$B(t+a)-B(a)$ is a Brownian motion

Question

Let $B(t)_{t\ge 0}$ be a Brownian motion and fix $a>0$.

I would like to prove that $W(t)=B(t+a)-B(a)$ is also a Brownian motion.

So:

$1)$ $W(0)=B(a)-B(a)=0$ a.e.

2) \begin{align*} \DeclareMathOperator{\Cov}{Cov} \Cov(W(s),W(t)) &=\Cov(B(s+a)-B(a),B(t+a)-B(a)) \\&=\Cov(B(s+a),B(t+a)) \\&=\min(s+a,t+a) \\ &=\min(s,t). \end{align*}

$3)$ $t\mapsto W(t)$ is continuous as a composed of continuous functions.

$4)$ I need to check that $W$ is gaussian with mean zero.

To prove that $W$ is gaussian it's equivalent to $(W(t_1),\ldots,W(t_n))$ to be a gaussian vector. I cannot figure out how can I prove that.

I tried with the characteristic function $$\phi_W(u)=\exp({u^tB(a))}\phi_{B(t+a)}(u).$$

Any help ?

Answer 1

Fix $0 \leq t_1 \leq \ldots \leq t_n$, $n \in \mathbb{N}$. The vector

$$\begin{pmatrix} B(a) \\ B(t_1+a) \\ \vdots \\ B(t_n+a) \end{pmatrix}$$

is Gaussian, and so is

$$\begin{pmatrix} W(t_1) \\ \vdots \\ W(t_n) \end{pmatrix} = \begin{pmatrix} -1 & 1 & 0 & 0&0& \ldots & 0& 0 \\ -1 & 0 & 1 & 0& 0&\ldots & 0 &0 \\ -1 & 0 & 0 & 1 & 0 & \ldots & 0 & 0 \\ \vdots & \vdots & & & & & \\ -1 & 0 & 0 & 0 & 0 & \ldots& 0 & 1\end{pmatrix} \cdot \begin{pmatrix} B(a) \\ B(t_1+a) \\ \vdots \\ B(t_n+a) \end{pmatrix}$$

as a linear combination of Gaussian random vector.