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
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}

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) }.

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


\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)}

Unfortunately, this Fourier transform may not be easily inverted. However, 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.

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