# Why predictable processes?

So far I have seen two approaches for a theory of stochastic integration, both based on $L^2$-arguments and approximations. One dealt with a standard Brownian motion as the only possible integrator and admitted integrands to be progressively measurable processes satisfying certain integrability conditions. The other theory I saw generalized the ideas of the first one to continuous semimartingales as integrators, but it "only" admitted predictable integrands (I write "only" as predictability implies progressive measurability). Skimming through the literature I get the impression that using predictable integrands is quite common. Is there a deeper reason why on "restricts" oneself to this class of processes?

• In my point of view, these processes are the analogues of step functions when one deals with measurability - on a product space - which are just elaborate rectangles in a sense. But I would be interested in a reasoning why these type of objects work well when building convergence-based machinery. Just came to mind that one of my classes had $C^*$ algebras in them (kinda uncovering the reason why my analysis classes had the same feel about them), and maybe it would apply to this situation as well – Ákos Somogyi Nov 16 '15 at 8:29
• @zhoraster Could you elaborate on that? E.g., p. 411, Classical Potential Theory and Its Probabilistic Counterpart by J. L. Doob seems to say the predictability does imply the progressively measurability and I wish to reconcile. – shall.i.am Apr 21 '16 at 12:23
• @shall.i.am, yes, this comment was rubbish, sorry. – zhoraster Apr 21 '16 at 14:24

Being predictable (as in being measurable with respect to the predictable $\sigma$-algebra) always implies being progressive (as in being measurable with respect to the progressive $\sigma$-algebra), but the other implication only holds in special cases, such as when e.g. the underlying filtration is the usual completion of the filtration generated by a Brownian motion.
There are good reasons for restricting yourself to predictable integrands. In general, stochastic integral processes with e.g. progressive integrands and local martingale integrators will not be local martingales (generally, such integrals will be difficult to define at all except for simple progressive integrands). For an example of this, let $N$ be a standard Poisson process and let $M_t = N_t - t$. Then $M$ is a local martingale (in fact, a true martingale).
We then have $$\int_0^t N_s d M_s =\int_0^t N_{s-} dM_s +\int_0^t \Delta N_s d M_s \\ =\int_0^t N_{s-} dM_s + \sum_{0<s\le }(\Delta N_s)^2 =\int_0^t N_{s-} dM_s + N_t.$$ Here, the first term is a local martingale (because it's the integral of a left-continuous and adapted, hence predictable, process, with respect to a local martingale), while the second term is not a local martingale. Therefore, the integral $\int_0^t N_s d M_s$ is not a local martingale. In essence, the reason for this is that $N$ is progressive (because it is cadlag and adapted) but not predictable.
One of the main properties which makes it possible to obtain the local martingale property for stochastic integrals of predictable processes with respect to local martingales is the following result: If $X$ is cadlag and predictable, then the jumps of $X$ can be covered with a countable sequence of predictable stopping times. And predictable stopping times plays well with martingales.