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 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 maringales are Ito processes and vice versa?

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. Representation theorem for local martingales

Also, these two blog posts are much of the basis for my question, because I think the answer might be contained somewhere within, but I am having difficulty putting everything together, because no hypotheses seem to be consistently used throughout.

https://almostsure.wordpress.com/2010/04/20/time-changed-brownian-motion/ https://almostsure.wordpress.com/2010/04/13/levys-characterization-of-brownian-motion/

See Theorem 1 here.

Theorem 1 Any continuous local martingale $X$ with $X_0 = 0$ is a continuous time-change of standard Brownian motion (possibly under enlargement of the probability space).

More precisely, there is a Brownian motion $B$ with respect to a filtration $\{\mathcal{G}\}_{t\ge 0}$ such that, for each $t \ge 0$, $\omega \mapsto [X]_t(\omega)$ is a $\mathcal{G}$ stopping time and $X_t = B_{[X]_t}$.

1. Is the converse of this theorem true?

2. What is the set of stochastic processes whose quadratic variation processes are (or just have sample paths which are) the cumulative distribution functions of measures absolutely continuous with respect to Lebesgue measure on $[0, \infty)$?

(Presumably this set includes the continuous local martingales, perhaps all local martingales, but is this a strict inclusion? Does it even include all continuous local martingales?)

At a very minimum: does any (continuous) local martingale have quadratic variation process which is not absolutely continuous with respect to Lebesgue measure?

3. Would we be able to prove Ito's isometry for every member of this set/class (from 2.)?

4. If 3. is correct, then is that the reason why (continuous) local martingales are the largest possible class (up to a finite variation process) of (good) stochastic integrators?

(I.e. why semimartingales are the largest possible class of good stochastic integrators, since every semimartingale is equal to a local martingale plus a process of finite variation, the integral of the latter can be computed pathwise, whereas the integral of the former requires the same type of machinery which one must apply on Brownian motion.)

EDIT: Theorem 5 here suggests some progress towards answering this question is possible.

Theorem 5 Let $X$ be a local martingale such that $[X]$ has absolutely continuous sample paths. Also, suppose that there exists at least one Brownian motion defined on the filtered probability space.

Then $X=X_0+\int \xi\ dB$ for a Brownian motion $B$ and a $B-$integrable process $\xi$.

This result looks tantalizingly close to answering my question, but I'm not sure how exactly to interpret it correctly (second exercise underneath the theorem by Dambis, Dubins-Schwarz).

If $(M_t)_{t\ge 0}$ is a continuous local martingale such that $M_0=0$ and $\langle M \rangle_{\infty}=+\infty$ (if the quadratic variation increases without bound), then there exists a Brownian motion $(B_t)_{t \ge 0 }$ such that for every $t \ge 0$, $M_t = B_{\langle M \rangle_t}$.

  • 1
    $\begingroup$ @ William : Her are my two cents, trying to reformulate the implication you are looking for, given a family stopping time $(\tau_t)_{t>0}$ with respect to a filtration under which it is possible to construct a BM $B$ equipped with a measure $P$, with the property that a.s. paths $(\tau_t)(\omega)$ are continuous, increasing, then does $X_t=B_{\tau_t}$ is a local martingale?A sufficient condition is that the $\tau_t$ being u.i., because in that case you can apply optimal stopping theorem without further ado and get a true martingale. For the general case well I don't know. Best regards $\endgroup$
    – TheBridge
    Jun 21, 2016 at 13:18
  • 1
    $\begingroup$ @Willaim : Regarding theorem 5, it is another representation theorem but the same problem arises, you only have one way, and not equivalence unless mistaken. Best regardds $\endgroup$
    – TheBridge
    Jun 21, 2016 at 13:20
  • $\begingroup$ @TheBridge wait I'm confused -- do you mean continuous sample paths, or absolutely continuous with respect to Lebesgue measure (i.e. the functions are the integrals of some density)? $\endgroup$ Jun 21, 2016 at 14:16
  • 1
    $\begingroup$ I mean almost surely, every path is continuous with respect to the probability space that supports the filtration, the sopping times and the Brownian motion. Best regards $\endgroup$
    – TheBridge
    Jun 21, 2016 at 15:46
  • 1
    $\begingroup$ But you can modify the reformulation to replace continuity by absolute continuity (in an almost sure way), but observe that in the "theorem 1" it is stated "a continuous time-change". best regards $\endgroup$
    – TheBridge
    Jun 21, 2016 at 16:11


You must log in to answer this question.