Lecture 1. The Paul Levy’s stochastic area formula

When studying functionals of a Brownian motion, it may be useful to embed this functional into a larger dimensional Markov process.

Consider the case of the Levy area
$S_t=\int_0^t B^1_s dB^2_s-B^2_s dB^1_s,$
where $B_t=(B^1_t,B^2_t)$ , $t \ge 0$ , is a two dimensional Brownian motion started at 0. We can write
$S_t=\int_{B[0,t]} \alpha$
where $\alpha=xdy-ydx$ . Since $d\alpha =2 dx \wedge dy$ , we interpret $S_t$ as (two times) the algebraic area swept out in the plane by the Brownian curve up to time $t$ . The process $(S_t)_{t \ge 0}$ is not a Markov process in its own natural filtration. However, if we consider the 3-dimensional process
$X_t=(B^1_t,B^2_t,S_t),$
then $X_t$ is solution of a stochastic differential equation
$dX^1_t =dB^1_t$

$dX^2_t =dB^2_t$

$dX^3_t =-X^2_t dB^1_t+X^1_t dB^2_t$

As a consequence $X_t$ is a Markov process with generator

$L=\frac{1}{2} (X^2+Y^2)$
where $X,Y$ are the following vector fields
$X=\frac{\partial}{\partial x}-y \frac{\partial}{\partial z}$
$Y=\frac{\partial}{\partial y}+x \frac{\partial}{\partial z}.$
Observe that the Lie bracket

$[X,Y]=XY-YX =2 \frac{\partial}{\partial z}.$
Thus, for every $x \in \mathbb{R}^3$ , $(X(x),Y(x), [X,Y](x))$ is a basis of $\mathbb{R}^3$ . From the celebrated Hormander’s theorem, this implies that for every $t > 0$ the random variable $X_t$ has a smooth density with respect to the Lebesgue measure of $\mathbb{R}^3$ . In particular $S_t$ also has smooth density whenever $t>0$ . We are interested in an expression for this density. The first idea is to reduce the complexity of the random variable $X_t$ by making use of symmetries.

Lemma: Let $r_t=\| B_t \|=\sqrt{ (B^1_t)^2 +(B^2_t)^2 }$ , $t \ge 0$ . Then, the couple

$(r_t , S_t)_{t \ge 0}$
is a Markov process with generator

$\mathcal{L}=\frac{1}{2r} \frac{\partial}{\partial r}+\frac{1}{2} \frac{\partial^2}{\partial r^2}+\frac{1}{2}r^2 \frac{\partial^2}{\partial s^2}$

Proof:
From Ito’s formula, we have

$dr_t =\frac{dt}{2 r_t}+\frac{B^1_t dB^1_t+B^2_t dB^2_t}{ \sqrt{ (B^1_t)^2 +(B^2_t)^2 }}$

$dS_t=r_t \frac{B^1_t dB^2_t-B^2_t dB^1_t}{ \sqrt{ (B^1_t)^2 +(B^2_t)^2 }}.$

Since the two processes

$\beta_t=\int_0^t \frac{B^1_s dB^1_s+B^2_s dB^2_s}{ \sqrt{ (B^1_s)^2 +(B^2_s)^2 }}$

$\gamma_t=\int_0^t \frac{B^1_s dB^2_s-B^2_s dB^1_s}{ \sqrt{ (B^1_s)^2 +(B^2_s)^2 }},$

are two independent Brownian motions, the conclusion easily follows. $\square$

We are now ready to prove the celebrated Levy’s area formula.

Theorem: For $t>0$ and $x \in \mathbb{R}^2$ , and $\lambda >0$

$\mathbb{E}\left( e^{i\lambda S_t} | B_t=x\right)=\frac{\lambda t}{\sinh \lambda t} e^{-\frac{\|x\|^2}{2t}(\lambda t \coth \lambda t -1) }.$

Proof:
First, we observe that by rotational symmetry of the Brownian motion $(B_t)_{t \ge 0}$ , we have

$\mathbb{E}\left( e^{i\lambda S_t} | B_t=x\right)=\mathbb{E}\left( e^{i\lambda S_t} | \| B_t \| =\| x \| \right).$
Then, according to the previous lemma,

$\mathbb{E}\left( e^{i\lambda S_t} | \| B_t \| =\| x \| \right)=\mathbb{E}\left( e^{i\lambda \gamma_{\int_0^t r_s^2 ds}} | r_t =\| x \| \right),$
where $\gamma_t$ is a Brownian motion independent from $r$ . We deduce

$\mathbb{E}\left( e^{i\lambda S_t} | B_t=x\right)=\mathbb{E}\left( e^{-\frac{\lambda^2}{2} \int_0^t r_s^2 ds} | r_t =\| x \| \right),$
As we have seen, $r_t$ solves a stochastic differential equation

$dr_t =\frac{dt}{2 r_t}+d\beta_t,$
where $\beta$ is a one-dimensional Brownian motion.
One considers then the new probability

