Show that Y is a Gaussian process

Question

I have a Moving Average of order $q$. That is to say $MA(q)$.

Answer 1

Possibly the fastest way is to use the direct definition.

Consider the set $$\{Y_{t_1},...,Y_{t_n}\}$$

We need to show that every linear combination of this set is Gaussian random variable. Take now such linear combination and denote the chosen indecies as in set $A$

$$\sum_{k\in A} \lambda_k Y_{t_k}$$

But what do we have here? A sum of sum of Gaussian variables times constants. And we know that is Gaussian by basic theorem. Hence we are done.

EDIT: To use the theorem that sums of Gaussians are Gaussian, we need that the Gaussians are independent. Let's show why that is. Consider $Y_{t_i},Y_{t_j}$ and show that the sum is indeed Gaussian using the moment generating function. Say that $t_i\leq t_j=t_i+n$

$$E(e^{s(Y_{t_i}+Y_{t_i+n})})=E(e^{s(\sum_{k=0}^q\theta_kZ_{t_i-k}+\sum_{k=0}^q\theta_kZ_{t_i+n-k})})$$

$$=E(e^{s(\sum_{k=0}^{q-n-1}Z_{t_i+n+k}(\theta_k+\theta_{k+n})+\sum_{k=0}^{\min(n-1,q)}\theta_{k} Z_{t_i+k}+\theta_{q-n+1+k} Z_{t_i+q+1+k})})$$

$$=\prod_{k=0}^{q-n-1} E(e^{s(Z_{t_i+n+k}(\theta_k+\theta_{k+n}))})\prod_{k=0}^{\min(n-1,q)}E(e^{s(\theta_{k} Z_{t_i+k})}) \prod_{k=0}^{\min(n-1,q)}E(e^{s(\theta_{q-n+1+k} Z_{t_i+q+1+k})})$$

And this is a product of MGTs of Gaussian variables, hence the sum is Gaussian. It remains to note that sums of $Y$ are always in this form.