end{aligned}

Note that the above calculation lists out the three possible paths to go from state 0 to state 2 in two steps – from state 0 to state 0 and then from state 0 to state 2, from state 0 to state 1 and then from state 1 to state 2 and from state 0 to state 2 and then from state 2 to state 2. Looking at the calculation more closely, the above calculation is basically the first row of $\mathbf{P}$ (the row corresponding to state 0) multiplying the third column of $\mathbf{P}$ (the column corresponding to state 2).

$P_{02}^2= \left[\begin{array}{ccc} 0.6 & 0.2 & 0.2 \\ \end{array}\right] \left[\begin{array}{c} 0.2 \\ 0.2 \\ 0.3 \end{array}\right]= \left[\begin{array}{c} 0.22 \end{array}\right]$

By the same idea, the following gives the other two 2-step transition probabilities.

\displaystyle \begin{aligned}P_{12}^2&=\sum \limits_{k=0}^2 P_{1k} \ P_{k2} \\&=P_{10} \ P_{02}+P_{11} \ P_{12}+P_{12} \ P_{22} \\&=0.3 \times 0.2+0.5 \times 0.2+0.2 \times 0.3 \\&=0.22 \end{aligned}

\displaystyle \begin{aligned}P_{22}^2&=\sum \limits_{k=0}^2 P_{2k} \ P_{k2} \\&=P_{20} \ P_{02}+P_{21} \ P_{12}+P_{22} \ P_{22} \\&=0.4 \times 0.2+0.3 \times 0.2+0.3 \times 0.3 \\&=0.23 \end{aligned}

As discussed, the above two calculations can be viewed as the sum of all the possible paths to go from the beginning state to the end state (conditioning on the interim state in the middle) or as a row in the transition probability $\mathbf{P}$ multiplying a column in $\mathbf{P}$. The following shows all three calculations in terms of matrix calculation.

$P_{02}^2= \left[\begin{array}{ccc} 0.6 & 0.2 & 0.2 \\ \end{array}\right] \left[\begin{array}{c} 0.2 \\ 0.2 \\ 0.3 \end{array}\right]= \left[\begin{array}{c} 0.22 \end{array}\right]$

$P_{12}^2= \left[\begin{array}{ccc} 0.3 & 0.5 & 0.2 \\ \end{array}\right] \left[\begin{array}{c} 0.2 \\ 0.2 \\ 0.3 \end{array}\right]= \left[\begin{array}{c} 0.22 \end{array}\right]$

$P_{22}^2= \left[\begin{array}{ccc} 0.4 & 0.3 & 0.3 \\ \end{array}\right] \left[\begin{array}{c} 0.2 \\ 0.2 \\ 0.3 \end{array}\right]= \left[\begin{array}{c} 0.23 \end{array}\right]$

The view of matrix calculation will be crucial in understanding Chpan-Kolmogorov equations discussed below. To conclude the example, consider the 3-step transition probability $P_{02}^3$. We can also condition on the first step. The chain goes from state 0 to the next step (3 possibilities) and then goes from that state to state 2 in two steps.

\displaystyle \begin{aligned} P_{02}^3&=\sum \limits_{k=0}^2 P_{0k} \ P_{k2}^2 \\&=P_{00} \ P_{02}^2+P_{01} \ P_{12}^2+P_{02} \ P_{22}^2 \\&=0.6 \times 0.22+0.2 \times 0.22+0.2 \times 0.23 \\&=0.222 \end{aligned}

The example shows that the calculation of a 3-step transition probability is based 2-step probabilities. Thus longer length transition probabilities can be built up from shorter length transition probabilities. However, it pays to focus on a more general framework before carrying out more calculation. The view of matrix calculation demonstrated above will help in understanding the general frame work.

Chapman-Kolmogorov Equations

The examples indicate that finding $n$-step transition probabilities involve matrix calculation. Let $\mathbf{Q}_n$ be the $n$-step transition probability matrix. The goal now is to have a systematic way to compute the entries in the matrix $\mathbf{Q}_n$. The computation is based on the Chapman-Kolmogorov equations. These equations are:

$\displaystyle (3) \ \ \ \ \ P_{ij}^{n+m}=\sum \limits_{k=0}^\infty P_{ik}^n \times P_{kj}^m$

for $n, m \ge 0$ and for all $i,j$. For a finite-state Markov chain, the summation in (3) does not go up to infinity and would have an upper limit. The number $P_{ij}^{n+m}$ is the probability that the chain will be in state $j$ after taking $n+m$ steps if it is currently in state $i$. The following gives the derivation of (3).

\displaystyle \begin{aligned} P_{ij}^{n+m}&=P[X_{n+m}=j \lvert X_0=i] \\&=\sum \limits_{k=0}^\infty P[X_{n+m}=j, X_n=k \lvert X_0=i]\\&=\sum \limits_{k=0}^\infty \frac{P[X_{n+m}=j, X_n=k, X_0=i]}{P[X_0=i]} \\&=\sum \limits_{k=0}^\infty \frac{P[X_n=k,X_0=i]}{P[X_0=i]} \times \frac{P[X_{n+m}=j, X_n=k, X_0=i]}{P[X_n=k,X_0=i]} \\&=\sum \limits_{k=0}^\infty P[X_n=k \lvert X_0=i] \times P[X_{n+m}=j \lvert X_n=k, X_0=i] \\&=\sum \limits_{k=0}^\infty P[X_n=k \lvert X_0=i] \times P[X_{n+m}=j \lvert X_n=k] \ * \\&=\sum \limits_{k=0}^\infty P_{ik}^n \times P_{kj}^m \end{aligned}

Here’s the idea behind the derivation. The path from state $i$ to state $j$ in $n+m$ steps can be broken down into two paths, one from state $i$ to an intermediate state $k$ in the first $n$ transitions, and the other from state $k$ to state $j$ in the remaining $m$ transitions. Summing over all intermediate states $k$ yields the probability that the process will go from state $i$ to state $j$ in $n+m$ transitions.

The entries in the matrix $\mathbf{Q}_n$ can be then computed by (3). There is an interesting an useful interpretation of (3). The following is another way to state the Chapman-Kolmogorov equations:

$(4) \ \ \ \ \ \mathbf{Q}_{n+m}=\mathbf{Q}_n \times \mathbf{Q}_m$.

A typical element of the matrix $\mathbf{Q}_n$ is $P_{ik}^n$ and a typical element of the matrix $\mathbf{Q}_m$ is $P_{kj}^m$. Note that $P_{ik}^n$ with $k$ varying is a row in $\mathbf{Q}_n$ (the row corresponding to the state $i$) and that $P_{kj}^n$ with $k$ varying is a column in $\mathbf{Q}_m$ (the column corresponding to the state $j$).

$\left[\begin{array}{cccccc} P_{i0}^n & P_{i1}^n & \cdots & P_{ik}^n & \cdots & P_{iw}^n \\ \end{array}\right] \left[\begin{array}{c} P_{0j}^m \\ \text{ } \\ P_{1j}^m \\ \vdots \\ P_{kj}^m \\ \vdots \\ P_{wj}^m \end{array}\right]$

The product of the above row and column is the transition probability $P_{ij}^{n+m}$.

Powers of the One-Step Transition Probability Matrix

Let $\mathbf{P}$ be the one-step transition probability matrix of a Markov chain. Let $\mathbf{Q}_n$ be the $n$-step transition probability matrix, which can be computed by using the Chapman-Kolmogorov equations. We now derive another important fact.

The $n$-step transition probability matrix $\mathbf{Q}_n$ is obtained by multiplying the one-step transition probability matrix $\mathbf{P}$ by itself $n$ times, i.e. $\mathbf{Q}_n$ is the $n$th power of $\mathbf{P}$.

This fact is important in terms of calculation of transition probabilities. Compute $\mathbf{P}^n$, the $n$th power of $\mathbf{P}$ (in terms of matrix calculation). Then $P_{ij}^n$ is simply an entry in $\mathbf{P}^n$ (in the row that corresponds to state $i$ and in the column that corresponds to state $j$). If the matrix size is large and/or $n$ is large, the matrix multiplication can be done using software or by using an online matrix calculator (here’s one matrix calculator).

Of course, the above fact is also important Markov chain theory in general. Almost all mathematical properties of Markov chains are at root based on this basic fact.

We can establish this basic fact using an induction argument. Clearly $\mathbf{Q}_1=\mathbf{P}^1=\mathbf{P}$. Suppose that the fact is true for $n-1$. Based on (4), $\mathbf{Q}_{n}=\mathbf{Q}_{n-1} \times \mathbf{Q}_1$. Continuing the derivation: $\mathbf{Q}_{n}=\mathbf{Q}_{n-1} \times \mathbf{Q}_1=\mathbf{P}^{n-1} \times \mathbf{P}=\mathbf{P}^n$.

The Row Vector and the Column Vector

As demonstrated in the preceding section, $\mathbf{P}^n$, the $n$th power of the $\mathbf{P}$, is precisely the $n$-step transition probability matrix. Let’s examine the matrix $\mathbf{P}^n$ more closely. Suppose that the Markov chain has a state space $\left\{0,1,2,\cdots,w \right\}$. The following shows the matrix $\mathbf{P}^n$ with special focus on a generic row and a generic column.

