Let $0<p=1-q,r<1$ be fixed probabilities, and $X_i,i=1,2,\dots$ be i.i.d. random variables with the following distribution: $$ X_i= \begin{cases} 1,&\text{with probability }rp,\\ 0,&\text{with probability }1-r,\\ -1,&\text{with probability }rq. \end{cases} $$ Define $$ S_0=0,\quad S_n=\sum_{i=1}^nX_i\quad(n>0). $$ This is a random walk that only moves with probability $r$, and stays put with probability $1-r$ at each step. Define the hitting time of level $1$ as usual: $$ \tau^+:=\inf\{n>0\:S_n=1\}. $$
Prove, for $0<s<1$, $$ P^+(s)=\frac{1-(1-r)s-\sqrt{((1-r)s-1)^2-4r^2pqs^2}}{2rqs}.\label{eq1}\tag{1} $$
Give a probabilistic interpretation of the form of $\eqref{eq1}$ of $P^+(s)$ of $\tau^+$ using the Geometric$(r)$ generating function $$ Q(s)=\frac{rs}{1-(1-r)s}. $$
I have successfully proved equation $\eqref{eq1}$, but I am stuck for the second part. To me, I would expect that if we set $q=0$, that would basically turn $\tau^+$ into a Geometric$(r)$ variable, since it stays at level $0$ with probability $1-r$, and it hits level $1$ with probability $r$, and it is the waiting time until we hit level $1$. However, when I substitute $q=0,p=1$ into my expression for $P^+(s)$, it does not return what I want it to. It seems like the reasonable way to geto to $Q(s)$ from $P^+(s)$ is to actually substitute in $q=1,p=0$, and then to take the reciprocal of this function. However, this does not make any intuitive sense to me, and I can't figure out what this would represent in terms of the random walk hitting level $1$. Why would we take the reciprocal? Why would we set it so that the random walk can only go down, rather than setting it so that it can only go up? Can anybody make sense of this? It seems like a contradiction between the two generating functions. Can anybody motivate why I would substitute in $q=1,p=0$ and then take the reciprocal? The reciprocal part seems especially confusing to me, because $$ \mathbb{E}\left(\frac1{s^{\tau^+}}\right)=\mathbb{E}(s^{-\tau^+}) $$ would seem to suggest to me that reciprocating the generating function is possibly related to considering the negative of that random variable, but how could we have the negative of a waiting time? Any help would be sincerely appreciated!