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It is well known that the Laplacian is the infinitesimal generator of a Brownian Motion, that is, $$ \lim_{t \to 0} \frac{E[f(x+B_t)-f(x)]}{t}= \Delta f(x). $$

Is it true that for the Fractional Brownian Motion $B_H$ with hurst parameter $H \in (0,1)$, that is, the continuous-time Stochastic Process with stationary gaussian increments, mean $0$ and covariance $$ \mathbb{E}[(B_H(t)-B_H(s))^2]= |t|^{2H} + |s|^{2H} - |t-s|^2H $$ has the infinitesimal generator to be a fractional Laplacian $-(-\Delta)^{\alpha}$for some $\alpha \in (0,\infty)$? If so, what is the relation between $H$ and $\alpha$?

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    $\begingroup$ Actually the fractional Laplacian is for $\alpha \in (0,1)$ the infinitesimal generator of so-called stable Lévy processes. $\endgroup$
    – saz
    Sep 4, 2018 at 18:26
  • $\begingroup$ Which is the case right? As $B_{H}(t) \sim t^{H}B_{H}(1) $. $\endgroup$
    – Kernel
    Sep 4, 2018 at 18:55
  • $\begingroup$ Moreover, the case $\alpha=1$ gives the $2$-stable process (Brownian Motion). Is there a scaling missing? I understand you are talking about the case $\alpha \in (0,1)$, but I’m just impressed with the gap. $\endgroup$
    – Kernel
    Sep 4, 2018 at 18:58
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    $\begingroup$ The fractional Brownian motion is not a Lévy process (unless $H=1/2$). Lévy processes have independent and stationary increments. They can be uniquely characterized by their so-called characteristic exponent. The $\alpha$-stable isotropic Lévy process ($\alpha \in (0,2)$) is uniquely characterized by its characteristic exponent $\psi(\xi) := |\xi|^{\alpha}$, and the associated infinitesimal generator equals, when restricted to the smooth compactly supported functions, the fractional Laplacian $-(-\Delta)^{\alpha/2}$. $\endgroup$
    – saz
    Sep 4, 2018 at 19:03
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    $\begingroup$ The standard reference for Lévy processes is Sato's book ("Lévy processes and infinitely divisible distributions"); you can also take a look at these lecture notes by Schilling, in particular Chapter 6 (and Example 6.5). $\endgroup$
    – saz
    Sep 4, 2018 at 19:12

1 Answer 1

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As saz points out, the fractional Laplacian is not the generator of fractional Brownian motion. In fact, since fractional Brownian motion is not a Markov process when $H \neq 1/2$, it does not even make sense to talk about a generator for it (e.g., the limit of the above difference quotient will simply be zero or $\infty$ for $H \neq 1/2$).

Instead, the fractional Laplacian for $\alpha \in (0,1]$ is the generator for the (radially symmetric) $2\alpha$-stable Levy process, i.e., the Levy process associated to the infinitely divisible probability distribution whose characteristic function is given by $\phi_{2\alpha}(t) = e^{-|t|^{2\alpha}}$. When $\alpha \neq 1$, one should notice that $-(-\Delta)^{\alpha}$ is a non-local differential operator, and this is perfectly embodied by the fact that the $2\alpha$-stable Levy process will have jumps (i.e., it is not a continuous path like Brownian motion when $\alpha \neq 1$).

With that being said, the fractional Laplacian actually does have a different kind of relationship to frational Brownian motion. Specifically, since $(B_H(t))_{t \in [0,1]}$ is a Gaussian variable in $C[0,1]$, it has a Cameron-Martin space which encodes its covariance structure. In the case of fractional Brownian motion with Hurst parameter $H$, the Cameron martin space is the Sobolev space $$W_0^{H+1/2,2}([0,1]) := \{ f \in C[0,1]: f(0)=0, \;\;\langle f, (-\Delta)^{H+1/2}f \rangle_{L^2[0,1]}<+\infty\}.$$ Since Sobolev spaces extend to all real parameters (corresponding to the exponent on the fractional Laplacian), this means that fractional Brownian motion can be naturally generalized to all $H \in \Bbb R$ and can also be generalized to $d$-dimensional indexing space, i.e., $\{B_H(t)\}_{t \in [0,1]^d}$ is the Gaussian random variable on $\mathcal D'([0,1]^d)$ whose Cameron-Martin space is $W^{H+d/2,2}_0([0,1]^d)$.

In the case $H=-d/2$, this is just white noise on $[0,1]^d$. In the case that $H=0$ and $d=2$, this object (called the Gaussian Free Field) is conformally invariant and a Fields medal has even been awarded for a detailed study of its geometry. These fractional fields may be coupled for $H<H'$ by the identity $$B_H = (-\Delta)^{(H’-H)/2}B_{H’}.$$In particular fractional BM may be obtained from ordinary BM just by applying a fractional Laplacian to the latter. Take a look at this nice survey paper for more information (and more general types of boundary conditions) on these fractional Gaussian fields.

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  • $\begingroup$ As a remark, the fractional Laplacian in the expression above needs to be interpreted with Dirichlet boundary condition on $[0,1]^d$. $\endgroup$
    – shalop
    Sep 18, 2018 at 22:30
  • $\begingroup$ sorry for digging up an old answer, but I was wondering about the following: $\endgroup$ Nov 25, 2021 at 13:58
  • $\begingroup$ we know that $dX = V_t(X_t)dt + dB_t$ for $B$ being a Brownian motion, is related to the PDE $\partial_t u + 1/2 \Delta u = f$, and $dX = V_t(X_t) dt + dL_t$, where $L$ is a Levy process with $\partial_t u + 1/2 \Delta^s u = f$. Then, if we take the fBm $B^H$ and a stochastic DE $dX = V_t(X_t) dt + dB^H_t$, does there exist PDE related in the similar way? $\endgroup$ Nov 25, 2021 at 14:00
  • $\begingroup$ I don’t think any such pde exists for H not equal to 1/2. The reason you have pde representation for Brownian motion and levy process is again because they have markov property. Without that you lose a lot. $\endgroup$
    – shalop
    Nov 29, 2021 at 1:55

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