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Stationary distribution for a doubly stochastic matrix.

Jason

New member
Feb 5, 2012
28
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?
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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
 

Jason

New member
Feb 5, 2012
28
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?