Why shocks are independent with weighted sum of normal process

Question

I am doing a problem and got stuck by the definition of "normal process". The problem is stated as follows:

Suppose $e_t = \sum_{j}^{\infty}\theta^j Y_{t - j} $ and assume that $Y_t$ is a normal process. Show that $e_t$ are normal and independent.

For the first part of proving $e_t$ being normal, it's kinda easy. We can show $e_t$'s MGF converges to a Gaussian's MGF.

For the independence part, we can simply prove $e_t$ are uncorrelated. But what's the definition of "normal process" here? Thanks for any guidance.

Answer 1

Probably "normal model" means a process in which every (finite) linear combination of $(Y_t)_{t\,\in\, \text{index set}}$ with constant (i.e. non-random) coefficients has a Gaussian distribution. The main point of the exercise is probably intended to be to prove that $e_t$ are independent.

Suppose $Y_1\sim\operatorname N(0,1)$ and $Y_2 = X Y_1$ where $X=\pm1$ each with probability $1/2.$ Then $Y_1,Y_2$ are each normally distributed and uncorrelated by NOT independent and $Y_1+Y_2$ is not normally distributed. And let $Y_3= \begin{cases} Y_1 & \text{if } |Y|>c, \\ Y_2 & \text{if } |Y|<c. \end{cases} \qquad$ Then for a certain value of $c>0,$ $Y_1,Y_3$ are normally distributed and uncorrelated but NOT independent and $Y_1+Y_3$ is not normally distributed. The point of these examples is that separately normal does not imply jointly normal, i.e. does not imply that this is a normal process. "Normal model" is the same as "jointly normal", i.e. every constant linear combination is normally distributed. When random variables are jointly normally distributed, then uncorrelatedness is as strong as mutual (not just pairwise) independence.