Self similarity of fractional Brownian motion

Question

The fractional Brownian motion with Hurst Paramter $H\in(0,1)$ is a Gaussian Process $\{X(t),t\ge0\}$ with mean $0$ and covariance function
$$\gamma(t,s)=1/2(|t|^{2h}+|s|^{2H}-|t-s|^{2H}).$$

I want to show that the fractional Brownian motion with Hurst parameter $H \in (0,1)$ is self-similar, that means for every fixed $c>0$ the processes $\{X(t),t\ge0\}$ and $\{c^{-H}X(ct),t\ge0\}$ have the same finite dimensional distributions.

I showed that $E[X(t)]=E[c^{-H}X(ct)]$ and $\operatorname{Cov}[X(t),X(s)]=\operatorname{Cov}[x^{-H}X(ct),c^{-H}X(cs)]$.

Does this mean they are equal in distribution? And therefore they have the same finite dimensional distributions which implies self similarity?

Or do I also have to show that $\{c^{-H}X(ct),t\ge0\}$ fulfills the $4$ properties of a Brownian motion? I mean $W(t)=0$, independent and stationary increments, Gaussian with mean $0$ and variance $t$, and continuous paths?

Answer 1

Recall that Gaussian distributions are uniquely characterized by their mean and covariance:

If $U:=(U_1,\ldots,U_n)$ and $V:=(V_1,\ldots,V_n)$ are Gaussian random vectors then $U=V$ in distribution if, and only if,

• $\mathbb{E}(U_j)=\mathbb{E}(V_j)$ for all $j=1,\ldots,n$
• $\text{cov}(U_i,U_j) = \text{cov}(V_i,V_j)$ for all $i,j=1,\ldots,n$.

Now let $0 \leq t_1 < \ldots < t_n$ be arbitrary times and set $$U := (X(t_1),\ldots,X(t_n)) \quad \text{and} \quad V := (c^{-H} X(ct_1),\ldots,c^{-H} X(ct_n))$$

for the fractional Brownian motion $(X(t))_{t \geq 0}$. Since $(X(t))_{t \geq 0}$ and $(c^{-H} X(ct))_{t \geq 0}$ are Gaussian processes, we know that $U$ and $V$ are Gaussian random vectors. Moreover, you have shown that the random vectors have identical mean vector and covariance matrix. Applying the above result we get $U=V$ in distribution. Since $0 \leq t_1 < \ldots < t_n$ are arbitrary, this shows that $(X(t))_{t \geq 0}$ and $(c^{-H} X(ct))_{t \geq 0}$ have the same finite-dimensional distributions.