8/3/2019 Set3 Markov Chains
1/50
1
Chapter 17Markov Chains
8/3/2019 Set3 Markov Chains
2/50
2
Description
Sometimes we are interested in how a random variablechanges over time.
The study of how a random variable evolves over time
includes stochastic processes.
An explanation of stochastic processes in particular, a type
of stochastic process known as a Markov chain is included.
We begin by defining the concept of a stochastic process.
8/3/2019 Set3 Markov Chains
3/50
3
5.1 What is a Stochastic Process?
Suppose we observe some characteristic of a system at discretepoints in time.
Let Xt be the value of the system characteristic at time t. In
most situations, Xt is not known with certainty before time t
and may be viewed as a random variable. A discrete-time stochastic process is simply a description of
the relation between the random variables X0, X1, X2 ..
Example: Observing the price of a share of Intel at the
beginning of each day Application areas: education, marketing, health services,
finance, accounting, and production
8/3/2019 Set3 Markov Chains
4/50
4
A continuous time stochastic process is simplythe stochastic process in which the state of the
system can be viewed at any time, not just at discrete
instants in time.
For example, the number of people in a supermarket
tminutes after the store opens for business may be
viewed as a continuous-time stochastic process.
8/3/2019 Set3 Markov Chains
5/50
5
5.2 What is a Markov Chain?
One special type of discrete-time is called a MarkovChain.
Definition: A discrete-time stochastic process is a
Markov chain if, fort= 0,1,2 and all statesP(Xt+1 = it+1|Xt = it, Xt-1=it-1,,X1=i1, X0=i0)
=P(Xt+1=it+1|Xt = it)
Essentially this says that the probability distribution
of the state at time t+1 depends on the state at time
t(it) and does not depend on the states the chain
passed through on the way to it at time t.
8/3/2019 Set3 Markov Chains
6/50
6
In our study of Markov chains, we make furtherassumption that for all states i andj and all t,P(Xt+1
=j|Xt = i) is independent oft.
This assumption allows us to writeP(Xt+1 =j|Xt = i)=pij wherepijis the probability that given the system
is in state i at time t, it will be in a statej at time t+1.
If the system moves from state i during one period tostatej during the next period, we call that a
transition from i toj has occurred.
8/3/2019 Set3 Markov Chains
7/50
7
Thepijs are often referred to as the transition
probabilities for the Markov chain.
This equation implies that the probability law
relating the next periods state to the current state
does not change over time. It is often called the Stationary Assumption and
any Markov chain that satisfies it is called a
stationary Markov chain.
We also must define qi to be the probability that the
chain is in state i at the time 0; in other words,
P(X0=i) = qi.
8/3/2019 Set3 Markov Chains
8/50
8
We call the vectorq= [q1, q2,qs] the initial
probability distribution for the Markov chain.
In most applications, the transition probabilities are
displayed as ans xstransition probability matrixP. The transition probability matrixPmay be written
as
=
ssss
s
s
ppp
ppp
ppp
P
21
22221
11211
8/3/2019 Set3 Markov Chains
9/50
9
For eachI
We also know that each entry in thePmatrix must benonnegative.
Hence, all entries in the transition probability matrix
are nonnegative, and the entries in each row must
sum to 1.
=
=
=sj
j
ijp1
1
8/3/2019 Set3 Markov Chains
10/50
10
The Gamblers Ruin Problem
At time 0, I have $2. At times 1, 2, , I play a game in
which I bet $1, with probabilities p, I win the game,and with probability 1 p, I lose the game. My goal isto increase my capital to $4, and as soon as I do, thegame is over. The game is also over if my capital is
reduced to 0. Let Xt represent my capital position after the time t game (if
any) is played
X0, X1, X2, . May be viewed as a discrete-time stochastic
process
8/3/2019 Set3 Markov Chains
11/50
11
The Gamblers Ruin Problem
$0 $1 $2 $3 $4
P =
100000100
0010
0001
00001
pp
pp
pp
8/3/2019 Set3 Markov Chains
12/50
12
5.3 n-Step Transition Probabilities
A question of interest when studying a Markov chain is: If aMarkov chain is in a state i at time m, what is the probability
that n periods later the Markov chain will be in statej?
