markov chain a, b and c are three towns. each year: 10% of the residents of a move to b 30% of the...

MARKOV CHAIN

A, B and C are three towns. Each year:10% of the residents of A move to B30% of the residents of A move to C20% of the residents of B move to A30% of the residents of B move to C20% of the residents of C move to A20% of the residents of C move to B

Everybody else stays put. No one dies. No one is born.

A

B

C

A, B, and C arethree towns.

10 people live in A

15 people live in C

20 people live in B

A

B

C

Each year10% of the residents of Amove to B

.10

and20% of the residents of A move to C

.20

A

B

C

.10

.20

Each year20% of the residents of Bmove to A

.20

and30% of the residents of B move to C

.30

A

B

C

.10

.20

.20

.30

Each year20% of the residents of Cmove to A

.20

and20% of the residents of C move to B

.20

A

B

C

.10

.20

.20

.30

.20

.20

A

B

C

.10

.20

.20

.30

.20

.20

A

B

C

.10

.20

.20

.30

.20

.20

BA

C

.10

.20

.20

.30.20

.20

14 14

17

AB

C

1020

15

This year

Next year

Where will everyone be in 10 years?

A

B

C

.10

.20

.20

.30

.20

.20

AB

C

.10

.20

.20

.30.20

.20

An+1 = .7An

A

C

.10

.20

.20

.30.20

.20

An+1 = .7An + .2Bn + .2Cn

B

A

C

.10

.20

.20

.30.20

.20

An+1 = .7An + .2Bn + .2Cn

Bn+1 = .1An + .5Bn + .2Cn

B

A

C

.10

.20

.20

.30.20

.20

An+1 = .7An + .2Bn + .2Cn

Bn+1 = .1An + .5Bn + .2Cn

Cn+1 = .2An + .3Bn + .6Cn

B

A

C

.10

.20

.20

.30.20

.20

An+1 = .7An + .2Bn + .2Cn

Bn+1 = .1An + .5Bn + .2Cn

Cn+1 = .2An + .3Bn + .6Cn

n

n

n

C

B

A

6.3.2.

2.5.1.

2.2.7.

B

An+1 = .7An + .2Bn + .2Cn

Bn+1 = .1An + .5Bn + .2Cn

Cn+1 = .2An + .3Bn + .6Cn

n

n

n

C

B

A

6.3.2.

2.5.1.

2.2.7.

Note: .7+.1+.2 = 1. This accounts for 100% of those in town A in year n

Cn+1 = .2An + .3Bn + .6CnCn+1 = .2An + .3Bn + .6Cn

Note: .2+.5+.3 = 1. This accounts for 100% of those in town B in year n

Note: .2+.2+.6 = 1. This accounts for 100% of those in town C in year n

n

n

n

n

n

n

C

B

A

C

B

A

6.3.2.

2.5.1.

2.2.7.

1

1

1

nn vMv 1

04

13

22

34

03

12

23

02

12

01

vMvMvMvMv

vMvMvMv

vMvMv

vMv

n

n

n

n

n

n

C

B

A

C

B

A

6.3.2.

2.5.1.

2.2.7.

1

1

1

nn vMv 1

0vMv nn

Because the sum of the numbers in each column of M is 1,

1 is an eigenvalue for M, and

An eigenvector belonging to 1 will describe stable proportions.

The null space of 1I – M = the null space of

6.3.2.

2.5.1.

2.2.7.

100

010

001

-

4.3.2.

2.5.1.

2.2.3.

=

4.3.2.

2.5.1.

2.2.3. The null space of =

13

8

14

Suppose there are this year 140 people in A, 80 people in B, and 130 people in C. How many will be in each town next year?

130

80

140

6.3.2.

2.5.1.

2.2.7.

?