$\mathbf{P}^n= \left[\begin{array}{ccccccc} P_{0,0}^n & P_{0,1}^n & \cdots & P_{0,k}^n & \cdots & P_{0,w}^n \\ P_{1,0}^n & P_{1,1}^n & \cdots & P_{1,k}^n & \cdots & P_{1,w}^n \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ P_{i,0}^n & P_{i,1}^n & \cdots & P_{i,k}^n & \cdots & P_{i,w}^n \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ P_{w,0}^n & P_{w,1}^n & \cdots & P_{w,k}^n & \cdots & P_{w,w}^n \\ \end{array}\right]$

Now look at the row corresponding to the state $i$ and call it $R_i$. Also look at the column corresponding to the state $k$ and call it $C_k$. They are separated from the matrix $\mathbf{P}^n$ below.

$R_i=\left[\begin{array}{cccccc} P_{i,0}^n & P_{i,1}^n & \cdots & P_{i,k}^n & \cdots & P_{i,w}^n \\ \end{array}\right] \ \ \ \ \ \text{row vector}$

$C_k=\left[\begin{array}{c} P_{0,k}^n \\ \text{ } \\ P_{1,k}^n \\ \vdots \\ P_{i,k}^n \\ \vdots \\ P_{w,k}^n \end{array}\right] \ \ \ \ \ \text{column vector}$

The row vector $R_i$ is a conditional distribution. It gives the probabilities of the process transitioning (after $n$ transitions) into one of the states given the process starts in state $i$. If it is certain that the process begins in state $i$, then $R_i$ gives the probability function for the random variable $X_n$ since $P[X_n=k]=P_{i,k}^n$.

On the other hand, if the initial state is not certain, then the column vector $C_k$ gives the probabilities of the process transitioning (after $n$ transitions) into state $k$ from any one of the possible initial states. Thus a column from the matrix $\mathbf{P}^n$ gives the probability of the process in a given state at the $n$th period from all possible initial states. The following gives the probability distribution for $X_n$ (the state of the Markov chain at time $n$).

$\displaystyle (5) \ \ \ \ \ P[X_n=k]=\sum \limits_{i=0}^\infty P_{i,k}^n \times P[X_0=i]$

The calculation in (5) can be viewed as matrix calculation. If we put the probabilities $P[X_0=i]$ in a row vector, then the product of this row vector with the column vector $C_k$ would be $P[X_n=k]$. The following is (5) in matrix multiplication.

$(6) \ \ \ \ \ P[X_n=k]=\left[\begin{array}{cccccc} P(X_0=0) & P(X_0=1) & \cdots & P(X_0=j) & \cdots & P(X_0=w) \\ \end{array}\right] \left[\begin{array}{c} P_{0k}^n \\ \text{ } \\ P_{1k}^n \\ \vdots \\ P_{jk}^n \\ \vdots \\ P_{wk}^n \end{array}\right]$

Even though the matrix calculation for $P_{ij}^n$ should be done using software, it pays to understand the orientation of the matrix $\mathbf{P}^n$. In the preceding section, we learn that a row of $\mathbf{P}^n$ is the conditional distribution $X_n \lvert X_0$ (the state of the process at time $n$ given the initial state). In the preceding section, we also learn that a column in the matrix $\mathbf{P}^n$ gives the probabilities of the process landing in a fixed state $k$ from any one of the initial states.

The Chapman-Kolmogorov equations in (3) tells us that an entry in the matrix $\mathbf{P}^{n+m}$ is simply the product of a row in $\mathbf{P}^n$ and a column in $\mathbf{P}^m$. This observation makes it possible to focus just on the transition probability that is asked in a given problem rather than calculating the entire matrix. For example, to calculate an entry $P_{ij}^2$ of the matrix $\mathbf{P}^2$, multiply a row of $\mathbf{P}$ (the row corresponding to the state $i$) and a column of $\mathbf{P}$ (the row corresponding to state $j$). There is no need to calculate the entire matrix $\mathbf{P}^2$ if the goal is just one entry of $\mathbf{P}^2$.

To calculate an entry of $\mathbf{P}^n$, there is a considerable amount of variation in multiplication approaches. For example, multiple a row of $\mathbf{P}$ and a column of $\mathbf{P}^{n-1}$ or multiple a row of $\mathbf{P}^{n-1}$ and a column of $\mathbf{P}$. Or multiple a row of $\mathbf{P}^2$ and a column of $\mathbf{P}^{n-2}$.

If a row of transition probabilities multiplies a transition matrix, the result is a row in a higher transition probability matrix. For example, if a row in $\mathbf{P}$ multiplies the matrix $\mathbf{P}^n$, the result is a row in $\mathbf{P}^{n+1}$. More specifically, the first row in $\mathbf{P}$ multiplying the matrix $\mathbf{P}$ gives the first row of $\mathbf{P}^2$. The first row of $\mathbf{P}$ multiplying the matrix $\mathbf{P}^2$ gives the first row in $\mathbf{P}^3$, etc.

On the other hand, when a transition probability matrix multiplies a column of transition probabilities, the result is a column in a higher transition probability matrix. For example, the matrix $\mathbf{P}^n$ multiplies a column in $\mathbf{P}$, the result is a column in $\mathbf{P}^{n+1}$. More specifically, when the matrix $\mathbf{P}$ multiples the first column in the matrix $\mathbf{P}$, the result is the first column of $\mathbf{P}^2$. When the matrix $\mathbf{P}^2$ multiples a column in the matrix $\mathbf{P}$, the result is the first column in $\mathbf{P}^3$, etc.

Examples

We now give some examples on how to calculate transition probabilities.

Example 1
Consider a Markov chain with the following transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.6 & 0.2 & 0.2 \cr 1 & 0.3 & 0.5 & 0.2 \cr 2 & 0.4 & 0.3 & 0.3 \cr } \qquad$

Determine $P[X_3=1 \lvert X_0=0]$ and $P[X_5=2 \lvert X_2=0]$.

Of course, the most straightforward way would be to calculate $\mathbf{P}^3$. Then $P[X_3=1 \lvert X_0=0]=P_{01}^3$ would be the (0,1)th entry of $\mathbf{P}^3$ and $P[X_5=2 \lvert X_2=0]=P_{02}^3$ would be the (0,1)th entry of $\mathbf{P}^3$. In fact, it is a good practice to use an online matrix calculator for this task. Doing so produces the following marrices.

$\mathbf{P}^2 = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.50 & 0.28 & 0.22 \cr 1 & 0.41 & 0.37 & 0.22 \cr 2 & 0.45 & 0.32 & 0.23 \cr } \qquad \ \ \ \ \ \ \mathbf{P}^3 = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.472 & 0.306 & 0.222 \cr 1 & 0.445 & 0.333 & 0.222 \cr 2 & 0.458 & 0.319 & 0.223 \cr } \qquad$

From the matrix $\mathbf{P}^3$, we see that $P_{01}^3=0.306$ and $P_{02}^3=0.222$. Note that $P_{02}$ is calculated in the introductory example earlier.

Example 2
For the transition matrix $\mathbf{P}$ in Example 1, use the ideas discussed in the section Additional Comments on Chapman-Kolmogorov to compute the first row and the first column in the transition matrix $\mathbf{P}^4$.

$\left[\begin{array}{ccc} 0.6 & 0.2 & 0.2 \\ \end{array}\right] \left[\begin{array}{ccc} 0.472 & 0.306 & 0.222 \\ 0.445 & 0.333 & 0.222 \\ 0.458 & 0.319 & 0.223 \end{array}\right] = \left[\begin{array}{ccc} 0.4638 & 0.314 & 0.2222 \end{array}\right]= \left[\begin{array}{ccc} P_{00}^4 & P_{01}^4 & P_{02}^4 \end{array}\right]$

$\left[\begin{array}{ccc} 0.472 & 0.306 & 0.222 \\ 0.445 & 0.333 & 0.222 \\ 0.458 & 0.319 & 0.223 \end{array}\right] \left[\begin{array}{c} 0.6 \\ 0.3 \\ 0.4 \end{array}\right] = \left[\begin{array}{c} 0.4638 \\ 0.4557 \\ 0.4597 \end{array}\right]= \left[\begin{array}{c} P_{00}^4 \\ P_{10}^4 \\ P_{20}^4 \end{array}\right]$

As discussed in earlier, the first row of $\mathbf{P}^4$ is obtained by multiplying the first row of $\mathbf{P}$ with the matrix $\mathbf{P}^3$. The first column of $\mathbf{P}^4$ is obtained by multiplying the matrix $\mathbf{P}^3$ with the first column of the matrix $\mathbf{P}$.

Example 3
A particle moves among states 0, 1 and 2 according to a Markov process with the following transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.6 & 0.3 & 0.1 \cr 1 & 0.3 & 0.3 & 0.4 \cr 2 & 0.3 & 0.2 & 0.5 \cr } \qquad$

Let $X_n$ be the position of the particle after the $n$th move. Suppose that at the beginning, the particle is in state 1. Determine the probability $P[X_2=k]$ where $k=0,1,2$.

