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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

by abetusk 1mo ago

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 1mo ago
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 1mo ago
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-a-markov-chain-right-result https://math.stackexchange.com/questions/1040972/entropy-of-...
- chermi 1mo ago
  That is the entropy rate. If I'm understanding your original question correctly, you were asking about the standard equilibrium-defining thermodynamic entropy?
  - abetusk 1mo ago
    I'm not sure I had a well formed question. I saw the post title, didn't get a clarification in the article and started searching. The entropy rate seems like a pretty natural definition of "entropy of a Markov chain", no? It's not exactly this but it's similar to "start at state i, end on state j (maybe in n steps?), what is the number of bits I need to send over the wire to tell you what path was taken". What does the entropy of the raw stationary distribution give you? Is the entropy rate related to the entropy of the stationary distribution (the thermodynamic entropy?)?
    - chermi 1mo ago
      I was probably projecting. I like both path and equilibrium entropies, but my assumption I guess was that when people hear entropy in physics they're usually thinking of the equilibrium-defining entropy. That is, the value that is maximized as the system relaxes to equilibrium, the vanilla thermodynamic entropy.
chermi 1mo ago
Find the equilibrium probabilities from first eigenvector, use definition of entropy like you did. If it's not ergodic then you get subsystems with their own entropies. Second eigenvalue iirc basically characterizes dominant relaxation time. This is all if you mean the equilibrium entropy of the underlying system, not the entropy rate.
bandrami 1mo ago
Anything with a probability mass or distribution function has entropy; the state space of Markov chains just tends to get combinatorially big as the chain's memory increases.
kgwgk 1mo ago
> 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? That's essentially what he's doing there. He makes the calculation for N=8 but for very large N the result converges to N times that entropy.

by abetusk 1mo ago

ssivark 1mo ago

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 1mo ago

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-a-markov-chain-right-result https://math.stackexchange.com/questions/1040972/entropy-of-...

chermi 1mo ago
That is the entropy rate. If I'm understanding your original question correctly, you were asking about the standard equilibrium-defining thermodynamic entropy?
- abetusk 1mo ago
  I'm not sure I had a well formed question. I saw the post title, didn't get a clarification in the article and started searching. The entropy rate seems like a pretty natural definition of "entropy of a Markov chain", no? It's not exactly this but it's similar to "start at state i, end on state j (maybe in n steps?), what is the number of bits I need to send over the wire to tell you what path was taken". What does the entropy of the raw stationary distribution give you? Is the entropy rate related to the entropy of the stationary distribution (the thermodynamic entropy?)?
  - chermi 1mo ago
    I was probably projecting. I like both path and equilibrium entropies, but my assumption I guess was that when people hear entropy in physics they're usually thinking of the equilibrium-defining entropy. That is, the value that is maximized as the system relaxes to equilibrium, the vanilla thermodynamic entropy.

chermi 1mo ago

Find the equilibrium probabilities from first eigenvector, use definition of entropy like you did. If it's not ergodic then you get subsystems with their own entropies. Second eigenvalue iirc basically characterizes dominant relaxation time. This is all if you mean the equilibrium entropy of the underlying system, not the entropy rate.

bandrami 1mo ago

Anything with a probability mass or distribution function has entropy; the state space of Markov chains just tends to get combinatorially big as the chain's memory increases.

kgwgk 1mo ago

> 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? That's essentially what he's doing there. He makes the calculation for N=8 but for very large N the result converges to N times that entropy.