I was wondering whether the following sum of a Brownian motion $\{W_t, t\geq 0\}$ is also a Gaussian Process: $$ X_t:= W_t +W_{t/2} $$ My attempt:
Choose $t_1,\ldots,t_n$ and $v_1,\ldots,v_n$ randomly, if we can show $v_1X_{t_1}+\ldots+v_nX_{t_n}$ is normally distributed, we are done and know that $X_t$ is a Gaussian Process.
To show this, we want to rewrite $v_1X_{t_1}+\ldots+v_nX_{t_n}$ as some sum of independent increments of the Brownian Motion $W_t$, since we know that the sum of independent normals is once again normal.
Note that the set of all points for which we evaluate $W_t$ will be: $\{t_{1/2},t_1,t_{3/2},\ldots,t_{\lfloor{n/2}\rfloor},t_{n/2},t_{\lfloor{n/2}\rfloor+1},\ldots,t_{n-1},t_n\}$. Thus you can just rewrite the sum $v_1X_{t_1}+\ldots+v_nX_{t_n}$ as the sum of independent normals and therefore it is distributed normally and $X_t$ is a Gaussian process.
Thoughts
Is my approach correct? I feel as if I am missing something and my result is not correct. Is there a better way of approaching questions with regard to whether some transformation of Brownian Motions is a Gaussian Process?
If you know that Brownian motion $(W_t)_{t \geq 0}$ is a Gaussian process, i.e. that $\sum_{j=1}^m c_j W_{s_j}$ is Gaussian for any $c_j \in \mathbb{R}$ and $s_1,\ldots,s_m>0$, then there is an easier way to prove the assertion:
Take $0<t_1 < \ldots<t_n$ and $d_1,\ldots,d_n \in \mathbb{R}$, then
$$\sum_{j=1}^n d_j X_{t_j} = \sum_{j=1}^n d_j W_{t_j} + \sum_{j=1}^n d_j W_{t_j/2}. \tag{1}$$
By renaming/-labeling, we can write the sum on the right-hand side in the form $\sum_{j=1}^{2n} c_j W_{s_j}$ for suitable coefficients $c_j$ and times $s_j>0$. Since $(W_t)_{t \geq 0}$ is a Gaussian process, it thus follows that the right-hand side (hence, also the left-hand side) of $(1)$ is Gaussian.