Since the initial state of the process is certain, $P[X_2=k]=P[X_2=k \lvert X_0=1]=P_{1k}^2$. Thus the problem is to find the second row of $\mathbf{P}^2$.

$\left[\begin{array}{ccc} 0.3 & 0.3 & 0.4 \\ \end{array}\right] \left[\begin{array}{ccc} 0.6 & 0.3 & 0.1 \\ 0.3 & 0.3 & 0.4 \\ 0.3 & 0.2 & 0.5 \end{array}\right] = \left[\begin{array}{ccc} 0.39 & 0.26 & 0.35 \end{array}\right]= \left[\begin{array}{ccc} P_{10}^2 & P_{11}^2 & P_{12}^2 \end{array}\right]$

Example 4
Consider a Markov chain with the following transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.3 & 0.2 & 0.5 \cr 1 & 0.1 & 0.3 & 0.6 \cr 2 & 0.4 & 0.3 & 0.3 \cr } \qquad$

Suppose that initially the process is equally likely to be in state 0 or state 1. Determine $P[X_3=0]$ and $P[X_3=1]$.

To find $P[X_3=0]$, we need the first column of $\mathbf{P}^3$. To find $P[X_3=1]$, we need the second column of $\mathbf{P}^3$. Using an online calculator produces $\mathbf{P}^3$.

$\mathbf{P}^3 = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.288 & 0.269 & 0.443 \cr 1 & 0.283 & 0.270 & 0.447 \cr 2 & 0.295 & 0.273 & 0.432 \cr } \qquad$

The following calculation gives the answers.

$P[X_3=0]=\left[\begin{array}{ccc} 0.5 & 0.5 & 0 \\ \end{array}\right] \left[\begin{array}{c} 0.288 \\ 0.283 \\ 0.295 \end{array}\right]=0.2855$

$P[X_3=1]=\left[\begin{array}{ccc} 0.5 & 0.5 & 0 \\ \end{array}\right] \left[\begin{array}{c} 0.269 \\ 0.270 \\ 0.273 \end{array}\right]=0.2695$

$P[X_3=2]=1-P[X_3=0]-P[X_3=1]=0.445$

Example 5
Consider a Markov chain with the following transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.6 & 0.2 & 0.2 \cr 1 & 0.2 & 0.2 & 0.6 \cr 2 & 0 & 0 & 1 \cr } \qquad$

At time 0, the Markov process is in state 0. Let $T=\text{min}\left\{n \ge 0:X_n=2 \right\}$. Note that state 2 is called an absorbing state. The Markov process will eventually reach and be absorbed in state 2 (it stays there forever whenever the process reaches state 2). Thus $T$ is the first time period in which the process reaches state 2. Suppose that the Markov process is being observed and that absorption has not taken place. Then we would be interested in this conditional probability: the probability that the process is in state 0 or state 1 given that the process has not been absorbed. Determine $P[X_4=0 \lvert T>4]$.

The key idea here is that for the event $T>4$ to happen, $X_4=0$ or $X_4=1$. Thus

$P[T>4]=P[X_4=0 \text{ or } X_4=1]=P[X_4=0]+P[X_4=1]=P_{00}^4+P_{01}^4$.

Note that since the initial state is certain to be state 0, $P[X_4=0]=P_{00}^4$ and $P[X_4=1]=P_{01}^4$. Using an online calculator gives the following matrix.

$\mathbf{P}^4 = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.1856 & 0.0768 & 0.7376 \cr 1 & 0.0768 & 0.032 & 0.8912 \cr 2 & 0 & 0 & 1 \cr } \qquad$

The following gives the desired result.

$\displaystyle P[X_4=0 \lvert T>4]=\frac{P_{00}^4}{P_{00}^4+P_{01}^4}=\frac{0.1856}{0.1856+0.0768}=0.7073$

Thus if absorption has not taken place in the 4th period, the process is in state 0 about 71% of the time.

Example 6
Continue with the Markov chain in Example 5. Determine $P[T=4]$.

Note that $P[T=4]=P[T>3]-P[T>4]$. The probability $P[T>4]$ is already determined using $\mathbf{P}^4$. To determine $P[T>3]$, we need the top row of $\mathbf{P}^3$.

$\mathbf{P}^3 = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.272 & 0.112 & 0.616 \cr 1 & 0.112 & 0.048 & 0.84 \cr 2 & 0 & 0 & 1 \cr } \qquad$

Thus $P[T>3]=P_{00}^3+P_{01}^3=0.272+0.112=0.384$. Thus $P[T=4]=0.384-0.2624=0.1216$. Thus about 12% of the time, absorption takes place in the 4th period about 12% of the time.

Exercises

We give a few practice problems to reinforce the concepts and calculation discussed here. Further practice problems are in a companion blog (here and here).

Exercise 1
Consider a Markov chain with the following transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.1 & 0.2 & 0.7 \cr 1 & 0.2 & 0.2 & 0.6 \cr 2 & 0.6 & 0.1 & 0.3 \cr } \qquad$

Determine these conditional probabilities.

• $P[X_3=2 \lvert X_1=1]$
• $P[X_3=2 \lvert X_0=1]$
• $P[X_4=2 \lvert X_0=1]$

Exercise 2
A particle moves among states 1, 2 and 3 according to a Markov process with the following transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 1 & 2 & 3 \cr 1 & 0 & 0.6 & 0.4 \cr 2 & 0.6 & 0 & 0.4 \cr 3 & 0.6 & 0.4 & 0 \cr } \qquad$

Let $X_n$ be the position of the particle after the $n$th move. Determine the probability $P[X_3=1 \lvert X_0=1]$ and $P[X_4=1 \lvert X_0=1]$.

Exercise 3
Consider a Markov chain with the following transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.1 & 0.2 & 0.7 \cr 1 & 0.9 & 0.1 & 0 \cr 2 & 0.1 & 0.8 & 0.1 \cr } \qquad$

Suppose that initially the process is equally likely to be in state 0 or state 1 or state 2. Determine the distribution for $X_3$.

Exercise 4
Continue with Example 5 and Example 6. Work these two examples assuming that the Markov chain starts in state 1 initially.

$\text{ }$

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Exercise 1

• $P[X_3=2 \lvert X_1=1]=P_{12}^2=0.13$
• $P[X_3=2 \lvert X_0=1]=P_{12}^3=0.16$
• $P[X_4=2 \lvert X_0=1]=P_{12}^4=0.1473$

Exercise 2

• $P[X_3=1 \lvert X_0=1]=P_{11}^3=0.24$
• $P[X_4=1 \lvert X_0=1]=P_{11}^4=0.456$

Exercise 3

• $\displaystyle P[X_3=0]=\frac{1.076}{3}=0.3587$
• $\displaystyle P[X_3=1]=\frac{1.013}{3}=0.3377$
• $\displaystyle P[X_3=2]=\frac{0.911}{3}=0.3037$

Exercise 4

• $\displaystyle P[X_4=0 \lvert T>4]=\frac{P_{10}^4}{P_{10}^4+P_{11}^4}=\frac{0.0768}{0.0768+0.0.032}=0.70588$
• $P[T=4]=P[T>3]-P[T>4]=0.16-0.1088=0.0512$

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$\copyright$ 2017 – Dan Ma

# A beginning look at transition probabilities

The previous post introduces the notion of Markov chains, more specifically discrete Markov chains. This and the next post focus on calculating transition probabilities.

Suppose that $\left\{X_n: n \ge 0 \right\}$ is a Markov chain with transition probability matrix $\mathbf{P}$. Elements of $\mathbf{P}$ are the one-step transition probabilities $P_{ij}$. As the Markov process moves through the states over time, the probabilities in the matrix $\mathbf{P}$ shows how likely the process will transition to state $j$ in the next time period if the process is currently in state $i$. Elements of $\mathbf{P}$ are conditional probabilities:

$(1) \ \ \ \ \ P_{ij}=P[X_{n+1}=j \lvert X_n=i]$

The transition probabilities described in (1) are independent of the time period $n$. These are called stationary transition probabilities. Markov chains with stationary transition probabilities are called stationary Markov chains or time-homogeneous Markov chains. One immediate goal is to develop the $n$-step transition probabilities $P_{ij}^n$ from the one-step transition probabilities $P_{ij}$. This is to be developed in the next post. This post focuses on basic calculations with $P_{ij}$.

Examples of Conditional Probabilities

We use examples to demonstrate conditional probabilities that can be calculated from the one-step transition probabilities $P_{ij}$. The calculation is based on the following transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 \cr 0 & 0.6 & 0.2 & 0.2 \cr 1 & 0.3 & 0.5 & 0.2 \cr 2 & 0.4 & 0.3 & 0.3 \cr } \qquad$

Example 1
Determine the following conditional probabilities:

• $P[X_1=0, X_2=1 \lvert X_0=0]$
• $P[X_2=0, X_3=1 \lvert X_1=0]$
• $P[X_1=2, X_2=1,X_3=0 \lvert X_0=1]$

Note that the first two are identical since the transition probabilities are stationary. To calculate the first and third probabilities, it is a matter of linking up the one-step transition probabilities from one state to the next.

