Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If I want to simulate Brownian motion in the Euclidean space I can simulate it by a point that is moving a distance $\epsilon$ in an arbitrary direction then it randomly choose a new direction and moves a distance $\epsilon$ again and so on. The smaller the $\epsilon$ the closer the simulation will be to the real Brownian motion.

How can I simulate Brownian motion in the hyperbolic space (Poincare Disk model for instance)? Does the same work here where I replace the Euclidean distance by the hyperbolic distance? My intuition is yes but when I did the simulation the random walk do not seem to be transient but it should be!

share|improve this question
1  
The same method should work. It's a little tricky to find the points which are distance $\epsilon$ away from a given point in the disk model. One way is to translate a circle from the center. –  Douglas Zare Apr 5 '11 at 20:26
1  
Yes, that's correct. And it is transient, but how can you tell this from running a simulation for finite time? Maybe you need to run it longer? –  George Lowther Apr 5 '11 at 22:51
    
I do not have the reputation to comment but shouldn't be the steps be Gaussian random variables of zero mean and variance $\epsilon$ instead of steps of constant length? Does it makes a difference? –  user9126 Apr 5 '11 at 23:18
1  
It makes no difference in the limit as the step size goes to zero (and the number of steps goes to infinity). –  George Lowther Apr 6 '11 at 0:08
    
@user9126: please do not use answers to comment. –  Qiaochu Yuan Apr 6 '11 at 20:35

3 Answers 3

Taking the Poincaré disk model for hyperbolic space as being formed by all complex numbers with modulus smaller than 1, the distance between two points can be calculated as

$$d(z_1,z_2)=\tanh^{-1}\left|\frac{z_1-z_2}{1-z_1\bar{z_2}}\right| \; .$$

The set of points at equal distance $\varepsilon=\tanh(d(z,z_0))$ from a fixed point $z_0$ can be shown to be represented by a circle in the Poincaré disk model with center

$$\frac{(1-\varepsilon^2)z_0}{1-\varepsilon^2|z_0|^2}$$

and radius

$$\frac{\varepsilon(1-|z_0|^2)}{1-\varepsilon^2|z_0|^2} \; .$$

But this is not the end of the story, you can't just pick points uniformly on this circle. What you want is to pick points such that you can arrive from $z_0$ to them such that the direction of departure from $z_0$ was chosen uniformly. This will induce a particular distribution on that circle.

To figure out this distribution, it is necessary to compute the intersection of an h-line* through $z_0$ for an arbitrary direction and the circle we just computed. From that, we can get the sought distribution. I'll update this later if I get to computing that part.

EDIT: Following the suggestion by @Douglas Zare, this can also be achieved by picking a point at a distance $\varepsilon$ from the origin uniformly, where the deformation of the Poincaré disk model won't be a problem since h-lines passing through the origin are all just diameters. All we have to do then is translate the point with the translation that takes the origin to the point $z_0$ which can be done with a Möbius transformation

$$z \mapsto w=\frac{z+z_0}{\bar{z}_0z+1} \; .$$

This is also easily implemented computationally. Here's how I coded it in R:

    # Brownian motion on Poincaré disk (no comments about crappy code plz kthx)

epsilon=0.01;
z<-0;
path<-c(z);
for (t in 1:10000) { 
    jitter=runif(1,0,1);
    dz=epsilon*complex(1,cos(2*pi*jitter),sin(2*pi*jitter));
    z<-(z+dz)/(Conj(dz)*z+1);
    path<-c(path,z);
}

plot(Re(path),Im(path),type="l",col="blue",asp=1,xlim=c(-1,1),ylim=c(-1,1));
curve(sqrt(1-x^2),-1,1,col="red",add=TRUE);
curve(-sqrt(1-x^2),-1,1,col="red",add=TRUE);

And here is a picture of a run of the code Brownian motion simulation on Poincaré Disk

*By h-line, I mean a line in the hyperbolic plane, which will be either a diameter, either a circle segment orthogonal on the boundary of the Poincaré disk.

share|improve this answer
1  
Rather than choosing from the appropriate distribution, wouldn't it be better to choose a point on a circle of distance $\epsilon$ around the origin (which can conveniently be done uniformly) and then 'translate' the chosen point by $z_0$? (As Douglas Zare suggests in his comment) –  Steven Stadnicki Apr 18 '11 at 23:16
    
Sure, but where's the fun in that? Mind you, I'll edit my post to reflect that. –  Raskolnikov Apr 19 '11 at 7:52

IIRC, there a discussion of Brownian motion in Feller (An Introduction To Probability Theory and it's Applications).

share|improve this answer
    
Is it a discussion about Brownian motion in non-Euclidean spaces? –  Raskolnikov Apr 11 '11 at 21:02
    
Euclidean only, I believe, but it's been 30 years since I read it. –  phv3773 Apr 15 '11 at 15:09

I think that the start will be to define what is a Brownian motion on certain subset of $\mathbb{R}^3$ in this case.

The key property of Brownian motion $B_t$ is that it has independent increments, i.e. $B_t-B_s$ is independent of $B_s$. Hence one way of simulating Brownian motion is by summing up independent random variables. For two-dimensional Brownian motion this is achieved by generating $n$ random vectors $X_i$ and then taking

$$\xi_t=\frac{1}{\sqrt{n}}\left(\sum_{i=1}^{[nt]}X_i+(nt-[nt])X_{[nt]+1}\right),$$

where $[\cdot]$ is whole part of a real number. The functional central limit theorem ensures that $\xi_t$ converges to $B_t$ as $T$ goes to infinity, so take large $T$ and you will get a simulation of $B_t$. It is possible also to take only the partial sums $\sum_{i=1}^{[nt]}X_i$, the end result will be the same.

Now to apply this approach to a certain space, you need, as I said before the definition of the Brownian motion is needed in said space and the corresponding functional central limit theorem. My own impression when writing my thesis (which was about multi-parameter Wiener processes) was that functional central limit theorems go hand in hand with the brownian motion, i.e. if you can define Brownian motion, you can prove the FCLT.

Before though delving into literature I would advise to find a way how to generate random sample from the space you are intend to work in. This might be the most difficult part in simulation.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.