# 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.

$$\newcommand{\e}{\varepsilon}$$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 $$X$$.
• @Semmah you're very welcome! The linear algebra helps with discovering the solution, but now that we know that there's a space of oscillations that all lead to the same $X$ we can just jump to that and describe it element-wise – jld May 19 at 21:41