1
$\begingroup$

Let $(X_n)_{n \in \mathbb{N}_0}$ be a sequence of independent and identically distributed random variables with $$\mathbb{P}(X_1 = 1) = \mathbb{P}(X_1 = -1) = \frac{1}{2}.$$ Define $S_t = \sum\nolimits_{k=1}^t X_k$ for $t\in \mathbb{N}$ and $S_0 = 0$

Show that $\sigma := \sup\{t \in \mathbb{N}_0 : S_t =1\}$ and $\tau := \inf\{t \in \mathbb{N}_0 : S_t =1\}$ are stopping times with respect to the natural filtration of $(X_n)$

I need to show that $\{\sigma = t\} \in F_t$ where $F_t$ is $\sigma(X_0, ... , X_t)$

I wanted to write $\{\sigma = t\}$ differently, but I don't see how this can be done.

Thanks for any help.

$\endgroup$
  • $\begingroup$ Are you sure that this is what you have to prove? I doubt that $\sigma$ is a stopping time. $\endgroup$ – saz Oct 30 '18 at 11:59
  • $\begingroup$ Hmm.. is it because $\sigma=\infty$ a.s.? I.e., for each $t<\infty$, $\{\sigma\le t\}$ is null, so it's trivially measurable w.r.t. $\mathcal F_t$? $\endgroup$ – AddSup Oct 30 '18 at 14:43
1
$\begingroup$
  1. $\tau$ is a stopping time: At time $t$ it is known if $\{\tau=t\}$ occurred. $$ \{\tau=t\}=\{S_0\neq 1,\dots,S_{t-1}\neq 1,S_t=1\}.$$

  2. $\sigma$ is not a stopping time: At time $t$ you can not know if $\{\sigma=t\}$ occurred. $$ \{\sigma=t\}=\{S_t=1,S_{t+1}\neq 1,S_{t+2}\neq 1\dots\}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.