Let $S(A)$ be the total amount of time spent in $A\subset \mathbb{Z}$ by a discrete-time asymmetric random walk on $\mathbb{Z}$. The transition probabilities are $P(i,i+1)=a,P(i,i-1)=b,P(i,i)=1-a-b$ where $0<a<b$.
Then there is a result that says $\mathbb{E}_0S(0)=\frac{1}{b-a}$ where the subscript means that the Markov chain starts at $0$ and $S(0)$ is the total time spent by the walk at $0$. I am not sure why this is true?
I know that this would be transient but I'm confused about the total time spent part and how to incorporate that into the expectation calculation? For example, if the total time spent is $1$, I don't know how to find the corresponding probability.
I got a different answer. I think the number of visits to $0$ is a geometric random variable with success probability being the probability of returning to $0$. Thus $\mathbb{E}_0S(0)=1/p$ where $p$ is the aforementioned probability.
We can figure out the probability of visiting $0$ again by using the gambler’s ruin starting from $1$ and that $0$ is hit before $n$, then let $n$ go to infinity. This probability is $1-\frac{1-\frac{a}{b}}{1-\left(\frac{a}{b}\right)^n}$ which goes to $\frac{a}{b}$ as $n\to \infty$. The probability of going from $0$ to $1$ is $b$ and thus $p=b\times \frac{a}{b}=a$ and $\mathbb{E}_0S(0)=\frac{1}{a}$ which is different from what was claimed in the book.
Is anything wrong in my argument?