OK, no, I'm wrong. The entropy of a Markov chain with stationary distribution v is [0]:

    -\sum v_i p_{i,j} \log(p_{i,j})

That is, the "entropy" of the transition matrix modified by the stationary distribution.

[0] https://math.stackexchange.com/questions/1040972/entropy-of-...

▲ chermi an hour ago | parent [-]

That is the entropy rate. If I'm understanding your original question correctly, you were asking about the standard equilibrium-defining thermodynamic entropy?

	▲	abetusk 32 minutes ago \| parent [-]
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