\displaystyle \begin{aligned} P[X_1=0, X_2=1 \lvert X_0=0]&=P[X_1=0 \lvert X_0=0] \times P[X_2=1 \lvert X_1=0] \\&=P_{00} \times P_{01} \\&=0.6 \times 0.2=0.12 \end{aligned}

$\displaystyle P[X_2=0, X_3=1 \lvert X_1=0]=P[X_1=0, X_2=1 \lvert X_0=0]=0.12$

\displaystyle \begin{aligned} P[X_1=2, X_2=1,X_3=0 \lvert X_0=1]&=P[X_1=2 \lvert X_0=1] \times P[X_2=1 \lvert X_1=2] \times P[X_3=0 \lvert X_2=1] \\&=P_{12} \times P_{21} \times P_{10} \\&=0.2 \times 0.3 \times 0.3=0.018 \end{aligned}

Example 2
Example 1 shows that the probability of a chain moving from along a path conditional on a starting state is simply the product of the transition probabilities on that path. We now calculate unconditional probability of a path. For this to be possible, we need to assume a distribution for the initial state.

Assume that $P[X_0=0]=0.4$, $P[X_0=1]=0.3$ and $P[X_0=2]=0.3$. Determine the following unconditional probabilities:

• $P[X_0=0,X_1=0,X_2=1]$
• $P[X_1=0,X_2=0,X_3=1]$
• $P[X_0=1,X_1=2, X_2=1,X_3=0]$

To calculate these unconditional probabilities, we multiply the transition probabilities on the path along with the probability of the first state in the path.

\displaystyle \begin{aligned} P[X_0=0,X_1=0,X_2=1]&=P[X_0=0] \times P[X_1=0 \lvert X_0=0] \times P[X_2=1 \lvert X_1=0] \\&=P[X_0=0] \times P_{00} \times P_{01} \\&=0.4 \times 0.6 \times 0.2=0.048 \end{aligned}

To calculate the second probability, we need to first compute $P[X_1=0]$. This can be done by conditioning on the initial state $X_0$.

\displaystyle \begin{aligned} P[X_1=0]&=P[X_0=0] \times P_{00}+ P[X_0=1] \times P_{10} +P[X_0=2] \times P_{20} \\&=0.4 \times 0.6 +0.3 \times 0.3+0.3 \times 0.4=0.45 \end{aligned}

\displaystyle \begin{aligned} P[X_1=0,X_2=0,X_3=1]&=P[X_1=0] \times P[X_2=0 \lvert X_1=0] \times P[X_3=1 \lvert X_2=0]\\&=P[X_1=0] \times P_{00} \times P_{01} \\&=0.45 \times 0.6 \times 0.2 \\&=0.054 \end{aligned}

The following gives the third probability.

\displaystyle \begin{aligned} P[X_0=1,X_1=2, X_2=1,X_3=0]&=P[X_0=1] \times P[X_1=2 \lvert X_0=1] \times P[X_2=1 \lvert X_1=2] \times P[X_3=0 \lvert X_2=1] \\&=P[X_0=1] \times P_{12} \times P_{21} \times P_{10} \\&=0.4 \times 0.2 \times 0.3 \times 0.3 \\&=0.0072 \end{aligned}

Note that each unconditional probability in this example is obtained by multiplying the corresponding conditional probability in Example 1 with the probability of the conditioned event.

Example 3
Sometimes we know for certain that the Markov chain starts in a given state. Then the unconditional probability of a path is simply the conditional probability of that path (conditioned on the initial state). Determine the three probabilities in Example 2 given that the chain starts in state 0 in the first two probabilities and the chain starts in state 1 in the third probability. The probabilities are:

• $P[X_0=0,X_1=0,X_2=1]$
• $P[X_1=0,X_2=0,X_3=1]$
• $P[X_0=1,X_1=2, X_2=1,X_3=0]$

For the first probability, $P[X_0=0,X_1=0,X_2=1]=P[X_0=0] \times P[X_1=0,X_2=1 \lvert X_0=0]$. With $P[X_0=0]=1$, $P[X_0=0,X_1=0,X_2=1]=P[X_1=0,X_2=1 \lvert X_0=0]=0.12$, the answer from Example 1.

For the second probability, we need to first calculate $P[X_1=0]$. Since $P[X_0=0]=1$, $P[X_1=0]=P[X_0=0] \times P_{00}=0.6$. Thus $P[X_1=0,X_2=0,X_3=1]=P[X_1=0] \times P_{00} \times P_{01}=0.6 \times 0.6 \times 0.2=0.072$, which is 0.6 times the corresponding conditional probability in Example 1.

Similarly, the third probability is

\displaystyle \begin{aligned} P[X_0=1,X_1=2, X_2=1,X_3=0]&=P[X_1=2, X_2=1,X_3=0 \lvert X_0=1] \\& =0.018\end{aligned}

Conditional Probabilities of Paths in Markov Chains

The conditional probabilities in the above examples are not difficult to do as long as the basic idea of conditional probabilities and the Markov property are understood. First, the idea of conditional probability. The idea is that the probability of the intersection of two events can be obtained by conditioning on one of the events. For example, given two events $A$ and $B$, we have

$(2a) \ \ \ \ \ P[A \cap B]=P[A] \times P[B \lvert A]$

$\displaystyle (2b) \ \ \ \ \ P[B \lvert A]=\frac{P[A \cap B]}{P[A]}$

The probability of the intersection of two events is the product of a conditional probability of one event given another event times the probability of the event being conditioned. Relation (2b) is the probability of the event $B$ given that event $A$ has occurred. Of course, the conditioned event has to have positive probability, i.e. $P[A]>0$.

Suppose that the event $A$ is $X=x$, the discrete random variable $X$ taking on the value $x$ and that the event $B$ is $Y=y$. Then we have the following relations.

$(3a) \ \ \ \ \ P[X=x,Y=y]=P[X=x] \times P[Y=y \lvert X=x]$

$\displaystyle (3b) \ \ \ \ \ P[Y=y \lvert X=x]=\frac{P[X=x,Y=y]}{P[X=x]}$

We can also write the conditional probability of $X=x$ conditioning on the event $Y=y$. Suppose we have a path in a Markov chain, $X_0=x_0,X_1=x_1,X_2=x_2,\cdots,X_n=x_n$. This is the event that the chain starts in state $x_0$ initially, then move to state $x_1$ in time 1 and so on and at time $n$ the chain is in state $x_0$. Conditioning on the initial state, consider the following conditional probability.

$\displaystyle (4) \ \ \ \ \ P[X_1=x_1,X_2=x_2,\cdots,X_n=x_n \lvert X_0=x_0]=\frac{P[X_0=x_0,X_1=x_1,X_2=x_2,\cdots,X_n=x_n]}{P[X_0=x_0]}$

Relation (4) can be further simplified since the random variables $X_0=x_0,X_1=x_1,X_2=x_2,\cdots,X_n=x_n$ satisfy the Markov property.

\displaystyle \begin{aligned} (5) \ \ \ \ \ P[X_1=x_1,X_2=x_2,\cdots,X_n=x_n \lvert X_0=x_0]&=P[X_1=x_1 \lvert X_0=x_0] \times P[X_2=x_2 \lvert X_1=x_1] \\& \ \ \ \times \cdots \times P[X_n=x_n \lvert X_{n-1}=x_{n-1}] \\&=P_{x_0,x_1} \times P_{x_1,x_2} \times \cdots \times P_{x_{n-1},x_n} \end{aligned}

Relation (5) says that the probability of a path in a Markov chain conditioning on the initial state is simply the product of the transition probabilities along the path. This is derived using the Markov property. Relation (5) can be established by an induction argument on the subscript $n$, i.e. the length of a path.

Clearly $P[X_1=j \lvert X_0=i]=P_{ij}$. This is the fundamental assumption in Markov chains. Now consider $P[X_1=j,X_2=k \lvert X_0=i]$.

\displaystyle \begin{aligned} P[X_1=j,X_2=k \lvert X_0=i]&=\frac{P[X_0=i,X_1=j,X_2=k]}{P[X_0=i]} \\&=\frac{P[X_0=i,X_1=j,X_2=k]}{P[X_0=i,X_1=j]} \times \frac{P[X_0=i,X_1=j]}{P[X_0=i]} \\&=P[X_2=k \lvert X_0=i,X_1=j] \times P[X_1=j \lvert X_0=i] \\&=P[X_2=k \lvert X_1=j] \times P[X_1=j \lvert X_0=i] \\&=P_{jk} \times P_{ij} \\&=P_{ij} \times P_{jk} \end{aligned}

Note that $P[X_2=k \lvert X_0=i,X_1=j]=P[X_2=k \lvert X_1=j]$ in the above derivation. This is Markov property in action since the probability of the future state depends only on the state immediately preceding it. Now that relation (5) is true for a path of length 2, the following shows how it is derived for length 3.

\displaystyle \begin{aligned} P[X_1=j,X_2=k,X_3=l \lvert X_0=i]&=\frac{P[X_0=i,X_1=j,X_2=k,X_3=l]}{P[X_0=i]} \\&=\frac{P[X_0=i,X_1=j,X_2=k,X_3=l]}{P[X_0=i,X_1=j,X_2=k]} \times \frac{P[X_0=i,X_1=j,X_2=k]}{P[X_0=i]} \\&=P[X_3=l \lvert X_0=i,X_1=j,X_2=k] \times P[X_1=j,X_2=k \lvert X_0=i] \\&=P[X_3=l \lvert X_2=k] \times P[X_1=j,X_2=k \lvert X_0=i] \\&=P_{ij} \times P_{jk} \times P_{kl} \end{aligned}

Note that going from the third line to the fourth line the Markov property is used. Relation (5) for length 2 is used at the end. The inductive proof keeps going and relation (5) is true for any path of arbitrary length.

