I saw today that $(B_t^1,B_t^2)$ is a brownian motion on $mathbb R^2$ iff $B_t^i$ a independent Brownian motion on $mathbb R$. So, if for example, $(B_t,B_t)$ won’t be a Brownian motion on $mathbb R^2$ if $(B_t)$ is a Brownian motion on $mathbb R$. This make sense, because it won’t be normal distributed on $mathbb R^2$. Nevertheless, I was wondering : does $(B_t,B_t)$ will be a Brownian motion on ${(x,x)mid xinmathbb R}$ ? Or, does $(cos(B_t),sin(B_t))$ will be a Brownian motion on the circle ?

I think it could be yes for both, however, what it mean to be normal distributed on ${(x,x)mid xinmathbb R}$ or on the circle.

Mathematics Asked on November 21, 2021

1 AnswersWelcome to the world of stochastic differential geometry. You are, of course, spot on that, in order to make sense of a Brownian Motion on some subspace of $mathbb{R}^2$, you need to find some feature of the usual definition (in $mathbb{R}^d$) that you can generalise.

So how do we usually define Brownian Motion? Well, we usually define it as a process with independent, jointly normal increments. This already looks slightly bad, because we would have to make sense of the normal distribution on some arbitrary space (which you would likely end up characterising through Fourier theory, which requires you to be in a group setting). However, it is also just true that you can characterise Brownian Motion as a process $W_t$ with continuous sample paths and independent, stationary increments, that is $W_t-W_s$ has the same law as $W_{t-s}-W_0$ and for any partition $s_0<t_0<...<s_j<t_j<s_{j+1}<t_{j+1}<...<t_n$, we have that $(W_{t_j}-W_{s_j})_{0leq jleq n}$ is independent.

Hence, if you have some notion of subtraction, i.e. you're in the setting of a (topological) group (such as the diagonal under addition or the circle under complex multiplication), you can generalise this to say that a (left-)Brownian Motion is a continuous process $W_t$ such that $W_s^{-1}W_t$ is identical in law to $W_{t-s}$ and for any partition $(W_{s_j}^{-1} W_{t_j})_{0leq jleq n}$ is independent.

It's clear that $(B_t,B_t)$ fits this definition and could therefore be considered Brownian Motion on the diagonal (indeed, for an arbitrary finite dimensional vector space $V$, pick an isomorphism $varphi: mathbb{R}^{dim(V)} to V$ and then if $B_t$ is Brownian Motion in $mathbb{R}^{dim(V)}$, then $varphi(B_t)$ would fit the above definition of Brownian Motion in $V$).

Now what about $(sin(B_t),cos(B_t))=e^{iB_t}$ (it's easier to work in complex form)? Well, $(e^{iB_s})^{-1} e^{i B_t}=e^{i(B_t-B_s)}$ so independence and stationarity for the increments of this process simply follows from those properties of Brownian Motion in $mathbb{R}$.

Indeed, we could play this game over and over for any (topological) group $G$ and some group homomorphism $varphi:mathbb{R}^nto G$. It is perhaps most natural in the setting where $varphi$ is surjective and in some sense minimal.

There is a more general notion still of Brownian Motion on manifolds, but that's rather involved. You can find material on it in Elton Hsu's *Stochastic Differential Geometry*, which has basic stochastic calculus as a prerequisite. One would probably also be best off having read Stroock & Varadhan's *Multi-dimensional Diffusion Processes* first, since Hsu refers to them and mimics their proofs a lot.

Answered by WoolierThanThou on November 21, 2021

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