1. Lecture 12.5 – Additional Issues Concerning Discrete-Time Markov Chains
Topics
Review of DTMC
Classification of states
Economic analysis
First-time passage
Absorbing states

2. Discrete-Time Markov Chain
A stochastic process { Xn } where n  N = { 0, 1, 2, } is called a discrete-time Markov chain if
Pr{ Xn+1 = j | X0 = k0, , Xn-1 = kn-1, Xn = i }
= Pr{ Xn+1 = j | Xn = i }  transition probabilities
for every i, j, k0, , kn-1 and for every n.
The future behavior of the system depends only on the
current state i and not on any of the previous states.

3. Stationary Transition Probabilities
Pr{ Xn+1 = j | Xn = i } = Pr{ X1 = j |X0 = i } for all n
(They don’t change over time)
We will only consider stationary Markov chains.
The one-step transition matrix for a Markov chain
with states S = { 0, 1, 2 } is
where pij = Pr{ X1 = j | X0 = i }

4. Classification of States
Accessible: Possible to go from state i to state j (path exists in the network from i to j).
Two states communicate if both are accessible from each other. A system is irreducible if all states communicate.
State i is recurrent if the system will return to it after leaving some time in the future.
If a state is not recurrent, it is transient.

5. Classification of States (continued)
A state is periodic if it can only return to itself after a fixed number of transitions greater than 1 (or multiple of a fixed number).
A state that is not periodic is aperiodic.
a. Each state visited every 3 iterations
b. Each state visited in multiples of 3 iterations

6. Classification of States (continued)
An absorbing state is one that locks in the system once it enters.
This diagram might represent the wealth of a gambler who begins with $2 and makes a series of wagers for $1 each.
Let ai be the event of winning in state i and di the event of losing in state i.
There are two absorbing states: 0 and 4.

7. Classification of States (continued)
Class: set of states that communicate with each other.
A class is either all recurrent or all transient and may be either all periodic or aperiodic.
States in a transient class communicate only with each other so no arcs enter any of the corresponding nodes in the network diagram from outside the class. Arcs may leave, though, passing from a node in the class to one outside.

8. Illustration of Concepts
Example 1
Every pair of states communicates, forming a single recurrent class; however, the states are not periodic.
Thus the stochastic process is aperiodic and irreducible.

9. Illustration of Concepts
Example 2
States 0 and 1 communicate and for a recurrent class.
States 3 and 4 form separate transient classes.
State 2 is an absorbing state and forms a recurrent class.

10. Illustration of Concepts
Example 3
Every state communicates with every other state, so we have irreducible stochastic process.
Periodic?
Yes, so Markov chain is irreducible and periodic.

11. Classification of States
Example .6 .7 1 2 .4 3 4 .5 .4 .5 .5 .3 .8 .1 5 .2

12. A state j is accessible from state i if pij(n) > 0 for some n > 0.
In example, state 2 is accessible from state 1
& state 3 is accessible from state 5
but state 3 is not accessible from state 2.
States i and j communicate if i is accessible from j and j is accessible from i.
States 1 & 2 communicate; also
states 3, 4 & 5 communicate.
States 2 & 4 do not communicate
States 1 & 2 form one communicating class.
States 3, 4 & 5 form a 2nd communicating class.

13. If all states in a Markov chain communicate
(i.e., all states are members of the same communicating class)
then the chain is irreducible.
The current example is not an irreducible Markov chain.
Neither is the Gambler’s Ruin example which
has 3 classes: {0}, {1, 2, 3} and {4}.
First Passage Times
Let fii = probability that the process will return to state i
(eventually) given that it starts in state i.
If fii = 1 then state i is called recurrent.
If fii < 1 then state i is called transient.

