Consider the following Markov chain on $\mathbb{Z}$ with transition probabilities given by $p(i,j)=1/2$ if $j=-1$, $p(i,j)=1/2$ if $j=i+1$ and $0$ otherwise. I want to find the probability $\mathbb{P}_{0}(T_{-1}=n)$ for each $n\geq 1$, where $T_{-1}$ is the first time the chain hits $\{-1\}$.
I found that $\mathbb{P}_{0}(T_{-1}=1)=1/2$. However, I have troubles to find $\mathbb{P}_{0}(T_{-1}=n)$ for $n\geq 2$. By the Chapman-Kolmogorov equation, it is reasonable that $\mathbb{P}_{0}(T_{-1}=2)=p^{2}(0,-1)=0$. What about for general $n$?
I will appreciate any help.
The only way for the event $\{T_{-1}=n\}$ to occur is if the the first $n-1$ steps of the process are $i\to (i+1)$, and then the $n^{th}$ step is $(n-1)\to 0$. The probability of this occurring is $(1/2)^n$.
By the way, you erred in applying the Chapman-Kolmogorov equation. The correct way is $$ P_0(T_{-1}=2)=\sum_j p(0,j)p(j,-1) $$ Since the only $f$ for which $p(0,j)$ is nonzero is $j=1$, $P_0(T_{-1}=2)=p(0,1)p(1,-1)=1/4$. In general, you need to sum over all possible paths the Markov process can take, but since your process is so simple, there always only a single path from $0$ to $-1$ which has length $n$ and does not hit $-1$ prematurely.