- Thread starter
- #1

$\boldsymbol\pi P=\boldsymbol\pi$, 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?

- Thread starter Jason
- Start date

- Thread starter
- #1

$\boldsymbol\pi P=\boldsymbol\pi$, 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?

- Jan 26, 2012

- 890

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.

$\boldsymbol\pi P=\boldsymbol\pi$, 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?

CB

- Thread starter
- #3

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?