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.


1 Answer 1


Intuitively speaking, you generally have an event for which you do not know when it occurs (the time of the occurrence of the event is random), but you do know that it will occur at some point in the future. The event is a stopping time if you do know at any point in time whether the event has occurred or not.

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.

  • 2
    $\begingroup$ :-) 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. $\endgroup$
    – nbbo2
    Feb 20, 2023 at 11:15

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