MHB Stationary distribution for a doubly stochastic matrix.

AI Thread Summary
The discussion revolves around finding the stationary distribution vector for a doubly stochastic matrix. A user notes that while they can compute the stationary distribution for a stochastic matrix using the equation $\boldsymbol\pi P=\boldsymbol\pi$, they struggle to find resources for doubly stochastic matrices. Another participant points out that the limiting distribution for a doubly stochastic matrix is the uniform distribution, expressed as $\boldsymbol\pi=\left(1/n,...,1/n\right)$ for an n x n matrix. This uniform distribution arises because both the rows and columns of a doubly stochastic matrix sum to 1, eliminating the need for solving equations. The conversation highlights the unique properties of doubly stochastic matrices in relation to their stationary distributions.
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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