The conditional probability in (4) by itself may not be special. With the additional assumption of Markov property, the conditional probability can be made to be a “chain” of one-step transition probabilities.

Using the idea of (3a), the unconditional probability of a path is:

\displaystyle \begin{aligned} (6) \ \ \ \ \ P[X_0=x_0,X_1=x_1,X_2=x_2,\cdots,X_n=x_n]&=P[X_0=x_0] \times P_{x_0,x_1} \\& \ \ \times P_{x_1,x_2} \times \cdots \times P_{x_{n-1},x_n} \end{aligned}

Relation (6) says that the unconditional probability of a path in a Markov chain is simply the conditional probability of the path conditioning on the initial state times the probability of the initial state.

More Examples

The earlier examples demonstrate relation (5) and relation (6). We now work a few more examples.

Example 4
There is a slightly different way to look at one unconditional probability $P[X_1=0,X_2=0,X_3=1]$ in Example 2. The first state in this path is $X_1=0$. We can condition on the initial state $X_0$.

$\displaystyle P[X_1=0,X_2=0,X_3=1]=\sum \limits_{k=0}^2 P[X_0=k,X_1=0,X_2=0,X_3=1]$

Then every one of the unconditional probability within the summation sign can be computed according to the idea in (6). Note that Example 2 gives the distribution of the initial state.

\displaystyle \begin{aligned} P[X_1=0,X_2=0,X_3=1]&=P[X_0=0,X_1=0,X_2=0,X_3=1]+ \\& \ \ P[X_0=1,X_1=0,X_2=0,X_3=1]+P[X_0=2,X_1=0,X_2=0,X_3=1] \\&=P[X_0=0] \times P_{00} \times P_{00} \times P_{01} +\\&\ \ P[X_0=1] \times P_{10} \times P_{00} \times P_{01}+P[X_0=2] \times P_{20} \times P_{00} \times P_{01} \\&=0.4 \times 0.6 \times 0.6 \times 0.2+\\&\ \ 0.3 \times 0.3 \times 0.6 \times 0.2 + 0.3 \times 0.4 \times 0.6 \times 0.2\\&=0.054 \end{aligned}

In calculating probabilities in a Markov chain, conditioning on the first step is a useful technique that will be used often in the subsequent posts.

Example 5
Consider a Markov chain with the following transition probability matrix.

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

State 3 is called an absorbing state since $P_{33}=1$. In other words, when the chain reaches state 3, it will never leave. Suppose that initially the Markov chain is in state 0 50% of the time, in state 1 30% of the time and in state 2 20% of the time. Calculate the probability $P[X_1=0,X_2=1,X_3=0,X_4=2]$.

We can condition on the initial state as in Example 4.

\displaystyle \begin{aligned} P[X_1=0,X_2=1,X_3=0,X_4=2]&=\sum \limits_{k=0}^2 P[X_0=k,X_1=0,X_2=1,X_3=0,X_4=2] \\&=P[X_0=0,X_1=0,X_2=1,X_3=0,X_4=2]+ \\&\ \ P[X_0=1,X_1=0,X_2=1,X_3=0,X_4=2]+\\&\ \ P[X_0=2,X_1=0,X_2=1,X_3=0,X_4=2] \\&=P[X_0=0] \times P_{00} \times P_{01} \times P_{10} \times P_{02}+ \\& \ \ P[X_0=1] \times P_{10} \times P_{01} \times P_{10} \times P_{02}+\\&\ \ P[X_0=2] \times P_{20} \times P_{01} \times P_{10} \times P_{02} \\&=0.0048\end{aligned}

Example 6
This is Example 1 in this previous post.

When a binary digit, 0 or 1, is transmitted through a communication system, it passes through several stages. At each stage, there is a probability $p$ that the digit is transmitted in error. Let $X_n$ be the digit that is the output at the $n$th stage. Then $\left\{X_n: n \ge 0 \right\}$ is a two-state Markov chain with the following transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & \text{ } & 1 \cr 0 & 1-p & \text{ } & p \cr \text{ } & \text{ } & \text{ } &\text{ } \cr 1 & p & \text{ } & 1-p \cr }$

Determine the following:

• If the digit to be sent is 0, determine the probability that no error occurs up to the second stage.
• If the digit to be sent is 0, determine the probability that a correct message is received at stage 2 through stage 4.

The first probability is $P[X_1=0,X_2=0 \lvert X_0=0]$. This is $P_{00} \times P_{00}=(1-p)^2$. To calculate the second probability, we can condition on the state in time 1. We first calculate $P[X_1=0]$ and $P[X_1=1]$. Since we know for certain that the initial state is 0, $P[X_1=0]=P_{00}=1-p$ and $P[X_1=1]=P_{01}=p$.

\displaystyle \begin{aligned} P[X_2=0,X_3=0,X_4=0]&=P[X_1=0,X_2=0,X_3=0,X_4=0]+P[X_1=1,X_2=0,X_3=0,X_4=0] \\&=P[X_2=0,X_3=0,X_4=0 \lvert X_1=0] \times P[X_1=0]+ \\&\ \ P[X_2=0,X_3=0,X_4=0 \lvert X_1=1] \times P[X_1=1] \\&=P_{00} \times P_{00} \times P_{00} \times (1-p)+ \\&\ \ P_{10} \times P_{00} \times P_{00} \times p\\&=(1-p)^4+p^2 \times (1-p)^2 \end{aligned}

Remarks

The calculation in this post involves very basic calculation with one-step transition probabilities, i.e. the elements of the transition probability matrix given in a Markov chain. The $n$-step transition probabilities are key to the development of the theory of Markov chains. The next post begins the discussion of Chapman-Kolmogorov equations.

Practice problems are in this companion blog.

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$\copyright$ 2017 – Dan Ma

# Introducing Markov Chains

This post continues the discussion of the previous two posts – the previous post that looks at random walks informally through graphs and the previous post that defines random walks more formally. This post introduces the notion of Markov chains.

Stochastic Process

A stochastic process is a collection of random variables $\left\{X_t: t \in T \right\}$. Each element of the collection is a random variable $X_t$. The index $t$ is often regarded as time. Thus we refer to $X_t$ the state of the process at time $t$. Thus a stochastic process is a family of random variables that describes the evolution through time of some physical process.

The set $T$ is called the index set of the stochastic process. When $T$ is a countable set, e.g. $T=\left\{0,1,2,3,\cdots \right\}$, the stochastic process is said to be a discrete-time process. When the index set $T$ is an interval of the real number line, the stochastic process is said to be a continuous-time process. For example, $\left\{X_n: n=0,1,2,\cdots \right\}$ is a discrete-time stochastic process indexed by the non-negative integers whereas $\left\{X_t: t \ge 0 \right\}$ is a continuous-time stochastic process indexed by the non-negative real numbers. The state space of a stochastic process is defined as the set of all possible values that can be taken by the random variables $X_t$.

In the present discussion, we focus on discrete-time stochastic processes $\left\{X_n: n=0,1,2,\cdots \right\}$ where the state space $S$ is finite or countably infinite. When the state space is finite, the process is said to be a finite-state stochastic process.

We also assume that the state space $S$ consists of integers. Thus $S$ is either $\mathbb{N}=\left\{0,1,2,3,\cdots \right\}$ (the natural numbers) or $\mathbb{Z}=\left\{\cdots,-3,-2,-1,0,1,2,3,\cdots \right\}$ (the integers), or some subset of $\mathbb{N}$ or $\mathbb{Z}$.

If we know nothing about $X_n$ other than the fact that they are random variables, then there is not much we can say about the stochastic process $\left\{X_n: n=0,1,2,\cdots \right\}$. To make the discussion more meaningful, it is necessary to impose some additional structure on these random variables.

The simplest structure we can assume is that the random variables $X_n$ are independent random variables. This would be a good model for phenomena that can be regarded as random experiments in which the future states of the process are independent of past and present states. However, in many situations the assumption of independence is often unjustified. In many stochastic processes that arise in practice, past and present states can influence the future states.

Markov Chains

We consider stochastic processes that possess the following property: the future states of the process (conditional on both past and present states) depends only upon the present state, not on the past states leading up to the present state. This property is called the Markov property. Any stochastic process that possesses the Markov property is called a Markov chain. More specifically, to satisfy the Markov property, given the present state $X_n$ and the past states $X_{n-1},X_{n-2},\cdots,X_2,X_1,X_0$, the conditional distribution of the future state $X_{n+1}$ depends only on the present state $X_n$. Even more specifically, to satisfy the Markov property, the process must satisfy the following requirement:

$\displaystyle (1) \ \ \ \ P[X_{n+1}=j \lvert X_n=i,X_{n-1}=x_{n-1},\cdots,X_0=x_0]=P[X_{n+1}=j \lvert X_n=i]$

for all possible states $x_0,x_1, \cdots,x_{n-1},i,j$ and all non-negative integers $n$. Thus under the Markov property, the conditional distribution of the future state $X_{n+1}$ depends only on the present state $X_n$ and that all the states preceding the present state have no influence on the future state.

