Invertibility of a Moving Average Process (MA) Let us consider a moving average process given by : $$X_t = \epsilon_t + \epsilon_{t-1}$$
Is it possible to show that this process is not invertible by expressing $\epsilon_t$ in terms of $X_t
, X_{t−1}, X_{t−2}, \dots$ ? If so, how can this be shown? Any solutions would be vastly appreciated.
 A: $\newcommand{\e}{\varepsilon}$Let's suppose we have $X=(X_1,X_2,X_3)$ so
$$
A\e = \begin{bmatrix}1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}
\begin{bmatrix}\e_0 \\ \e_1 \\ \e_2 \\ \e_3\end{bmatrix} = 
\begin{bmatrix}X_1 \\ X_2 \\ X_3\end{bmatrix} = X.
$$
The system $A\e = X$ is underdetermined since $A$ is rank 3 and not square with rank 4, so there is no unique inverse. By rank-nullity $A$ has a one dimension null space spanned by $v = (1, -1, 1, -1)^T$ so, given a realization of $X$, we can perturb $\e$ along this subspace and still arrive at the same realization.
If the white noise process is indexed by $\mathbb Z$, so it doesn't just start at $0$, nothing changes: we'll have
$$
\begin{bmatrix}X_t \\ X_{t-1} \\ X_{t-2} \\ \vdots \end{bmatrix} = \begin{bmatrix}1 & 1 & 0 & 0 & 0 &\dots \\ 0 & 1 & 1 & 0 & 0 &\dots \\  0 &0&1&1&0&\dots \\ \vdots & &&\ddots&&\dots\end{bmatrix}\begin{bmatrix}\e_t \\ \e_{t-1} \\ \e_{t-2} \\ \vdots\end{bmatrix}
$$
and we can still perturb $\e$ in the same way. For example, suppose we create a new process $\tilde \e$ with $\tilde\e_t = \e_t + (-1)^t$. Then the $\pm 1$s cancel and we get the same realization of $X$.
