Home > Documents > Markov Chains. Summary Markov Chains Discrete Time Markov Chains Homogeneous and non-homogeneous...

# Markov Chains. Summary Markov Chains Discrete Time Markov Chains Homogeneous and non-homogeneous...

Date post: 16-Dec-2015
Category:
Author: lillian-jackson
View: 262 times
Embed Size (px)
of 55 /55
Markov Chains
Transcript Markov Chains Summary

Markov Chains Discrete Time Markov Chains

Homogeneous and non-homogeneous Markov chains

Transient and steady state Markov chains

Continuous Time Markov Chains Homogeneous and non-homogeneous Markov

chains Transient and steady state Markov chains Markov Processes

Recall the definition of a Markov Process The future a a process does not depend on its past, only on its

present

1 1 0 0

1 1

Pr | ,...,

Pr |

k k k k

k k k k

X t x X t x X t x

X t x X t x

Since we are dealing with “chains”, X(t) can take discrete

values from a finite or a countable infinite set. For a discrete-time Markov chain, the notation is also

simplified to

1 1 0 0 1 1Pr | ,..., Pr |k k k k k k k kX x X x X x X x X x

Where Xk is the value of the state at the kth step Chapman-Kolmogorov Equations

Define the one-step transition probabilities

1Pr |ij k kp X j X ik Clearly, for all i, k, and all feasible transitions from state i

1ijj i

p k

xi

x1

xR

… xj

k u k+n

Pr |,ij k n kp X j X ik k n

Define the n-step transition probabilities Chapman-Kolmogorov Equations

xi

x1

xR

… xj

k u k+n

1

Pr | , Pr |,R

ij k n u k u kr

p X j X r X i X r X ik k n

Using total probability

Pr | , Pr |k n u k k n uX j X r X i X j X r

Using the memoryless property of Marckov chains

1

, , , ,R

ij ir rjr

p p p k u k nk k n k u u k n

Therefore, we obtain the Chapman-Kolmogorov Equation Matrix Form

Define the matrix

We can re-write the Chapman-Kolmogorov Equation

Choose, u = k+n-1, then

,, ijp k k nk k n H

, , ,k k n k u u k n H H H

, , 1 1,

, 1 1

k k n k k n k n k n

k k n k n

H H H

H P

One step transition probability

Forward Chapman-Kolmogorov Matrix Form

Choose, u = k+1, then

, , 1 1,

1,

k k n k k k k n

k k nk

H H H

P H

One step transition probability

Backward Chapman-Kolmogorov Homogeneous Markov Chains

The one-step transition probabilities are independent of time k.

1 or Pr |ij k kp X j X ik P P

Even though the one step transition is independent of k, this does not mean that the joint probability of Xk+1 and Xk is also independent of k Note that

1 1Pr , Pr | Pr

Pr

k k k k k

ij k

X j X i X j X i X i

p X i Example

Consider a two processor computer system where, time is divided into time slots and that operates as follows At most one job can arrive during any time slot and this can

happen with probability α. Jobs are served by whichever processor is available, and if

both are available then the job is given to processor 1. If both processors are busy, then the job is lost When a processor is busy, it can complete the job with

probability β during any one time slot. If a job is submitted during a slot when both processors are

busy but at least one processor completes a job, then the job is accepted (departures occur before arrivals).

Describe the automaton that models this system. Describe the Markov Chain that describe this model. Example: Automaton

Let the number of jobs that are currently processed by the system by the state, then the State Space is given by X= {0, 1, 2}.

Event set: a: job arrival, d: job departure

Feasible event set: If X=0, then Γ(X)= a If X= 1, 2, then Γ(Χ)= a, d.

State Transition Diagram

0 1 2

a

- / a,d

a

-d d / a,d,d

dd Example: Alternative Automaton

Let (X1,X2) indicate whether processor 1 or 2 are busy, Xi= {0, 1}. Event set:

a: job arrival, di: job departure from processor i Feasible event set:

If X=(0,0), then Γ(X)= a If X=(0,1) then Γ(Χ)= a, d2. If X=(1,0) then Γ(Χ)= a, d1. If X=(0,1) then Γ(Χ)= a, d1, d2.

