Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.