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What is the most computationally efficient way to compute the stationary distribution of a reversible finite-state continuous-time Markov process?

Assume that the process is irreducible and the rate matrix is dense. All eigenvalues will be real. A single eigenvalue of the rate matrix will be zero and all of the other eigenvalues will be negative. The stationary distribution is given by the entrywise squares of the unit left-eigenvector corresponding to the eigenvalue that is zero.

I hope this question is mathematically interesting enough because of how much structure is known about the rate matrix, and perhaps this structure can be used cleverly.

Following a comment mentioning detailed balance I guess you can compute the stationary distribution $\pi$ given rate matrix $R$ as something like $\pi = \frac{v}{\sum_i{v_i}}$ where $v = (1, \frac{R_{1,2}}{R_{2,1}}, \frac{R_{1,3}}{R_{3,1}}, ..., \frac{R_{1,n}}{R_{n,1}})$ as long as you don't divide by zero anywhere. Is there a better way to do it?

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    $\begingroup$ detailed balance $\endgroup$ – Anthony Quas Jan 13 '12 at 1:39

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