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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.

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$\newcommand{\e}{\varepsilon}$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 $X$.

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  • $\begingroup$ I never thought of invertibility in that way. nice intuition and thanks for enlightenment. $\endgroup$ – mlofton May 19 at 16:54
  • $\begingroup$ @mlofton thanks! $\endgroup$ – jld May 19 at 19:12
  • $\begingroup$ Thanks for this answer! It has very interesting intuition, but I was wondering if it is possible for you to represent this in non-matrix form, since I am trying to format a solution without any reference to linear algebra! $\endgroup$ – Semmah May 19 at 20:22
  • $\begingroup$ @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 $\endgroup$ – jld May 19 at 21:41

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