Can all Local Martingales Be Represented using Only Brownian Motion and Finite Variation Processes?

This is a cross-post of my unanswered (more than a week) question on Math.SE. Since it covers topics from my graduate-level course on stochastic processes, I thought it might be appropriate to try to ask it here.

Prove or Disprove: A process $(X_t)_{t \ge 0}$ is a (continuous) local martingale if and only if it can be represented in the form: $$\int_0^t \xi dB = \large B_{\int_0^t \xi_s^2 ds}$$ where the quadratic variation $\langle X \rangle_t=\int_0^t \xi_s^2 ds$ for some increasing finite variation process $\xi$ (whose paths are therefore absolutely continuous with respect to Lebesgue measure).

In particular, a (continuous) local martingale is a deterministic time change of Brownian motion if and only if its quadratic variation is a deterministic function (and absolutely continuous with respect to Lebesgue measure).

Basically I want to know how much the theory of stochastic integration/local martingales can be reduced to the study of Brownian motion and finite variation processes only.

Is this the same thing as saying thing that all local martingales are Ito processes and vice versa?

EDIT: See here for my latest attempt to solve this problem using Stephan Sturm's answer below.

Related: this question covers a lot of similar ground, although it does not address how that specific representation can be made to relate to time changes. https://math.stackexchange.com/questions/1457270/representation-theorem-for-local-martingales

This question on MathOverflow seems somewhat similar in spirit: The only continuous martingales with stationary increments are Brownian motions

Also, and most importantly, these two blog posts by George Lowther (who answered the above linked MO question) are the primary basis for my question, because I think the answer might be contained somewhere within the two posts, but I am having difficulty putting everything together.

Motivation: Basically I am trying to devise as simple of an introduction to the concept of local martingale as possible for my study guide for stochastic integration, to be found here.

I find the most commonly used characterization, as a localization of a martingale, to be unsatisfactory for several reasons.

First, it affords none of the intuition developed from studying martingales in the discrete-time case, since in the discrete time case, local and regular martingales coincide. Hence, for someone learning for the first time, it is not immediately clear why the two concepts shouldn't coincide in continuous time, or why localization of processes should be at all interesting, when it is not at all interesting in discrete time.

Second, the commonly used intuition that "a martingale is just an integrable local martingale" is incorrect, as Rick Durrett found out the hard way after publishing the first edition of his textbook on stochastic calculus. In fact the necessary and sufficient condition (Theorem 2.5 on p. 41) is $$\mathbb{E}\left( \underset{0 \le s \le t}{\sup} |X_s| \right) < \infty$$ which is a lot less elegant than would be desired.

Third, and along the lines of the second objection, as noted on p. 123 in Revuz and Yor, local martingales are much more general than martingales, not just "a little more", making the name "local martingale" something of a misnomer. Revuz and Yor list at least two integrability classes of stochastic processes, $(D)$ and $(DL)$ on p. 124 which one presumably would think of before thinking of the concept of local martingale.

Fourth and finally, local martingales as a concept were introduced in order to generalize the integral with respect to Brownian motion developed by Ito, not to generalize martingales, so I think it would be more appropriate to characterize them in terms of stochastic integration. Both my professor, who mentioned Kunita and Watanabe's work, as well as Revuz and Yor in their remark preceding Theorem 1.8 on p.124, state that local martingales as a concept were introduced in order to advance the theory of stochastic integration.

By the Bachteler-Dellacherie theorem, which tells us that the most general class of "good" stochastic integrands is equal to a local martingale up to (plus) a finite variation process, if we were able to characterize local martingales entirely in terms of Brownian motion and finite variation processes, I think it would be an impressive conceptual achievement. Maybe not a practical one, but a conceptually elegant one, which is what interests me at the moment.

First, a martingale is always only specified with respect to a filtration, and so is thus a local martingale. You do not specify any filtration in your problem, so I assume you mean the natural filtration of the local martingale (i.e., the smallest filtration w.r.t. which $X$ is a local martingale).

Second, your prove/disprove statement does not consist of one, but of multiple claims. I will try to disentangle them first before discussing their correctness. I will refer throughout to the the text Continuous Martingales and Brownian Motion which you mention, too.

(1) Let $X$ be a continuous local martingale, then there exists a Brownian motion $B$ and a predictable process $\xi$ such that $X_t = B_{\int_0^t \xi_s^2 \, ds}$ for all $t \in [0, \infty)$.

(2) Let $X$ be a continuous local martingale, then there exists a Brownian motion $\tilde{B}$ and a predictable process $\tilde{\xi}$ such that $X_t = \int_0^t \tilde{\xi}_s \, d\tilde{B}_s$ for all $t \in [0, \infty)$.

(3) $\xi$ and $\tilde{\xi}$ as well as $B$ and $\tilde{B}$ agree in an appropriate sense, say, are versions of each other.

(4) The quadratic variation of $\langle X \rangle_t$ is then given by $\int_0^t \xi^2_s \, ds$ (resp. $\int_0^t \tilde{\xi}^2_s \, ds$).

(5) It holds true that $\xi$ is increasing and of finite variation.

(6) The statement (3) holds true at least in the case when $\xi$ is is deterministic.

Lets now asses the claims in detail:

(1) This statement holds true as long as $\langle X \rangle_\infty = \infty$, it is called Dambis -- Dubins-Schwarz theorem (cf. R-Y, Thm. V.1.6). Without the assumption this is not necessarily true in the natural filtration (as it might be not rich enough to support a Brownian motion). However, it is possible to enlarge the probability space resp. the filtration such that this statement is true in the larger space (cf. R-Y, Thm V.1.7).

