# Why do we need optional stopping theorem?

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?

• $M_\tau (\omega) = M_{\tau ( \omega) }(\omega)$, 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 $\tau$. Sep 22, 2015 at 21:23
• 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 $\tau$. Sep 22, 2015 at 21:29

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.
• 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. Sep 22, 2015 at 21:44