A question concerning terminal time

Question

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.

Answer 1

$\newcommand\om\omega\newcommand\th\theta$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.