State Transition Diagram

a

- / a,d1

a

-

d2 d1

d1,d2

-

d1

a,d2

a,d1,d2

00

10

11

01 Example: Markov Chain

For the State Transition Diagram of the Markov Chain, each transition is simply marked with the transition probability

0 1 2

p01

p11

p12

p00p10

p21

p20

p22

00 1p 01p 02 0p

10 1p 11 1 1p 12 1p

220 1p 2

21 2 1 1p 2

22 21 1p Example: Markov Chain

Suppose that α = 0.5 and β = 0.7, then,

0 1 2

p01

p11

p12

p00p10

p21

p20

p22

0.5 0.5 0

0.35 0.5 0.15

0.245 0.455 0.3ijp

P State Holding Times

Suppose that at point k, the Markov Chain has transitioned into state Xk=i. An interesting question is how long it will stay at state i.

Let V(i) be the random variable that represents the number of time slots that Xk=i.

We are interested on the quantity Pr{V(i) = n}

1 1

1

1 1

Pr Pr , ,..., |

Pr | ,...,

Pr ,..., |

k n k n k k

k n k n k

k n k k

V n X i X i X i X ii

X i X i X i

X i X i X i

1

1 2

2 1

Pr |

Pr | ...,

Pr ,..., |

k n k n

k n k n k

k n k k

X i X i

X i X X i

X i X i X i State Holding Times

This is the Geometric Distribution with parameter pii. Clearly, V(i) has the memoryless property

1

1 2

2 1

Pr Pr |

Pr | ...,

Pr ,..., |

k n k n

k n k n k

k n k k

V n X i X ii

X i X X i

X i X i X i

1Pr 1 nii iiV n p pi

1 2

2 3

3 1

1 Pr |

Pr | ,...,

Pr ,..., |

ii k n k n

k n k n k

k n k k

p X i X i

X i X i X i

X i X i X i State Probabilities

An interesting quantity we are usually interested in is the probability of finding the chain at various states, i.e., we define

Pri kX ik For all possible states, we define the vector

0 1, ...k k k π Using total probability we can write

1 1Pr | Pr

1

i k k kj

ij jj

X i X j X jk

p k k

In vector form, one can write

1k k k π π P 1k k π π POr, if homogeneous Markov Chain State Probabilities Example

Suppose that

1 0 00 π

Find π(k) for k=1,2,…

with

0.5 0.5 0

0.35 0.5 0.15

0.245 0.455 0.3

P

0.5 0.5 0

1 0 0 0.35 0.5 0.15 0.5 0.5 010.245 0.455 0.3

π

Transient behavior of the system: MCTransient.m In general, the transient behavior is obtained by solving the difference equation

1k k π π P Classification of States

Definitions State j is reachable from state i if the probability to go

from i to j in n >0 steps is greater than zero (State j is reachable from state i if in the state transition diagram there is a path from i to j).

A subset S of the state space X is closed if pij=0 for every iS and j S

A state i is said to be absorbing if it is a single element closed set.

A closed set S of states is irreducible if any state jS is reachable from every state iS.

A Markov chain is said to be irreducible if the state space X is irreducible. Example

Irreducible Markov Chain

0 1 2

p01 p12

p00p10

p21

p22

Reducible Markov Chain p01 p12

p00p10

p14

p224

p23

p32

p33

0 1 2 3

Absorbing State

Closed irreducible set Transient and Recurrent States

Hitting Time

Recurrence Time Tii is the first time that the MC returns to state i.

Let ρi be the probability that the state will return back to i given it starts from i. Then,

0min 0 : ,ij kT k X i X j

1

Pri iik

T k

The event that the MC will return to state i given it started

from i is equivalent to Tii < ∞, therefore we can write

1

Pr Pri ii iik

T k T

A state is recurrent if ρi=1 and transient if ρi<1 Theorems

If a Markov Chain has finite state space, then at least one of the states is recurrent.

