Consider a Simple Random Walk, $(X_n)_n≥0$, on $\mathbb{Z}$ where we have the following:
$ P_{ij} = \begin{cases} p&\text{if i = j + 1}\\ q=1-p&\text{if i = j - 1}\\ 0&\text{otherwise}\\ \end{cases} $
Suppose that $1 > p > q > 0$. We know that the process $(X_n)_{n≥0}$ is transient for all states $i ∈ Z$.
QUESTION: For state $m ∈ I$, let $H^m := inf\{n ≥ 0 : X_n = m\}$. Suppose that $m < 0$. Then show that for $m < i ∈ I$ $$h_i:= P_i(H^m < ∞) = a^{f(i,m)}$$ where $a$ is a constant and $f(i, m)$ is a function of $i$ and $m$.
So far I have: I have looked around on here and through other resources and I can't find anything to help me solve this question rigorously. I have thought about this problem and as $$h_i = ph_{i+1} + qh_{i-1}. $$ This recurrence relation with assuming $h_i = \lambda^i$ produces a solution of the form $$ h_i = A + B(\frac{q}{p})^i$$
As $h_m$ is the probability of hitting state $m$ given we start at $m$, I have concluded that $h_m = 1$.
However, I am having trouble coming up with a second boundary condition to solve for A and B. Any help would be appreciated. As q < p, this is an asymmetric random walk on Z. Please note the state space is all of the integers and that $m < i$ and $m < 0$.
I believe as q < p this leads to $h_i = (\frac{q}{p})^{i-m}$ but I cannot get there.