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The answer here makes sense to me, in the sense that the stochastic integral does not preserve the local martingale property if the integrand is not predictable, but I am confused as to why. I have been taught the following definition of the stochastic integral:

We first define the quadratic variation for a RCLL martingale, $M$, to be $$ Q(M,t) \equiv \lim_{n \rightarrow \infty} \sum_{i=1}^\infty(M_{\sigma_{i+1} \wedge t}-M_{\sigma_{i} \wedge t})^2$$ where $\sigma_0 \equiv 0$ and $$ \sigma_{i+1} \equiv \inf \left(t > \sigma_i : |M_t - M_{\sigma_i}| \ge 2^{-n} \text{ or } |M_{t-} - M_{\sigma_i}| \ge 2^{-n} \right) $$ This limit is shown to converge uniformly on compact sets almost surely from first principles in this excellent pedagogical paper. The limit is shown to be non-decreasing, RCLL, etc. Now for any bounded RCLL martingale $M$, and any (predictable, but where does this argument rely on predictability?) process $H$ such that $$\mathbb{E} \left(\int_0^\infty H_s^2 Q(M,ds) \right) < \infty $$ we define the following mapping from the Hilbert space of bounded RCLL martingales to $\mathbb{R}$:

$$J(N) \equiv \mathbb{E} \left(\int_0^\infty H_s Q(M,N,ds) \right),$$

where $Q(M,N,t)$ is the quadratic covariation $$Q(M,N,t) \equiv \frac{1}{2} \Big(Q(M+N,t) - Q(M,t) - Q(N,t) \Big).$$

Since $J$ is a linear map, Riesz representation gives us a unique element $L$ which is in the space of bounded RCLL martingales which is such that $$\textbf{(1)} \qquad \qquad J(N) = \mathbb{E}(L_{\infty} N_{\infty})$$ and we define this $L$ to be the stochastic integral of $H$ with respect to $M$.

I know that it is typical (as in Protter's text) to define the stochastic integral first and then obtain its quadratic variation afterward, hence my confusion. Since $L$ is by construction a RCLL martingale, this seems to contradict the notion that lack of predictability of $H$ is a problem with respect to preserving the (local) martingale property of the stochastic integral, so my questions are as follows: where in this argument is the predictability (or frankly, because my knowledge is so utterly and depressingly poor, adaptedness) of $H$ used? What am I missing/where does everything fall apart?

Any and all help in understanding this would be massively appreciated, and I thank you for even just reading through this ''mini -essay''.