The conditional probabilities $P[X_{n+1}=j \lvert X_n=i]$ in (1) are called transition probabilities of the Markov chain. In the present discussion, we consider only the Markov chains in which the transition probabilities $P[X_{n+1}=j \lvert X_n=i]$ are independent of the current period $n$. Such Markov chains are called time-homogeneous Markov chains or stationary Markov chains. So the additional assumption of time-homogeneity means that the probability of transitioning into state $j$ from state $i$ is identical regardless of where the process is in the time scale (at the beginning in the process or later in the process).

With the transition probability $P[X_{n+1}=j \lvert X_n=i]$ independent of $n$, we can denote this probability by the states $i$ and $j$ only:

$\displaystyle (2) \ \ \ \ P_{ij}=P[X_{n+1}=j \lvert X_n=i]$

The probability $P_{ij}$ represents the probability that the process, starting in state $i$, will enter into state $j$ in the next period. Since the process must transition into one of the states in the state space in the next time period, $P_{ij}$, with $i$ fixed and $j$ varying, must represent a probability distribution. Thus $P_{i0}+ P_{i1}+\cdots=1$. Here we assume that the state space is the set of all non-negative integers.

The probabilities $P_{ij}$ are called one-step transition probabilities since they give the probabilities of transitioning from one state to another state in one period of time. The one-step transition probabilities are usually expressed in a matrix or a transition diagram. The following shows what a transition probability matrix looks like.

$\mathbf{P}= \left[\begin{array}{cccccc} P_{0,0} & P_{0,1} & P_{0,2} & \cdots & P_{0,m} \\ P_{1,0} & P_{1,1} & P_{1,2} & \cdots & P_{1,m} \\ P_{2,0} & P_{2,1} & P_{2,2} & \cdots & P_{2,m} \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ P_{m,0} & P_{m,1} & P_{m,2} & \cdots & P_{m,m} \\ \end{array}\right]$

The above transition matrix is for a finite-state Markov chain with state space being $\left\{0,1,\cdots,m \right\}$. The probabilities in each row sum to 1. If the process is currently in state $i$, the row with the first index as $i$ (in this case the $(i+1)$st row) will give the probabilities of transitioning into all the states. If the state space is countably infinite, then each row would have infinitely many terms and there would be infinitely many rows.

In many situations, it is helpful to have a column label and row label to specify the states in the Markov chain. For example, a particle moves through the states 0, 1, 2, 3 and 4 according to a Markov chain described by the following transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 \cr 0 & 0 & 0.6 & 0.2 & 0.1 & 0.1 \cr 1 & 0.5 & 0 & 0.2 & 0.2 & 0.1 \cr 2 & 0.4 & 0.2 & 0 & 0.1 & 0.3 \cr 3 & 0.2 & 0.3 & 0.2 & 0 & 0.3 \cr 4 & 0.6 & 0.2 & 0.1 & 0.1 & 0 \cr } \qquad$

In the above matrix, $P_{31}=0.3$, which means that when the Markov process is in state 3, it moves to state 1 with probability 0.3. As discussed earlier, the probabilities in a given row sum to 1, which makes sense since the probabilities in a row describes all possible moves when the process is a given state.

When it is helpful to do so, the transition probability matrices will have a row label and a column label. If the row label and column label are not given, we assume that the rows and columns are listed in ascending order (recall that the states are integers). The transition probability matrices can be displayed using square brackets or parentheses.

Examples

The key to working with Markov chains is the transition probabilities $P_{ij}$. These probabilities allow us describe the random transitions through the states over time. The remainder of this post gives several examples of Markov chains focusing on transition probability matrices.

Example 1
When a binary digit, 0 or 1, is transmitted through a communication system, it passes through several stages. At each stage, there is a probability $p$ that the digit is transmitted in error. Let $X_n$ be the digit that is the output at the $n$th stage. Then $\left\{X_n: n \ge 0 \right\}$ is a two-state Markov chain with the following transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & \text{ } & 1 \cr 0 & 1-p & \text{ } & p \cr \text{ } & \text{ } & \text{ } &\text{ } \cr 1 & p & \text{ } & 1-p \cr }$

Some typical problems might be:

• If the digit to be sent is 0, determine the probability that no error occurs up to the second stage.
• If the digit to be sent is 0, determine the probability that a correct message is received at stage 2.
• If the digit to be sent is 0, determine the probability that a correct message is received at stage $n$ where $n \ge 1$.

Example 2
Let’s consider the example of gambler’s ruin with finitely many states. Suppose that a gambler makes a series of one-unit bets against the house. For the gambler, the probabilities of winning and losing each bet are $p$ and $1-p$, respectively. Whenever the capital reaches zero, the gambler is in ruin and his capital remains zero thereafter. On the other hand, if the capital of the gambler increases to $m$, a large level of capital, the gambler quits playing. Let $X_n$ be the capital of the gambler after the $n$th bet. Then $\left\{ X_n: n=0,1,2,\cdots \right\}$ is a random walk, as discussed in the previous post. Since the future capital $X_{n+1}$ depends only on the current level of capital $X_n$ (up one unit or down one unit), this is also a Markov chain. The following is the transition probability matrix.

$\mathbf{P} = \bordermatrix{ & 0 & 1 & 2 & 3 & 4 & \cdots & m \cr 0 & 1 & 0 & 0 & 0 & 0 & \cdots & 0 \cr 1 & 1-p & 0 & p & 0 & 0 & \cdots & 0 \cr 2 & 0 & 1-p & 0 & p & 0 &\cdots & 0 \cr 3 & 0 & 0 & 1-p & 0 & p &\cdots & 0 \cr \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \cr m & 0 & 0 & 0 & 0 & 0 &\cdots & 1 } \qquad$

The above is an $(m+1) \times (m+1)$ matrix. The states are of the process are $0,1,2,\cdots,m$. The states $0$ and $m$ are absorbing states since the process would stay there and not leave once the process arrive at these two states. Note that $P_{00}=1$ and $P_{mm}=1$. State 0 would be the state of ruin. State $m$ would mean the gambler becomes fabulously rich (if $m$ is large).

For the rows in between, the transition probabilities are $P_{i,i-1}=1-p$ and $P_{i,i+1}=p$. In other words, from state $i$, the process would go to state $i+1$ with probability $p$ and would go down to $i-1$ with probability $1-p$. More specifically, if $m=5$, the following is the transition probability matrix.

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

Assuming that the gambler has initial capital $d$ which is positive but small in comparison to $m$. For example, $d=10$ and $m=1000$. Some interesting questions are:

• If the staring capital is 10, what is the probability that the gambler’s capital will be 5 after 8 plays of the game?
• What is the probability of ruin for the gambler, i.e. reaching state 0?
• What is the mean time until reaching the absorbing states (ruin or fabulously rich)?

Example 3 (Restless Mouse)
A maze has eight areas as shown below.

The rat moves through the areas in the maze at random. That is, if an area has exits to $w$ areas, the rat moves to each of these $w$ areas with probability $1/w$. Let $X_n$ be the area the mouse is located in after the $n$th move. Then $\left\{X_n: n \ge 0 \right\}$ is a Markov chain with the following transition probability matrix.

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

Here’s some interesting questions about this Markov chain. Suppose that area 8 contains food and area 6 has a device for an electric shock.

• Given that the rat is placed in area 1 at the beginning, what is the probability that the rat will reach area 8 in 4 moves, in general in $n$ moves?
• Given that the rat is placed in area 1 at the beginning, what is the mean number of moves for the rat to reach area 8 or area 6?
• Given that the rat is placed in area 1 at the beginning, what is the probability that the rat will reach the food area before the electric shock area?

Next Steps

The concept of Markov chains has been introduced and illustrated with examples of random walks (in earlier posts) and other examples in this post. All the properties of Markov chains (in this case time-homogeneous Markov chains) are derived from the transition probability matrix. One of the most urgent tasks is to develop a way to calculation the transition probabilities. Recall that the elements of the transition probability matrix $\mathbf{P}$ are the one-step transition probabilities $P_{ij}$, which gives the probability of going from state $i$ into state $j$. The natural next step is to calculate the $n$-step transition probabilities $P^n_{ij}$, which is the probability that the process will in state $j$ in the $n$th period given the process is initially in state $i$. The next post is a beginning look at transition probabilities. The post that follows the next post is on Chapman-Kolmogorov equations, a systematic way of calculating transition probabilities using matrix multiplication.

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$\copyright$ 2017 – Dan Ma

# More on random walks

This post continues the discussion of the previous post on random walks. The previous post looks at random walks graphically. This post introduces the notion of random walks more formally.

