 Solution

Let $X$ be a coarse-grained configuration, with restricted partition function $Z_X$ and free energy $F(X)=-k_{B}T\log Z_X$. Its Boltzmann distribution is $\pi (X)\propto e^{-\beta F(X)}$. Microscopic reversibility pairs every equilibrium microscopic history with a history read backwards, including reversal of all time-odd variables. Summing this identity over the microscopic histories representing a given coarse-grained path preserves it; a Markov assumption is not needed for this argument.

Write ${\mathbb{P}}_F$ for the probability density of a specified history conditional on its starting configuration $X_1$, and ${\mathbb{P}}_B$ for its physically reversed history conditional on starting at $\Theta X_2$, where $\Theta$ reverses time-odd variables. Equilibrium time-reversal symmetry givesSince $F(\Theta X)=F(X)$, the desired conditional path relation isFor a reversible Markov chain, this also follows by multiplying detailed balance ratios $P_{ij}/P_{ji}={\pi }_j/{\pi }_i$ along the history: the intermediate equilibrium weights telescope.

The conditioning is essential. If the starting equilibrium weights are included in the path probabilities, the forward and reversed probabilities are equal, rather than differing by the displayed exponential. For example, a two-state continuous-time Markov chain with $F_1=0$, $F_2=\Delta >0$, rates $k_{12}=e^{-\beta \Delta }$ and $k_{21}=1$ has equal equilibrium-weighted probabilities for the two reversed single-jump histories. Their conditional rate ratio is instead $e^{-\beta \Delta }$. Normalized bridges conditioned on both endpoints also have a different normalization from the path weights used above. The PDF prints ${\mathbb{F}}_B$ in the denominator while its explanatory text defines ${\mathbb{P}}_B$; the latter is the intended notation. For the scalar order parameter used subsequently, $\Theta \varphi =\varphi$.