qp212223
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    Lots of stuff. In this answer we can see in action a RCLL martingale and a non predictable integrand who together make for a very interesting stochastic integral. – Kurt G. Feb 19 '24 at 23:56
  • Thanks for another example! I understand that this proof of existence is wrong. However, the problem I have is that I still have no idea where I have implicitly used predictability of $H$ in this isometry construction of the stochastic integral. Are you able to see where specifically one has to use this? – qp212223 Feb 20 '24 at 00:06
  • The answer I linked to shows that the answer to your title is a clear no. On the other hand: It does not contradict the preservation of the (local) martingale property. The proof I gave there is pretty self contained. It shows that $\int_0^tN_s,dM_s$ is not even a local martingale. In contrast: $\int_0^tN_{s-},dM_s$ (predictable integrand) is one. I have read your post only quickly an it is late at night. What in a nutshell is your confusion? – Kurt G. Feb 20 '24 at 00:24
  • My confusion is not about whether or not $H$ needs to be predictable. I know from the example I linked in my post, AND from yours, that $H$ needs to be predictable in order for the stochastic integral to be a local martingale. I am asking where, specifically, in the "proof" I have given, is there a breakdown in logic (the questions I have are in bold and are italicized). I put proof in quotations because I KNOW it must be wrong due to the counterexamples. In particular, where in the construction of the stochastic integral defined by $L$ in my post rely on the predictability of $H$? – qp212223 Feb 20 '24 at 15:04
  • What is $N$ in $J(N),?$ And why is the space of bounded rcll martingales a Hilbert space? – Kurt G. Feb 20 '24 at 16:56
  • $N$ is a bounded RCLL martingale (the mapping is from said Hilbert space to $\mathbb{R}$, as stated in the question). It is not difficult to show that $\mathcal{M} \equiv$ the set of bounded RCLL martingales is a Hilbert space with inner product $\mathbb{E}(M_\infty N_\infty)$ for $M, N \in \mathcal{M}$. The fact that this induces a norm is obvious. The fact that the space is complete with this norm is a consequence of Doob's inequality. I can write the full proof out if you want me to (or I'll try and find a good resource for you). – qp212223 Feb 20 '24 at 17:39
  • Ok. I can probably look up that Hilbert space stuff myself. So your stochastic integral depends on the choice of $N,?$ Might be a bit awkward. It also looks like the only integrators $M$ that are allowed have to be bounded (rcll) martingales. There might be a few but Brownian motion or the compensated Poisson process are not among them. – Kurt G. Feb 20 '24 at 18:11
  • @KurtG. The stochastic integral does not depend on the choice of $N$; that is the point of using the Reisz representation theorem. I don't think the boundedness of $M$ is important in this argument, but if it is, then the argument would be to extend to unbounded martingales by stopping. This is the stochastic integral construction done in Revuz and Yor's book. – user6247850 Feb 20 '24 at 21:59
  • Why does he integral not depend on the choice of $N,?$ If I take another $N$ called $N'$ you need to show that $$ \mathbb E[L_\infty N_\infty]=\mathbb E[L'\infty N'\infty] $$ implies $L=L'$ I think. – Kurt G. Feb 21 '24 at 10:06
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    No. $J(N') = \mathbb{E}(L_\infty N'\infty) \ne \mathbb{E}(L\infty N_\infty) = J(N)$. $J(\cdot)$ is a linear *function* that takes a bounded RCLL martingale as input and outputs a real number. Since the space of RCLL bounded martingales is a Hilbert space under the inner product I mentioned, we are using the following (elementary) theorem to claim the existence of such an $L$: https://en.wikipedia.org/wiki/Riesz_representation_theorem. I will bounty this question at some point in the hope that someone else answers. – qp212223 Feb 21 '24 at 14:21
  • I looked up how Revuz & Yor define stochastic integrals. They do not mention Riesz but it must be the same representation theorem. Their setup is slightly different from yours and I find it better to understand. That is not the point. It is rarely mentioned in the literature, but what we want from a stochastic integral $\int_0^t H_s,dM_s$ is that it should depend only on the restriction of $H$ to $[0,t]\times\Omega$ and this should be a $({\cal B}[0,t]\otimes {\cal F}_t)$-measurable random variable. – Kurt G. Feb 22 '24 at 17:38
  • This property is called progressive measurability of $H,.$ It is weaker than predictable. R&Y mention that their integrands satisfy just this condition. And if they were right or left continuous and adapted they would automatically then be progressively measurable. – Kurt G. Feb 22 '24 at 17:38
  • My setting here is not identical to that of Revuz and Yor (as the title of their book suggests, they handle continuous martingales, where progressive measurability suffices). My question is not about what is or is not required to define a stochastic integral for RCLL martingales which preserves local martingale property. I know that it is written in many sources that it suffices to have $H$ predictable and such that $\int_0^t H^2_s d[M]_s < \infty$ almost surely for every $t \ge 0$. My question is *where in the argument provided to construct such an integral is the predictability used?* – qp212223 Feb 22 '24 at 19:21
  • You mix up the integrator (continuous martingale) with the integrand. Predictability is *not* used. It is sufficiently general and implies progressive measurability. – Kurt G. Feb 22 '24 at 19:34
  • I am not mixing up the integrator and integrand. In almost all contexts, $H$ is assumed to be predictable (for example, see https://encyclopediaofmath.org/wiki/Stochastic_integral, or open almost any textbook on the subject). I want to know why. Perhaps the reason for this is because the stochastic integral $L$ defined in my question does not equal the limit of Riemann-like sums if $H$ is not predictable (I know they are equal for $H$ left continuous). I do not know if this is the answer. I am looking for said answer. I don't understand how to make this more clear to you at this point. – qp212223 Feb 22 '24 at 19:59
  • You are free to start that bounty, or even post it on MO, to get someone else's opinion. I am personally glad that I had to think this through one more time. So summarize: all that integral construction stuff (be it Riesz, $L^2$, simple integrands, or whatever) does not seem to need a lot of measurability requirements on the integrand. Progressive is the minimum reasonable requirement to make the integral meaningful in the sense of carrying information that is not from the future. I don't know who sharpened this to predictable. – Kurt G. Feb 22 '24 at 20:34
  • Perhaps it was again the Strasbourg school in the Bourbaki days of that theory. – Kurt G. Feb 22 '24 at 20:34

