Average Value of Area Closed by Brownian Motion

Question

Two dimensional brownian motion will intersect its own path infinitly many times. What is the average value of area, closed by curve during an intersection in brownian motion?

Answer 1

One way to define the "enclosed area" for a curve $\mathbf{r}(t)$ in the $x$-$y$ plane of duration $T$, with $\mathbf{r}(0)=\mathbf{r}(T)$, is via the socalled algebraic area $A=\tfrac{1}{2}\int_0^T (\mathbf{r}\times\dot{\mathbf{r}})\cdot\hat{z}\,dt$. (The dot indicates the time derivative and $\hat{z}$ is a unit vector perpendicular to the plane.) A loop traversed in counterclockwise direction contributes a positive amount to $A$, clockwise propagation contributes a negative amount. The probability distribution of $A$ for Brownian motion is known [1,2,3,4],

$$P(A)=\frac{\pi}{4DT}\cosh^{-2}\left(\frac{\pi A}{2DT}\right).$$

The two-dimensional diffusion constant is defined by the mean square displacement,

$$E(|\mathbf{r}(T)-\mathbf{r}(0)|^2)=4DT.$$

The average of $A$ itself is zero (because positive and negative contributions cancel), the average of $|A|$ is

$$E(|A|)= \frac{2DT\log 2}{\pi}.$$

[1] P. Lévy, Le mouvement Brownien plan (1940), and later references here.

[2] D.C. Khandekar and F.W. Wiegel, Distribution of the area enclosed by a plane random walk (1988).

[3] B. Duplantier, Areas of planar Brownian curves (1989).

[4] A. Comtet, J. Desbois, and S. Ouvry, Winding of planar Brownian curves (1990).

Answer 2

There is a beautiful connection between the area swept out by Brownian motion and the Dirichlet function:

$L(s)=\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^s}$.

Proposition: Let $A_t$ be the area swept out by the Brownian motion up to time $t$. Then for every $\alpha \ge 0$ $$\mathbb{E}(|A_t|^\alpha)=\frac{4 \Gamma(1+\alpha)}{\pi^{1+\alpha}} L(1+\alpha)t^\alpha$$

Proof: Let $B_t=(B^1_t,B^2_t)$ be a two-dimensional Brownian motion. The algebraic area enclosed by $B$ up to time $t$ is given by

$A_t=\frac{1}{2} \int_0^t B_s^1 dB_s^2-B_s^2dB_s^1$

where the integral is understood as a Ito integral. It is an easy exercise to check that

$A_t=\frac{1}{2} \beta_{\int_0^t \rho_s^2 ds}$

where $\rho_t=\| B_t \|$ and $\beta$ is a Brownian motion independent from $\rho$.

Thus we have,

$\mathbb{E}(| A_t |^\alpha)=\frac{1}{2^\alpha} \mathbb{E}\left(\left| \beta_{\int_0^t \rho_s^2 ds} \right|^\alpha \right)$

Using the Brownian scaling property and the independence, we obtain

$\mathbb{E}(| A_t |^\alpha)=\frac{1}{2^\alpha} \mathbb{E}\left(| \beta_{1} |^\alpha\right) \mathbb{E}\left(\left( \int_0^t \rho_s^2 ds\right)^{\alpha/2}\right)=\frac{t^\alpha}{2^\alpha} \mathbb{E}\left(| \beta_{1} |^\alpha\right) \mathbb{E}\left(\left( \int_0^1 \rho_s^2 ds\right)^{\alpha/2}\right)$

Now, it is well-known that for every $\lambda \ge 0$

$\mathbb{E}\left(e^{-\frac{\lambda^2}{2} \int_0^1 \rho_s^2 ds}\right)=\frac{1}{\cosh \lambda} .$

We can deduce from this Laplace transform the following Mellin transform formula

$\mathbb{E}\left(\left( \int_0^1 \rho_u^2 du\right)^{s}\right)=\Gamma(1+s)2^{1+s} \left(\frac{2}{\pi} \right)^{2s+1}L(1+2s)$

The result follows then easily.