2
$\begingroup$

For martingale,optional stopping theorem states:

Let $(M_n)_{n\in \mathbb{N}}$ be adapted with $M_n\in L^1$ for all $n$ and if $(M_n)_{n\in \mathbb{N}}$ is a martingale, then $E[M_T]=E[M_0]$, for all bounded stopping time T.

So, why do we need this theorem? If $(M_n)_{n\in \mathbb{N}}$ is a martingale, then by its property $E[M_n]=E[M_i]=E[M_0], \forall i<n$. So $E[M_T]=E[M_0]$ will be implied.

Have I misunderstood something here?

$\endgroup$
2
  • 1
    $\begingroup$ $M_\tau (\omega) = M_{\tau ( \omega) }(\omega)$, so the $n$ depend on the $\omega$. So you cannot say $\mathbb{E}M_n = \mathbb{E}M_0$ implies $\mathbb{E} M_\tau = \mathbb{E} M_0$ for a random bounded stopping time $\tau$. $\endgroup$
    – Hetebrij
    Sep 22, 2015 at 21:23
  • $\begingroup$ Thanks for your comment! Can you be more specific? What I understand is that since the stopping time must take some value of $n$, for what holds for $n$ must hold for $\tau$. $\endgroup$
    – Cancan
    Sep 22, 2015 at 21:29

1 Answer 1

1
$\begingroup$

The stopping time $\tau$ can take multiple values of $n$, so we have, when $\tau$ is bounded by $k$ that $$\mathbb{E} M_\tau = \mathbb{E} \sum_{n=1}^k M_n \mathbf{1}_{\{\tau = n\}}= \sum_{n=1}^k \mathbb{E} M_n \mathbf{1}_{\{\tau = n\}}.$$ And we do not know whether $\sum_{n=1}^k \mathbb{E} M_n \mathbf{1}_{\{\tau = n\}} = \mathbb{E} M_0$ without the optional stopping theorem.

$\endgroup$
4
  • $\begingroup$ Last question: the last equality really looks alright for me :D Can you Specify a case where this equality doesn't hold without optional stopping theorem? $\endgroup$
    – Cancan
    Sep 22, 2015 at 21:42
  • 1
    $\begingroup$ I suppose not, since the last equality is true for bounded stopping times $\tau$ and martingales $M$ by the optional stopping theorem. I mean that if you haven't encountered the optional stopping theorem yet, you cannot say whether the equality is right or not for any bounded stopping time and any martingale. But of course equality holds as the Optional stopping theorem is true, regardless if you have heard of it or not. $\endgroup$
    – Hetebrij
    Sep 22, 2015 at 21:44
  • $\begingroup$ Oh, yea. I think I just got confused before. Now I got it! Thanks! $\endgroup$
    – Cancan
    Sep 22, 2015 at 21:49
  • $\begingroup$ The St. Petersburg paradoxon (en.wikipedia.org/wiki/St._Petersburg_paradox for example) might help you to understand the value of the OST. $\endgroup$ Sep 23, 2015 at 17:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .