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). $
I would be grateful if anyone could illuminate me about this particular stochastic process. Thank you.
Maurice
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, $dX(t) = -Y(t) dW(t) - \frac12 X(t) dt$, $dY(t) = X(t) dW(t) -\frac12 Y(t)dt$.