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This is a follow-up to my previous question on MathOverflow.

Is there a way to combine the Dambis-Dubins-Schwarz theorem and the Martingale Representation Theorem to get the following result?

Let $(M_t, t\ge 0)$ be a continuous $(\mathcal{F}_t)_{t \ge0}$ local martingale, then there exists some Brownian motion $B$ and some progressively measurable process $\xi$ on some probability space such that $$M_t = \int_0^t \xi_s dB_s = \large{ B_{\int_0^t \xi_s^2 ds}} \quad \forall t \in [0,\infty)$$

In particular it would follow that $\langle M \rangle_t=\int_0^t \xi_s^2 ds$ (by Ito's Isometry?).

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This appears to be correct (at least for continuous local martingales such that $\langle M_t \rangle_{\infty}=\infty$), provided the following claim is true:

Claim If $(M_t, t\ge 0)$ is a continuous local $(\mathcal{F}_t)_{t \ge 0}$ martingale, then it is also a continuous local $(\mathcal{F}_{\tau_u})_{u \ge0}$ martingale, where $\tau_u:=\inf\{t: \langle M \rangle_t > u\}$.

The truth of this claim might be a direct corollary of the Dambis-Dubins-Schwarz Theorem, although I am not entirely certain of this.

Let's assume we have a continuous $(\mathcal{F}_t)_{t \ge 0}$ local martingale $(M_t, t\ge 0)$ such that $\langle M \rangle_{\infty}=\infty$. Then it is also a $(\mathcal{F}_{\tau_u})_{u \ge 0}$ local martingale, which by the Dambis-Dubins-Schwarz Theorem equals $B_{\langle M \rangle_t}$ for a standard Brownian motion on the same probability space $B_u$ whose natural filtration is $(\mathcal{F}_{\tau_u})_{u \ge 0}$. Then it follows from the martingale representation theorem that $(M_{\tau_u}, u\ge0)$ (the "same" continuous local martingale as before, but using the coarser filtration $(\mathcal{F}_{\tau_u})_{u \ge 0}$ as opposed to the original filtration) can be written as $M_{\tau_u}=M_0 + \int_0^u \xi dB$ for a predictable, $B$-integrable process $\xi$. It then follows (by Ito's Isometry?) that $\langle M \rangle_t = \int_0^t \xi_s^2 ds$.

(But why is this integral even necessarily defined? Just because $\xi$ is predictable doesn't necessarily mean that it is of finite variation, hence I don't see how we can necessarily conclude that the pathwise Lebesgue-Stieltjes integral is applicable, unless we can somehow guarantee that $\xi^2$ is a finite variation process, which seems dubious to me.)

In summary

Local Martingales as Both Time Changes of and Integrals of Brownian Motion Let $(M_t, t\ge 0)$ be a continuous local $(\mathcal{F}_t)_{t \ge 0}$ martingale. Define $\tau_u:=\inf\{t: \langle M \rangle_t > u \}$. Then there exists a standard $(\mathcal{F}_{\tau_u})_{u \ge 0}$ Brownian motion $(B_u, u \ge 0)$ and a predictable $B$-integrable process $\xi$ such that $$\langle M \rangle_t = \int_0^t \xi_s^2 ds$$ and $$M_{\tau_u} = B_{\langle M \rangle_t}=B_{\int_0^t \xi_s^2 ds}= M_0 + \int_0^{\tau_u} \xi dB.$$


According to Professor Marc Yor, we have:

Dambis-Dubins-Schwarz Theorem If $(M_t, t\ge 0)$ is a continuous local $(\mathcal{F}_t)_{t \ge 0}$ martingale such that $\langle M \rangle_{\infty}=\infty$, then the process $$B_u=M_{\tau_u}, \quad \tau_u=\inf\{t:\langle M \rangle_t > u\}$$ is a standard $(\mathcal{F}_{\tau_u})_{u \ge 0}$ Brownian motion, and $M$ may be represented as: $$M_t=B_{\langle M \rangle_t}, \quad t \ge 0$$

I believe that the validity of the representation $M_t=\beta_{\langle M \rangle_t}$ must have as a corollary that $M_t$ is a continuous local $(\mathcal{F}_{\tau_u})_{u \ge 0}$ martingale, although again I am not 100% sure about this.

According to George Lowther (note that his notation is not the same as Marc Yor's), we have:

Martingale Representation Theorem Let $B$ be a standard Brownian motion defined on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and $( \mathcal{F}_t )_{t \ge 0}$ be its natural filtration. Then every local $(\mathcal{F}_t)_{t \ge 0}$ martingale $(M_t,t \ge 0)$ can be written as $$M_t=M_0 + \int_0^t \xi dB$$ for a predictable, $B$-integrable process $\xi$.

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