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?

1$\begingroup$ 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? $\endgroup$ – Douglas Zare Apr 9 '15 at 16:09

$\begingroup$ What is the meaning of "Area With Multiplicity" in your comment? $\endgroup$ – Mohammad Ghiasi Apr 9 '15 at 16:20

$\begingroup$ 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 figure8 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 $1$? $\endgroup$ – Douglas Zare 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).$$
The twodimensional diffusion constant is defined by the mean square displacement,
$$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).

5$\begingroup$ This is a bit subtle for Brownian motion because the time derivative doesn't exist. Basically, the nowheredifferentiability 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. $\endgroup$ – Nate Eldredge Apr 9 '15 at 21:29

$\begingroup$ thanks for the caveat, I added pertinent references to Lévy and later work. $\endgroup$ – Carlo Beenakker 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 twodimensional Brownian motion. The algebraic area enclosed by $B$ up to time $t$ is given by
$A_t=\frac{1}{2} \int_0^t B_s^1 dB_s^2B_s^2dB_s^1$
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 wellknown 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.

$\begingroup$ 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. $\endgroup$ – Carlo Beenakker Apr 14 '15 at 21:49

$\begingroup$ 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. $\endgroup$ – Fabrice Baudoin Apr 15 '15 at 0:51

$\begingroup$ 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"?) $\endgroup$ – Carlo Beenakker Apr 15 '15 at 0:55

$\begingroup$ 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. $\endgroup$ – Fabrice Baudoin Apr 15 '15 at 1:10

1$\begingroup$ @Captain Darling, since $A_t$ is non negative, $\mathbb{E}(A_t)=0$ would imply $A_t=0$ almost surely, which is not possible. $\endgroup$ – Fabrice Baudoin May 17 '15 at 16:13