Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The matter is in the title.

Is there a means to define the white noise process in Lie group. A basic definition link text

Question: Can we replace $\mathbb{R}^n$ by a Lie group?

In fact, I would like for studying the stochastic differential equation

\begin{align} X_t^{x} &=x + \sum_{k=1}^m \int_0^t \mathcal{X}_k (X_s^{x}) dB^k_s & ( *) \end{align}

For $H > \frac{1}{2}$,

  1. $\mathcal{X}_k$'s are $C^{\infty}$-bounded vector fields on $\mathbb{R}^m$ which I would like to remplace it by a lie group $G$.

  2. $B$ is a $m$-dimensional fractional Brownian motion taking its values ​​in $G$.

In Hida space $\mathcal{S}^*$, equation (*) becomes $$ d X_t^{x} =\sum_{k=1}^m \mathcal{X}_k (X_t^{x})\diamondsuit B^k_t dt \qquad (**) $$ whith $X_0^x=x$. In order to exploit the underlying Lie algebra should be established white noise on the Lie group.

Thanks having taken bother to read this post.

share|improve this question
See Rogers and Williams, vol. 2 –  Steve Huntsman Dec 21 '11 at 16:37
I'll second Steve Huntsman's comment. But, what is H? –  George Lowther Dec 21 '11 at 17:34
$H\in ]\frac{1}{2}, 1[$ –  Jonas Dec 21 '11 at 20:17
@alabair: ok, so that is the range of values that H can take. But, you never said what H means. Is it the Hurst index of fractional Brownian motion? So the question is a bit more complicated than just regular Brownian motion, and is not covered by Rogers and Williams. If you're asking about fractional Brownian motion on a Lie group, does that mean that you are already familiar with regular BM and stochastic differential equations on the group? Also, why specifically $H > 1/2$? Have I understood you correctly? I think the question should provide more information. –  George Lowther Dec 21 '11 at 22:09
add comment

Your Answer


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

Browse other questions tagged or ask your own question.