Operations on Stopping Times

Closed operations

If $T_1$ and $T_2$ are nonnegative integer-valued stopping times, then:

$\min(T_1,T_2)$ is a stopping time, because

$$ \{\min(T_1,T_2)\le n\}=\{T_1\le n\}\cup\{T_2\le n\}. $$

$\max(T_1,T_2)$ is a stopping time, because

$$ \{\max(T_1,T_2)\le n\}=\{T_1\le n\}\cap\{T_2\le n\}. $$

In discrete time, $T_1+T_2$ is a stopping time. For example,

$$ {T_1+T_2=n}

\bigcup_{k=0}^n {T_1=k}\cap{T_2=n-k}. $$

For finite nonnegative integer-valued stopping times, $T_1T_2$ is also a stopping time. For $n\ge1$,

$$ {T_1T_2\le n}

{T_1=0}\cup{T_2=0}\cup \bigcup_{\substack{k,l\ge1\kl\le n}} \bigl({T_1=k}\cap{T_2=l}\bigr). $$

The union is finite and every term belongs to $\mathcal F_n$. If extended values such as $+\infty$ are allowed, specify the convention for products such as $0\cdot\infty$ before applying this statement.

An operation that is not closed

Even if $T_1\ge T_2$, the difference $T_1-T_2$ need not be a stopping time. The difference measures elapsed time from a random reference time; deciding $\{T_1-T_2=n\}$ at absolute time $n$ may require information after time $n$.

Useful consequence

The second smallest of three stopping times is still a stopping time, because

$$ T_{(2)}

\max\bigl(\min(T_1,T_2),\min(T_1,T_3),\min(T_2,T_3)\bigr). $$