14. If pii = 1 then state i is called an absorbing state.
Above example has no absorbing states
States 0 & 4 are absorbing in Gambler’s Ruin problem.
The period of a state i is the smallest k > 1 such that
all paths leading back to i have a length that is
a multiple of k;
i.e., pii(n) = 0 unless n = k, 2k, 3k, . . .
If a process can be in state i at time n or time n + 1
having started at state i then state i is aperiodic.
Each of the states in the current example are aperiodic

15. Example of Periodicity - Gambler’s Ruin
States 1, 2 and 3 each have period 2.
1 1-p 0 p 0 0
2 0 1-p 0 p 0
p 0 p
If all states in a Markov chain are
recurrent, aperiodic, & the chain is irreducible
then it is ergodic.

16. Existence of Steady-State Probabilities
A Markov chain is ergodic if it is aperiodic and allows the attainment of any future state from any initial state after one or more transitions. If these conditions hold, then
For example,
State-transition network 1 3 2
Conclusion: chain is ergodic.

17. Economic Analysis Two kinds of economic effects:
those incurred when the system is in a specified state, and
those incurred when the system makes a transition from one state to another.
The cost (profit) of being in a particular state is represented by the m-dimensional column vector
where each component is the cost associated with state i.
The cost of a transition is embodied in the m  m matrix where each component specifies the cost of going from state i to state j in a single step.

18. Expected Cost for Markov Chain
Expected cost of being in state i :
Let C = (c1, cm)T
ei = (0, 0, 1, 0, 0) be the ith row of the m  m identity matrix, and
fn = a random variable representing the economic return associated with the stochastic process at time n.
Property 3: Let {Xn : n = 0, 1, . . .} be a Markov chain with finite state space S, state-transition matrix P, and expected state cost (profit) vector C. Assuming that the process starts in state i, the expected cost (profit) at the nth step is given by 
E[fn(Xn) |X0 = i] = eiP(n)C.

19. Additional Cost Results
What if the initial state is not known?
Property 5: Let {Xn : n = 0, 1, . . .} be a Markov chain with finite state space S, state-transition matrix P, initial probability vector q(0), and expected state cost (profit) vector C. The expected economic return at the nth step is given by
  E[fn(Xn) |q(0)] = q(0)P(n)C.
Property 6: Let {Xn : n = 0, 1, . . .} be a Markov chain with finite state space S, state-transition matrix P, steady-state vector π, and expected state cost (profit) vector C. Then the long-run average return per unit time is given by
  SiS πici = πC.

20. Insurance Company Example
An insurance company charges customers annual
premiums based on their accident history
in the following fashion:
No accident in last 2 years: $250 annual premium
 Accidents in each of last 2 years: $800 annual premium
 Accident in only 1 of last 2 years: $400 annual premium
Historical statistics:
If a customer had an accident last year then they have a 10% chance of having one this year;
If they had no accident last year then they have a 3% chance of having one this year.

21. (N, N) (N, Y) (Y, N) (Y, Y) (N, N) 0.97 0.03 0 0 (N, Y) 0 0 0.90 0.10
Problem: Find the steady-state probability and the long-run average annual premium paid by the customer.
Solution approach: Construct a Markov chain with four states: (N, N), (N, Y), (Y, N), (Y,Y) where these indicate (accident last year, accident this year).
(N, N) (N, Y) (Y, N) (Y, Y)
(N, N)
(N, Y)
(Y, N)
(Y, Y)
P =

22. State-Transition Network for Insurance Company
.90
.03
.90
.03
Y, N
Y, Y
N, N
N, Y
.97
.10
.97
.10
This is an ergodic Markov chain.
All states communicate (irreducible)
Each state is recurrent (you will return, eventually)
Each state is aperiodic

23. Solving the Steady–State Equationsm
i=1
(N,N) = (N,N) (Y,N)
(N,Y) = (N,N) (Y,N)
(Y,N) = (N,Y) (Y,Y)
(N,N) + (N,Y)+(Y,N) + (Y,Y) = 1
pj = å pipij, j = 0,…,m
å pj = 1, pj  0,  j m
j =1
Solution:
(N,N) = 0.939, (N,Y) = 0.029, (Y,N) = 0.029, (Y,Y) = 0.003
& the long-run average annual premium is
0.939* * * *800 = 260.5

