Stationary distribution for a doubly stochastic matrix.

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SUMMARY

The stationary distribution vector $\boldsymbol\pi$ for a doubly stochastic matrix can be determined using the same principle as for a stochastic matrix, specifically $\boldsymbol\pi P=\boldsymbol\pi$ with the constraint that $\pi_1+\pi_2+\ldots+\pi_k=1$. For doubly stochastic matrices, the stationary distribution is uniformly distributed across the state space, represented as $\boldsymbol\pi=\left(1/n,...,1/n\right)$ for an $n\times n$ matrix. This uniform distribution arises because both the rows and columns of the matrix sum to 1, negating the need for solving additional equations.

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  • Familiarity with linear algebra concepts, particularly matrix multiplication
  • Knowledge of uniform distributions in probability theory
  • Basic understanding of Markov chains and their stationary distributions
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Jason4
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I can find the stationary distribution vector $\boldsymbol\pi$ for a stochastic matrix $P$ using:

$\boldsymbol\pi P=\boldsymbol\pi$, where $\pi_1+\pi_2+\ldots+\pi_k=1$

However, I can't find a textbook that explains how to do this for a doubly stochastic (bistochastic) matrix. Could somebody show me how?
 
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Jason said:
I can find the stationary distribution vector $\boldsymbol\pi$ for a stochastic matrix $P$ using:

$\boldsymbol\pi P=\boldsymbol\pi$, where $\pi_1+\pi_2+\ldots+\pi_k=1$

However, I can't find a textbook that explains how to do this for a doubly stochastic (bistochastic) matrix. Could somebody show me how?

Doubly stochastic matrices are a subset of right stochastic matrices so the same method should work, unless you have something else in mind for the stationary distribution.

CB
 
Well my notes say:

The limiting distribution for a doubly stochastic is the uniform distribution over the state space, i.e.

$\boldsymbol\pi=\left(1/n,...,1/n\right)$ for an $n\times n$ matrix.

So I assume that if both the columns and the rows sum to 1, you don't have to solve any equations.

Why is this?
 

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