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To find the limiting probability you solve the systems of equations:

  • $\vec{\pi}=P\vec{\pi}$
  • $\Sigma \pi_j = 1$

and my teacher told us "you could rewrite this as matrices". Having just completed a first intro course in Linear Algebra, I'm curious on how you would do that. Guess it's a matter of simple algebra.

jacob
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1 Answers1

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$P$ is already a matrix ($n \times n$). Your first equation is $(P - I) \vec{\pi} = \vec{0}$ where $I$ is the $n \times n$ identity matrix. Your second is $(1,\ldots,1) \vec{\pi} = 1$. So you get the matrix-vector equation $A \vec{\pi} = \vec{b}$ where $A$ is the $(n+1)\times n$ matrix whose first $n$ rows are $P - I$ and last row is all $1$'s, and $\vec{b}$ consists of $n$ $0$'s and then a $1$.

The system is redundant, since the sum of the first $n$ rows of $A$ is $0$ (do you see why?); you can leave out one of those rows to get an $n \times n$ system.

Robert Israel
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  • "do you see why" I think I do. It's because P represents a markov chain, thus the sum of its rows are 1. If you subtract $I$ then the sum is 0. – jacob Apr 09 '14 at 18:08
  • So I should put A in rref, and then calculate $rref(A)^{-1} \cdot b$ ? – jacob Apr 11 '14 at 12:11
  • rref is Reduced Row Echelon Form – jacob Apr 11 '14 at 13:19
  • Use whatever method you use to solve $A x = b$. – Robert Israel Apr 11 '14 at 14:28
  • Well, I need to find a row that is OK to throw out ("redundant" as you call it). And by doing gauss eliminations, I would be able to see what row is linearly dependent (i.e. redundant). Correct? – jacob Apr 11 '14 at 15:19