Proof Verification?

Question

Let $(B_t)_{t\geq0}$ be a standard Brownian motion. We define the stochastic process $(X_t)_{0\leq t\leq1}$ with $X_t = B_t - tB_1$. To show that the stochastic process $(X_t)_{0\leq t\leq1}$ is a Gaussian process, is it sufficient to state that because $(B_t)_{t\geq0}$ is a Gaussian process (since it is a standard Brownian motion), and $X_t$ is normally distributed (since $B_t$ is normally distributed and a linear combination of the form $aB_t+b$ ($a,b \in \mathbb{R}$) is also normally distributed), $(X_t)_{0\leq t\leq1}$ is a Gaussian process?

Answer 1

The correct argument is as follows: $ \sum\limits_{k=1}^{n} c_kX_{t_k}=\sum\limits_{k=1}^{n} (B_{t_k}-t_kB_1)$ which is a linear combination of $B_{t_1},B_{t_2},...B_{t_n}, B_1$, hence Gaussian. The argument in the second case is similar.