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The Entropy of a Markov Chain
81 points by surprisetalk
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5 comments
abetusk
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So how does one calculate the entropy of a Markov Chain? Is it actually specified? If so, it seems buried.
The Markov chain provided as an example has the edge labels swapped (np should be qp and qp should be np). Regardless, what is the entropy of the example provided?
The problem with Markov chains is that states are dependent, so simply cataloguing states now violates the basic entropy calculation as neighboring states are now dependent on each other.
If the Markov chain is ergodic then maybe you can talk about the entropy of the stationary distribution? Then it's just $-\sum p_i lg(p_i)$ of the stationary distribution probabilities?
The article alludes to how entropy evolves. In the context of ergodic Markov chains, this is related to the size of the second eigenvalue?
ssivark
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Here's a thought: once could derive the spectrum of the Markov transition matrix, and assign an entropy to each of the eigenvectors. The dominant eigenvector (highest entropy) would be the ergodic / stationary distribution, but it seems likely that each successive eigenvector would have a little less entropy. One could initialize the system in a "localized" state (very low entropy) and study the thermalization process as each of the low-entropy eigen-components decay away (exponentially, with rates proportional to the corresponding eigenvalue of the transition matrix) finally leaving the system in the high-entropy stationary distribution. The balance between the eigenvalues (exponential rates) and the entropies of respective eigenvectors would characterize the rate of entropy production (at different times) in the Markov chain!
abetusk
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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})
[0] https://math.stackexchange.com/questions/1040972/entropy-of-...