I'm doing the point b) of exercise number $4$, taken from here .
More precisely,
[...] let $(X_t)_{t \in [0,1]}$ be a Brownian bridge. Let $Z$ be a $N(0,1)$ -distributed random variable. Show that $\tilde{B} = X_t + tZ$ is a Brownian motion for $t \in [0,1]$.
To show it's a Brownian motion, I need to check that the process $\tilde{B}$ is gaussian, with zero mean, and that the $Cov(\tilde{B}_s, \tilde{B}_t) = \min(s,t)$.
My big problem is that I can't even show it's Gaussian and it seems to me that $Z$ should be independent from the process $(X_t)_{t \geq 0}$.
Indeed, to check it's Gaussian, I fix a partition of $[0,1]$, $0 \leq t_1 \leq t_2 \leq \ldots t_n=1$ and consider the random vector $$(X_{t_1} - t_1 Z, \ldots ,X_{t_n} - t_nZ)$$
I can see that it is a linear mapping of the vector $(X_{t_1}, \ldots, X_{t_n},Z)$ but I don't know if this random vector is Gaussian. I just know that the first $n$ components, namely $(X_{t_1}, \ldots, X_{t_n})$ form a Gaussian random vector (because $(X_t)_t$ is a Brownian bridge), but I think that if I add $Z$ to the vector, without independence assumption, it's no more Gaussian.
Am I right or am I missing something?