the forward experiments. The probability distribution of trajectory $X'_n$ with backward protocol is given by $$P[X'_N | \Lambda_N^\dagger] = \prod_{k=0}^{N-1} P[x'_k | X'_k, \lambda_k^\dagger] P_0^\dagger [x'_0] =: P^\dagger[X'_N],
\label{prob_b1}$$ where we write $P[X'_N | \Lambda_N^\dagger]$ as $P^\dagger[X'_N]$ for simplicity. Correspondingly, $$P[X'_N | x'_0, \Lambda_N^\dagger] = \prod_{k=0}^{N-1} P[x'_k | X'_k, \lambda_k^\dagger] =: P^\dagger [X'_N | x'_0] .
\label{prob_b2}$$

In special cases, the backward trajectory $X'_N$ is equal to the time-reversal of the forward trajectory $X_N$. Let $x^\ast$ be the time-reversal of phase-space point $x$. For example, if $x = (\bm r, \bm p)$ with $\bm r$ and $\bm p$ being the position and the momentum respectively, we have $x^\ast := (\bm r, - \bm p)$. The time-reversal of trajectory $X_n$ is then given by $X_n^\dagger := (x_N^\ast, x_{N-1}^\ast, \cdots, x_{N-n}^\ast)$. With notation $x_n^\dagger := x_{N-n}^\ast$, we write $X_n^\dagger = (x_0^\dagger, x_1^\dagger, \cdots, x_n^\dagger)$. By substituting $x'_n = x_n^\dagger$ to Eqs. (\[prob\_b1\]) and (\[prob\_b2\]), we obtain the probability of realizing a backward trajectory under the backward protocol as $$P^\dagger [X_N^\dagger ] = \prod_{k=0}^{N-1} P[x_k^\dagger | X_k^\dagger, \lambda_k^\dagger] P_0^\dagger [x_0],$$ where the conditional probability under initial $x_0^\dagger$ is given by $$P^\dagger [X_N^\dagger | x_0^\dagger] = \prod_{k=0}^{N-1} P[x_k^\dagger | X_k^\dagger,
\lambda_k^\dagger].$$ We note that 