If state i is recurrent and state j is reachable from state i then, state j is also recurrent.

If S is a finite closed irreducible set of states, then every state in S is recurrent. Positive and Null Recurrent States

Let Mi be the mean recurrence time of state i

A state is said to be positive recurrent if Mi<∞. If Mi=∞ then the state is said to be null-recurrent.

Theorems If state i is positive recurrent and state j is reachable

from state i then, state j is also positive recurrent. If S is a closed irreducible set of states, then every

state in S is positive recurrent or, every state in S is null recurrent, or, every state in S is transient.

If S is a finite closed irreducible set of states, then every state in S is positive recurrent.

1

Pri ii iik

M E T k T k Example

p01 p12

p00p10

p14

p224

p23

p32

p33

0 1 2 3

Recurrent State

Transient States Positive

Recurrent States Periodic and Aperiodic States

Suppose that the structure of the Markov Chain is such that state i is visited after a number of steps that is an integer multiple of an integer d >1. Then the state is called periodic with period d.

If no such integer exists (i.e., d =1) then the state is called aperiodic.

Example1 0.5

0.5

0 1 2

1

Periodic State d = 2

0 1 0

0.5 0 0.5

0 1 0

P Recall that the probability of finding the MC at state i after the kth step is given by

Pri kX ik 0 1, ...k k k π An interesting question is what happens in the “long run”,

i.e., limi kk

Questions: Do these limits exists? If they exist, do they converge to a legitimate

probability distribution, i.e., How do we evaluate πj, for all j.

1i

This is referred to as steady state or equilibrium or stationary state probability Recall the recursive probability 1k kπ π P

If steady state exists, then π(k+1) π(k), and therefore the steady state probabilities are given by the solution to the equations

If an Irreducible Markov Chain the presence of periodic states prevents the existence of a steady state probability

Example: periodic.m

π πP and 1ii

0 1 0

0.5 0 0.5

0 1 0

P 1 0 00 π THEOREM: In an irreducible aperiodic Markov chain consisting of positive recurrent states a unique stationary state probability vector π exists such that πj > 0 and

1limj jk

j

kM

where Mj is the mean recurrence time of state j

The steady state vector π is determined by solving

π πP and 1ii

Ergodic Markov chain. Birth-Death Example

1-p 1-p

pp

1-p

p

0 1 i

p1 0

0 1

0 0

p p

p p

p

P

Thus, to find the steady state vector π we need to solve

π πP and 1ii Birth-Death Example

0 0 1p p In other words

1 1 , 1, 2,...1j j j p jp

1 0

1 p

p

Solving these equations we get2

2 0

1 p

p

In general 0

1j

j

p

p

Summing all terms we get

0 00 0

1 11 1

i i

i i

p p

pp Birth-Death Example

Therefore, for all states j we get

0

1i

i

p

p

If p<1/2, then 0, for all j j

0

1 1j i

ji

p p

p p

All states are transient

0

10

2 1

i

i

p pp p

If p>1/2, then 12 1

, for all j

j

ppj

pp

All states are positive recurrent Birth-Death Example

If p=1/2, then

0

1i

i

p

p

0, for all j j

All states are null recurrent Reducible Markov Chains

In steady state, we know that the Markov chain will eventually end in an irreducible set and the previous analysis still holds, or an absorbing state.

The only question that arises, in case there are two or more irreducible sets, is the probability it will end in each set

Transient Set T

Irreducible Set S1

Irreducible Set S2 Transient Set T

Reducible Markov Chains

Suppose we start from state i. Then, there are two ways to go to S. In one step or Go to r T after k steps, and then to S.

