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Having read this link on math stackexchange, I would like to submit to your wisdom the following questions.

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

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

  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.1.

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1 Answer 1

up vote 3 down vote accepted

Your construction has been carried out for much more general situations. See

Baudoin, F., Coutin, L. Operators associated with a stochastic differential equation driven by fractional Brownian motions, Stoch. Proc. Appl. 117, 5, 550–574, 2007. ArXiv: math/0509511.

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