14
$\begingroup$

Although my resarch focuses on PDEs (optimal transport, these days), I am currently trying to learn stochastic calculus and integration. I am just beginning in this topics, but I was wondering: why do we mainly integrate with respect to martingales? Typically $$ Y_t=\int_0^s X_s \,dH_s $$ seems to require that $H_s$ be a martingale, as far as I can tell after skimming through a few lecture notes and textbooks.

I'm aware that one can extend stochastic integration to semi-martingales (i.e. processes that are the sum of a martingale plus a "reasonable" process locally of bounded variations $H_s=\tilde H_s+B_s$): is this just a slight technical variation on the concept of martingale (which I should then think of as the sole and typical object w.r.t. which one can integrate), or are semi-martingales really very general processes and should I believe that one can essentially integrate w.r.t to (almost) ANY process?

I'm not interested in highly technical details (such as "if $H_s$ is not a martingale then this precise term is not square-integrable in the discrete construction of the stochastic integral", or whatever), I'm more focused on the big picture and I'd like a "philosophical/heuristic" explanation if possible. For example, I believe that one can think of martingales as processes that "do not see the future", which makes a whole lot of sense to me at least when it comes to what randomness should mean in real life. So what I'm wondering is: does it make sense to integrate w.r.t. processes that might be able to predict the future (some kind of non-locality in time, but why not after all?), or does that not make sense for some deep reason? If not, what is the said deep reason?

PS: feel free to migrate to MSE, although I do think that this is a fair research-level question.

$\endgroup$
12
  • 3
    $\begingroup$ "Can't see the future" is the property of being adapted (or perhaps progressively measurable). Martingales have the much stronger property of being a "fair game", whereas semimartingales allow for some "drift" which can bias the game one way or another. There are certainly theories of stochastic calculus that extend beyond semimartingales and even beyond adapted processes; see for instance the Skorokhod anticipating stochastic integral. $\endgroup$ Commented Feb 17, 2018 at 18:59
  • 1
    $\begingroup$ @MateuszKwaśnicki: Sure, adaptedness is only a relevant concept with respect to a specified filtration. The fact that a process is adapted to its own filtration, or any other, is not so relevant for that process as an integrator. This issue is relevant instead for the integrand and whether it is adapted to the integrator, which asks whether the integrand can see the future of the integrator. I guess my comment came out looking misleading that way. But I agree that to focus on adaptedness for this question would be barking up the wrong tree. $\endgroup$ Commented Feb 17, 2018 at 20:08
  • 7
    $\begingroup$ @leo monsaingeon: People do integrate with respect to non-semi-martingales, for example, with respect to fractional Brownian motion. I think the main point of rough paths theory is about extending the Itô calculus to certain non-martingales, but, honestly, I do not know this area at all. $\endgroup$ Commented Feb 17, 2018 at 20:09
  • 11
    $\begingroup$ @leomonsaingeon The claim that semimartingales are precisely the "good integrators" is known as the Bichteler-Dellacherie theorem, which can be found somewhere in Protter's book (or here: arxiv.org/abs/1201.1996 ). The idea of this theorem is as follows: There's no question as to how to define the integral of a simple process X with respect to any other process H. A "good integrator" is then defined as a process H for which the map from simple process X to integral $\int X dH$ is continuous in a suitable (and rather weak) sense. $\endgroup$
    – Dan
    Commented Feb 18, 2018 at 13:38
  • 1
    $\begingroup$ It‘s hard to give purely philosophical reason for all of that. Predictability of the integrand has to do with causality and avoiding circularity. For example, if the integrator is a counting process (cadlag version), then the integrand will determine the jump sizes of the integral. If the integrand jumps at the same time as the integrator, the proper jump size is the value of the integrand just before the jump. Therefore we always want to take predictable versions as integrands. $\endgroup$
    – S.Surace
    Commented Mar 2, 2018 at 17:29

1 Answer 1

9
$\begingroup$

I am aware that this is a really old post, but I feel it is a very important question. My personal favourite answer to this question lies in the theory of rough paths. I will give both a technical answer and a heuristic/qualitative answer, as my view of maths is that both should be present and linked as tightly as possible.

The main problem when trying to define a stochastic integral in the usual Riemann Stiltjes way is the fact that the squared increments $(M_{t+s} - M_{s})^2$ of Brownian motion, and martingales in general - are not of order $o(t)$. In other words, the paths of the integrator are too rough.

Recalling how Riemann Stiltjes integration works, the key observation is that the above increments of second order and higher are of order $o(t)$, and thus vanish in the limit as the mesh size goes to $0$. This is not the case for martingales.

However, this problem only occurs on a pathwise level. The orthogonality of martingale increments leads to the miraculous fact that the conditional expectations of all the problematic second order terms vanish, so long as the integrand is adapted (this is why adaptedness is such a crucial assumption). This leads to the well definedness of the Riemann-Stiltjes limits in $L^2$ sense, and this is none other than the Ito integral. This also explains why typically Ito integrals are defined as a limit $L^2$ or in probability instead of almost sure limits.

Now, the observation of rough paths theory is that we do not need the full martingale property! We simply need the conditional expectations of the higher increments to be of order $o(t)$ in the limit. In proving this theorem, we gain a finer understanding of why martingale integration works to begin with, and how the Ito theory may be naturally extended.

All of this is made precise in the so called Stochastic Sewing Lemma in rough path theory. I highly recommend the exposition in Hairer’s A Course in Rough Paths, where the Stochastic Sewing Lemma is found as Proposition 4.19.

On a more philosophical level, what the above is saying is this - adapted processes cannot see into the future. Martingales are fair in the future given the present. These two aspects together give rise to an “averaging effect” that allows the stochastic integral to be well defined despite the roughness of paths. I like to think of it as the integrand being “constant” from the point of view of the future martingale increments. If the integrand were not adapted, it could conspire with the martingale paths to ruin the fairness, and hence the averaging property.

$\endgroup$
5
  • $\begingroup$ "the paths of the integrator are too rough." Does this mean that you view the integral as a path integral over the path of the stochastic process? I'm new to this, so please excuse the basic questions. And what does it mean for Ito integrals to be defined in $L^2$ rather than pathwise? $\endgroup$
    – 900edges
    Commented Sep 19 at 15:54
  • $\begingroup$ @900edges Ah, so the precise meaning is that if we try to view it as a usual deterministic Riemann-Stiltjes integral for each value of the random parameter $\omega \in \Omega$, the integral fails to exist in the usual sense. $\endgroup$
    – Nate River
    Commented Sep 19 at 20:11
  • $\begingroup$ @900edges And uh, I think I should have said that the Ito integrals are defined as limits in $L^2$ or probability, rather than almost sure limits. $\endgroup$
    – Nate River
    Commented Sep 19 at 20:13
  • 1
    $\begingroup$ Great answer @Nate, precisely the kind of insight I was looking into. (BTW: sorry for coming back here so late!) $\endgroup$ Commented Sep 20 at 7:33
  • $\begingroup$ I’m glad it was understandable! @leomonsaingeon $\endgroup$
    – Nate River
    Commented Sep 20 at 7:38

You must log in to answer this question.

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