Define

Irreducible Set S

i

rs1

sn

0Pr | , 1, 2,...i kX S X i kS Reducible Markov Chains

Next consider the general case for k=2,3,…

1 0Pr |X S X i ijj S

p

1 1 1 0Pr , ..., |k k kX S X r T X r T X i

1 1 1 0

1 0

Pr , ...,| ,

Pr |

k k kX S X r T X r T X i

X r T X i

i ij r irj S r T

p pS S

First consider the one-step transition

1 1 1Pr , ...,|k k k irX S X r T X r T p r irpS Continuous-Time Markov Chains

In this case, transitions can occur at any time Recall the Markov (memoryless) property

1 1 0 0

1 1

Pr | ,...,

Pr |

k k k k

k k k k

X t x X t x X t x

X t x X t x

where t1 < t2 < … < tk

Recall that the Markov property implies that X(tk+1) depends only on X(tk) (state memory)

It does not matter how long the state at X(tk) (age memory).

The transition probabilities now need to be defined for every time instant as pij(t), i.e., the probability that the MC transitions from state i to j at time t. Transition Function

Define the transition function Pr | , ,ijp X j X i s ts t t s

The continuous-time analogue of the Chapman-Kolmokorov equation is

Using the memoryless property

,

Pr | , Pr |

ij

r

p s t

X j X r X i X r X it u s u s

Pr | Pr |,ijr

p X j X r X r X is t t u u s Define H(s,t)=[pij(s,t)], i,j=1,2,… then

, , , , s u ts t s u u t H H H Note that H(s, s)= I Transition Rate Matrix

Consider the Chapman-Kolmogorov for s ≤ t ≤ t+Δt

, , ,s t t s t t t t H H H

Subtracting H(s,t) from both sides and dividing by Δt

, ,, , s t t t ts t t s t

t t

H H IH H

Taking the limit as Δt0

,

,s t

s t tt

H

H Q

where the transition rate matrix Q(t) is given by

0

,lim

t

t t tt

t

H I

Q Homogeneous Case

In the homogeneous case, the transition functions do not depend on s and t, but only on the difference t-s thus

,ij ijp ps t t s

It follows that

,s t t s H H P

and the transition rate matrix

0 0

,lim lim , constant

t t

t t t tt

t t

H I H IQ Q

Thus

1 if with 0

0 if ij

i jt pti jt

PP Q tet QP State Holding Time

The time the MC will spend at each state is a random variable with distribution

1 iiG et

where

i j

j i

Explain why… Transition Rate Matrix Q.

Recall that t

tt

P

P Q

ijir rj

r

p tp qt

t

First consider the qij, i ≠ j, thus the above equation can be written as

ij

ii ij ir rjr i

p tp q p qt t

t

Evaluating this at t = 0, we get that

0

ijij

t

p t qt

pij(0)= 0 for all i ≠ j

The event that will take the state from i to j has exponential residual lifetime with rate λij, therefore, given that in the interval (t,t+τ) one event has occurred, the probability that this transition will occur is given by Gij(τ)=1-exp{-λijτ}. Transition Rate Matrix Q.

0

ijij

pq

Since Gij(τ)=1-exp{-λijτ}.

0

ij

ijije

In other words qij is the rate of the Poisson process that activates the event that makes the transition from i to j.

Next, consider the qjj, thus

ijij jj ir rj

r j

p tp q p qt t

t

Evaluating this at t = 0, we get that

0

ijjj

t

p t qt

0

1 ij jjt

p qtt

Probability that chain leaves state j Transition Rate Matrix Q.

0ijj

q Note that for each row i, the sum

The event that the MC will transition out of state i has exponential residual lifetime with rate Λ(i), therefore, the probability that an event will occur in the interval (t,t+τ) given by Gi(τ)=1-exp{- Λ(i)τ}.

0

ijjq e ii Transition Probabilities P.

1Pr |ij k kP X j X i

Recall that in the case of the superposition of two or more Poisson processes, the probability that the next event is from process j is given by λj/Λ.

