Proof of Brownian Scaling Relation by Durrett Below is the proof of the Brownian scaling relation given by Durrett's Probability Theory and Examples. However, in the proof, I don't see how we get the equivalence of the distributions simply from independent increments, i.e. if $t_0<t_1<\cdots <t_n$, then $B(t_0),B(t_1)-B(t_0),\dots, B(t_n)-B(t_{n-1})$ are independent, as given by Durrett. I would greatly appreciate it if anyone could help me with this.



 A: It seems that the idea is to prove that both processes are Gaussian with the same mean and covariance. The Brownian motion is Gaussian thanks to,in particular, the independent increment properties. 
Let $t>0$, $X_s=B_{st}$ and $Y_s=\sqrt{t}B_{s}$ $s\geq0$, two processes.
We chose $0<s_1<s_2..<s_n$, therefore we have to prove that $U_n=(X_{s_1},...,X_{s_n})$ and $V_n=(Y_{s_1},...,Y_{s_n})$ are Gaussian.
if we define $t_k=ts_k$ for $k\in\{1,..,n\}$, we have that $U_n=(B_{t_1},...,B_{t_n})$.
let $(\lambda_1,...,\lambda_n) \in \mathbb{R^n}$ , $U_n$ is Gaussian if
 $$Z=\sum_{k=0}^{n}{\lambda_kX_{s_k}}$$ or
$$Z=\sum_{k=0}^{n}{\lambda_kB_{t_k}}$$ is normally distributed.
We can always find ${(\bar{\lambda_1},...,\bar{\lambda_n})\in \mathbb{R^n}}$ such as 
$$Z=\sum_{k=0}^{n}{\bar{\lambda_k}(B_{t_k}-B_{t_{k-1}})}$$
by rearranging the "$\{B_{t_k}\}$".
No doubt that Z is Gaussian as it is a linear combination of a Brownian motion increments, which  are mutually independent and normally distributed
We can conclude that $B$ is a Gaussian process, and finally $X$ as well. Its mean is $(0,...,0)$, and the covariance matrix is $Cov(X_{s_k},B_{s_l})=Cov(B_{t_k},B_{t_l})=min(t_k,t_l)=tmin(s_k,s_l) $for $k\in\{1,..,n\}$ and for $l\in\{1,..,n\}$
The second case,
let $(\lambda_1,...,\lambda_n) \in \mathbb{R^n}$ , $V_n$ is Gaussian if
 $$Z=\sum_{k=0}^{n}{\lambda_kY_{s_k}}$$
or
$$Z=\sum_{k=0}^{n}{\bar{\lambda_k}B_{s_k}}$$  is normally distributed, where $\bar{\lambda_k}=\lambda_k\sqrt{t}$
Z is normally distributed because we prove that $B$ is a Gaussian process. The mean of $V_n$ is obviously $(0,...,0)$, and the covariance matrix $Cov(Y_{s_k},Y_{s_l})=\sqrt{t}\sqrt{t}Cov(B_{s_k},B_{s_l})=tmin(s_k,s_l)$
$U_n$ and $V_n$ are Gaussian with the same mean and same covariance matrix, therefore they have the same distribution, and subsequently $X$ and $Y$ too. 