This probability will be independent ofm, so we may write
P(Xm+n =j|Xm = i) =P(Xn =j|X0 = i) =Pij(n)wherePij(n) is called the n-step probability of a transition
from state i to statej.
Forn > 1,Pij(n) = ijth element ofPn
Pij(2) is the (i, j)th element of matrix P2 = P1 P1
Pij(n) is the (i, j)th element of matrix Pn = P1 Pn-1
8/3/2019 Set3 Markov Chains
13/50
13
The Cola Example
Suppose the entire cola industry produces only twocolas.
Given that a person last purchased cola 1, there is a
90% chance that their next purchase will be cola 1.
Given that a person last purchased cola 2, there is an
80% chance that their next purchase will be cola 2.1. If a person is currently a cola 2 purchaser, what is the probability
that they will purchase cola 1 two purchases from now?
2. If a person is currently a cola 1 a purchaser, what is the
probability that they will purchase cola 1 three purchases from
now?
8/3/2019 Set3 Markov Chains
14/50
14
The Cola Example
We view each persons purchases as a Markov chainwith the state at any given time being the type of cola
the person last purchased.
Hence, each persons cola purchases may be
represented by a two-state Markov chain, where State 1 = person has last purchased cola 1
State 2 = person has last purchased cola 2
If we define Xn to be the type of cola purchased by aperson on hernth future cola purchase, then X0, X1,
may be described as the Markov chain with the
following transition matrix:
8/3/2019 Set3 Markov Chains
15/50
15
The Cola Example
We can now answer questions 1 and 2.
1. We seekP(X2
= 1|X0
= 2) =P21
(2) = element 21 of
P2:
=
21
80.20.
10.90.
2
1ColaCola
Cola
ColaP
=
=
66.34.
17.83.
80.20.
10.90.
80.20.
10.90.2P
8/3/2019 Set3 Markov Chains
16/50
16
The Cola Example
Hence,P21(2) =.34. This means that the probability is .34
that two purchases in the future a cola 2 drinker will
purchase cola 1.
2. We seekP11(3) = element 11 ofP3:
Therefore,P11(3) = .781
=
== 562.438.
219.781.
66.34.
17.83.
80.20.
10.90.
)(
23PPP
8/3/2019 Set3 Markov Chains
17/50
17
Many times we do not know the state of the Markov chain at time0. Then we can determine the probability that the system is in state i
at time n by using the reasoning.
Probability of being in statej at time n
where q=[q1, q2, q3].
Hence, qn = qopn = qn-1p
Example, q0
= (.4,.6) q1= (.4,.6)
q1 = (.48,.52)
)(1
nPq ij
si
i
i=
=
=
80.20.
10.90.
8/3/2019 Set3 Markov Chains
18/50
18
To illustrate the behavior of the n-step transitionprobabilities for large values ofn, we have computed
several of the n-step transition probabilities for the
Cola example.
This means that for large n, no matter what the initial
state, there is a .67 chance that a person will be a cola
1 purchaser.
8/3/2019 Set3 Markov Chains
19/50
19
5.4 Classification of States in a
Markov Chain
To understand the n-step transition in more detail, weneed to study how mathematicians classify the states
of a Markov chain.
The following transition matrix illustrates most ofthe following definitions. A graphical representation
is shown in the book (State-Transition diagram)
=
2.8.000
1.4.5.00
07.3.000005.5.
0006.4.
P
8/3/2019 Set3 Markov Chains
20/50
20
Definition: Given two states ofi andj, a path from i
toj is a sequence of transitions that begins in i andends inj, such that each transition in the sequence
has a positive probability of occurring.
Definition: A statej is reachable from state i if
there is a path leading from i toj. Definition: Two states i andj are said to
communicate ifj is reachable from i, and i is
reachable fromj.
Definition: A set of states Sin a Markov chain is a
closed set if no state outside ofSis reachable from
any state in S.
8/3/2019 Set3 Markov Chains
21/50
21
Definition: A state i is an absorbing state ifpij
=1.
