0
$\begingroup$

I posted this question on mathstack a couple of weeks ago and even with 100 bounty on it Ive not been able to get any feedback. Hence I tought Id try posting it here.

https://math.stackexchange.com/questions/2950955/is-an-sde-really-equal-to-an-integral-equation-or-is-it-rather-its-integral-t

Ive been told and been reading in some textbooks on SDE's that an SDE or stochastic differential really is an integral equation. In other words, that

$ dX= \beta dt + \sigma dW$ $\,$ "really means" $\,$ $X_{t}= X_{0} +\int_{0}^{t} \beta dt +\int_{0}^{t} \sigma dW$

However I am getting the impression that this is not the case and that $dX$ really is a measure type object which we in turn can "take the integral off", giving us the latter above. This in turn implies that the SDE really has a meaning maybe not as a differental equation in the usual sense but as something else. This impression was obtained for instance from comments in the following post,

https://math.stackexchange.com/questions/1647732/if-dx-t-x-t-dt-db-t-why-does-e-tdx-t-e-t-x-t-dt?noredirect=1&lq=1

In particular the discussion involving $dX=dY$, $AdX=AdY$ and then taking the integral of that.

With that background I ask the following;

Is an SDE then really equal to an integral equation? This would be the same as saying or declaring rather that $\mu= d\mu$, which is fine if one sticks to that convention, but that dosnt seam to be the case in some calculations with differentials.

I also wonder if it is possible to make sense of the $dX$ if one uses more advanced tools? Similar to the case of $dx$ in a ODE, that indeed justify some "informal" operations. My impression is that in general the $dx$ can be tought of as a measure, a differential form etc..

Found these;

https://math.stackexchange.com/questions/2850431/are-sdes-really-differential?rq=1

https://math.stackexchange.com/questions/930578/problem-with-understading-mixed-integration?rq=1,

https://math.stackexchange.com/questions/523724/why-do-people-write-stochastic-differential-equations-in-differential-form?rq=1

that looks like similar questions, but which are not answering my question.

Update

While reading Revus and Yors book Continuous Martingales and Brownian Motion I noticed that they denote the equation $e_{x}(f,g)$ for drift and diffusion coefficents $f$ and $g$.

It thus looks like they deliberately avoid what seem to be convention in other books to give the meaning to the differential version to be equal to the integral version.

I suppose the reason for this is to not get into the whole mess that I got into, but that just a guess on my part.

Any thoughts on this would also be great to hear.

Update 2

Comments on why this question is so hard to answer, which seams to be the case ,would also be great. After all it has 8 upvotes and has not been closed due to being regarded as nonsense. Someone with experience must have tought about this before since it is a rather elementary matter.

$\endgroup$
  • 1
    $\begingroup$ I'm not an expert in these matters, but I think the situation is this: while the 'differential version' is {\em really} just a compact way of writing the integral equation, it can (as the respondent in the second link says) "serve as a guide to deducing one integral equation from another". The 'differentials' one encounters in connection with the Ito integral (and in measure theory) are not the same as the `real' ones encountered in differential geometry. $\endgroup$ – DCM Nov 4 '18 at 11:30
  • $\begingroup$ @DCM right, measures and 1-forms are not the same. However as one sees from one of the linked questions comments, it looks like we run into troubles applying Ito formula if we following the injunction of some authors. If we multiply our object from the left for instance we end up with wierd things if we follow their injunction to interpretation explicitly. $\endgroup$ – Maxed Nov 4 '18 at 11:40
  • $\begingroup$ @DCM this is the post which comments I refer to math.stackexchange.com/questions/1647732/… and here is the book where the statement is th.if.uj.edu.pl/~gudowska/dydaktyka/Oksendal.pdf. The "interpretation" is on page 21-22 and the exercise on page 73-74 in particular 5.4 ii. The pages refer to the book not the pdf. $\endgroup$ – Maxed Nov 4 '18 at 11:47
  • 2
    $\begingroup$ I am not sure I completely understand the question. Both differential and integral notation are short-hands for certain $L^2$ limits. As with any notation, there are certain rules for manipulating these objects that give correct answers. If you take things too far and interpret them in ways that are not consistent with the rules, you may get the wrong answers. But maybe there is something deeper in here.. $\endgroup$ – S.Surace Nov 4 '18 at 15:07
  • $\begingroup$ @S.Surace do you have a reference for the $dX$ being an $L^2$ limit? $\endgroup$ – Maxed Nov 4 '18 at 20:23
1
$\begingroup$

This is not really an answer as the question is well-posed. This one rather is a collection of some possible answers. And for the better, in my opinion, you should develop more what you want. For example, which type of representation of $dW$ do you want? Which properties do you wish it to have?

Let $(W_t)$ Weiner process, and $(F_t)$ its canonic filtration.

Answer 1 $dW$ as the couple $\left( (W_t)_{t \ge 0}, \left(\mathcal{F}_t\right)_{t \ge 0} \right)$

Answer 2 $dW$ as an unbounded linear operator of the space of all $\left(\mathcal{F}_t\right)_{t \ge 0}$ (continuous) local martingale.

Answer 3 For fixed $T$, $dW$ as a transition kernel of the couple $( M_T, \int_{0}^T M_tdW_t)$, for $(M_t)$ is a martingale. In some senses, it means that $dW$ can be as a measurable function into space of probability measure.

Answer 4: $dW$ define a path integral for a class of 'lovely' $(F_t)$-progressively measurable process.

Answer 5: However, you can't define a path integral (in some sense) between $dW$ and an arbitrary process For the first counter with this idea, you can reference yourself to the link https://math.stackexchange.com/questions/1212111/could-someone-explain-rough-path-theory-more-specifically-what-is-the-higher-o

$\endgroup$
  • $\begingroup$ So you are saying that $dX$ does exist in itself and does not nesseserly have to mean the integral of it? As the author in the book from the comment suggests. $\endgroup$ – Maxed Nov 5 '18 at 12:08
  • $\begingroup$ How do you define 'exist in itself' ? $\endgroup$ – Taro NGUYEN Nov 5 '18 at 12:18
  • $\begingroup$ If you really do want to find an answer, you should put more work on your question. I saw that in your previous question in math stackexchange, you also faced the same problem. $\endgroup$ – Taro NGUYEN Nov 5 '18 at 12:26
  • $\begingroup$ K ill see what I can do $\endgroup$ – Maxed Nov 5 '18 at 12:46

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.