# Average Value of Area Closed by Brownian Motion

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?

• In what sense do you mean area? If you count the area with multiplicities, the expectation does not exist. See mathoverflow.net/questions/130532/…. Do you count areas where the winding number is 0, but there is no path to infinity disjoint from the Brownian path? Apr 9 '15 at 16:09
• What is the meaning of "Area With Multiplicity" in your comment? Apr 9 '15 at 16:20
• The topological multiplicity is the winding number of the curve around the point. If you take something like $(cost(t),sin(t))$ for $t\in[0,4\pi]$, this winds around the origin twice. If you take a figure-8 knot, $4_1$, or the knots $6_1$ or $7_2$, with their standard minimum crossing diagrams, there is a bounded region about which the knot has $0$ winding number. Would you want to count that area as $0$ or $1$? Apr 9 '15 at 17:03

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).

• This is a bit subtle for Brownian motion because the time derivative doesn't exist. Basically, the nowhere-differentiability of Brownian motion prevents you from sensibly determining whether a loop is being traversed clockwise or counterclockwise, since you don't know which direction it's going at any given moment. The usual solution is to replace the Riemann integral with a stochastic integral, giving you what's known as the Levy area process. I don't have access to the paper you link right now, so I'm not certain if their model is equivalent. Apr 9 '15 at 21:29
• thanks for the caveat, I added pertinent references to Lévy and later work. Apr 10 '15 at 6:29

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.

• hmm, I'm a bit confused: the average of $|A|$ would give $L(2)$ which is the Catalan number, which seems to disagree with the answer I found in the literature. Apr 14 '15 at 21:49
• The formula you give is for the Brownian loop, not for the Brownian motion. The Brownian loop is the Brownian motion conditioned to go back at 0 at time T. Unfortunately, the Fourier transform of the area swept out by the Brownian motion can not be inverted simply and the distribution of A is not known. I found the above formula for the Brownian motion quite cute and it was too long for a comment, so I put it as an answer. Apr 15 '15 at 0:51
• thanks for the clarification; if the path is not closed, wouldn't the definition of "enclosed area" depend on the choice of origin? (I'm thinking of a straight line, what area does it "enclose"?) Apr 15 '15 at 0:55
• Yes indeed, it depends on the starting point. The computation I do is for the Brownian motion B started at 0 and run up to time t. I then look at the algebraic area between the segment [0,B_t] and the curve up to time t. So the formula is $A_t=1/2 \int_0^t B_s \times dB_s$ where the integral is understood as a Ito integral. Apr 15 '15 at 1:10
• @Captain Darling, since $|A_t|$ is non negative, $\mathbb{E}(|A_t|)=0$ would imply $A_t=0$ almost surely, which is not possible. May 17 '15 at 16:13