Definition: A state i is a transient state if there exists a statej that isreachable from i, but the state i is not reachable from statej.
Definition:If a state is not transient, it is called a recurrent state.
Definition:A state i is periodic with period k > 1 if k is the smallestnumber such that all paths leading from state i back to state i have alength that is a multiple of k. If a recurrent state is not periodic, it isreferred to as aperiodic.
If all states in a chain are recurrent, aperiodic, and communicate witheach other, the chain is said to be ergodic.
The importance of these concepts will become clear after the next two
sections.
8/3/2019 Set3 Markov Chains
22/50
22
5.5 Steady-State Probabilities and
Mean First Passage Times
Steady-state probabilities are used to describe thelong-run behavior of a Markov chain.
Theorem 1: LetPbe the transition matrix for an
s-state ergodic chain. Then there exists a vector = [1 2 s] such that
=
s
s
s
nn
P
21
21
21
lim
8/3/2019 Set3 Markov Chains
23/50
23
Theorem 1 tells us that for any initial state i,
The vector = [1 2 s] is often called the
steady-state distribution, orequilibriumdistribution, for the Markov chain. Hence, they are
independent of the initial probability distribution
defined over the states
jijn
nP =
)(lim
8/3/2019 Set3 Markov Chains
24/50
24
Transient Analysis & Intuitive
Interpretation
The behavior of a Markov chain before the steadystate is reached is often call transient (or short-run)
behavior.
An interpretation can be given to the steady-state
probability equations.
This equation may be viewed as saying that in thesteady-state, the flow of probability into each state
must equal the flow of probability out of each state.
=jk
kjkjjj pp )1(
8/3/2019 Set3 Markov Chains
25/50
25
Steady-State Probabilities
The vector= [1, 2, . , s ] is often known as the steady-state
distribution for the Markov chain For large n and all i,
Pij(n+1) Pij(n) j
In matrix form = P
For any n and any i,
Pi1(n) + Pi2(n) + + Pis(n) = 1
As n, we have 1 + 2 + . + s = 1
=
=
=
=
=+
s
kkjkj
kj
sk
k ikij
p
PnPnP
1
1)()1(
8/3/2019 Set3 Markov Chains
26/50
26
An Intuitive Interpretation of Steady-State
Probabilities
Consider
Subtracting jpjj from both sides of the above equation, wehave
Probability that a particular transition enters state j =probability that a particular transition leaves state j
=
=s
kkjkj p
1
=jk
kjkjjj pp )1(
f S d S b bili i i
8/3/2019 Set3 Markov Chains
27/50
27
Use of Steady-State Probabilities in
Decision Making
In the Cola Example, suppose that each customermakes one purchase of cola during any week.
Suppose there are 100 million cola customers.
One selling unit of cola costs the company $1 toproduce and is sold for $2.
For $500 million/year, an advertising firm guarantees
to decrease from 10% to 5% the fraction of cola 1
customers who switch after a purchase.
Should the company that makes cola 1 hire the firm?
8/3/2019 Set3 Markov Chains
28/50
28
At present, a fraction 1 = of all purchases are cola1 purchases, since:
1 = .901+.2022 = .101+.802
and using the following equation by 1 + 2 = 1
Each purchase of cola 1 earns the company a $1profit. We can calculate the annual profit as
$3,466,666,667 [2/3(100 million)(52 weeks)$1].
The advertising firm is offering to change theP
matrix to
=80.20.
05.95.1P
8/3/2019 Set3 Markov Chains
29/50
29
ForP1, the steady-state equations become1 = .951+.2022 = .051+.802
Replacing the second equation by 1 + 2 = 1 andsolving, we obtain 1=.8 and2 = .2.
Now the cola 1 companys annual profit will be
$3,660,000,000 [.8(100 million)(52 weeks)$1-($500million)].
Hence, the cola 1 company should hire the ad
agency.
