Lecture 23. Martingales as a time changed Brownian motion

In the previous Lecture, we proved that any martingale which is adapted to a Brownian filtration can be written as a stochastic integral. In this section, we prove that any martingale can also be represented as a time changed Brownian motion. To prove this fact, we give first first a characterization of the Brownian motion which is interesting in itself. In this section, we denote by (\mathcal{F}_t)_{t \ge 0} a filtration that satisfies the usual conditions.

Proposition: (Levy’s characterization theorem) Let (M_t)_{t \ge 0} be a continuous local martingale such that M_0=0 and such that for every t \ge 0, \langle M \rangle_t =t. The process (M_t)_{t \ge 0} is a standard Brownian motion.

Proof. Let N_t=e^{i\lambda M_t +\frac{1}{2}\lambda^2 t}. By using Itō’s formula, we obtain that for s \le t, N_t =N_s +\int_s^t N_u dM_u.
As a consequence, the process (N_t)_{t \ge 0} is a martingale and, from the above equality we get
\mathbb{E}\left( e^{i\lambda (M_t-M_s)} \mid \mathcal{F}_s \right)=e^{-\frac{1}{2}\lambda^2 (t-s)}.
The process (M_t)_{t \ge 0} is therefore a continuous process with stationary and independent increments such that M_t is normally distributed with mean 0 and variance t. It is thus a Brownian motion \square

The next proposition shows that continuous martingales behave in a nice way with respect to time changes.

Proposition Let (C_t)_{t \ge 0} be a continuous and increasing process such that for every t \ge 0, C_t is a finite stopping time of the filtration (\mathcal{F}_t)_{t \ge 0}. Let (M_t)_{t \ge 0} be a continuous martingale with respect to (\mathcal{F}_t)_{t \ge 0}. The process (M_{C_t})_{t \ge 0} is a local martingale with respect to the filtration (\mathcal{F}_{C_t})_{t \ge 0}. Moreover \langle M_C \rangle =\langle M \rangle_C.

Proof. . By using localization, we may assume C to be bounded. According to the Doob’s stopping theorem, we need to prove that for every bounded stopping time T of the filtration (\mathcal{F}_{C_t})_{t \ge 0}, we have \mathbb{E}( M_{C_T})=0. But C_T is obviously a bounded stopping time of the filtration (\mathcal{F}_t)_{t \ge 0} and thus from Doob’s stopping theorem we have \mathbb{E}( M_{C_T})=0. The same argument shows that M_C^2- \langle M \rangle_C \square

Exercise. Let (C_t)_{t \ge 0} be an increasing and right continuous process such that for every t \ge 0, C_t is a finite stopping time of the filtration (\mathcal{F}_t)_{t \ge 0}. Let (M_t)_{t \ge 0} be a continuous martingale with respect to (\mathcal{F}_t)_{t \ge 0} such that M is constant on each interval [C_{t-},C_t]. Show that the process (M_{C_t})_{t \ge 0} is a continuous local martingale with respect to the filtration (\mathcal{F}_{C_t})_{t \ge 0} and that \langle M_C \rangle =\langle M \rangle_C.

We can now prove the following nice representation result for martingales.

Theorem. ( Dambis, Dubins-Schwarz) Let (M_t)_{t \ge 0} be a continuous martingale such that M_0=0 and \langle M \rangle_\infty =+\infty. There exists a Brownian motion (B_t)_{t \ge 0}, such that for every t \ge 0,
M_t =B_{\langle M \rangle_t}.

Proof. Let C_t =\inf \{ s \ge 0, \langle M \rangle_s >  t \}. (C_t)_{t \ge 0} is a right continuous and increasing process such that for every t \ge 0, C_t is a finite stopping time of the filtration (\mathcal{F}_t)_{t \ge 0} and M is obviously constant on each interval [C_{t-},C_t]. From the previous exercise the process (M_{C_t})_{t \ge 0} is a local martingale whose quadratic variation is equal to t. From Levy’s characterization theorem, it is thus a Brownian motion \square

Exercise. Show that if (M_t)_{t \ge 0} is a continuous local martingale such that M_0=0 and \langle M \rangle_\infty =+\infty, there exists a Brownian motion (B_t)_{t \ge 0}, such that for every t \ge 0, M_t =B_{\langle M \rangle_t}.

Exercise.
Let (u_t)_{t \ge 0} be a continuous adapted process and let (B_t)_{t \ge 0} be a Brownian motion. Show that for every T \ge 0, the process \left( \int_0^t u_s dB_s\right)_{0\le t \le T} has \frac{1}{2}-\varepsilon Holder paths, where 0 < \varepsilon \le \frac{1}{2}.

The study of the planar Brownian is deeply connected to the theory of analytic functions. The fundamental property of the Brownian curve is that it is a conformal invariant. The following proposition is easily proved as a consequence of Itō’s formula and of the Dambins-Dubins-Schwarz theorem. By definition, a complex Brownian motion is a process (B_t)_{t \ge 0} in the complex plane that can be decomposed as B_t =B^1_t +i B^2_t where B^1 and B^2 are independent Brownian motions.

Proposition.(Conformal invariance of the planar Brownian motion) Let (B_t)_{t \ge 0} be a complex Brownian motion and f : \mathbb{C} \to \mathbb{C} be an analytic function. Then
f(B_t)=f(0)+\int_0^t f'(B_s) dB_s.
As a consequence, there exists a complex Brownian motion (\beta_t)_{t \ge 0} such that
f(B_t)=f(0)+\beta_{\int_0^t | f'(B_s)|^2 ds}.

To study the complex Brownian motion, it is useful to look at it in polar coordinates. It leads to the so-called skew-product decomposition of the complex Brownian motion.

Proposition. Let (B_t)_{t \ge 0} be a complex Brownian motion started at z \neq 0. There exists a complex Brownian motion (\beta_t)_{t \ge 0} such that
B_t=z \exp{\left(  \beta_{\int_0^t \frac{ds}{\rho_s^2} }\right)}, where \rho_t =| B_t |.

Proof. The proof is let as an exercise to the reader. The main idea is to prove, by using Itō’s formula, that B_t=z \exp\left( \int_0^t \frac{dB_s}{B_s}  \right), and then to used the Dambins-Dubins-Schwarz theorem \square

Exercise. In the previous proposition, show that the process (\rho_t)_{t \ge 0} is independent from the process (\mathbf{Im} ( \beta_t))_{t \ge 0}.

You will find below a video of a talk by Pr. Marc Yor concerning quadratic functionals of the planar Brownian motion. The talk was given at the University of Bristol in December 2008 for a special event.

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2 Responses to Lecture 23. Martingales as a time changed Brownian motion

  1. Gábor Pete says:

    In the proof of Dambis, Dubins-Schwarz, C_t is NOT necessarily continuous (e.g., let M_t be a BM, staying put for exponential times at the rings of an exponential clock). Hence the proof here is very much incomplete, isn’t it?

  2. Thanks, this is now corrected

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