Handout #8 Markov Chains

 

Suppose there is a random sequence, X, such that:

 

X = {a, b, �, N }

 

That corresponds to a time interval [0, t].

X = {

a,

b,

�,

N }

 

�

�

 

�

 

 

 

 

 

 

So,

Xt=0 = X0 = ����������������

Xt=1 = X1 = ����������������

M

Xt=t = Xt = �����������������

 

[Note:� The values that Xt take on are finite or countably infinite.]

 

Then, the Markov Chain Property is:

P[Xt-1 = j | X0 = a, X1 = b, K, Xt = i ] = P[Xt-1 = j | Xt = i ] = qij ( t )

= the conditional probability that the Markov Chain is at time t in state j,

given that at time (t-1) it was in state i.

 

qij ( t ) is also called the transition probability

 

What does this mean?

 

P [ future states | past & present states ] is independent of the past events and depends only on the present state of the probability pj( t + 1 ) {which represents the random value Xt+1 takes on in state j}.

 

 

 

 


Or, in matrix and vector representation,

p( t + 1 ) = p( t ) Q( t )

where,

p( t ) is an N-dimensional row vector with elements pj( t ), for j = 1, 2, �, N

Q( t ) is an N x N matrix with the elements qij( t ) and is called the Markov transition probability matrix at time t.

 

Stationary or Homogeneous

 

If all qij( t ) are independent of time t, then,

p( t + 1 ) = p( t ) Q

Q and Q( t ) must follow the properties:

qij( t ) � 0, for all i, j and t = 0, 1, 2, �

 

Example:� State Transition Diagram, page 83 of handout

 

States:� { 1, 2, 3 }

Transitions:

q12 = 1, q23 = 1/2, q33 = 1/2, q32 = 1/2, q22 = 1/4, q21 = 1/4

 

So,

 

q11

q12

q13

Q =

q21

q22

q23

 

q31

q32

q33

 

 

 

 

 

 

Q =

 

 

 

 

 

 

 

 


Nonstationary:

p( t ) = p( 0 ) Q( 0 ) Q( 1 ) � Q( t � 1 )

 

Stationary:

p( t ) = p( 0 ) Qt

 

If we let t = s + n, then:

p( s + n ) = p( 0 ) Qs Qn = p( s ) Qn

 

Then,

From magic, we end up with:

P( z ) = p( 0 ) [ I � Qz ]-1, where I is the identity matrix

 

Asymptotic Behavior:

 

 

 

 

length of time = "first passage of time"

(in terms of number of transitions)

 

when i = j

 

 

 

 

 

 

length of time = "recurrence time" for state i

 

 

Let be the probability that the first passage time from state i to state j is equal to n.

 

 

 

 

Then,

is the probability that starting in state i, the system reaches state j.

 

If:

fij �������������������� :� may never reach j

fij �������������������� :� probability distribution of the first passage of time

 

If i = j, �is the probability that starting in state i, system reaches state j (so if i = j, system gets back to i).

 

If:

fii �������������������� :� transient state

fii �������������������� :� recurrent state � probability distribution of recurrence time

 

Since they are just probability distributions, then we know they have mean values!

 

 

So, whenever fij = 1,

 

 

 

Look at Figure 2.10.

 

 

 

 

 

 

Difficult to find

 

i

j

n =

1

2

3

4�

1

1

0

1(1/4) = ��

1(1/4)(1/4) =

�

 


The mean though is easy to find:

 

etc.

 

When j = i,� mii = expected recurrence time

 

If:

mii ������������������ , then recurrent state i is a null-recurrent state

mii ������������������ , then recurrent state i is a positive-recurrent state or regular-recurrent state

 

Periodic state:� A state is periodic if n/t is not integer divisible (t is the greatest integer with this property) and ��In other words, a return to state i is impossible unless it occurs in t, 2t, 3t,� steps.

 

Ergodic state:� A state that is neither null nor periodic.

 

Absorbing state:� A state where once in, one can never leave.

 

Reachable state: , for n � 1.� If a state is reachable, then the state can communicate.

 

If all states in a state diagram can communicate, then the chain is irreducible.

 

So (page 88)

�         States of an irreducible chain are either all recurrent or all transient.

�         If one is periodic with period t, then all states are periodic with period t.

 

Question:� Does a system converge to some limiting distribution?

 

Assume an aperiodic, irreducible system is stable.� Let,

Then, by page 83,

p( t + 1 ) = p( t ) Q

p = p Q

A probability distribution p satisfying this equation is called a stationary distribution.� See Theorem 2.9.2., page 89.

 

An irreducible aperiodic Markov Chain has:

�         At most one stationary distribution

�         And either

1.                  All states are transient or all are null, so there's no stationary distribution.

2.                  All states are ergodic, so there's a unique stationary distribution.