Hitting times by Brownian motion

Question

[Edited] Suppose that $A$ is a (Borel) measurable set and $X$ is an Ito diffusion, i.e., $dX_{t}=\mu(X_t)dt+\sigma(X_t)dB_t$.

Consider a hitting time $\tau_A$ of the given set $A$ by the process $X$: $\tau_A:=\inf\{t\geq0:X_t\in A\}$.

Also, let $\tau_\bar{A}:=\inf\{t\geq0:X_t\in \bar{A}\}$ where $\bar{A}$ is the closure of $A$.

Now, my question is: Is $\tau_\bar{A} = \tau_A$ (a.s.)? I think this must be true, but I do not find any reference for this... In case my conjecture is incorrect, any idea on reasonable conditions (either on $A$ or $X$) under which this is true?

I'd really appreciate your help!

---

Based on the comments below, suppose that $X_t$ is an Ito diffusion process and let me refine my question as follows:

1. Let $X_0 = x$. If the process $X$ comes back to $x$ infinitely often in any time interval $(0,\epsilon)$, $\epsilon>0$, then can we say that my assertion above ($\tau_\bar{A} = \tau_A$ (a.s.)) is true? My intuition is that if $X$ has this property, then along with the Markov property of $X$, for any given $t$, a sample path $X_{s>t}$ always "zig-zagged" around $X_t$ for any $s>t$. So, given any set $A$, as soon as $X$ hits its boundary point of $A$ (which belongs to the closure of $A$), it should also hit $A$ in arbitrarily small time $\epsilon>0$. Does my argument make sense?

2. If #1 is correct, then the follow-up question is: under what condition on $\mu(\cdot)$ and $\sigma(\cdot)$ an Ito diffusion $X$ has this property? As a canonical example, does a Brownian motion with drift ($\mu(\cdot)=\mu$ and $\sigma(\cdot)=\sigma$) have this property?

Any idea or suggestion for the reference would be really appreciated!

Answer 1

I'm very new in this field, but I'm giving a try anyways. So please be critical and let me know if something is wrong (We are here to learn so...).

The assertion is not true in general, indeed we have a couple of counterexamples.

A simple counterexample. Let $\mu\equiv 1$ and $\sigma\equiv 0$. Define $$A:=(0,1)$$ Then we have for $X_0=1$, the explicit unique solution for $X_t$ given by $$X_t=1+t$$ In this particular case $$\tau_A=\infty, \ \ \text{ and }\ \ \ \tau_A^-=0.$$

---

We might add extra conditions to get that $\tau_A=\tau_A^-$ a.s., I guess we can cook up the following simple proposition which is inspired by Saz' comment. I also add some thoughts at the end. But before we do that we need some Lemmas. Let us fix the filtered probability space $(\Omega,\mathcal F,\mathcal F_t,\mathbb P)$.

Lemma 1. For the case $X=B$ we have $$\tau_A=\tau_A^- \ \ \ \text{ a.s. }$$

Proof. This Lemma is a generalization of what Saz proved here in this answer. The proof should go more or less the same so it will be omitted.

///

Now we say the same if we consider time-changed Brownian motions.

Lemma 2. Let $(T(t))_{t\geq 0}$ be a strictly increasing and continuous stochastic process such that $T(0)=0$ and $\lim_{t\to\infty}T(t)=\infty$. Then for $X_t=B_{T(t)}$ we have $$\tau_A=\tau_A^- \ \ \ \text{ a.s. }$$

Proof. Since $t\mapsto T(t)(\omega)$ for some fixed $\omega\in\Omega$ is increasing and continuous we have the existence of increasing and continuous "inverse random variable" $t\mapsto T^{-1}(t)(\omega)$. The domain of $T$ and $T^{-1}$ (as a function of $t$) is $[0,\infty)$. We have \begin{align} \tau_A&=\inf\{t\geq 0 \ :\ B_{T(t)}\in A\}\\ &=\inf\{t\geq 0\ : \ s=T(t), \ B_s\in A\}\\ &=T^{-1}(\inf\{s\geq 0\ : \ B_s\in A\}) \\ &=T^{-1}(\inf\{s\geq 0\ : \ B_s\in \overline{A}\}) \\ &=...\\ &=\tau_A^{-1} \end{align} In the third line we have used continuity and monotonicity of $T^{-1}$ and the line before the dots is due to Lemma 1. In the dots we did the steps above with $\overline{A}$ to get the last line.

///

Now we can finally state the proposition.

Proposition. Assume that $\mu$ and $\sigma$ are Lipschitz continuous so that a unique solution $(X_t)_{t\geq 0}$ exists for $X_0=x\in\mathbb R$. Moreover assume that $\sigma$ is bounded away from zero, i.e. there is $c>0$ such that $|\sigma|\geq c$. Finally assume that $$Z_t=\exp\left(-\int^t_0 \frac{\mu(X_s)}{\sigma(X_s)}\,dB_s+\frac 1 2 \int^t_0\frac{\mu(X_s)^2}{\sigma(X_s)^2}\,ds \right)$$ is a $\mathcal F_t$-martingale. Then we have that $\tau_A=\tau_A^-$ a.s..

Proof. Sinze $Z_t$ is a martingale, we can define a probability measure $\mathbb Q$, that is equivalent to $\mathbb P$, by $$\frac{d\mathbb Q}{d\mathbb P}|_{\mathcal F_t}=Z_t$$ Applying Girsanov gives us that $$W_t=B_t+\int^t_0\frac{\mu(X_s)}{\sigma(X_s)}\,ds$$ is a BM w.r.t. $\mathbb Q$. We also verify w.r.t. $\mathbb Q$ we can write $X$ as $$X_t=x+\int^t_0\sigma(X_s) \,dW_s$$ Now we notice that $$\langle X\rangle_t=\int^t_0 \sigma(X_s)^2\,ds\geq \int^t_0 c^2\,ds=tc^2$$ Therefore $\langle X\rangle_\infty=\infty$. By the Theorem of Dubins-Schwarz, there exists a BM w.r.t. $\mathbb Q$, say $\tilde W$, such that $$X_t=x+\tilde W_{\langle X\rangle_t}$$ Now take $T(t)=\langle X\rangle_t$. Apply Lemma 2 with this $T(t)$ and $A'=A-\{x\}$ to get the proof done.

---

Some remarks/thoughts.

• The proposition shows that, in particular, a BM with drift satisfies the property.
• There are lot of ways to show that $Z_t$ is a martingale, maybe Novikov's or Kazamaki are applicable.
• Thanks to Saz who told me to use Dubins-Schwarz Theorem to mke the proposition applicable to a larger class of Ito-diffusions.