Entropy production on a finite biased ring

The total rate leaving every state is $a+b$. Each state also receives total stationary flow $(a+b)\mathrm{/}N$ from its two neighbors, so the uniform row vector satisfies $\pi Q=0$. By [F2], starting the process in $\pi$ makes it stationary. The condition $N\ge 3$ ensures $i+1$ and $i-1$ are distinct states.

By [F1] and [F4], almost surely every jump is clockwise or counterclockwise. A clockwise jump $i\to i+1$ contributes $\log (({\pi }_{i}a)\mathrm{/}({\pi }_{i+1}b))=\log (a\mathrm{/}b)$, while a counterclockwise jump contributes $\log (({\pi }_{i}b)\mathrm{/}({\pi }_{i-1}a))=-\log (a\mathrm{/}b)$. Summing these almost-sure increments gives $S_T=(N_{+}-N_{-})\log (a\mathrm{/}b)$ almost surely, including the no-jump case when both counts are zero.

Detailed balance on an edge requires ${\pi }_{i}a={\pi }_{i+1}b$, which under the uniform law is equivalent to $a=b$. If $a=b$, this equality holds on every edge; nonedges have zero rate in both directions. Hence detailed balance holds exactly in the symmetric-rate case.

By [F3] and stationarity from step 1.1, $\mathbb{E}[N_{+}]={\sum }_i\mathbb{E}[N_{i,i+1}]={\sum }_i{\int }_0^T\frac{1}{N}adt=aT,\mathbb{E}[N_{-}]={\sum }_i\mathbb{E}[N_{i,i-1}]={\sum }_i{\int }_0^T\frac{1}{N}bdt=bT\mathrm{.}$ Thus step 1.2 gives $\mathbb{E}[S_T]=(aT-bT)\log (a\mathrm{/}b)=T(a-b)\log (a\mathrm{/}b)$. Since $\log$ is strictly increasing by [F6], $a-b$ and $\log (a\mathrm{/}b)$ have the same sign, so their product is nonnegative.

The ring is irreducible because both rates are positive and repeated clockwise jumps connect every pair of states. It has positive stationary law $\pi$ and symmetric support because every edge has both rates positive. Therefore [F5] applies and gives the stated exact finite-time fluctuation relation, with no assumption of a probability density for $S_T$.