I am not able to understand the reasoning behind the load-balance equation of the birth-death processes (Markov chain) . The equation is $\pi_ib_i = \pi_{i+1}d_{i+1}$ , where the symbol have their usual meanings ie ( $b_i$ is the probability of going from state $i$ to $i+1$ and $d_{i+1}$ is the probability of coming from state $i+1$ to state $i$ ) .
What the book says :
For a birth-death process, the balance equations can be substantially simplified. Let us focus on two neighboring states, say, $i$ and $i+1$. In any trajectory of the Markov chain, a transition from $i$ to $i+1$ has to be followed by a transition from $i + 1$ to $i$, before another transition from $i$ to $i + 1$ can occur. Therefore, the frequency of transitions from $i$ to $i + 1$, which is $π_ib_i$, must be equal to the frequency of transitions from $i + 1$ to $i$, which is $π_{i+1}d_{i+1}$. This leads to the local balance equations.
If I simply go on and write a load balance equation at node $i$ the equation must be $$\pi_i = \pi_{i-1}b_{i-1} + \pi_{i+1}d_{i+1}+\pi_i(1-b_{i-1}-d_{i+1})$$ but if I go for solving these I can never reach the earlier the simple equation given earlier.
I don't understand why the frequencies should be same . Suppose I have more probability of going from $i$ to $i+1$ and less of returning then how can they be balanced in the long run . I know my reasoning is wrong but I am not understanding the fallacy . Please help me here .
PS: I am studying markov chains for the first time
Thanks in advance
The book suggests to look at what goes through an edge. Indeed, at equilibrium, since the only way to go from $\{j\mid j\leqslant i\}$ to $\{j\mid j\geqslant i+1\}$ or from $\{j\mid j\geqslant i+1\}$ to $\{j\mid j\leqslant i\}$ is to use the edge $\{i,i+1\}$, the fluxes through this edge in the direction $i\to i+1$ and in the direction $i+1\to i$ must compensate, that is, $$ \pi_ib_i = \pi_{i+1}d_{i+1}.\tag{$\ast$} $$ Your idea, on the other hand, is to look at what leaves, what arrives and what stays at a node. Unfortunately, there is a mistake in your computations, and this approach leads actually to $$ \pi_i = \pi_{i-1}b_{i-1} + \pi_{i+1}d_{i+1}+\pi_i(1-b_{i}-d_{i}), $$ the last term on the RHS counting what stays at node $i$. Usually though, one equates what leaves and what arrives at a node, in which case one gets the equivalent identity $$ \pi_{i-1}b_{i-1} + \pi_{i+1}d_{i+1}=\pi_i(b_{i}+d_{i}).\tag{$\dagger$} $$ And now the Grand Finale: the systems $(\ast)$ and $(\dagger)$ are in fact equivalent... hence each yields the correct stationary distribution $$ \pi_i=\pi_0\cdot\prod_{j=1}^i\left(\frac{b_{j-1}}{d_j}\right), $$ for the unique positive factor $\pi_0$ which makes that the mass of $(\pi_i)$ is $1$.