My question is from the book Introduction to Probability Models, 10th edition, by Sheldon Ross.
Here is a example in the book. Consider a Markov chain with states $0, 1,\cdots , n$ having $P_{0,1} = 1, P_{i,i+1} = p, P_{i,i−1} = q = 1 − p, 1 \leq i < n$.
and suppose that we are interested in studying the time that it takes for the chain to go from state $0$ to state $n$.
One of the approaches the author uses in the book is to construct a Markov Chain. The method is as following:
let $N_i$ denote the number of additional transitions that it takes the chain when it first enters state $i$ until it enters state $i + 1$. By the Markovian property, it follows that these random variables $N_i, i = 0, . . . , n − 1$ are independent. Also, we can express $N_{0,n}$, the number of transitions that it takes the chain to go from state $0$ to state $n$, as $$ N_{0,n} = \sum_{i=0}^{n-1}N_i$$ Letting $\mu_i = E[N_i]$ we obtain, upon conditioning on the next transition after the chain enters state $i$, that for $i = 1, . . . , n − 1$
$$\mu_i = 1 + E[\text{number of additional transitions to reach } i + 1|\text{chain to } i − 1]q$$.
My question is why the author uses 1 instead of $p$ in the last equation.
To me, it sounds more reasonable that $$\begin{align} \mu_i &= E[\text{number of additional transitions to reach } i + 1 \text{ from } i|\text{chain to } i + 1]p \\& \quad+ E[\text{number of additional transitions to reach } i + 1\text{ from } i|\text{chain to } i − 1]q \\ &= p + E[\text{number of additional transitions to reach } i + 1|\text{chain to } i − 1]q \end{align}$$