8/3/2019 Set3 Markov Chains
30/50
30
Inventory Example A camera store stocks a particular model camera that can be ordered
weekly. Let D1, D2, represent the demand for this camera (the number
of units that would be sold if the inventory is not depleted) during the firstweek, second week, , respectively. It is assumed that the Dis areindependent and identically distributed random variables having a Poissondistribution with a mean of 1. Let X0 represent the number of cameras onhand at the outset, X1 the number of cameras on hand at the end of week 1,X2 the number of cameras on hand at the end of week 2, and so on.
Assume that X0 = 3. On Saturday night the store places an order that is delivered in time for the
next opening of the store on Monday. The store using the following order policy: If there are no cameras in stock,
3 cameras are ordered. Otherwise, no order is placed. Sales are lost when demand exceeds the inventory on hand
I E l
8/3/2019 Set3 Markov Chains
31/50
31
Inventory Example
Xt is the number of Cameras in stock at the end of week t (as defined
earlier), where Xt represents the state of the system at time t
Given that Xt = i, Xt+1 depends only on Dt+1 and Xt (Markovian property)
Dt has a Poisson distribution with mean equal to one. This means that
P(Dt+1 = n) = e-11n/n! for n = 0, 1,
P(Dt = 0 ) = e-1 = 0.368
P(Dt = 1 ) = e
-1
= 0.368 P(Dt = 2 ) = (1/2)e
-1 = 0.184
P(Dt 3 ) = 1 P(Dt 2) = 1 (.368 + .368 + .184) = 0.08
Xt+1 = max(3-Dt+1, 0) if Xt = 0 and Xt+1 = max(Xt Dt+1, 0) if Xt 1, for t =
0, 1, 2, .
I t E l (O St ) T iti
8/3/2019 Set3 Markov Chains
32/50
32
Inventory Example: (One-Step) Transition
Matrix P
03
= P(Dt+1
= 0) = 0.368
P02 = P(Dt+1 = 1) = 0.368
P01 = P(Dt+1 = 2) = 0.184
P00 = P(Dt+1 3) = 0.080
44434241
34333231
23221211
04030201
3
2
1
0
3210
pppp
pppp
pppp
pppp
8/3/2019 Set3 Markov Chains
33/50
33
Inventory Example: Transition Diagram
0 1
2 3
8/3/2019 Set3 Markov Chains
34/50
34
Inventory Example: (One-Step) Transition
Matrix
368.368.184.080.3
0368.368.264.2
00368.632.1
368.368.184.080.0
3210
8/3/2019 Set3 Markov Chains
35/50
35
Transition Matrix: Two-Step
P(2) = PP
165.300.286.249.3097.233.319.351.2
233.233.252.283.1
165.300.286.249.0
3210
)2( =P
8/3/2019 Set3 Markov Chains
36/50
36
Transition Matrix: Four-Step
P(4) = P(2)P(2)
164.261.286.289.3171.263.283.284.2
166.268.285.282.1
164.261.286.289.0
3210
)4( =P
8/3/2019 Set3 Markov Chains
37/50
37
Transition Matrix: Eight-Step
P(8) = P(4)P(4)
166.264.285.286.3166.264.285.286.2
166.264.285.286.1
166.264.285.286.0
3210
)8(=P
8/3/2019 Set3 Markov Chains
38/50
38
Steady-State Probabilities
The steady-state probabilities uniquely satisfy the following
steady-state equations
0 = 0p00 + 1p10 + 2p20 + 3p30
1 =
0p01 +
1p11 +
2p21 +
3p31 2 = 0p02 + 1p12 + 2p22 + 3p32
3 = 0p03 + 1p13 + 2p23 + 3p33
1 = 0 + 1 + 2 + 3
1
......s2,1,0,jfor
0
0
=
==
=
=
s
jj
s
iijij p
8/3/2019 Set3 Markov Chains
39/50
39
Steady-State Probabilities: Inventory Example
0 = .0800 + .6321 + .2642+ .0803
1 = .1840 + .3681 + .3682 + .1843
2 = .3680 + .3682 + .3683
3 = .3680 + .3683
1 = 0 + 1 + 2 + 3
0 = .286, 1 = .285, 2 = .263, 3 = .166
The numbers in each row of matrix P(8) match the
corresponding steady-state probability
8/3/2019 Set3 Markov Chains
40/50
40
Mean First Passage Times
For an ergodic chain, let mij = expected number oftransitions before we first reach statej, given that we
are currently in state i; mij is called the mean first
passage time from state i to statej. In the example, we assume we are currently in state i.
