# 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

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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.

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Thank you, but if I am not wrong in this way you assume discrete time process (which I already proved). I want to prove it for continues time.. –  user39097 Dec 22 '12 at 11:23
See Edit.   –  Did Dec 22 '12 at 11:57
OK, now I understand what you meant. I have a silly (sorry) question: from where the equation $\mathbb P(X_{t_1}=x,X_{t_2}=x')=\pi(x)\left(\mathrm e^{(t_2-t_1)Q}\right)(x,x')$ comes from? BTW, in order to generalize it to any n I can just use induction, right? –  user39097 Dec 23 '12 at 9:21
This is the definition of a Markov process with transition kernel $Q$ such that the distribution of $X_{t_1}$ is $\pi$. –  Did Dec 23 '12 at 10:04
Sorry but what is YOUR definition of a Markov process? –  Did Dec 23 '12 at 10:34