1 Answers1

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As mentioned in Revuz-Yor pg.120, the main reason that we require $H_{s}$ to be progressively measurable, is because we want the integral against an adapted increasing process $A_{s}$

$$H\cdot A:=\int H_{s}dA_{s}$$

to remain adapted and in turn preserve martingale too. So as mentioned in (2.2) Theorem part (a), if we don't have that $H\cdot A$ is a martingale, then we can't do the uniqueness argument i.e. we want

$$\langle L-L',L-L'\rangle=0$$

to give us $L=L'$. Then they also use Riesz-representation theorem to get a representative.

So in the above construction, one needs to add some measurability assumption for $H$ in order to ensure that we get a martingale back.

This is also shown here in lemma 2 https://almostsuremath.com/2010/01/03/the-stochastic-integral/ using the monotone class theorem.

See also Le Gall Brownian Motion, Martingales, and Stochastic Calculus Proposition 5.3 and Theorem 5.4, in particular on pg. 100

To be precise, we should here say “equivalence classes of elementary processes” (recall that $H$ and $H'$ are identified if $E\int (H_s-H'_s)^{2}d\langle M\rangle_{s}=0$).

So since work over this quotient space, we have to make sure that the representative we obtain is well-defined even if we switch to some other equivalent element.

In fact, the Bichteler-Dellacherie theorem and here say that for semimartingales this the most general construction possible.

Theorem 1 (Bichteler-Dellacherie) For a cadlag adapted process $X$, the following are equivalent.

  • $X$ is a semimartingale.

  • For each ${t\geq 0}$, the set given by $$\displaystyle \left\{\int_0^t\xi\,dX\colon\xi{\rm\ is~ simple~ predictable },\ \vert\xi\vert\le1\right\} $$ is bounded in probability.

  • $X$ is the sum of a local martingale and an FV process.

