Finding long term Markov behaviors

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Homework Help Overview

The discussion revolves around finding the long-term behavior of a Markov system represented by a transition matrix and initial conditions. The original poster attempts to determine the limiting distribution of the system as time approaches infinity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the nth power of the transition matrix and the implications for the long-term behavior of the system. Questions arise regarding the correct method for matrix multiplication and the concept of equilibrium equations.

Discussion Status

Some participants have pointed out potential errors in the original poster's approach to calculating the transition matrix's power. Guidance has been offered regarding the need for vector-matrix multiplication and the formulation of equilibrium equations to find the limiting probabilities.

Contextual Notes

There is mention of a potential lack of resources in the original poster's textbook regarding Markov chains, which may be contributing to the confusion in understanding the concepts discussed.

Lolsauce
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Homework Statement



Find the long-term behavior of the Markov system described by the following matrix with initial conditions below: (i.e. find T^infinity)

Find x_infinity

Homework Equations



x_0 = [.3 .3 .4]^TT =

[.5 .4 .1]
[.4 .3 .3]
[.1 .3 .6]

The Attempt at a Solution



So in my attempt to find x_infinity I know that x_n = ( T^n)(x_0)

This means:

[.5^n .4^n .1^n] * [.3 .3 .4]^T
[.4^n .3^n .3^n]
[.1^n .3^n .6^n]

Which I solved to be
[ .3*.5^n .3*.4^n .4*.1^n]
[ .3*.4^n .3*.3^n .4*.3^n]
[ .3*.1^n .3*.3^n .4*.6^n]

But as n approaches infinity all the terms get smaller and all go to zero. Does this seem correct so far?

That would mean:

x_infinity = ( 0 0 0 )^T ? I tried this on my webassign, but it is not correct. Any tips or help in pointing out my mistakes would be great, thanks!
 
Last edited:
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Your expression for T^n is wrong. T^n = nth power of T in the sense of matrix multiplication. For example, T^2 = T*T =[[.42,.35,.23],[.35,.35,.31],[.23,.31,.46]], etc. You need to perform vector-matrix multiplications, or else solve the "equilibrium equations" to find the long-run limiting probabilities.

RGV
 
Ray Vickson said:
Your expression for T^n is wrong. T^n = nth power of T in the sense of matrix multiplication. For example, T^2 = T*T =[[.42,.35,.23],[.35,.35,.31],[.23,.31,.46]], etc. You need to perform vector-matrix multiplications, or else solve the "equilibrium equations" to find the long-run limiting probabilities.

RGV

Oh I see how I was doing the expression of T^n was wrong. What do you mean by vector-matrix multiplcation or "equilibrium equations"? Do you mean keep finding T^1,T^2,T^3, and see if it approaches a limit?
 
The state-probability row vector after n steps is x_n = x_0 * T^n, which can be written as x_{n-1}*T (a vector times a matrix). If the x_n approach a limiting distribution y, we must have y = y*T, and sum y(j) = 1. This is a simple system of linear equations, called the equilibrium or balance equations. I cannot believe that this material is not all in your textbook, but if not, Google "Markov chain" to see various relevant articles.

RGV
 

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