# Why is the stochastic integral only defined for predictable integrands?

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''.

• 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. Feb 19 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? Feb 20 at 0:06
• 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? Feb 20 at 0:24
• 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 $H$? Feb 20 at 15:04
• 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 $L$: en.wikipedia.org/wiki/Riesz_representation_theorem. I will bounty this question at some point in the hope that someone else answers. Feb 21 at 14:21

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.

• 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 Feb 23 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). Feb 25 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. Feb 25 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. Feb 25 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. Feb 25 at 22:45