Let $(B_t)_{t \geq 0}$ be a standard Brownian motion. The reflection principle tells us that $\sup_{u \leq t} B_s$ has the same distribution as $|B_t|$. Let $H_a$ be the first hitting time of $a$ by $|B_t|$, and $T_a$ the first hitting time of $a$ by $B_s$.

I believe that the first hitting time of $a$ by $B_s$ is equal to the first hitting time of $a$ by $\sup_{u \leq t} B_s$. Why then, is $\mathbb{E}[H_a]$ finite and $\mathbb{E}[T_a]$ infinite? Are the underlying distributions not equal in each case due to the reflection principle?



The reflection principle states that

$$M_t := \sup_{s \leq t} B_s \quad \text{and} \quad |B_t|$$

have the same distribution for each fixed $t$; it does not say that the processes $(M_t)_{t \geq 0}$ and $(|B_t|)_{t \geq 0}$ are equal in distribution. In fact, we cannot expect that the two processes are equal in distribution; this follows for instance from the fact that $(M_t)_{t \geq 0}$ has non-decreasing sample paths whereas the sample paths of $(|B_t|)_{t \geq 0}$ fail to be monotone.

Since the stopping times $T_a$ and $H_a$ clearly depend on the whole sample path of Brownian motion, the reflection principle cannot be used to deduce that $H_a$ and $T_a$ have the same distribution. (For instance if we want to know whether $T_a(\omega)=t$ for some $\omega$ and $t$, then it does not suffice to know the value $B_t(\omega)$; we need to know the path $(B_s(\omega))_{s \leq t}$.)

  • $\begingroup$ Thank you, this makes a lot of sense. $\endgroup$ – Daven May 26 '18 at 17:13
  • $\begingroup$ @Daven You are welcome. $\endgroup$ – saz May 26 '18 at 17:14

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