Are the stopping times $\tau_1 = \inf\{t \geq 0: X_t > a\}$ and $\tau_2 = \inf\{t \geq 0: X_t = a\}$ equivalent?

Question

Let $(X_t)_{t \in [0, T]}$ be a continuous stochastic process with paths which are a.s. continuous, the underlying space of which is irrelevant but is well defined. Let $a$ be a constant Define two stopping times $$\tau_1 = \inf\{t \geq 0: X_t > a\}$$ $$\tau_2 = \inf\{t \geq 0: X_t = a\}$$ Evidently, $X_{\tau_2} = a$. However, can we claim $X_{\tau_1} = a$ ? This "feels like" having something to do with continuity/topology but I cannot figure it out. Any help would be greatly appreciated.

Answer 1

Yes. Indeed, let $\omega \in \{X_{\tau_1}>a\}$. In particular, $X_{\tau_1(\omega )-h}\leq a$. Therefore, $$\lim_{h\to 0}X_{\tau_1(\omega )-h}=X_{\tau_1^-(\omega )}\leq a<X_{\tau_1(\omega )}.$$

Since $(X_{t})$ is continuous, $\mathbb P\{X_{\tau_1}>a\}=0$.

Answer 2

You need some assumption on $X_0$ for this. Assume that $X_0 <a$. Then you can apply the following elementary fact:

Let $f:[0,\infty) \to \mathbb R$ be a continuous function with $f(0) <a$. Then $\inf \{t\geq 0 : f(t) \geq a\}=\inf \{t\geq 0: f(t)=a\}$.

Proof of this fact is an easy consequence of intermediate value property of continuous functions.

Answer 3

In general they are not. Consider the constant process which is identically $a$ for all $t$. Then $\tau_2 = 0$ a.s. but $\tau_1 = +\infty$. (If you are working with the convention $\inf \emptyset = + \infty$)