Consider $\left\{X_n: n=0,1,2,\cdots \right\}$, which is a sequence of random variables indexed by the non-negative integers. By definition this is a stochastic process, to be precise a discrete stochastic process since the random variables are indexed by integers. The integers in the subscripts can be interpreted as time. Thus the random variables $X_n$ describe the states of some process over time. The stochastic processes discussed in the previous post, the future state $X_{n+1}$ of the process is obtained by adding one (with probability $p$) or subtracting one (with probability $1-p$) to the current state $X_n$. So a given state of the process $X_n$ is one higher or lower than the state in the previous period. The random walks in the previous post go up or down with equal chance, i.e. $p=0.5$. The following graph is from the previous post, which shows 5 simulations of such random walks.

Figure 1 – Examples of Random Walks in One Dimension

Random Walks – A General Description

We now give a more general description of the random walk. Instead of a random one-unit up or down move, the moves in the random walk are determined by a predetermined discrete distribution. Let $Y_1,Y_2,Y_3,\cdots$ be integer-valued random variables that are independent and identically distributed. Let $P[Y=y]$ be the common probability function. Let $X_0$ be an integer-valued random variable that is independent of the random variables $Y_n$. The random variable $X_0$ will be the initial position of the random walk. The random variables $Y_n$ are the increments (they are the amounts added to the stochastic process as time increases).

For $n \ge 1$, define $X_n=X_0+Y_1+\cdots+Y_{n-1}+Y_n$. Then $\left\{X_n: n=0,1,2,\cdots \right\}$ is called a random walk. Wherever the process is at any given time, the next step in the process is always a walk in a distance and direction that is dictated by the independent random variables $Y_n$. For convenience, we will call the random variables $Y_n$ the increments in the random walk since they tell the process how far (and what direction) to walk in the next step.

The state space is the set of all the values that can be taken on by the random variables $X_n$. In this case, the state space is either the set of all integers $\mathbb{Z}=\left\{\cdots,-3,-2,-2,0,1,2,3,\cdots \right\}$ or the set of all non-negative integers $\mathbb{N}=\left\{0,1,2,3,\cdots \right\}$ or some appropriate subset of $\mathbb{Z}$ or $\mathbb{N}$.

Given the current state of the process $X_n$, the next state $X_{n+1}$ is obtained by adding the random value $Y_{n+1}$ to the previous state $X_n$. If the random variables $Y_n$ are independent Bernoulli variables with $P(Y_n=1)=p$ and $P(Y_n=-1)=1-p$. Then each state $X_n$ is the result of an up or down move by one unit from the previous state. The examples in the previous post are based on this type of Bernoulli one-unit up and down move. In the general definition, the future state $X_{n+1}$, instead of being adjusted by a uniform up or down adjustment, is adjusted by the random variable $Y_{n+1}$.

One important task is to establish the probabilities of the transition from one state to another. For example, if the process is currently in state $i$, what is the probability that it will be in state $j$ in the next period? More specifically, given that the random walk is in state $i$, i.e. $X_n=i$, what is the probability that the process in in state $j$ in the next period, i.e. $X_{n+1}=j$? Such probabilities are called one-step transition probabilities.

The probabilities that we wish to obtain are $P[X_{n+1}=j \lvert X_n=i]$ for $n \ge 0$. Recall that $P[Y=y]$ is the common probability function for the random variables $Y_n$. Given states $i,j$, define $P_{ij}=P[Y=j-i]$.

When $X_n=i$, in order for $X_{n+1}=j$ to happen, the increment must be $Y_{n+1}=j-i$. Note that $P[Y_{n+1}=j-i]=P[Y=j-i]=P_{ij}$. Observe that this probability is independent of the time period $n$ since the $Y_n$ have a common distribution $Y$. Thus it is appropriate to use $P_{ij}$ to express this probability, i.e. there is no need to include the time $n$ in the subscript of $P_{ij}$. For all $n \ge 0$,

$P[X_{n+1}=j \lvert X_n=i]=P[Y_{n+1}=j-i]=P[Y=j-i]=P_{ij}$

The number $P_{ij}$ is the probability of the process transitioning from state $i$ to state $j$ in one step, which is evaluated based on the common probability function of the increments $Y_n$.

Now that we know the probability of the process going one state to another state in one step, other probability questions are: what is the probability of a path and what is the probability of going from state $i$ to state $j$ in $n$ steps? The probabilities in the second question are called $n$-step transition probabilities. The probability of a path is discussed here. The $n$-step transition probabilities are discussed here.

The first question has straightforward answers. For example, if the process starts in state $i_0$ at time 0, moves to state $i_1$ at time 1 and so on until reaching state $i_n$ at time $n$, then the following is the probability of this path:

\displaystyle \begin{aligned} P[X_0=i_0,X_1=i_1,\cdots,X_n=i_n]&=P[X_0=i_0,Y_1=i_1-i_0,\cdots,Y_n=i_n-i_{n-1}] \\&=P[X_0=i_0] \times P[Y_1=i_1-i_0] \times \cdots \times P[Y_n=i_n-i_{n-1}] \\&=P[X_0=i_0] \times P_{i_0,i_1} \times \cdots \times P_{i_{n-1},i_n} \end{aligned}

\displaystyle \begin{aligned} P[X_1=i_1,\cdots,X_n=i_n \lvert X_0=i_0]&=\frac{P[X_0=i_0,X_1=i_1,\cdots,X_n=i_n]}{P[X_0=i_0]} \\&=P_{i_0,i_1} \times \cdots \times P_{i_{n-1},i_n} \end{aligned}

Conditional on the initial state $X_0=i_0$, the probability of the process going through the path $X_1=i_1,\cdots,X_n=i_n$ is simply the product of the one-step transition probabilities.

If the path is observed to start at a time period other than time 0, conditional on the first state in the path, the probability of the process going through the path would be:

$\displaystyle P[X_{k+1}=i_1,\cdots,X_{k+n}=i_n \lvert X_k=i_0]=P_{i_0,i_1} \times \cdots \times P_{i_{n-1},i_n}$

Note that the transition probabilities $P_{ij}$ and the probabilities of paths are independent of the current period $n$. Random walks with such property are called time-homogeneous or stationary. Thus the probability of transitioning into state $j$ from state $i$ or the probability of transitioning through a path is identical regardless of where the process is in the time scale (at the beginning in the process or later in the process).

Specific Examples

For special cases of the random walks discussed above, let the increments $Y_n$ be defined by a Bernoulli random variable with $P[Y=1]=p$ and $P[Y=-1]=1-p$. Then the resulting random walk is a series of up and down moves as shown in Figure 1 above. The one-step transition probabilities are:

$\displaystyle P_{ij} = \left\{ \begin{array}{ll} \displaystyle p &\ \ \ \ \ \ j=i+1 \\ \text{ } & \text{ } \\ \displaystyle 1-p &\ \ \ \ \ \ j=i-1 \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ \text{otherwise} \end{array} \right.$

If the increments $Y_n$ are defined by the distribution where $P[Y=1]=p$, $P[Y=-1]=q$ and $P[Y=0]=1-p-q$, then the following gives the transition probabilities:

$\displaystyle P_{ij} = \left\{ \begin{array}{ll} \displaystyle p &\ \ \ \ \ \ j=i+1 \\ \text{ } & \text{ } \\ \displaystyle q &\ \ \ \ \ \ j=i-1 \\ \text{ } & \text{ } \\ \displaystyle 1-p-q &\ \ \ \ \ \ j=i \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ \text{otherwise} \end{array} \right.$

This random walk has three moves at any given state, an up move by one unit, a down move by one unit and a possibility that the process stays at the same state. Of course, the possibility for modeling is limitless. We can use other distributions for the common distribution of the increments $Y_n$, e.g. binomial distribution, Poisson, preferably having these distributions shifted in order to account for up moves and down moves.

Finite-State Random Walks

The random walks discussed above potentially have an infinite state space, i.e. the process can potentially reach far from the initial state without bounds. Depending on the common distribution of the increments $Y_1,\cdots,Y_n,\cdots$, the process can go up indefinitely to $+\infty$ or go down to $-\infty$. The random walks in Figure 1 have increments 1 and -1 and can reach states that are arbitrarily high or low, though it would take a great many steps to reach such high levels.

Any random walk with unbounded state space can be modified to have a finite state space. Pick two states that serve as boundaries, one lower and one upper. The states in between the boundary states are called interior states. Whenever the process reaches the upper boundary or the lower boundary, the process either stays there and not move any further or transition back into an interior state. The transition probabilities in the interior states remain the same as discussed above. If a boundary state is such that the process always stays there after reaching it, it is called an absorbing state.

A handy example of a finite-state random walk is the gambler’s ruin. Suppose that a gambler starts out with $d$ units in capital (in dollars or other monetary units) and makes a series of one-unit bets against the house. For the gambler, the probabilities of winning and losing each bet are $p$ and $1-p$, respectively. Whenever the capital reaches zero, the gambler is in ruin and his capital remains zero thereafter. On the other hand, if the capital of the gambler increases to $m$ where $d, then the gambler quits playing. Let $X_n$ be the capital of the gambler after the $n$th bet. Then $\left\{X_n: n=0,1,2,3,\cdots \right\}$ is a random walk. The starting state is $X_0=d$. The states $0$ and $m$ are absorbing states. The following gives the transition probabilities.

