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 $a,b\in\mathbb{Z}$.

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    $\begingroup$ Sorry, what is $\mathbb{R}_{++}$? $\endgroup$
    – Brian Tung
    Commented Sep 9, 2022 at 2:44
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    $\begingroup$ I haven't run through the details, but I imagine you can use the density of $\mathbb{Q}\subset \mathbb{R}$ to reduce to the integer case. $\endgroup$ Commented Sep 9, 2022 at 2:45
  • $\begingroup$ You want to ask this person where there might be research on that topic. He is kind of the expert on higher level crossings in a time series context. math.umd.edu/~bnk Or maybe he has some papers related to your problem. I didn't look at his website carefully recently but I looked at crossings once a while back and he was the big person in it. $\endgroup$
    – mark leeds
    Commented Sep 10, 2022 at 0:31
  • $\begingroup$ You say you can’t write a recursive equation, but why not? The escape probability starting from $x$ seems like a function of its values at $x+a$ and $x-b$, with some boundary behavior near $z$ and at $\infty$. I would expect some exponential ansatz would do the trick. $\endgroup$
    – mjqxxxx
    Commented Sep 14, 2022 at 7:46
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    $\begingroup$ My first reaction was that there should be some cute martingale solution to this problem similar to the regular gambler's ruin question (which does not particularly care whether whether the state space is finite or infinite). However, such a solution has been more elusive than I thought. Nice question! (+1) $\endgroup$ Commented Sep 14, 2022 at 13:40

1 Answer 1


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

$$ P(\tau=\infty) \geq 1 - \lambda^{x-z} $$


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