Self similarity of fractional Brownian motion

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?

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.