Thomas Kojar
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  • Thanks for you response! They do not use anything related to predictability to show the existence of $[M, M]t$, but I agree that you do need predictability of the integrand in order to justify their equation (1.3). My issue is that even for non-predictable $H$, the argument given in my question implies that there exists a RCLL martingale $L$ for which $$\mathbb{E}(L\infty N_\infty) = \mathbb{E}\left(\int_0^\infty H_s d[M, N]_s\right),$$ for any bounded RCLL martingale $N$. I know this must be wrong, but I am baffled as I cannot see the error in my argument, and am hoping you might be able to – qp212223 Feb 23 '24 at 17:45
  • Their statement (3.23) has nothing to do with any integrand to any stochastic integral. And your statement about predictability is false. Doob's maximal inequality holds for RCLL martingales (which are not necessarily predictable, obviously - take a Poisson process). Note that in this paper, the quadratic variation which is constructed is not predictable itself (with the same example, if $M$ is a Poisson process, $[M, M]_t = M_t$ in this paper). – qp212223 Feb 25 '24 at 21:49
  • @qp212223 I think I understand your issue now. The problem is that you didn't define a "stochastic integral" that people are interested in. People want to study $$\int \xi dX,$$ whereas you simply discuss the integral $\int \xi d[X]_t,$ which falls under RIemann-Stieltjes integration since $[X]_t$ is increasing. – Thomas Kojar Feb 25 '24 at 22:14
  • Unfortunately, I don't think you do understand my problem. In the case where $M$ is continuous, the martingale $L$ I defined in my question is most certainly the stochastic integral that people are interested in (see, for example, Revuz and Yor, where the integrand may in fact be progressively measurable since they deal with continuous martingales). I think there might be some confusion on your end based on the fact that for the Poisson process with rate 1, $M$, $M-t$ is a RCLL finite variation martingale, so the stochastic integral in that case is a Riemann-Stieltjes integral, obviously. – qp212223 Feb 25 '24 at 22:42
  • @qp212223 In Revuz-Yor, the main work goes towards constructing integrals of the form $$\int \xi dX,$$ that are L2-limits of $\sum \xi_{t_{i-1}} (X_{t_{i}}-X_{t_{i-1}})$. The integral you wrote is only in terms of the quadratic variation. Think of it as the difference between studying $\int \xi dB_{t}$ and $\int \xi d[B]_{t}=\int \xi dt$. I agree with you that defining the later doesn't need predictability. – Thomas Kojar Feb 25 '24 at 22:45
  • In Revuz and Yor, the stochastic integral is built in same way as my question and then shown to be equal to said $L^2$ limit when the process $\xi$ is predictable. In the case of Brownian motion, for example, $L$ *is* the stochastic integral $\int \xi_{t-}dB_t$ (and $\xi$, which is $H$ in my notation, need only be progressively measurable in that case). It is definitely *not* $\int \xi dt$. I don't think you understand the definition of $L$. $L$ is NOT an integral with respect to QV. It is the unique process satisfying $E(L_\infty N_\infty) = J(N)$ for all bounded RCLL martingales $N$. – qp212223 Feb 26 '24 at 16:31
  • @qp212223 One main issue is that you lose uniqueness. See updated answer. – Thomas Kojar Feb 26 '24 at 20:25
  • I am sorry to keep bothering you, but the $L$ I define as a candidate for the stochastic integral is unique by construction (see the Riesz representation theorem), so this is again not a problem. Revuz and Yor is define the stochastic integral as an element such that the condition of theorem (2.2) holds, mine is the same for continuous martingales but not for RCLL ones (for example, for a Poisson process $M$, both $M-[M] \equiv 0$ and $M-t$ are martingales). – qp212223 Feb 26 '24 at 21:57
  • @qp212223 Revuz-Yor use Riesz-representaiton theorem too. The problem is that as in L2-space is a quotient space. Here the quotient space is with respect to the cross-variation. See Le-Gall "stochastic integration" chapter where he talks about that in more detail. See updated answer for some links – Thomas Kojar Feb 26 '24 at 23:10
  • I am aware they also use Riesz representation. But they don't define the stochastic integral using the Riesz representation there. They define it as "a process $\tilde{L}$ such that $$\langle \tilde{L}, N\rangle_t = \int_0^t H d\langle M, N \rangle$$ for every continuous martingale $N$". Obviously $\tilde{L}$ is the same as my $L$ in the continuous martingale case. I am defining $L$ exclusively through Riesz representation (I do not claim there exists a unique process such that...condition holds). – qp212223 Feb 29 '24 at 17:32
  • @qp212223 well that is an issue though. We want our stochastic integral to be uniquely defined. – Thomas Kojar Feb 29 '24 at 17:36
  • I am defining it to be *the unique RCLL martingale* that is given by the Riesz representation theorem. I am not trying using the same definition out of Revuz and Yor (the definition that uses the property in the previous comment). I'm sorry if I'm missing something here. – qp212223 Feb 29 '24 at 17:48
