A stochastic Process where $X_{T_A}\not \in A$. Must it always be in $\bar A$?

Question

So, I was wondering how to define a stochastic process $X$, with continuous paths, such that $X_{T_A}\not \in A$, where $T_A=\inf\{t>0:X_t \in A\}$.

For this, I was thinking of $X$ mimicking the behaviour of $\sin(1/t)$...

Also, is it true that we always have $X_{T_A}\in \bar A$?

Answer 1

Yes, we will always have $X_{T_A} \in \overline A$ because $X$ has continuous paths. For a specific example, you could do something as simple as $X_t = t$ and $A = (1,2)$.

EDIT: To prove $X_{T_A} \in \overline A$, from the definition of $T_A$ we have that there exist $(t_n)$ decreasing to $T_A$ such that $X_{t_n} \in A$. Since $X$ has continuous paths, $(X_{t_n}) \rightarrow X_{T_A},$ and since $X_{T_A}$ is the limit of points in $A$ we have $X_{T_A} \in \overline A$.

Answer 2

On the condition that $T_A<\infty$.
By definition of $T_A$, there is a sequence of times $\tau_1,...,\tau_n$ such that:

• $X_{\tau_n} \in A$
• $\tau_n \longrightarrow T_A$

By the (right) continuity of $X$, $X_{T_A} = \lim X_{\tau_n}$.
By definition of $\overline{A}$, $\lim X_{\tau_n} \in \overline{A}$
Hence the conclusion.