Equivalence between detailed balance and time reversal I need help in proving the following popular claim
A continuous time and stationary Markov jump process obeys the detailed balance equations
$$
P(x)q(x,x') = P(x')q(x',x)
$$
where $q(\cdot,\cdot)$ is the state transition rates, and $P(\cdot)$ is the equilibrium distribution ($P_t(x) = P_r(X_t=x)\to P(x)$), if and only if he obeys the time reversal property.
Thank you 
 A: Hint: Write down $\mathbb P(X_{t_1}=x,X_{t_2}=x')$ and $\mathbb P(X_{t_1}=x',X_{t_2}=x)$ and equate them. Likewise, for every sequence $(x_k)_{1\leqslant k\leqslant n}$, write down $\mathbb P(X_{t_1}=x_1,X_{t_2}=x_2,\ldots,X_{t_n}=x_n)$ and $\mathbb P(X_{t_1}=x_n,X_{t_2}=x_{n-1},\ldots,X_{t_n}=x_1)$ and equate them.
For example, denoting by $\pi$ the stationary measure and $Q$ the matrix with off-diagonal entries $q(x,x')$ and row sums zero, for every $x\ne x'$,
$$
\mathbb P(X_{t_1}=x,X_{t_2}=x')=\pi(x)\left(\mathrm e^{(t_2-t_1)Q}\right)(x,x').
$$
When $t\to0$, $\mathrm e^{tQ}=I+tQ+o(t)$ hence, when $t_2\to t_1$ the RHS is
$$
\pi(x)(t_2-t_1)q(x,x')+o(t_2-t_1).
$$
This coincides with $\mathbb P(X_{t_1}=x',X_{t_2}=x)$ at the first order if and only if the function
$$
(x,x')\mapsto\pi(x)q(x,x')
$$
is symmetric with respect to $(x,x')$.
Edit: The OP alludes in the comments to the fact that $\mathbb P(X_{t+s}=b\mid X_t=a)=q(a,b)s+o(s)$ when $s\to0^+$. This leads once again to the same condition, as follows. Assume the distribution of $X_t$ is $\pi$, then
$$
\mathbb P(X_t=a,X_{t+s}=b)=\mathbb P(X_{t+s}=b\mid X_t=a)\mathbb P(X_t=a)=\pi(a)q(a,b)s+o(s),
$$
and this expression must be symmetric with respect to $(a,b)$ for every $(t,s)$, hence in particular when $s\to0^+$. QED.
