Given a Markov chain (finite state space) $X_1,X_2,...$ with transition matrix $P$ and initial distribution $\pi$, if they satisfy $\pi(x)P(x,y)=\pi(y)P(y,x)$, we say they satisfy detailed balance.
If the joint distribution of $(X_1,X_2,...,X_n)$ is identical to $(X_n,X_{n-1},...,X_1)$ for any $n \ge 1$, i.e. $\mathbb{P}\left\{ {{X_0} = {x_0},{X_1} = {x_1}, \ldots ,{X_n} = {x_n}} \right\} = \mathbb{P}\left\{ {{X_0} = {x_n},{X_1} = {x_{n - 1}}, \ldots ,{X_n} = {x_0}} \right\}$ for any realization $x_0,x_1,...,x_n$ and any $n$, then we say the Markov chain is reversible.
Detailed balance implies reversibility (subscripts are used to in the proof).
My question is, does time reversibility implies detailed balance? I think it is probably not true, can anyone help give a counter example? Thank you!
$\Bbb P[X_0=x_0,X_1=x_1]=\pi(x_0)P(x_0,x_1)$ and $\Bbb P[X_0=x_1,X_1=x_0]=\pi(x_1)P(x_1,x_0)$, so the $n=1$ case of reversibility already implies detailed balance.