$\displaystyle P_{ij} = \left\{ \begin{array}{ll} \displaystyle p &\ \ \ \ \ \ j=i+1,i \ne 0, i \ne m \\ \text{ } & \text{ } \\ \displaystyle 1-p &\ \ \ \ \ \ j=i-1,i \ne 0, i \ne m \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ i=0,j=0 \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ i=m,j=m \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ \text{otherwise} \end{array} \right.$

Whenever the gambler reaches the absorbing state of 0, the gambler is in ruin. If the process reaches the absorbing state of $m$, the gambler strikes gold. If the initial capital of the gambler is small relative to the casino, there is a virtually 100% chance that the gambler will be in ruin even if each bet has even odds (see here for the calculation). One interesting questions: on average how long does it take for the gambler to be in ruin? Such questions will be answered after necessary tools are built in subsequent posts.

Remarks

A random walk is a stochastic process $\left\{X_n: n=0,1,2,\cdots \right\}$ such that at each time point, the process walks a certain distance that is dictated by an independent and identically distributed sequence of random variables. In such a stochastic process, the future state depends only on the preceding state and not on the past states before the preceding state. For example, in the gambler’s ruin, the gambler’s capital at any given point depends only on the his capital in the previous play.

In a stochastic process $\left\{X_n: n=0,1,2,\cdots \right\}$, the property that the future state $X_{n+1}$ depends only on the preceding state $X_n$ and not on the past states $X_0,X_1,\cdots,X_{n-1}$ is called the Markov property. Any stochastic process with the Markov property is called a Markov chain. A random walk is a Markov chain. There are Markov chains that are not random walk. The notion of Markov chains is introduced in this post.

$\copyright$ 2017 – Dan Ma

# A walk down Random Lane

A stochastic process is a sequence of random variables $\left\{X_n: n=0,1,2,3,\cdots \right\}$. Since the random variables are indexed by the integers, this is a discrete stochastic process. The integers in the subscripts can also be interpreted as time. Suppose $X_0$ is the random value taken on by the process at time 0. Then $X_1$ is the random value taken on by the physical process at time 1 and so on. The notion of stochastic processes will be examined in more details in subsequent posts. This post looks at plots of some simple random walks, a particular type of stochastic processes. The next post gives a more formal definition of random walks.

Random Walk

One type of stochastic processes is a random walk, which can be regarded as a sequence of random steps (up or down for example) on some mathematical spaces such as the integers. For example, assume that $X_0$ is an integer (the initial position of the walk) and further assume that it is equally likely that $X_n$ is either $X_{n-1}-1$ or $X_{n-1}+1$. Then $\left\{X_n: n=0,1,2,3,\cdots \right\}$ is an example of a random walk. A colorful way to describe this random walk is that a drunkard takes steps along a path at random by taking one step to the left or one step to the right with equal chance. For convenience, we can assume that the initial position is $X_0=0$ (the origin). To demonstrate the random nature of the walk, we simulate such a random walk five times using random numbers generated by the function =Rand() in Excel.

Figure 1 – Examples of Random Walks in One Dimension

All 5 examples of the walk starts with $X_0=0$. One hundred steps are shown for each random walk. Each random move is basically the result of a coin toss. If the toss is a head, the step is to the right, which means adding one to the previous position (in the graph the path would go up by one step). If the coin toss is a tail, the step is to the left, which means subtracting one to the previous position (the graph would go down one step).

It is clear that the positions of the drunkard are initially close to the origin but as time goes by the drunkard may be far from the origin (either to the left or to the right). At the end of the 100 moves, the drunkard is about 10-12 steps from either side of the origin.

For each random walk in Figure 1, the random moves to the left or to the right (up or down in the plot) are generated using the function =Rand() in Excel. The following are the first 10 random values for the random walk in red in Figure 1.

0.762417629
0.475581918
0.916720504
0.455643747
0.918072861

0.141459931
0.871570535
0.813575225
0.092492375
0.823638012

If a random value is less than 0.5, we take that as a tail and if it is more than 0.5, we take that as a head. With $X_0=0$, the next 10 positions are 1, 0, 1, 0, 1, 0, 1, 2, 1, 2.

The random walks in Figure 1 have no restriction. The random variables $X_n$ can take on both positive and negative integers without bounds. The set of all values that can be taken by the random variables in the random walk is called the state space. So in these examples, the state space is $\mathbb{Z}=\left\{\cdots,-3,-2,-1,0,1,2,3,\cdots \right\}$, the set of all integers.

The random walks in Figure 1 have finite versions, i.e. the state space is a finite set. For example, the path the drunkard is on has barriers at both ends (10 steps away from the origin at either end). Whenever the drunkard reach 10 or -10, he stops. For example, his house is at step 10 and a bar is at step -10. Whenever he reaches the house or the bar, he stays there. The following is a graph of the same random walks with the 10-step limitatons.

Figure 2 – Examples of Random Walks in One Dimension with Absorbing States

Note that when the process reaches 10 or -10, it stays there from that point on. Such states are called absorbing states. The random walks in Figure 2 have two absorbing states 10 and -10. With minor adjustments, these random walks have a gambling interpretation.

Gambler’s Ruin

Suppose that a gambler starts out with $d$ units in capital (in dollars or other monetary units) and makes a series of one-unit bets against the house. For the gambler, the probabilities of winning and losing each bet are $p$ and $1-p$, respectively. Whenever the capital reaches zero, the gambler is in ruin and his capital remains zero thereafter. On the other hand, if the capital of the gambler increases to $m$ where $d, then the gambler quits playing. Let $X_n$ be the capital of the gambler after the $n$th bet. Then $\left\{X_n: n=0,1,2,3,\cdots \right\}$ is a random walk. The starting state is $X_0=d$. The states $0$ and $m$ are absorbing states.

Note the similarity between the random walk that describes the gambler’s ruin and the random walks in Figure 2. In gambler’s ruin, the capital $X_n$ is increased or decreased by one unit at random (probability $p$ for an increase and probability $1-p$ for a decrease). Whenever the process reaches 0, it stays at 0 thereafter (the gambler is in ruin and he stops playing). Similarly, whenever the process reaches $m$, the process remains at $m$ (the gambler reaches his goals and stops playing). The following is a graph of five simulations (100 plays each) with starting capital 10 units ($d=10$) and maximum target 100 units ($m=100$).

Figure 3 – Gambler’s Ruin – 5 Simulations, 100 Plays Each

Note the similarity in shapes between Figure 2 and Figure 3. The 100 random moves in Figure 3 are obtained by the same random numbers behind the random walks in Figure 2. Two of the simulations end in ruin (the purple at 36th play and the black at 38th play) fairly early in the process. At the end of 100 plays, the other three simulations are not yet in ruin, even though the random walk in green appears to be close to be in ruin. Let’s extend the random walks up to 500 plays of the gambling games.

Figure 4 – Gambler’s Ruin – 5 Simulations, 500 Plays Each

In all 500 plays of the game, no random walks have reached the maximum target of 100 units. In fact, the gambler is in ruin in four of the simulations, the purple and the black mentioned earlier along with the green and the red. The yellow random walk will likely be in ruin too if the game continues.

An alternative interpretation of the gambler’s ruin random walk is this. Assume that two players make a series of one-unit bets against each other and that the total capital between the two players is $m$ units with the first player having $d$ units in capital. Further suppose that the first player has probability $p$ of winning a bet and the second player has probability $1-p$ of winning a bet. Then the two gamblers play until one of them goes broke (is in ruin). Let $X_n$ be the capital of the first player after the $n$th play. Then $\left\{X_n: n=0,1,2,3,\cdots \right\}$ is the same random walk demonstrated in Figure 3 and Figure 4.

In Figures 3 and 4, the initial capital of the first player is 10 units and the total capital of the two players is 100 units. The probability of the first player winning all 100 units is 10/100 = 0.10 (see this post on gambler’s ruin). In other words, there is a 90% chance for the first player going broke! Whenever the gambler’s capital is small and is dwarfed by the other player (e.g. a casino), the probability of ruin for the gambler is all by certain. The 90% chance of ruin is based on the bets being even bets, which is not the case in the casino. Thus being in ruin in casino games is even more of a certainty.

Markov Chains

For a stochastic process $\left\{X_n: n=0,1,2,3,\cdots \right\}$ in general, there is no additional assumption on the random variables $X_n$. We can assume that $X_n$ are independent random variables. However, the independence assumption may not be realistic. In many applications, the future states may be dependent on the past and current states. Note that in the random walk examples discussed in this post, the future state $X_{n+1}$ is dependent on the preceding state $X_{n}$. In the gambler’s ruin example, the future capital $X_n$ is always one unit above or below the capital of the previous period. In the drunkard’s random walk, the future position is always one step to the left or to the right of the previous position. The past position preceding the immediately preceding period do not influence the future state.

The examples discussed in this post are excellent beginning examples to illustrate the stochastic processes $\left\{X_n: n=0,1,2,3,\cdots \right\}$ in which the future state $X_{n+1}$ depends only on the preceding state $X_n$ and in which the past states $X_0,X_1,\cdots,X_{n-1}$ do not influence the future states. This property is called the Markov property. Any stochastic process with the Markov property is called a Markov chain. A random walk is an example of a Markov chain. Random walks are discussed further in the next post. The notion of Markov chains is introduced in this post.

$\copyright$ 2017 – Dan Ma