$\mathbb{P}_{/ \mathcal{F}_t}^\lambda = \exp \left(- \lambda \int_0^t r_s d\beta_s -\frac{\lambda^2}{2} \int_0^t r_s^2 ds \right)\mathbb{P}_{/ \mathcal{F}_t},$
where $\mathcal{F}$ is the natural filtration of $\beta$ . Observe that

$\int_0^t r_s d\beta_s =\int_0^t r_s dr_s -\frac{t}{2} =\frac{1}{2} r_t^2 -t$

Therefore

$\exp \left( -\lambda \int_0^t r_s d\beta_s -\frac{\lambda^2}{2} \int_0^t r_s^2 ds \right)=e^{\lambda t} \exp \left( -\frac{\lambda}{2} r_t^2 -\frac{\lambda^2}{2} \int_0^t r_s^2 ds \right)$

In particular, one deduces that

$\exp \left( -\lambda \int_0^t r_s d\beta_s -\frac{\lambda^2}{2} \int_0^t r_s^2 ds \right) \le e^{\lambda t},$
which proves that $\exp \left( -\lambda \int_0^t r_s d\beta_s -\frac{\lambda^2}{2} \int_0^t r_s^2 ds \right)$ is a martingale. By using this change of probability, if $f$ is a bounded and Borel function, we have

$\mathbb{E}\left( f(r_t) e^{-\frac{\lambda^2}{2} \int_0^t r_s^2 ds} \right) =e^{-\lambda t} \mathbb{E}^\lambda \left(f(r_t) \exp \left( \frac{\lambda}{2} r_t^2 +\frac{\lambda^2}{2} \int_0^t r_s^2 ds \right) e^{-\frac{\lambda^2}{2} \int_0^t r_s^2 ds} \right) =e^{-\lambda t} \mathbb{E}^\lambda \left(f(r_t) \exp \left( \frac{\lambda}{2} r_t^2 \right) \right).$
Putting things together, we are thus let with the computation of the distribution of $r_t$ under the probability $\mathbb{P}^\lambda$ . From Girsanov’s theorem, the process

$\beta^\lambda_t=\beta_t +\lambda \int_0^t r_s ds$

is a Brownian motion under the probability $\mathbb{P}^\lambda$ . Thus

$dr_t =\left( \frac{1}{2 r_t} -\lambda r_t \right) dt+d\beta^\lambda_t.$

In law, this is the stochastic differential equation solved by $\| Y_t \|$ where

$dY_t =-\lambda Y_t dt +dB^\lambda_t, \quad Y_0=0.$

We deduce that $r_t$ is distributed as $\| Y_t \|$ , the norm of a two-dimensional Ornstein Uhlenbeck process with parameter $-\lambda$ . Since $Y_t$ is a Gaussian random variable with mean 0 and variance $\frac{1-e^{-2\lambda t}}{2\lambda} \mathbf{Id}$ , the conclusion follows from standard computations about the Gaussian distribution. $\square$

This formula is due to Paul Levy who originally used a series expansion of the Brownian motion. The proof we present here is due to Marc Yor.

The Levy’s area formula has several interesting consequences. First, when $x=0$ , we deduce that

$\mathbb{E}\left( e^{i\lambda S_t} | B_t=0\right)=\frac{\lambda t}{\sinh \lambda t}.$

This gives a formula for the characteristic function of the algebraic stochastic area within the Brownian loop with length $t$ . Inverting this Fourier transform yields

$\mathbb{P} \left( S_t \in ds | B_t=0 \right)=\frac{\pi}{2t} \frac{1}{\cosh^2 \left( \frac{\pi s}{t}\right)} ds.$

Next, integrating the Levy’s area formula with respect to the distribution of $B_t$ yields the characteristic function of $S_t$ :

$\mathbb{E}\left( e^{i\lambda S_t} \right)=\frac{1}{\cosh (\lambda t)}$

Inverting this Fourier transform yields

$\mathbb{P} \left( S_t \in ds \right)=\frac{\pi}{t} \frac{1}{\cosh \left( \frac{\pi s}{2t}\right)} ds.$

One may deduce from it the following formula (due to Biane-Yor): For $\alpha>0$ ,

$\mathbb{E} (|S_t|^\alpha)=\frac{2^{\alpha+2} \Gamma(1+\alpha)}{\pi^{1+\alpha}} L(1+\alpha) t^\alpha,$

where $L(s)=\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^s}$ is the Dirichlet function. This provides an unexpected and fascinating connection with the Riemann zeta function.

2 Responses to Lecture 1. The Paul Levy’s stochastic area formula

Pierre Bizeul says:

January 12, 2022 at 9:51 am

The fourier transform for S_t is easily inverted as 1/ch is a fixed point of the fourier transform

- Fabrice Baudoin says:
  
  January 12, 2022 at 1:38 pm
  
  I edited the post