Day 3 Markov Chains

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Day 3 Markov Chains. For some interesting demonstrations of this topic visit: http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/Tools/index.htm. Equations of the form: are called discrete equations because they only model the system at whole number time increments. - PowerPoint PPT Presentation

Transcript of Day 3 Markov Chains

Day 3 Markov Chains

For some interesting demonstrations of this topic visit: http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/Tools/index.ht

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Equations of the form:

are called discrete equations because they only model the system at whole number time increments

• Difference equation is an equation involving differences. We can see difference equation from at least three points of views: as sequence of number, discrete dynamical system and iterated function. It is the same thing but we look at different angle.

Difference Equations vs Differential Equations

Dynamical system come with many different names. Our particular interesting dynamical system is for the system whose state depends on the input history. In discrete time system, we call such system difference equation (equivalent to differential equation in continuous time).

Markov Matrices

.1 .01 .3 A = .2 .99 .3 .7 0 .4

Properties of Markov MatricesAll entries are ≥ 0All Columns add up to one Note: the powers of the matrix will maintain

these propertiesEach column is representing probabilities

[ ]Consider the matrix

Markov Matrices .1 .01 .3

A = .2 .99 .3

.7 0 .4

1 is a eigenvalue of all Markov Matrices

Why?

Subtract 1 down each entry in the diagonal.

Each column will then add to zero which means that the rows are dependent.

Which means that the matrix is singular

[ ]

Markov Matrices .1 .01 .3 A = .2 .99 .3 .7 0 .4 One eigenvalue is one all other eigenvalues have an

absolute value ≤ 1We are interested in raising A to some powers

If 1 is an eigenvector and all other vectors are less than 1 then the steady state is the eigenvector

Note: this requires n independent vectors

[ ]

Short cuts for finding eigenvectors -.9 .01 .3

A-I = .2 -.01 .3 det(A -1I)

.7 0 -.6

To find the eigenvector that corresponds to λ=1

Use < .6, ?? , .7> to get the last row to be zero.

Then use the top row to get the missing middle value.

<.6,33,.7>

[ ]

Applications of Markov Matrices

Markov Matrices are used to when the probability of an event depends on its current state.

For this model, the probability of an event must remain constant over time.

The total population is not changing over timeMarkov matrices have applications in

Electrical engineering

Applications of Markov Matricesuk+1 = Auk

Suppose we have two cities Suzhou (S) an Hangzhou (H) with initial condition at k=0, S = 0 and H = 1000

We would like to describe movement in population between these two cities.

us+1 = .9 .2 uS

uH+1 .1 .8 uH

Population of Suzhou and Hongzhou at time t+1

Column 1: .9 of the people in S stay there and .1 move to H Column 2: .8 of the people in H stay there are and .2 move to S

[ ] [ ][ ]Population of S and H at time t

Applications of Markov Matricesuk+1 = Auk

.

us+1 = .9 .2 uS

uH +1 .1 .8 uH

Find the eigenvalues and eigenvectors

[ ] [ ][ ]

Applications of Markov Matricesuk+1 = Auk

.

us +1 = .9 .2 uS

uH +1 .1 .8 uH

eigenvalues 1 and .7 (from properties of Markov Matrices and the trace)

Eigenvectors Ker (A-I), Ker (A-.7I)

A-I -.1 .2 Ker=<2,1> A-.7I = .2 .2 Ker=<1,-1>

.1 -.2 .1 .1

[ ] [ ][ ]

[ ] [ ]

Applicationsuk+1 = Auk

us+1 = .9 .2 uS

uH+1 .1 .8 uH

eigenvalue 1 eigenvector <2,1>eigenvalue .7 eigenvector <-1,1>This tells us about time and ∞λ=1 will be a steady state, λ=.7 will disappear as t→∞The eigenvector tells us that we need a ratio of 2:1The total population is still 1000 so the final population

will be 1000 (2/3) and 1000 (1/3)

[ ] [ ][ ]

Applications us+1 = .9 .2 uS

uH +1 .1 .8 uH

To find the amounts after a finite number of steps

Aku0 = c1(1) 2 + c2 (.7) k -1

1 1

Use the initial condition to solve for constants

0 = c1 2 + c2 -1 c1 =1000/3

1000 1 1 c2 = 2000/3

[ ] [ ][ ]

[ ][ ]

Initial condition at k=0, S = 0 and H = 1000

[ ][ ][ ]

k

Steady state for Markov Matrices

Every Markov chain will be a steady state.

The steady state will be the eigenvector for the eigenvalue λ=1

Applications of Markov matrices

Airlines - Markov matrices are used in creating networks for airlines to determine routes of planes.

The sum of the probabilities is 1 in each column because all planes in a given location go somewhere (includes possibly not moving)

Airlines want to create flight plans so they do not end up with too many planes in one part of the world and not enough in another.

More applications of Markov

Game theory – looking setting house rules for casinos ensuring casinos come out ahead

Economics- Economic mobility over generations

• http://www.facstaff.bucknell.edu/ap030/Math345LAApplications/MarkovProcesses.html

Homework (diff 3): review worksheet 8.1 3-6,8,9,13 eigenvalue review worksheet 1-5

"Genius is one per cent inspiration, ninety-nine per cent perspiration.“ Thomas Alva Edison

http://people.revoledu.com/kardi/tutorial/DifferenceEquation/

WhatIsDifferenceEquation.htm

http://www.math.duke.edu/education/ccp/materials/linalg/diffeqs/diffeq2.html

Fibonacci via matriceshttp://www.maths.leeds.ac.uk/applied/0380/fibonacci03.pdf

For More information visit: