Is stochastic process $X_t = \varepsilon_t - t\varepsilon_{t-1}$ Gaussian? Consider Gaussian white noise $\varepsilon_t$ with variance $\sigma^2$. Is the following stochastic process $$X_t = \varepsilon_t - t\varepsilon_{t-1}, \hspace{1cm} t \in \mathbb{Z},$$ a Gaussian process?

I can show that this process is not stationary (don't know if it helps anything). As $\varepsilon_t$ is Gaussian I would expect $X_t$ to be gaussian as well. Am I right?
 A: Let $\{t_1, \ldots, t_k\}$ be such that $t_1 < \cdots < t_k$.  There are two cases:

*

*$t_i < t_{i + 1} - 1$ for all $i = 1, \ldots, k - 1$. It then follows that
\begin{align}
\begin{bmatrix}
X_{t_1} \\
X_{t_2} \\
\vdots \\
X_{t_k}
\end{bmatrix} = 
\begin{bmatrix}
-t_1 & 1 &        &        &      &  \\
     &   & -t_2   & 1      &      &  \\
     &   & \cdots & \cdots &      &  \\
     &   &        &        & -t_k & 1
\end{bmatrix}_{k \times 2k}
\begin{bmatrix}
\varepsilon_{t_1 - 1} \\ 
\varepsilon_{t_1} \\
\varepsilon_{t_2 - 1} \\
\varepsilon_{t_2} \\
\vdots \\
\varepsilon_{t_k - 1} \\
\varepsilon_{t_k}
\end{bmatrix}. \tag{1}
\end{align}
By definition of $\{\varepsilon_t\}$, $[
\varepsilon_{t_1 - 1}, 
\varepsilon_{t_1},
\varepsilon_{t_2 - 1},
\varepsilon_{t_2},
\ldots,
\varepsilon_{t_k - 1},
\varepsilon_{t_k}
]' \sim N_{2k}(0, \sigma^2I_{(2k)})$.  Therefore, $(1)$ shows that $[X_{t_1}, \ldots, X_{t_k}]'$ has Gaussian distribution.


*$t_{i_j} = t_{i_j + 1} - 1$ for $j = 1, 2, \ldots, q$, where $1 \leq q \leq k - 1, 1 \leq i_1 < i_2 < \cdots < i_q \leq k - 1$.  In this case, there are $2k - q$ distinctive $\varepsilon_t$s, and the linear transformation $(1)$ can be modified accordingly, say $\mathbf{X} = A\mathbf{\varepsilon}$.  Still, by the same token, $[X_{t_1}, \ldots, X_{t_k}]'$ will have (non-degenerate) Gaussian distribution (since $\operatorname{rank}(A) = k < 2k - q$).
In summary, $\{X_t\}$ is indeed a Gaussian process.