  • @qp212223 Revuz Yor obtain the stochastic integral using Riesz-representation exactly as you do. The issue is that "uniqueness" in a quotient function space is only up to equivalence.. And as mentioned in Le Gall, the particular equivalence here is a.s. with respect to the quadratic variation. – Thomas Kojar Feb 29 '24 at 17:52
  • The difference is very, very, subtle. Theorem (2.2) in Revuz and Yor establishes the existence of a unique process satisfying the quadratic covariation condition given in the theorem (the QV condition is different from the condition I am defining $L$ with). They obtain existence via Riesz representation (Riesz representation obtains the existence of a unique martingale such that (2.1) holds - (2.1) in the RCLL case is not equivalent to the condition of the theorem, but it is in the continuous case, as the only continuous martingales with finite variation are constant). – qp212223 Mar 01 '24 at 20:41
  • Riez representation implies the existence of a unique RCLL martingale such that condition (1) of my post is true. So both existence and uniqueness are established this way. My question is why is this not ok? – qp212223 Mar 01 '24 at 20:43
  • @qp212223 As I mentioned in my post, there is a subtlety with the quotient space that your construction is not addressing. Indeed, Revuz-Yor apply RRT to get a Hilbert-unique representative. The problem is you also need well-posedness/definedness of that representative i.e. <L-L',L-L'>=0 implying L=L'. Without this there is a risk of not having a unique random variable as stochastic integral. But instead we are forced talking about a representative/stochastic-integral that is "unique" up to RCLL corrections i.e. $\int \xi dX+M$, for RCLL M as opposed to just $\int \xi dX$ as in Revuz-Yor. – Thomas Kojar Mar 01 '24 at 22:04
  • No, we are not. Since the element chosen via Riesz representation is the unique element which satisfies (1) and is a martingale, the representation is not unique up to any changes. It is, by definition, unique. There may be other stochastic processes $L'$ such that $[L-L', L-L'] = 0$ and satisfy (1), but they are not martingales (if they were, then by Riesz representation, they would be equal to $L$ and thus $[L-L', L-L'] = 0$). – qp212223 Mar 06 '24 at 14:27
  • @qp212223 Yes and that's why RY require the process $\xi$ to be predictable, in order that we get martingale $L$ and hence uniqueness. Your question was "why do we need predictable?" and that is exactly the subtle answer. – Thomas Kojar Mar 06 '24 at 18:29
  • RY do not require predictable. They require progressively measurable, since we are in the continuous case (definition 2.1 on p. 137). Bichteler-Dellacherie tells us that if we want to define the stochastic integral via limits of Riemann-type sums, we have to do it with predictable processes. My point is that if we define it via the alternative in (1), we can do so for non-predictable integrands, our integral satisfies the martingale property, and we get uniqueness by definition. This is not the same as RY's definition (except for continuous martingales ofc). – qp212223 Mar 07 '24 at 16:53
  • @qp212223 Sorry yes, I see that, I used the wrong term. But still in your construction you simply say "any process H". How does your construction satisfy the martingale property? There are many counterexamples of just picking H depend on the future. – Thomas Kojar Mar 07 '24 at 18:24
  • @qp212223 My point is simply that you cannot get martingale $\int H dX$ unless you make some assumptions for $H$ such as predictable or progressively measurable. Otherwise the $\int H dX$ will not be adapted and martingale. – Thomas Kojar Mar 07 '24 at 23:53
  • This is precisely why I am so confused. I understand there exist many problems if $H$ is arbitrary (via the examples in other posts). But the problem I have is that I cannot see, where the error in the construction itself is and I want to find that out (i.e. where in my post I am making a claim that is not true). – qp212223 Mar 13 '24 at 16:21
  • @qp212223 In your above comment you say " we define it via the alternative in (1), we can do so for non-predictable integrands, our integral satisfies the martingale property,". But how do you know your particular construction in (1) satisfies the martingale property? – Thomas Kojar Mar 13 '24 at 16:57
  • The space of bounded RCLL martingales is a Hilbert space. For any measurable process $H$ and any RCLL martingale $M$, $N \mapsto J(N)$ defines a linear mapping from this Hilbert space to $\mathbb{R}$. The Riesz representation theorem gives an element $L$, which is in the space of RCLL martingales, such that (1) is satisfied. I have no idea where this logic is broken. The only thing I can think of is that $L$ is indeed well defined, but is equal to the stochastic integral (the properly defined one as limit of Riemann type sums, etc.) of the predictable projection of $H$, or something like this. – qp212223 Mar 21 '24 at 15:05
  • @qp212223 In that case ok. The resulting element $L_{M,H}$ is simply some representative

    $$E[\Phi_{M,H}(N)]=<L_{M,H},N>,$$

    and since $H\cdot M$ is not a martingale (because we need predictability for H), that means $L_{M,H}$ is definitely not equal to $H\cdot M$. From the proof of Riesz-rep, it will just be an element from the subspace of RCLL-martingales N that are orthogonal to the kernel of $\Phi_{M,H}$.

     – Thomas Kojar Mar 21 '24 at 18:33