Stationary distribution for a doubly stochastic matrix.

[spitz]

Homework Statement

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?

 

[Ray Vickson]

Homework Statement

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?

What are you trying to do? Has somebody told you what the actual values are of \pi_1, \ldots, \pi_n, and you want to verify (or at least understand) the results? Or, do you not know what the solution is?

RGV

 

[spitz]

I know how to find the distribution vector for any r\times r stochastic matrix. I want to know why, if the matrix is doubly stochastic, you don't need to solve the system of equations and the distribution vector is just (1/r,\ldots ,1/r).

 

[Ray Vickson]

I know how to find the distribution vector for any r\times r stochastic matrix. I want to know why, if the matrix is doubly stochastic, you don't need to solve the system of equations and the distribution vector is just (1/r,\ldots ,1/r).

Without some qualifications the result is not true: consider P = nxn identity matrix. It is doubly-stochastic, but any row vector (v1, v2, ..., vn) satisfies the equation vP = v. If, however, we assume P corresponds to an irreducible chain, the result is true.

Take the case where P has at least two nonzero entries in each column. Suppose the row vector π does not have equal entries. Look at the jth equation \pi_j = \sum_{i} \pi_i p_{i,j}. Since the column entries sum to 1, this is a weighted average of [/itex] \pi_1, \ldots, \pi_n.[/itex] and since there are at least two nonzero entries in the column, we have \min(\pi_1,\ldots,\pi_n) < \pi_j < \max(\pi_1,\ldots, \pi_n). You ought to be able to derive a contradiction from this.

You still need to deal with cases where at least one column has only one entry (which would be 1.0), but which is, nevertheless, irreducible.

RGV

 

[Ray Vickson]

I know how to find the distribution vector for any r\times r stochastic matrix. I want to know why, if the matrix is doubly stochastic, you don't need to solve the system of equations and the distribution vector is just (1/r,\ldots ,1/r).

Without some qualifications the result is not true: consider P = nxn identity matrix. It is doubly-stochastic, but any row vector (v1, v2, ..., vn) satisfies the equation vP = v. If, however, we assume P corresponds to an irreducible chain, the result is true.

Take the case where P has at least two nonzero entries in each column. Suppose the row vector π does not have equal entries. Look at the jth equation \pi_j = \sum_{i} \pi_i p_{i,j}. Since the column entries sum to 1, this is a weighted average of \pi_1, \ldots, \pi_n. and since there are at least two nonzero entries in the column, we have \min(\pi_1,\ldots,\pi_n) < \pi_j < \max(\pi_1,\ldots, \pi_n). You ought to be able to derive a contradiction from this.

You still need to deal with cases where at least one column has only one entry (which would be 1.0), but which is, nevertheless, irreducible.

RGV