Why are Optional Stochastic Processes Important? I understand to some degree why adapted processes, progressive processes, and predictable processes are important. EDIT: I am referring only to the continuous time case, NOT discrete time.
But why do we care about optional processes?
What is significant about the optional sigma-algebra?
Are optional times and optional processes ever as important as stopping times and predictable processes?
(Also I know that given the "usual assumptions", specifically the second regarding a right-continuous filtration, that optional and stopping times are equivalent -- are optional and predictable processes equivalent?)
Finally, does the concept of optional processes have something to do with the "Optional Sampling Theorem"? The name always seemed somewhat arbitrary to me.
 A: Let's assume we are in the context of a filtered probability space $(\Omega,(\mathcal F_t),\Bbb P)$ satisfying the usual conditions.
Here are a few features of optionality that make it important.


*

*If $T$ is a stopping time and $A\in \mathcal F_T$ then there is a bounded optional process $Z$ such that $Z_T=1_A$ on $\{T<\infty\}$.

*If $Z$ is an optional process such that $Z_T=0$, $\Bbb P$-a.s. on $\{T<\infty\}$, for each stopping time $T$, then $\Bbb P[\sup_t|Z_t|=0]=1$.
(In this event $Z$ is said to be evanescent.)

*If $D\subset\Omega\times[0,\infty)$ is an optional set, and if $0<\delta<\Bbb P[\pi_\Omega(D)]$, where $\pi_\Omega(D)$ is the projection of $D$ unto $\Omega$, then there is a stopping time $T$ such that $\Bbb P[T<\infty]\ge\delta$ and $\{(\omega,T(\omega)): T(\omega)<\infty\}\subset D$.

*If $X$ is a bounded measurable process, there is an optional process $Z$ (unique up to evanescence) such that $\Bbb E[X_T\,|\,\mathcal F_T]=Z_T$, $\Bbb P$-a.s. on $\{T<\infty\}$, for each stopping time $T$. ($Z$ is the optional projection of $X$.)
The point is that for an optional process $Z$, properties of the random variables $Z_T$, as $T$ varies over the class of stopping times, lead to conclusions about the entire process $Z$.
An example showing that "progressive" is a weaker property. Consider the natural filtration of a standard Brownian motion $B$. Let $G$ denote the set of times $t\ge 0$ such that $B(t)=0$ but $B(t+s)\not=0$ for all sufficiently small $s$. It can be shown that $G$ is progressive. But the strong Markov property of $B$ implies that if $T$ is a stopping time, then $\Bbb P[T\in G]=0$. If $G$ were optional then we could conclude from point 2. above that $G$ was evanescent. But this is clearly not the case, so $G$ is not optional.
A: I think you've got the terminology mixed up a little. You have properties of a process (such as being adapted to a filtrartion, being progressively measurable, or being predictable), and then you have stopping and optional times (where the difference is, roughly speaking, merely that optional processes cannot answer questions regarding the "current time"). 
If you have a process and a stopping/optional time, you can also regard the stopped process (imagine a Brownian motion being klled after passing a certain barrier for example), which itself can be seen as a stochastic process by definining $X_{\tau \wedge t}$, where $\wedge$ should denote the minimum. So yes, to a certain degree, these stopped processes might be as important as the original processes, since they model "stopping whenever I have obtained a desired result".
Concerning the OST: yes, the OST exactly states that under certain conditions on your stopping time (which have to be fulfilled, see for example the St. Petersburg paradoxon), there is no strategy (i.e. stopping time $\tau$) which uses the information available to you (i.e. does not look into the future), that will give you any benefit on average, i.e. $\mathbb{E}(X_\tau) = \mathbb{E}(X_1)$ for a martingale $(X_n)_{n \in \mathbb{N}}$ and a stopping time $\tau$
