# Why is $T:=\inf\{t\geq0:B_t\leq at^p-b\}$ a stopping time?

How can I prove that $$T:=\inf\{t\geq0:B_t\leq at^p-b\}$$ is a stopping time w.r.t. a natural filtration of $$B$$, where $$B$$ is a $$BM$$, $$p>1/2$$ and $$a,b>0$$?

I can introduce a new random process, $$X_t:=e^{B_t-at^p}$$, for which $$T=\inf\{t\geq 0:X_t\leq e^{-b}\}$$.

I started: $$\{T\leq t\}= \{\exists s\leq t:X_s\leq e^{-b}\}.$$ Is this then equal to $$\cup_{s\leq t}\{X_s\leq e^{-b}\} = \cup_{s\leq t\cap \mathbb{Q}}\{X_s\leq e^{-b}\}?$$

Is it now enough for me to say, as $$X_s$$ is measurable on the given filtration, that a countable union is also measurable and that gives us the stopping time?

• Seems kosher to me, but I haven't studied this in ~4 years, so don't take my word for it. Commented Dec 9, 2018 at 1:23

Proposition. Let $$X=(X_t)_{t\geq0}$$ be an adapted process with values in a metric space $$(E, d)$$. Assume that the sample paths of $$X$$ are continuous, and $$F$$ be a closed subset of $$E$$. Then $$T_F=\inf\{t\geq0: X_t \in F\}$$ is a stopping time.
Clearly $$X_t$$ has continuous sample paths.  Since $$(-\infty, e^{-b}]$$ is closed, we can conclude that $$T$$ is a stopping time.
Your argument is not valid. LHS of the set theoretic identity you have written need not be contained in RHS. For example $$X_t \leq e^{-b}$$ does not imply that there exists $$s\leq t,s \in \mathbb Q$$ such that $$X_t \leq e^{-b}$$. Instead, consider $$\{T>t\}$$. You can write this as union over $$s \leq t, s \in \mathbb Q$$ of $$\{X_s>e^{-b}\}$$ and this proves that $$\{T>t\} \in \sigma \{B_u:u \leq t\}$$.
• Thank you for your answer. So if I understand correctly I can do a countable intersection, if I have $\{T>t\}$? Why is that so, what changes with the change of inequality? Commented Dec 10, 2018 at 20:28
• It is possible that $X_s >e^{-b}$ for all $s <t$ but yet $X_t \leq e^{-b}$. But if $X_t >e^{-b}$ then $X_s >e^{-b}$ for all $s$ sufficiently close to $t$ hence for some rational number $s <t$. Commented Dec 10, 2018 at 23:12