A terminal time $\tau$ is a stopping time satisfying $$\omega \in \{\tau(\omega) > t\} \text{ implies that } \tau(\omega) = t + \tau(\theta\_t\omega), $$ for all $t\ge 0$. Here $\theta\_t$ is the shift operator. I am wondering why the first hitting time is a terminal time but the second hitting time is not? Thanks.



A question concerning terminal time
$\newcommand\om\omega\newcommand\th\theta$Let $\tau$ be the first hitting time of a set $B\subseteq S$, so that for each "path" $\om\colon[0,\infty)\to S$ we have
$$\tau(\om)=\inf\{u\colon u\ge0,\om(u)\in B\}.$$
For each real $t\ge0$, the $t$-shifted path $\th\_t\om$ is defined by the formula
$$(\th\_t\om)(s):=\om(t+s)$$
for real $s\ge0$.


If now $\tau(\om)\ge t$ for some real $t\ge0$, then
$$
\begin{aligned}
\tau(\om)&=\inf\{u\colon u\ge t,\om(u)\in B\} \\ 
&=\inf\{t+s\colon t+s\ge t,\om(t+s)\in B\} \\ 
&=\inf\{t+s\colon s\ge0,(\th\_t\om)(s)\in B\} \\ 
&=t+\inf\{s\colon s\ge0,(\th\_t\om)(s)\in B\} \\
&=t+\tau(\th\_t\om).
\end{aligned}$$
So, $\tau$ is a terminal time.




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The second hitting time $\tau\_2$ of $B$ is defined by the formula
$$\tau\_m(\om)=\inf\{u\colon u>\tau\_1(\om),\om(u)\in B\},$$
where $\tau\_1$ is the first hitting time of $B$.
Suppose now a path $\om$ visited a set $B$ only at time moments $1$ and $3$, and choose $t=2$. Then $\tau\_2(\om)=3$, so that $\tau\_2(\om)<\infty$ and $\tau\_2(\om)>t$. However, $\tau(\th\_t\om)=\infty$ and therefore the condition $\tau(\om)=t+\tau(\th\_t\om)$ fails to hold. So, $\tau\_2$ is not a terminal time.


