Finite-time fluctuation identity for stationary path entropy

Since $X_0\sim \pi$ and $\pi Q=0$, [F3] gives $\mathbb{P}(X_t=x)={\pi }_x$ for every $x\in E$ and every $t\in [0,T]$. Thus the process is stationary on the whole interval under the stated initial law.

Fix a path with $n$ jumps, state sequence $x_0,\dots ,x_n$ and times $0<t_1<\cdots <t_n<T$. Its density is $f(\gamma )={\pi }_{x_0}\left({\prod }_{j=1}^n{q}_{x_{j-1}{x}_j}\right)\exp (-{\sum }_{j=0}^n{\lambda }_{x_j}(t_{j+1}-t_j)),$ where $t_0=0$ and $t_{n+1}=T$, by [F2]. The reversed path has state sequence $x_n,\dots ,x_0$ and jump times $T-t_n,\dots ,T-t_1$, and its holding-time exponent is the same because it spends exactly the same duration in each state. If any forward jump rate in the sequence is zero, symmetric support makes the corresponding reversed rate zero too, so both path densities vanish and the sequence contributes nothing. On every sequence with positive density, symmetric support makes all reversed jump rates positive. The ratio of forward to reversed density is therefore $\frac{f(\gamma )}{f({\gamma }^{†})}=\frac{{\pi }_{x_0}}{{\pi }_{x_n}}{\prod }_{j=1}^n\frac{q_{x_{j-1}{x}_j}}{q_{x_j{x}_{j-1}}}={\prod }_{j=1}^n\frac{{\pi }_{x_{j-1}}{q}_{x_{j-1}{x}_j}}{{\pi }_{x_j}{q}_{x_j{x}_{j-1}}}=e^{S_T(\gamma )}\mathrm{.}$ The middle equality telescopes the factors ${\pi }_{x_{j-1}}\mathrm{/}{\pi }_{x_j}$; for $n=0$ both densities are positive and their ratio is $1$.

For $n=0$, reversal fixes the no-jump path. For $n\ge 1$, time reversal sends $(t_1,\dots ,t_n)$ to $(T-t_n,\dots ,T-t_1)$. This is an affine bijection of the ordered time simplex; its linear part is minus a coordinate-reversal permutation matrix, whose determinant has absolute value $1$. Thus [F9] shows it preserves Lebesgue measure and transports each nonnegative density integral to the reversed-path integral. The map also reverses the finite state sequence and sends $S_T$ to $-S_T$. Thus $-B$ is Borel by [F8]. Integrating the density identity in step 1.2 over paths with $S_T\in B$, and using the measure-preserving change of variables, gives ${\mu }_T(B)={\int }_{\{S_T\in -B\}}{e}^{-S_T}d\mathbb{P}={\int }_{-B}{e}^{-u}{\mu }_T(du)\mathrm{.}$ The countable union over $n$ covers all paths by [F2]. Taking $B=\mathbb{R}$ gives $\mathbb{E}[e^{-S_T}]=1$. The same identity is equivalent, by the change of variable $u=-s$, to ${\mu }_T(ds)=e^s{\mu }_T(-ds)$ as an equality of measures.

There are finitely many directed edges, all nonzero forward and reverse rates are positive, and every ${\pi }_x>0$. Thus the absolute value of each jump increment in [F4] is bounded by a finite constant $C$; if there are no edges, set $C=0$. Hence $\mathrm{\mid }{S}_T\mathrm{\mid }\le C{N}_T$, and [F2] gives $\mathbb{E}\mathrm{\mid }{S}_T\mathrm{\mid }<\mathrm{\infty }$. For every $z>0$, the function $g(z):=z-1-\log z$ has $g(1)=0$ and derivative $g^{\mathrm{\prime }}(z)=1-1\mathrm{/}z$, negative on $(0,1)$ and positive on $(1,\mathrm{\infty })$; the mean value theorem [F6] gives $g(z)\ge 0$. Apply this pointwise with $z=e^{-S_T}$. The function $g(e^{-S_T})$ is nonnegative and integrable because $S_T$ is integrable and $\mathbb{E}[e^{-S_T}]=1$; its expectation is therefore nonnegative. By linearity [F7] and $\log (e^{-S_T})=-S_T$, $0\le \mathbb{E}[g(e^{-S_T})]=\mathbb{E}[e^{-S_T}]-1+\mathbb{E}[S_T]=\mathbb{E}[S_T]\mathrm{.}$ Therefore $\mathbb{E}[S_T]\ge 0$.