Suppose that state transitions occur at random points in time T1 < T2 <…< Tk <…

Let Xk be the state after the transition at Tk

Define

In this case, we have

,ijij

ii

qP i j

q

0iiP and Example

Assume a computer system where jobs arrive according to a Poisson process with rate λ.

Each job is processed using a First In First Out (FIFO) policy.

The processing time of each job is exponential with rate μ. The computer has buffer to store up to two jobs that wait

for processing. Jobs that find the buffer full are lost. Draw the state transition diagram. Find the Rate Transition Matrix Q. Find the State Transition Matrix P Example

The rate transition matrix is given by

a

d

0 1 2 3

a a

a

d d

0 0

0

0

0 0

Q

0 0 0

0 01

0 0

0 0 0

P

The state transition matrix is given by State Probabilities and Transient Analysis

Similar to the discrete-time case, we define Prj X jt t

In vector form 1 2, ,...t t t π

With initial probabilities 1 2, ,...0 0 0 π

Using our previous notation (for homogeneous MC)

0t t π π P 0teQπ

Obtaining a general solution is not easy!

Differentiating with respect to t gives us more “inside” d t

tdt

π

π Q

jjj j ij i

i j

d tq qt t

dt “Probability Fluid” view

We view πj(t) as the level of a “probability fluid” that is stored at each node j (0=empty, 1=full).

j

jj j ij ii j

d tq qt t

dt

Change in the probability fluid

inflowoutflow

ri

j

qij… qjr…

Inflow Outflow

jj jrr j

q q Often we are interested in the “long-run” probabilistic behavior of the Markov chain, i.e.,

limj jtt

As with the discrete-time case, we need to address the following questions Under what conditions do the limits exist? If they exist, do they form legitimate probabilities? How can we evaluate these limits?

These are referred to as steady state probabilities or equilibrium state probabilities or stationary state probabilities Theorem: In an irreducible continuous-time Markov Chain consisting of positive recurrent states, a unique stationary state probability vector π with

limj jtt

These vectors are independent of the initial state probability and can be obtained by solving

and 1jj

πQ = 0

0 jj j ij ii j

q qt t

0 Change

inflow

outflow ri

j

qij… qjr…

Inflow Outflow

Using the “probability fluid” view

0j t

dt Example

For the previous example, with the above transition function, what are the steady state probabilities

a

d

0 1 2 3

a a

a

d d

0 1 2 3

0 0

0

0

0 0

πQ 0

0 1 2 3 1

Solve Example

The solution is obtained

0 1 2 3 1

0 1 0 1 0

0 1 2 0 2

2 0

1 2 3 0 3

3 0

0 2 3

1

1 Birth-Death Chain

0 0

1 11 1

2 2 2

0

0

Q

Similarly to the previous example,

λ0

0 1 i

λ1 λi-1 λi

μ1μi μi+1

And we solve

πQ 00

1ii

and Example

The solution is obtained

0 0 1 1 0 01 0

1

0 0 1 2 21 1 0 0 12 0

1 2

In general

1 1 1 1 0j jj j j j j 0

1 01 1

...

...j

jj

Making the sum equal to 1

0 1

01 1

...1 1

...j

j j

Solution exists if

0 1

1 1

...1

...j

j j

S Uniformization of Markov Chains

In general, discrete-time models are easier to work with, and computers (that are needed to solve such models) operate in discrete-time

Thus, we need a way to turn continuous-time to discrete-time Markov Chains

Uniformization Procedure Recall that the total rate out of state i is –qii=Λ(i). Pick

a uniform rate γ such that γ ≥ Λ(i) for all states i. The difference γ - Λ(i) implies a “fictitious” event that

returns the MC back to state i (self loop). Uniformization of Markov Chains

Uniformization Procedure Let PU

ij be the transition probability from state I to state j for the discrete-time uniformized Markov Chain, then

if

if

ij

Uij

ijj i

qi j

Pq

i j

i

j

k

……

qij

qik

Uniformization i

j

k

……

ijq

ikq

iiq

Recommended