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Consider the following transition matrix

$$ T= \left[ {\begin{array}{cccc} \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6}\\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{2}{6}\\ \frac{1}{3} & \frac{1}{4} & \frac{3}{5}& 0\\ 0 & \frac{1}{4} & 0& \frac{3}{6}\\ \end{array} } \right] $$

of a Markov chain process $$P_{t+1}=TP_t$$

The steady state is given by

$$ P^*=\frac{1}{137} \left[ {\begin{array}{c} 33 \\ 36 \\ 50 \\ 18 \\ \end{array} } \right] $$

The detailed balance condition implies that $$\sum_i T_{ij}P_i^*=\sum_j T_{ji}P_j^*$$

since for example $T_{24}P^*_2\neq T_{42}P_4^*$, the chain process does not satisfy the detailed balance condition.

I want to understand if the system is in thermal equilibrium. Obviously the system is not reversible, since there is no detailed balance, but is it a sufficient condition?

can a system be irreversible and in thermal equilibrium?

If so, what is the mathematical condition that needs to be satisfied in order to be in thermal equilibrium?

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    $\begingroup$ Why are you dividing by the square root of 5209? Don't the entries in your equilibrium state need to sum to 1? $\endgroup$ – Aaron Stevens Jul 27 '18 at 10:50
  • $\begingroup$ @Aaron Stevens, that's correct, I've just normalized it with mathematica without paying attention. trying to trace my mistake now. $\endgroup$ – jarhead Jul 27 '18 at 10:56
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    $\begingroup$ Just divide by the sum of your entries :) Those numbers are correct. $\endgroup$ – Aaron Stevens Jul 27 '18 at 10:56
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According to the Wikipedia entry on detailed balance:

The detailed balance condition is stronger than that required merely for a stationary distribution; that is, there are Markov processes with stationary distributions that do not have detailed balance. Detailed balance implies that, around any closed cycle of states, there is no net flow of probability. For example, it implies that, for all a, b and c, $$P(a,b)P(b,c)P(c,a)=P(a,c)P(c,b)P(b,a)$$

If you mess around with your system, you will see that this does not hold for closed cycles between states 1, 2, and 4 and closed cycles involving all four states. Therefore, we have an equilibrium distribution without detailed balance.

This is not my area of expertise, but all we have here is that the probability distributions approach an "equilibrium value", but I am not sure if this necessarily means that the system itself needs to be at thermodynamic equilibrium. From the first page on this site https://www.stat.auckland.ac.nz/~fewster/325/notes/ch9.pdf

Note: Equilibrium does not mean that the value of $X_{t+1}$ equals the value of $X_t$ . It means that the distribution of $X_{t+1}$ is the same as the distribution of $X_t$

I imagine a circular body of water where there is a uniform whirlpool in the center. Even though we have an equilibrium probability distribution of where a water molecule will be, the system itself is not in equilibrium. This is not a perfect way to view this (as pointed out in the comments) since sometimes the system might not converge to the equilibrium distribution, but I still think it is a good picture here.

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    $\begingroup$ +1 good answer, but there's a subtlety, highlighted on math.SE. Example: the matrix $$T=\begin{pmatrix} 0 & 0.5 & 0 & 0.5 \\ 0.5 & 0 & 0.5 & 0 \\ 0 & 0.5 & 0 & 0.5 \\ 0.5 & 0 & 0.5 & 0 \end{pmatrix}$$ has stationary distribution $(0.25,0.25,0.25,0.25)$ and obeys detailed balance, but is not aperiodic. You can see that $T^n$ is different for odd and even $n$; therefore an arbitrary starting distribution may not converge to the "equilibrium" one. So I think the "no net flow" picture can be misleading! $\endgroup$ – user197851 Jul 27 '18 at 13:40
  • $\begingroup$ @LonelyProf Good point! Like I said, I am not too familiar with this stuff. I will make a note in the answer about my "example" $\endgroup$ – Aaron Stevens Jul 27 '18 at 14:09
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    $\begingroup$ The point about the question on math.SE is the textbook claim that the detailed balance condition was introduced to eliminate the danger of "limit cycles". The OP of that question doubted the logic of the argument. My initial thoughts were similar to yours, but now I doubt it too. Just to be clear: my comments relate to the final question of the OP here. Your analysis of the main question is fine, as one can confirm numerically by iterating the supplied $T$ several times: it converges to equilibrium. $\endgroup$ – user197851 Jul 27 '18 at 14:36

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