(2) This is the \textit{martingale representation theorem}. It is usually formulated in the context of Brownian filtrations, i.e., it is a priori assumed that $X$ is a local martingale with respect to a filtration generated by a Brownian motion. However, it holds true without this assumption as long as the measure generated by the quadratic variation process, $d\langle X \rangle_t$, is equivalent to the Lebesgue measure $dt$. If there is no equivalence between the measures but at least $d\langle X\rangle_t$ is absolutely continuous with respect to the Lebesgue measure, one can save again the situation by enlarging the probability space (cf. R-Y, V.3.8 and discussion thereafter).

Without absolute continuity the statement is false as can be seen on the following example. Let $C_t^i$ be copies of the Cantor function on the unit interval $[0,1)$ and define the extension to the positive halfline by

$$K_t = \sum_{i=0}^\infty i + C_{t-i}^i {{\mathchoice{1\mskip-4mu\mathrm l}{1\mskip-4mu\mathrm l} {1\mskip-4.5mu\mathrm l}{1\mskip-5mu\mathrm l}}}_{[i,i+1)}(t)$$

Let $W$ be a Brownian motion and define $X$ by $X_t = W_{K_t}$. This is clearly a martingale with quadratic variation $\langle X \rangle_t = K_t$. However, if $X$ could be written $X_t = \int_0^t \tilde{\xi}_s \, d\tilde{B}_s$, then we would have by (4) $\langle X \rangle_t = \int_0^t \tilde{\xi}^2_s \, ds$. However, this is always an absolutely continuous function whereas $K$ is not, therefore they cannot agree.

(3) From the restrictions above it is clear that this cannot hold true in general. But you can even find very simple counterexamples, e.g., set $\tilde{\xi} =1$. Then you have

$$\int_0^t (-1) \, d\tilde{B}_s = - \tilde{B}_t = B_t = B_{\int_0^t 1^2 \, ds} = B_{\int_0^t (-1)^2 \, ds}$$

So in this case $\tilde{B} = - B$ whereas $\xi$ is not uniquely determined (e.g., it could be $1$ or $-1$). This can be of course made worse using processes jumping between $1$ and $-1$. (4) This is true and in both cases (direct calculation using properties of Brownian motion and R-Y, Prop. IV.2.7 resp.).

(5) Every increasing process is of finite variation by definition. However, the claimed result is clearly not true, just think about an arbitrary integral with deterministic non- monotone integrand, i.e. $X_t = \int_0^t \sin{(s)} \, \, dW_s$ for some Brownian motion $W$. I assume this is simply a typo and you intended to claim that the quadratic variation itself is increasing (or rather: non-decreasing) and thus of bounded variation (which is evidently true, you just add more squares).

(6) No, the example from (3) or the Cantor function example from (2) are counterexamples.

Let me conclude with two remarks about the context:

(a) The idea to place stochastic calculus on change of time instead of stochastic integration has some interesting history. Actually, it predates Ito calculus and was the first form of stochastic calculus, developed by Wolfgang Döblin (Vincent Doblin) in his famous lettre scellée to the French Academy of Science, https://mathoverflow.net/a/100040/20026.

(b) I disagree with your statement that local martingale is a misnomer. Yes, a local martingale is much more general than a martingale. But it is intentionally not called generalized martingale but local martingale, as it is locally (i.e. up to an increasing sequence of stopping times) a martingales. Generalizations that resemble locally the original object are very often far more general than the original object. The gap between manifolds and Euclidean spaces is for sure much larger than that between true and local martingales.

• Yes you are correct that I meant the natural filtration; I apologize for not being more specific. Also I am a little confused; (1) do you mean local martingale instead of martingale in most of the places above? I am more interested in local martingales than martingales. (2) So the continuity of paths (i.e. the distinction between continuous local martingale and general local martingale) doesn't matter, but the absolute continuity of paths (a strictly weaker condition) (the difference between "absolutely continuous local martingales" and general local martingales) does matter? – Chill2Macht Jun 29 '16 at 15:12
• Also I want to upvote your post but I don't have enough reputation after posting this bounty. – Chill2Macht Jun 29 '16 at 15:13
• Also also to summarize, we can say that most, if not all, of the sub-claims are true if we add the two extra conditions: (1) we allow enlargements of the probability space, (2) the sample paths of the local martingale are absolutely continuous, (3) the measure generated by the quadratic variation process is absolutely continuous w.r.t. Lebesgue measure. Are (2) and (3) equivalent given (1)? (is this what you are saying in your answer of the 2nd sub-claim?) – Chill2Macht Jun 29 '16 at 15:23
• And finally, the sample paths of the quadratic variation are always absolutely continuous (but this is always true since the quadratic variation is always increasing and thus of finite variation)? But even if the sample paths of the quadratic variation are always absolutely continuous, that does not imply that the measure they generate is necessarily absolutely continuous w.r.t. Lebesgue measure? (I don't understand how the counterexample to the 2nd sub-claim is a local martingale, although I may be misremembering the theorem characterizing local martingales in terms of q.v. processes.) – Chill2Macht Jun 29 '16 at 15:26
• See also math.stackexchange.com/questions/1796863/… for more examples of my confusion regarding optimal properties of corresponding quadratic variation processes and suitability as stochastic integrators. – Chill2Macht Jun 29 '16 at 16:29