# Fractional Brownian motion as integral, mean zero

Let $H\in$ $]0,1[$. A fractional Brownian motion $\left(B_H(t)\right)_{t\geq 0}$ can be represented as $$X(t)={1\over C(H)}\int_\mathbb{R}\left((t-s)_+^{H-{1\over2}}-(-s)_+^{H-{1\over2}}\right)dB(s)$$ where $B(t)$ is the standard Brownian motion and $$C(H)=\left(\int_0^{\infty}\left((1+s)^{H-{1\over2}}-(-s)^{H-{1\over2}}\right)^2ds+{1\over 2H}\right)^{1\over2}$$

I would like to show that $\mathbb{E}[X(t)]=0$. My professor said that the zero mean can be explained by the symmetry of $X(t)$. What did he mean? How can he tell that $X(t)$ is symmetric? Can I state anything by knowing that $\mathbb{E}[X(t)^2]=t^{2H}$?

Thanks.

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Is there nobody who can help? –  Ichigo Jan 14 '13 at 16:18

If $(B_s)_{s\geqslant0}$ is a standard Brownian motion, so is $(\bar B_s)_{s\geqslant0}$, where $\bar B_s=-B_s$ for every $s$. Hence, if $$X_t=\int\varphi(s,t)\mathrm dB_s,\qquad \bar X_t=\int\varphi(s,t)\mathrm d\bar B_s,$$ for some suitable function $\varphi$, then the distributions of $X_t$ and $\bar X_t$ coincide. Since $\bar X_t=-X_t$, the proof is complete.