I'm not very familiar with the theory of Markov chains, and I'd like to learn how complicated the following problem actually is.
Let there be a discrete time Markov chain on $\mathbb{Z}$, where the random variable giving the position on the line is called $Z(t)$. The state is only allowed to change to the nearest neighbours, so the chain is defined by the transition probabilities:
\begin{align} P \{ Z(t+1)=n+1 \mid Z(t)=n \} &= q_{n} \\ P \{ Z(t+1)=n-1 \mid Z(t)=n \} &= 1-q_{n} \\ P \{ Z(t+1)=m \mid Z(t)=n \} &= 0 &\mathrm{if} \ m \notin \{n-1,n+1 \} \end{align}
$ \forall \ n \in \mathbb{Z}$. The point is that the transition probabilities $q_{n}$ are not necessarily the same everywhere.
On this system, there is a periodic symmetry. With period $N \in \mathbb{N}$, we have: $q_{n+k \cdot N} = q_{n} \ \ \forall k \in \mathbb{Z}$.
Now, let's define $p_{ij}^{(t)} = P\{Z(t)=j \mid Z(0)=i\}$ (everything is time-homogeneous) and the corresponding first-passage probability $f_{ij}^{(t)}$, the probability to first reach $j$ from $i$ after exactly $t$ timesteps. The probability to make at least one passage at all at location $j$ is given by:
\begin{equation} f_{ij} = \sum_{t=1}^{\infty} f_{ij}^{(t)} \end{equation}
Intuitively, the system should run forward if the forward probabilities $q_{n}$ are larger than the backward ones. More formally, I define $Q=\prod_{n=1}^{N} \frac{q_{n}}{1-q_{n}}$. For $Q>1$, I expect that:
\begin{align} f_{ij}&= 1 &\mathrm{if} & \ j>i \\ f_{ij}&<1 &\mathrm{if} & \ j \leq i \end{align}
Is there any way I can prove this? Help would be much appreciated.