Consider a DTMC with state space $S$ and transition matrix $P$. We say a probability distribution $(p_0,p_1,\ldots)$ satisfy the detailed balanced condition if for any state $i$ and $j$, $p_i P_{i,j} = p_j P_{j,i}$
Assume a Markov Chain has a tri-diagonal transition matrix, that is $P_{i,j} = 0$ whenever $|i-j| >1$. Prove that every stationary distribution for this chain satisfy the detailed balanced condition.
My trial: Suppose $(\pi_0,\pi_1\ldots,)$ is a stationary distribution, then the result trivially holds for any $i,j \in S$ such that $|i-j| \ge 2$ and $i = j$. Hence suppose $i = j+1$, then it suffices to show $\pi_{j+1} P_{j+1,j} = \pi_j P_{j,j+1}$, by the property of stationary distribution and tri-diagonal matrix we have $\pi_{j+1} = \pi_{j} P_{j,j+1} + \pi_{j+1} P_{j+1,j+1} + \pi_{j+2} P_{j+2,j+1}$. But then I got stuck, hope someone can help me, thanks!
Edit: I just found one proof which firstly prove the Global Balance Equation, that is for any subset $S \subset \{0,1,2,\ldots\}$ we have $\sum_{j \in S}\pi_j \sum_{i \not \in S}P_{ji} = \sum_{i \not \in S}\pi_i \sum_{j \in S}P_{ij}$. This equation is very easy to prove and apply it I can get the detailed balance equation very esaily.
Best Answer
If the transition matrix is tri diagonal, you can write $\pi_i = p_{i-1,i}\pi_{i-1} + p_{i,i}\pi_{i} + p_{i+1,i}\pi_{i+1}$ for all i, where $\pi$ is the stationary distribution.
So $(1 - p_{i,i})\pi_i = p_{i-1,i}\pi_{i-1} + p_{i+1,i}\pi_{i+1}$.
As $1 - p_{i,i} = p_{i,i-1} + p_{i,i+1}$ you can write this equation $p_{i,i-1} \pi_i - p_{i-1,i}\pi_{i-1} = p_{i+1,i}\pi_{i+1} - p_{i,i+1}\pi_i$.
So the sequence $u_i = p_{i,i-1} \pi_i - p_{i-1,i}\pi_{i-1}$ is constant equal to $c$ on state space $S$. Now two cases:
S is infinite : As $\sum p_{i,i-1} \pi_i < \infty$ and $\sum p_{i-1,i}\pi_{i-1} < \infty$ (because sum of positive terms smaller respectively than $\pi_i$ and $\pi_{i-1}$), $\sum_{i \in S} u_i$ is absolutely convergent. As $S$ is infinite, this means that $c = 0$.
S is finite : if $0$ si the left-most state, then $\pi_0 = p_{0,0}\pi_{0} + p_{1,0}\pi_{1}$. So $u_1 = 0$ and $u_i =0$ for all other $i$.