# On first and last zeros before t in a Brownian Motion

Suppose we have the following random variables, given a fixed $$t$$ we define the last zero before $$t$$ and the first zero after $$t$$:

\begin{align*} \alpha_t &= \sup\left\{ s\leq t: B(s) = 0 \right\}\\ \beta_t &= \inf\left\{ s\geq t: B(s) = 0 \right\}\ \end{align*}

Why $$\beta_t$$ is a stopping time but $$\alpha_t$$ is not?

Given the intuitive definition of a stopping time it makes much more sense in my head that the result would be the other way around.

In your example, you have a fixed time $$t$$. The event $$\alpha_t$$ is now describing the last occurrence of $$B(s)=0$$ before $$t$$. However, as you approach $$t$$ you will not know at any point if the last $$B(s)=0$$ before $$t$$ has occurred as there always could be another one coming up.
On the other hand, with $$\beta_t$$ you are now moving away from $$t$$ and you are waiting until you observe the first $$B(s)=0$$. As soon as you observe it, you will know that this was indeed the first $$B(s)=0$$ after $$t$$, as there could not have been another one before it.
• :-) Ha, ha, maybe $\alpha$ is like "the last bug in the code before the release date t". In my code I always think I have found the last bug and then suddenly I find another. Feb 20, 2023 at 11:15