How was induction used to arrived at this equation I am trying to understand the solution to a Markov chain model of the nodes of a wireless network and found a great discussion of it here.
What I do not get is the part where induction was involved:

The only way state $(i,k)$  can be reached, for  $1≤i≤m−1$ , and
$0≤k≤Wi−2$  is from the states  $(i−1,0)$  and  $(i,k+1)$ , from which
it is jumped to with probabilities  $\ \frac{p}{W_i}\ $  and $1$,
respectively, at each step. Therefore,
$b_{i,k} = \frac{p}{W_i}b_{i-1,0} + b_{i,k+1}$
for those values of $i$  and $k$. By induction on $k$, these equations
give
$b_{i,k}=\left(W_i - k-1\right)\frac{p}{W_i}b_{i-1,0}+ b_{i,W_i-1}$

Question: I do not get how this $b_{i,k}=\left(W_i - k-1\right)\frac{p}{W_i}b_{i-1,0}+ b_{i,W_i-1}$ was obtained by induction.
As a brief overview, the $b_{i,k}$ here is the stationary distribution of the chain defined as  $b_{i,k} = \lim_{t \rightarrow \infty} P\{s(t)=i, b(t)=k\}$. Each state in the chain is denoted as $(i,k)$ and $s(t)$ and $b(t)$ denote the processes incrementing $i$ and $k$, respectively. I cannot post images, but the Markov chain model is available here.
 A: The induction probably looks a bit unfamiliar since (a) it's finite induction rather than infinite, and (b) because of how everything's written you actually have to start from the other end of the sequence. You could re-index everything by making the substitution, say, $j = W_i - 1 - k$, but if we don't do that then we get something like this:
Base case: When $k = W_i - 2$,
$b_{i, k} = b_{i, W_i - 2} = \frac{p}{W_i} b_{i - 1, 0} + b_{i, W_i - 1} = (W_i - (W_i - 2) - 1)\frac{p}{W_i} b_{i - 1, 0} + b_{i, (W_i - 2)  +1} = (W_i - k - 1)\frac{p}{W_i} b_{i-1,0} + b_{i, k + 1}$
so the formula holds.
Induction hypothesis: Assume that the formula holds for some $1 \leq k \leq W_i - 2$. Then consider $k - 1$,
$\begin{eqnarray}b_{i, k - 1} & = & \frac{p}{W_i} b_{i - 1, 0} + b_{i, k} \\
& = & \frac{p}{W_i} b_{i - 1, 0} + (W_i - k - 1) \frac{p}{W_i} b_{i - 1, 0} + b_{i, W_i - 1} \\
& = & (1 + W_i - k - 1) \frac{p}{W_i} b_{i - 1, 0} + b_{i, W_i - 1} \\
& = & (W_i - (k - 1) - 1) \frac{p}{W_i} b_{i - 1, 0} + b_{i, W_i - 1}\end{eqnarray}$
which matches the formula, and so by induction it applies for all relevant $k$.
