Consider a 2-dimensional Wiener process $(W_t)_{t \in [0,1]}$. Color every area which is enclosed by the line parametrised by $W_t$ (this means that, when the Wiener process makes a loop and intersects itself you color the points of the plane inside the loop). What is the expectation value of the area colored in that way?

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    $\begingroup$ Have you considered running a simulation of this to observe the empirical result? $\endgroup$
    – Danra
    Dec 15, 2012 at 21:01

2 Answers 2


@Danra suggested running a simulation - so that's what I did.

  • Simulation of the Brownian motion: Since a 2-dimensional Brownian motions consists of two independent 1-dimensional Brownian motions, it suffices to simulate paths of a 1-dim. Brownian motion. To this end, I implemented the following algorithm (in R):

    (Lévy-Ciesielski) Let $J \geq 1$ the order of refinement. Initialize $b_0 := 0$. Generate $b_1 \stackrel{s}{\sim} N(0,1)$.
    $\quad$ For $j=0$ to $J-1$:
    $\quad \quad$For $\ell=0$ to $2^j-1:$
    $\quad \quad \quad$ Generate $y \stackrel{s}{\sim} N(0,1)$. Set $$b_{(2\ell+1)/2^{j+1}} = \frac{1}{2} (b_{\ell/2^j} + b_{(\ell+1)/2^j})+2^{- \left( \frac{j}{2}+1 \right)} \cdot y$$

    (René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, p. 320)levy-cieselski

  • Calculation of the enclosed area: First, I calculated the selfintersection-points of the given path. From this, we can determine polygons enclosed by the curve and approximately calculate the area of these polygons. (I did this part in Matlab, the tools "Fast and Robust Curve Intersections" and "Fast points-in-polygon test" were quite helpful.)

Here are some examples:path1path2
The area colored in red is the area detected as enclosed by the curve.

Altogether, I did 2000 simulations and obtained the following histogram and empirical cumulative distribution function:histogram cdf

Somehow, it looks a bit like an exponential distribution, but it doesn't fit properly. The average of the enclosed area is equal to $$0,0026$$

  • $\begingroup$ How many subdivisions are there in $2^{nd}$ graph? Is there a drift term in your random walk? The upper left corner of the $2^{nd}$ curve looks a little bit too straight for a zero drift Brownian motion... $\endgroup$ Mar 3, 2013 at 9:39
  • $\begingroup$ @achillehui I took $J=10$, i.e. there are $2^{10}$ subdivisions (for all simulations). There isn't a drift term. It would look "less straight" if one would choose a larger $J$. Perhaps, I could try this. (I had used the inpolygon-test implemented in Matlab (and this one is really slow if the polygons have a lot of vertices) before I found the faster implementation I mentioned above.) $\endgroup$
    – saz
    Mar 3, 2013 at 9:55

The probability that a point $x$ is enclosed by the Brownian path starting at $0$ at time $1$ depends only on the distance between $x$ and $0$. Call this probability $p(\|x\|)$, then the expected area $A$ enclosed by the Brownian path is $2\pi$ times the integral of $r\mapsto rp(r)$.

The Brownian path starting at distance $r$ from the origin and running during a time interval of length $1$ is the scaled version of the Brownian path starting at distance $1$ from the origin and running during a time interval of length $1/r^2$ hence $p(r)=\mathbb P(T\leqslant1/r^2)$, where $T$ denotes the first time the Brownian path starting from $1$ encloses the origin. Thus, $$ A=2\pi\int_0^{+\infty}\mathbb P(T\leqslant1/r^2)r\mathrm dr=\pi\cdot\mathbb E\left(\frac1T\right). $$ A skew-product representation of the planar Brownian motion indicates that the process of its angular component may be represented as $(B(U_t))_t$ where $(B(t))_t$ is a standard (linear) Brownian motion starting at $0$ and $U_t=\inf\{u\mid\int\limits_0^u\mathrm e^{2\beta}\geqslant t\}$ where $\beta$ is another (linear) Brownian motion starting from $0$ and independent of $B$. Thus, that $T\leqslant t$ implies that $|B|$ reaches $2\pi$ before time $U_t$, that is, that $\int\limits_0^\tau\mathrm e^{2\beta}\leqslant t$ where $\tau$ is the first hitting time of $2\pi$ by $|B|$.

In other words, $T\geqslant\int\limits_0^\tau\mathrm e^{2\beta}$ where $\tau$ is independent of $\beta$ and the distribution of $\tau$ is known. These observations might allow to deduce an upper bound of the value of $A$.

  • $\begingroup$ $T \le t$ is not equivalent to |B| reaching $2*\pi$ before time $U_t$ because the latter doesn't imply that the Brownian motion is self-intersecting. Consider for example a Brownian motion in the form of a spiral which winds several times around the origin without intersecting itself, thus enclosing no area. $\endgroup$
    – Dominik
    Mar 6, 2013 at 10:46
  • $\begingroup$ @Dominik Indeed. Post modified. Sorry about this (stupid) mistake and thanks for the correction. $\endgroup$
    – Did
    Mar 6, 2013 at 10:52

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