Fractional Brownian motion and Laplacian [closed]

Possible Duplicate:
Fractional Brownian motion and Laplacian

I would like to submit to your wisdom the following questions.

Is it possible, mutatis mutandis, repeat the same reasoning for a fractional Brownian motion?

More specifically, A real valued Gaussian process $B^H:={B^H(t)}{t\geq 0}$ on a probability space $(\Omega,\mathscr{F},\mathbb{P})$ is a fractional Brownian motion (fBm) with Hurst parameter $H \in(0,1)$ if for all $s,t\in \mathbb{R}+$ we have

    1)  $B^H(0)=0$,

2) $\mathbb{E}B^H(t)=0$,


3) $\operatorname{Cov}[B^H(t),B^H(s)]=\frac{1}{2} \left(t^{2 H}+s^{2 H}-|t-s|^{2 H}\right)$.

In addition, the It\^o formula for fBm is written as: $$f(B^H(t))= \displaystyle\int_0^t f'\left(B^H(s)\right)\, d B^H(s) + H \displaystyle\int_0^t f''\left(B^H(s)\right) s^{2H-1}\, ds.$$ Taking the expectation in both sides of the above equality, we obtain: $$\mathbb{E}\left[f(B^H(t))\right]= H \displaystyle\int_0^t \mathbb{E}\left[f''\left(B^H(s)\right)\right] s^{2H-1}\, ds.$$ I might continue as; changing the expectation by the conditional expectation $\mathbb{E}_x$ with respect to the event ${X_0=x}$ where $X(t)=B^H(t)+ x$, it follows: $$\mathbb{E}_x\left[f(X^H(t))\right]= H \displaystyle\int_0^t \mathbb{E}_x\left[f''\left(X^H(s)\right)\right] s^{2H-1}\, ds.$$ And if we put $$m(x,t; H)= \mathbb{E}_x\left[f\left(X^H(t)\right)\right].$$ We get: $$\displaystyle\frac{\partial}{\partial t} m(x,t; H)= H \, t^{2H-1}\displaystyle\frac{\partial^2}{\partial x^2 }m(x,t; H)$$ Knowledge that fBm is not a semimartingale nor a Markov process except for cases $H=\frac{1}{2}$. I have some doubts about the last deduction.

-
Since you seem to have repeated this question, presumably in an effort to fix some of the formatting, I'm voting to close this as a duplicate. – Yemon Choi Jul 17 2011 at 23:42