## 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