# Question on Brownian motion on unit circle.

I have been trying to investigate some stochastic processes that I find interesting. I came across the Brownian motion on the unit circle. For some reason I would have expected that it keeps "going around" on the same circle as the starting point. So, I wrote some Mathematica code that allows me to simulate the process. I tried to let the process run for a long time. Indeed, for a while, the process behaved as I expected but, once I let the time go for a very long time, I seem to notice that the process tends to move towards the origin. I simulated it using the following:

\begin{equation*}\begin{split} dX(t) &= - \frac{1}{2}X(t)\, dt - Y(t)\, dW(t)\\ dY(t) &= -\frac{1}{2}Y(t)\, dt + X(t)\, dW(t). \end{split}\end{equation*} with $X(0) = 1$ and $Y(0) = 1.$

I computed the average and, unless my calculations are wrong, the dynamic of the system seems to prove that over a long period of time it should move towards zero. Then, I wonder why do they call it Brownian motion on the unit circle?

Here is what my simulations show. The red dot is the starting point, the blue dot is the last point in my simulation. The first plot shows what happen for $t \in [0, 30];$ the second plot for $t \in [100, 200].$

If my computation is correct the mean for the process is $(1e^{-0.5t}, 1e^{-0.5t})$ and thus the simulations seem to be doing the correct thing to some extent, even if I am suprised how long it takes to perceive a definitive movement towards $(0, 0).$

Maurice

• they will not drift off from the circle – Nebo Alex Jul 18 '16 at 5:26
• That is what I think too. I suppose that in simulating the dynamics of diffusion processes over long periods of time, one has to be careful not to reduce the number of points simulated in each unit period below a certain threshold or the simulation could show the wrong thing. I went back and tried to redo the simulations with more points and indeed this time it did not drift towards the origin..... Thank you for your input. – Maurice Jul 18 '16 at 14:07
• i simulated it on sphere ,it didn't drift from sphere. – Nebo Alex Jul 18 '16 at 14:09
• The phenomenon you observed is genuine but it is due to the discretization steps inherent to every simulation algorithm, not to some theoretical property of the underlying stochastic process. To wit, considering the complex valued process $Z_t=X_t+iY_t$, and assuming to simplify things that you used a Euler scheme to simulate the stochastic differential equations in your post, one iterates the discrete step $$Z_{n+1}^h=Z_n^h+iZ_n^hU_n^h-\tfrac12hZ_n^h$$ where $h$ is small, $(U_n^h)$ is i.i.d. centered normal with variance $h$, and $Z_n^h$ approximates $Z_t$ at time $t=nh$. ... – Did Jul 26 '16 at 12:34
• ... Thus, the distance to the origin is such that $\log|Z_{n+1}^h|=\tfrac12R_n^h+\log|Z_n^h|$ with $$R_n^h=\log|1-\tfrac12h+iU_n^h|^2=\log(1-h+\tfrac14h^2+(U_n^h)^2).$$ Since $U_n^h$ is distributed like $\sqrt{h}U$ wth $U$ standard normal, $$E(R_n^h)=E(\log(1-h(1-U^2)+\tfrac14h^2)).$$ Expanding the logarithm up to the second order in $h$ and using $E(U^2)=1$ and $E(U^4)=3$, one gets $$E(R_n^h)=-\tfrac34h^2+o(h^2).$$ By the law of large numbers applied to the i.i.d. sequence $(R_n^h)_n$, this shows that $$|Z_n^h|=\exp\left(-\tfrac38nh^2+o(n)\right)$$ when $n\to\infty$, ... – Did Jul 26 '16 at 12:34

Since $X(0) = Y(0) = 1$, the process obviously cannot live on a unit circle. But it does live on a circle of radius $\sqrt{2}$.
In fact, it is easy to see that the solution to your system is given by $X(t) = \sqrt{2}\cos (W(t)+\frac\pi4)$, $Y(t) = \sqrt{2}\sin (W(t)+\frac\pi4)$. Indeed, $X(0)=Y(0)=1$ and, by Itô's lemma, $dX(t) = -Y(t) dW(t) - \frac12 X(t) dt$, $dY(t) = X(t) dW(t) -\frac12 Y(t)dt$.