The down-cross probability of a general random walk Let $\{X_t\}_{t=0,1,2\dots}$ be a random walk, defined by $X_0 = x$ and
$$
X_{t+1} = X_0 + \sum_{\tau=1}^t q_\tau,
$$
where $q_t$ is an iid random variable. We let
$$
q_t = \begin{cases}
a & \text{ w.p. }p\\
-b & \text{ w.p. }1-p
\end{cases},
$$
where $\mathbb{E}[q_t]>0$ and $a,b>0$. Thus, the state space is a continuum and this random walk has a positive drift.
For $z < x$, I want to compute the probability that $\{X_t\}$ does NOT down-cross $z$. Intuitively, this probability is positive, because the process has a positive drift. At least, I'd like to show formally that this probability is indeed positive.
I have no idea how to start. Resources I can find online or in textbooks deal with random walks on lattices, not on continuums. Since I cannot write a recursive equation for each state as in the countable state space case, I think I need to take a different approach from the case where $a,b\in\mathbb{Z}$.
 A: If you just want to show that the probability is positive, you can adapt the answer here, as follows.
The idea (as explained in Aaron Montmogery's comment) is to use a sort of de Moivre's martingale, by looking for a positive $\lambda$, $\lambda\neq 1$, such that $Y_n=\lambda^{X_n}$ is a martingale wrt the filtration $\cal F=({\cal F}_n)$ where ${\cal F}_n=\sigma(q_1,q_2,\ldots,q_n)$.
Now, $(Y_n)$ will be a martingale iff $\mathbb{E}(Y_{n+1}|{\mathcal{F}}_n)=Y_n$ , or equivalently ${\mathbb{E}}(\frac{Y_{n+1}}{Y_n}|{\mathcal{F}}_{n})=1$.
Now the LHS of this latter equality evaluates to $p\lambda^{a}+(1-p)\lambda^{-b}$ ; let's denote this expression by $f(\lambda)$. Then $f(1)=1$ and $f'(1)=(pa-(1-p)b)=\mathbb{E}[q_t]>0$. Thus, for small enough $\varepsilon$, we will have $f(1-\varepsilon)<1$. Since $\lim_{\lambda \to 0,\lambda >0}f(\lambda)=+\infty$, by the intermediate value theorem there is indeed some $\lambda\in (0,1-\varepsilon)$ such that $f(\lambda)=1$.
Let $\tau=\inf(n|X_n \lt z)$, (where we use the convention $\inf(\emptyset)=\infty$). Then $\tau$ is a stopping time wrt $\cal F$, and the probability you're looking for is $P(\tau=\infty)$.
Let $n\geq 1$ and $\tau_n=\min(\tau,n)$. Then $\tau_n$ is also a stopping time wrt $\cal F$, and further $|X_{\tau_n}|\leq |x|+n\max(a,b)$, and hence $|Y_{\tau_n}|\leq \exp((|x|+n\max(a,b))|\log(\lambda)|)$. We may therefore apply Doob's optional stopping theorem : $\mathbb{E}(Y_{\tau_n})=\mathbb{E}(Y_0)=\lambda^{x}$.
It follows that
$$
\begin{array}{lcl}
\lambda ^ x & = & \mathbb{E}(Y_{\tau_n}{\mathbf 1}_{\tau \leq n})+
\mathbb{E}(Y_{\tau_n}{\mathbf 1}_{\tau \gt n}) \\ 
&=&
\mathbb{E}(Y_{\tau}{\mathbf 1}_{\tau \leq n})+
\mathbb{E}(Y_{n}{\mathbf 1}_{\tau \gt n}) \\
&\geq& \mathbb{E}(\lambda^{z}{\mathbf 1}_{\tau \leq n})+
\mathbb{E}(\lambda^{X_n}{\mathbf 1}_{\tau \gt n}) \\
&=& \lambda^{z}P(\tau \leq n) +  \mathbb{E}(\lambda^{X_n}{\mathbf 1}_{\tau \gt n})
\end{array}
$$
We now take limits as $n\to \infty$. Because of the positive drift, the law of large numbers ensures that $X_n \to +\infty$ a.s., and hence $\lim_{n\to\infty}\mathbb{E}(\lambda^{X_n}{\mathbf 1}_{\tau \gt n})=0$ by the dominated convergence theorem. It follows that $P(\tau < \infty) \leq \lambda^{x-z}$, and hence
$$
P(\tau=\infty) \geq 1 - \lambda^{x-z}
$$
