= Solution
The difference between the two squared actions is $|\dot p+\Gamma\mu|^2-|-\dot p+\Gamma\mu|^2=4\Gamma\dot p\cdot\mu$. The common factor cancels, giving
$$
\boxed{\frac{P_F}{P_B}=\exp\left[-\frac{2\Gamma}{\sigma^2}\int\dot p\cdot\frac{\delta F}{\delta p}\,d\mathbf r\,dt\right]
=\exp\left[-\frac{2\Gamma}{\sigma^2}(F_2-F_1)\right].}
$$
The last equality uses the <functional chain rule> with the same midpoint convention as the <Onsager–Machlup functional>.
At equilibrium, <microscopic reversibility> requires $e^{-F_1/(k_BT)}P_F=e^{-F_2/(k_BT)}P_B$, equivalently $P_F/P_B=e^{-(F_2-F_1)/(k_BT)}$. Comparing the coefficients of an arbitrary free-energy change gives the <model A fluctuation-dissipation relation>
$$
\boxed{\sigma^2=2\Gamma k_BT.}
$$
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