Then with probabilitypij, it will take one transition to
go from state i to statej. Fork j, we next go with
probabilitypik to state k. In this case, it will take an
average of 1 + mkj transitions to go from i andj.
8/3/2019 Set3 Markov Chains
41/50
41
This reasoning implies
By solving the linear equations of the equationabove, we find all the mean first passage times. It can
be shown that
+=jk
kjikij mpm 1
i
iim1=
8/3/2019 Set3 Markov Chains
42/50
42
For the cola example, 1=2/3 and2 = 1/3
Hence, m11 = 1.5 andm22 = 3
m12 = 1 + p11m12= 1 + .9m12
m21 = 1 + p22m21 = 1 + .8m21
Solving these two equations yields,
m12 = 10 andm21 = 5
S l i f St d St t P b biliti
8/3/2019 Set3 Markov Chains
43/50
43
Solving for Steady-State Probabilities
and Mean First Passage Times on the
Computer Since we solve steady-state probabilities and mean
first passage times by solving a system of linear
equations, we may use LINDO to determine them. Simply type in an objective function of 0, and type
the equations you need to solve as your constraints.
8/3/2019 Set3 Markov Chains
44/50
44
5.6 Absorbing Chains
Many interesting applications of Markov chainsinvolve chains in which some of the states are
absorbing and the rest are transient states.
This type of chain is called an absorbing chain.
To see why we are interested in absorbing chains we
consider the following accounts receivable example.
A R i bl E l
8/3/2019 Set3 Markov Chains
45/50
45
Accounts Receivable Example The accounts receivable situation of a firm is often
modeled as an absorbing Markov chain. Suppose a firm assumes that an account is
uncollected if the account is more than three months
overdue.
Then at the beginning of each month, each account
may be classified into one of the following states: State 1 New account
State 2 Payment on account is one month overdue
State 3 Payment on account is two months overdue
State 4 Payment on account is three months overdue
State 5 Account has been paid
State 6 Account is written off as bad debt
8/3/2019 Set3 Markov Chains
46/50
46
Suppose that past data indicate that the followingMarkov chain describes how the status of an account
changes from one month to the next month:
100000
010000
3.7.0000
06.4.000
05.05.00
04.006.0New
1 month
2 months
3 monthsPaid
Bad Debt
New 1 month 2 months 3 months Paid Bad Debt
8/3/2019 Set3 Markov Chains
47/50
47
To simplify our example, we assume that after threemonths, a debt is either collected or written off as a
bad debt.
Once a debt is paid up or written off as a bad debt,
the account if closed, and no further transitions
occur.
Hence, Paid or Bad Debt are absorbing states. Since
every account will eventually be paid or written offas a bad debt, New, 1 month, 2 months, and 3 months
are transient states.
8/3/2019 Set3 Markov Chains
48/50
48
A typical new account will be absorbed as either acollected debt or a bad debt.
What is the probability that a new account will
eventually be collected?
To answer this questions we must write a transition
matrix. We assumes m transient states and m
absorbing states. The transition matrix is written in
the form of mcolumns
10
RQs-m rows
m rows
s-mcolumns
P =
8/3/2019 Set3 Markov Chains
49/50
49
The transition matrix for this example is
Thens =6, m =2, and Q andR are as shown.
100000
010000
3.7.0000
06.4.000
05.05.00
04.006.0New
1 month
2 months
3 months
Paid
Bad Debt
New 1 month 2 months 3 months Paid Bad Debt
Q R
8/3/2019 Set3 Markov Chains
50/50
50
1. What is the probability that a new account willeventually be collected? (.964)
2. What is the probability that a one-month overdue
account will eventually become a bad debt? (.06)
3. If the firms sales average $100,000 per month,
how much money per year will go uncollected?
From answer 1, only 3.6% of all debts are
uncollected. Since yearly accounts payable are$1,200,000 on the average, (0.036)(1,200,000) =
$43,200 per year will be uncollected.