# Fractional Brownian motion, selfsimilar

Let $0<H<1$. A real-valued Gaussian process $\left(B_H(t)\right)_{t\geq 0}$ is called fractional Brownian motion (fBm) if $\ \mathbb{E}[B_H(t)]=0$ and $$\mathbb{E}[B_H(t)B_H(s)]={\mathbb{E}[B_H(1)^2]\over2}\left(t^{2H}+s^{2H}-|t-s|^{2H}\right).$$

I would like to prove that a fBm is selfsimilar with parameter $H$, and I found something in the literature:

For any $a>0$ we have \begin{eqnarray*} \mathbb{E}[B_H(at)B_H(as)] &=& {\mathbb{E}[B_H(1)^2]\over2}\left((at)^{2H}+(as)^{2H}-(a|t-s|)^{2H}\right)\\ &=& a^{2H}\mathbb{E}[B_H(t)B_H(s)]\\ &=& \mathbb{E}\left[\left(a^HB_H(t)\right)\left(a^HB_H(s)\right)\right] \end{eqnarray*} Since all processes here are mean zero Gaussian, this equality in covariance implies that $B_H(at)\overset{d}{=}a^HB_H(t)$.

I cannot really follow the conclusion of the last sentence. Any help please.

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Gaussian processes are "uniquely" determined by their mean functions and covariance functions. The last sentence is saying that both $B_H(a t)$ and $a^H B_H(t)$ have zero mean functions, and equal covariance functions, hence they are equal in distribution. – Sasha Dec 10 '12 at 19:23
@Sasha Thanks for the explanation. – Ichigo Dec 10 '12 at 20:05