24. Markov Chain Add-in Matrix

25. Economic Data and Solution

26. Transient Analysis for Insurance Company

27. First Passage Times Let ij = expected number of steps to transition
from state i to state j
If the probability that we will eventually visit state j
given that we start in i is less than 1, then
we will have ij = +.
For example, in the Gambler’s Ruin problem,
20 = + because there is a positive probability
that we will be absorbed in state 4 given that we
start in state 2 (and hence visit state 0).

28. Computations for All States Recurrent
If the probability of eventually visiting state j given
that we start in i is 1 then the expected number
of steps until we first visit j is given by
mij = 1 + å pirmrj, for i = 0,1, , m–1
rj
We go from i to r in the first step with probability pir and it takes mrj steps from r to j.
It will always take
at least one step.
For j fixed, we have linear system in m equations and m unknowns mij , i = 0,1, , m–1.

29. First-Passage Analysis for Insurance Company
Suppose that we start in state (N,N) and want to find
the expected number of years until we have accidents
in two consecutive years (Y,Y).
This transition will occur with probability 1, eventually.
For convenience number the states
(N,N) (N,Y) (Y,N) (Y,Y)
Then, 03 = 1 + p00  p01 13 + p0223
13 = 1 + p10  p11 13 + p1223
23 = 1 + p20  p21 13 + p2223

30. First-Passage Computations
(N, N) (N, Y) (Y, N) (Y, Y)
(N, N)
(N, Y)
(Y, N)
(Y, Y) 1 2 3
states
Using P =
03 =  13
13 = 23
23 =  13
So, on average it takes years to transition
from (N,N) to (Y,Y).
Note, 03 = 23. Why? Note, 13 < 03.
Solution: 03 = 343.3, 13 = 310, 23 = 343.3

31. First Passage Probabilities

32. Game of Craps
Probability of win = Pr{ 7 or 11 } = = 0.223
Probability of loss = Pr{ 2, 3, 12 } = = 0.112

33. First Passage Probabilities for Craps

34. Absorbing States An absorbing state is a state j with pjj = 1.
Given that we start in state i, we can calculate the
probability of being absorbed in state j.
We essentially performed this calculation for the
Gambler’s Ruin problem by finding
P(n) = (pij(n) ) for large n.
But we can use a more efficient analysis
like that used for calculating first passage times.

35. Let 0, 1, . . . , k be transient states and
k + 1, , m – 1 be absorbing states.
Let qij = probability of being absorbed in state j given that we start in transient state i.
Then for each j we have the following relationship
qij = pij +  pirqrj , i = 0, 1, , k k
r = 0
Go directly to j Go to r and then to j
For fixed j (absorbing state) we have k + 1 linear equations in k + 1 unknowns, qrj , i = 0, 1, , k.

36. Absorbing States – Gambler’s Ruin
Suppose that we start with $2 and want to calculate the
probability of going broke, i.e., of being absorbed in state 0.
We know p00 = 1 and p40 = 0, thus
q20 = p20 + p21 q10 + p22 q p23 q30 (+ p24 q40)
q10 = p10 + p11 q10 + p12 q p13 q
q30 = p30 + p31 q10 + p32 q p33 q
where
P =
1 1-p 0 p 0 0
p 0 p 0
p 0 p

37. Solution to Gambler’s Ruin Example
Now we have three equations with three unknowns.
Using p = 0.75 (probability of winning a single bet)
we have
q20 = q q30
q10 = q20
q30 = q20
Solving yields q10 = 0.325, q20 = 0.1, q30 = 0.025
(This is consistent with the values found earlier.)

38. What You Should Know About The Mathematics of DTMCs
How to classify states.
What an ergodic process is.
How to perform economic analysis.
How to compute first-time passages.
How to compute absorbing probabilities.