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Stationary distribution for a doubly stochastic matrix.

  1. Mar 12, 2012 #1
    1. The problem statement, all variables and given/known data

    I can find the stationary distribution vector [itex]\boldsymbol\pi[/itex] for a stochastic matrix [itex]P[/itex] using:

    [itex]\boldsymbol\pi P=\boldsymbol\pi[/itex], where [itex]\pi_1+\pi_2+\ldots+\pi_k=1[/itex]

    However, I can't find a textbook that explains how to do this for a doubly stochastic (bistochastic) matrix. Could somebody show me how?
     
  2. jcsd
  3. Mar 12, 2012 #2

    Ray Vickson

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

    RGV
     
  4. Mar 12, 2012 #3
    I know how to find the distribution vector for any [itex]r\times r[/itex] 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 [itex](1/r,\ldots ,1/r)[/itex].
     
  5. Mar 12, 2012 #4

    Ray Vickson

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    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 [itex] \pi_j = \sum_{i} \pi_i p_{i,j}.[/itex] 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 [itex] \min(\pi_1,\ldots,\pi_n) < \pi_j < \max(\pi_1,\ldots, \pi_n).[/itex] 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
     
  6. Mar 12, 2012 #5

    Ray Vickson

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    Science Advisor
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    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 [itex] \pi_j = \sum_{i} \pi_i p_{i,j}.[/itex] 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 [itex] \min(\pi_1,\ldots,\pi_n) < \pi_j < \max(\pi_1,\ldots, \pi_n).[/itex] 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
     
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