For standard Brownian motion $B$, define stopping time $T_1:=\inf\{t>0: B_t = 3\}$ and $T_2:=\inf\{t>0: B_t = -3\}$ and $T_3 := \min\{T_1, T_2\}$.

  1. Can I say that $T_3 = \inf\{t>0, B_t \in \{-3, 3\}\}$?
  2. In addition, how to relate $T_1$ and $T_2$ in terms of their distributions? I would imagine some easy relationship between their moment generating function or Laplace transform since $-B$ is also a Brownian motion. Hence, $T_2$ is the counterpart of $T_1$ for $-B$. Is there any available result for this? Thank you!
  1. Yes, this one is very simple and has nothing to do with probability: $$ \min(\inf\{t: B_t = a\}, \inf\{t: B_t = -a\}) = \inf\{t: B_t = a \text{ or } B_t = -a\}$$

  2. Indeed, as in probability law for stochastic processes $(B_t) = (-B_t)$, $T_1$ and $T_2$ have the same distribution:

$$ Ef(\inf\{t: B_t = a\}) =^* \inf\{t: -B_t = a\} = \inf\{t: B_t = -a\} $$

where you use the fact that $(B_t) = (-B_t)$ in the $*$ed equality.

  • $\begingroup$ maybe in another question? ;) this is quite long to be explained just in a comment. But just a hint: it is based on rescaling properties of the Brownian motion. $\endgroup$ – mookid Oct 18 '14 at 23:31
  • $\begingroup$ you are welcome. did you see my edit? $\endgroup$ – mookid Oct 18 '14 at 23:35
  • $\begingroup$ the lsat comment. $\endgroup$ – mookid Oct 18 '14 at 23:39
  • $\begingroup$ exactly.${{{}}}$ $\endgroup$ – mookid Oct 18 '14 at 23:48
  • $\begingroup$ no, I don't think so. $\endgroup$ – mookid Oct 19 '14 at 0:00

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