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 $a,b\in\mathbb{Z}$.
Best Answer
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
$$ P(\tau=\infty) \geq 1 - \lambda^{x-z} $$