Independent increment and gaussian increments process is a gaussian process

Question

Lemma: An independent increments process $(X_t)_{t∈R^+}$ such that $X_0 = 0$ with Gaussian increments (i.e. for all s < t, $X_t − X_s$ has Gaussian law) is a Gaussian process.

Proof:

So what I want to prove is that $(X_{t_1},...,X_{t_n})$ is a multivariate normal. I see that i can rewrite $X_{t_j}=\sum_{k=1}^j(X_{t_k}-X_{t_{k-1}})$ therefore i know that each $X_{t_j}$ is a sum of independent increment and then is Normal.

Now, how can I conclude that $(X_{t_1},...,X_{t_n})$ is multivariate?

Using the definition of multivariate i should show that $$\sum_{i=1}^na_iX_{t_j}=\sum_{i=1}^na_i\sum_{k=1}^j(X_{t_k}-X_{t_{k-1}})$$ is normal. But how could i do that? because in this case $\sum_{i=1}^na_iX_{t_j}$ the $X_{t_j}$ seems to me that is not independent among each element, so I cannot conclude that is a sum of independen normal (and from here conclude that is normal)

Please some help, I thought a lot about this problem.

Answer 1

Without loss of generality assume that $t_1 \leq t_2\leq ...\leq t_n$. If $Y_1,Y_2,...,Y_n$ are independent normal then the vector $TY$ is normal for any linear transformation $T: \mathbb R^{n} \to \mathbb R^{m}$. Take $T(y_1,y_1,...,y_n)=(y_1,y_1+y_2,...,y_1+y_2,+...+y_n)$ and $Y_i=X_{t_i}-X_{t_{i-1}}$ ($t_0=0$).

Answer 2

The $Z_j:= X_{t_j}-X_{t_{j-1}}$'s are all independent where $t_0:=0$, hence jointly normal. Now rearranging the terms in the right hand side will give you $$ \sum_k b_k Z_k$$ with $b_k= a_1 + a_2 + ... + a_k$.

P.s. the j's should be i's in your sums.