0
$\begingroup$

Consider two independent unidimensional Brownian motion $w_1$ and $w_2$. What is the polar form of $(w_1,w_2)$?

If $r(t)$ and $\phi(t)$ satisfy $(w_1,w_2) = r(t)(\cos(\phi(t)),\sin(\phi(t)))$, how to get an equation for $\phi$?

The equation for $r(t)$ can easily be obtained using Ito’s rule, that is $$ d r(t) = \frac{1}{r(t)} d t + d \tilde w(t). $$ where $\tilde w$ is a Brownian motion.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

You have $d\varphi(t)=\frac{1}{r(t)}d\check{w}_t$, where $\check{w}_t$ is another Brownian motion, independent of $\tilde{w}_t$. This can be calculated, e.g., by applying Ito's formula to $\varphi=\arctan(w_2/w_1).$ Perhaps an even easier way is to start with these $r(t),\varphi(t)$ and check that $r(t)\cos\varphi(t)$ and $r(t)\sin\varphi(t)$ are independent Brownian motions.

An even simpler way is to use the theory of Markov processes, i.e., view this process as a Markov process with generator $\Delta=\partial^2_x+\partial^2_y,$ which in polar coordinates is $\frac{1}{r}\partial^2_\varphi+\frac{1}{r}\partial_r+\partial^2_r.$

Improve this answer
$\endgroup$
2
  • $\begingroup$ Since the codomain of arctan is $(-\frac{\pi}{2},\frac{\pi}{2})$, it seems to me that some information gets lost. I’ll use the generator. Thanks. $\endgroup$
    – happy and healthy
    Commented Dec 17, 2023 at 3:36
  • $\begingroup$ If I am not mistaken, a square is missing in front of the second order partial derivative w.r.t. $\varphi$. $\endgroup$
    – happy and healthy
    Commented Dec 17, 2023 at 7:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .