Let $M$ denote the Markov chain on states $\{0, 1, 2, ...\}$ with absorbing state $0$. For $i \geq 1$, let the transition probabilities be $p$ for $(i, i-1)$ and $1-p$ for $(i, i+1)$. Further, assume $p > 1/2$, thus, there is a drift towards $0$. I am interested in the mean absorbing/hitting time in state $0$, when started at state $n > 0$, denoted as $\mathbb{E}\{h(n)\}$. Due to the drift towards $0$, this should easily be determinable by using the linearity of the expectation on each single step?
More precise, the expected change ('increment') $\Delta$ on each single step of a walk that did not yet reach $0$, is $\mathbb{E(\Delta)} = (-1) \cdot p + 1 \cdot (1-p) = 1 - 2p < 0$. Thus, 'on average' we make progress $|1 - 2p|$ towards $0$ on each step, so we should hit $0$ after $h := \frac{n}{|1-2p|}$ many time steps, since $n + \mathbb{E}\{h \cdot \Delta\} = n + h \mathbb{E}\{\Delta\} = n - n = 0$. Thus, it should hold that $\mathbb{E}\{h(n)\} = h$.
I suppose that there is something wrong with this easy argument, since I fail to find a formal proof for it; I cannot adopt the hitting time definition to this. So my two questions are:
(1) Is this approach right or wrong, and how can $\mathbb{E}\{h(n)\}$ be determined correctly?
(2) If this argument is wrong, shouldn't it at least provide an upper bound on the hitting time?
Could somebody please help on this?
Hints:
(1) Show that $\mathbb E(h(n))=n\mathbb E(h(1))$.
(2) Show that $\mathbb E(h(1))=1+(1-p)\mathbb E(h(2))$.
(3) Conclude that, indeed, $\mathbb E(h(n))=n/(2p-1)$ for every $n\geqslant1$.