= Steady-state asymptotic fluctuation theorem
{title2=$I(-s)-I(s)=s$}
A <steady state> does not by itself give every fluctuating observable an <entropy production> symmetry. If the dimensionless total <entropy production> $\Sigma_t$ has the same-law finite relation and a <large deviation principle> with <rate function> $I$ for $\Sigma_t/t$, then, on its common domain,
$$
p_t(\Sigma_t/t\approx s)\asymp e^{-tI(s)},\qquad
I(-s)-I(s)=s.
$$
The first expression denotes leading exponential scaling; the second follows by taking the scaled <logarithm> of the probability ratio. A finite-state, autonomous <Markov process> with time-even states, positive bidirectional rates and a positive <stationary distribution> supplies a standard setting for a finite total-<entropy production> relation. The stochastic system <entropy> is then bounded, so its endpoint contribution does not change the long-time <rate function> of bath <entropy production>. <Stochastic thermodynamics, fluctuation theorems, and molecular machines by Udo Seifert>[Seifert, 2012, §3.3.2, p. 22, Eq. (65); §4.4.3, p. 28, Eqs. (93)–(95): finite stationary total entropy and endpoint-conditioned medium entropy]{ref}
Unbounded state spaces can have endpoint fluctuations on the scale relevant to a <large deviation principle>. Dropping system <entropy>, changing the measured observable to <heat>, or changing the reverse preparation therefore requires a fresh argument. The exact finite <integral fluctuation theorem>, a same-law finite relation, and an asymptotic <rate function> symmetry are distinct claims.
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