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Given a non-invertible MA model

$$Y_t = e_t - \theta_1e_{t-1} - \theta_2e_{t-2}$$

where $\theta_1$ and $\theta_2$ are provided (known) parameters, for which values of these parameters can I take a single difference (making this ARIMA) to yield an invertible ARIMA? And how would I do it?

That is to say, how can I show (for given values of $\theta_1$ and $\theta_2$) whether the resulting $W_t$ where $W_t=Y_t-Y_{t-1}$ is invertible?

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An MA(q) process has the characteristic equation: $$1-\theta_1 x-\theta_2 x^2 - \theta_3 x^3 - ... - \theta_q x^q$$ According to Time Series Analysis with Applications in R, the MA process is invertible "if and only if the roots of the MA characteristic equation exceed 1 in modulus."

The first difference of your MA(2) process is: $$ \begin{aligned}\nabla Y_t &=e_t-\theta_1 e_{t-1} -\theta_2 e_{t-2} - (e_{t-1}-\theta_1 e_{t-2} -\theta_2 e_{t-3}) \\ &= e_t - (\theta_1+1)e_{t-1}-(\theta_2-\theta_1)e_{t-2}-(-\theta_2)e_{t-3} \end{aligned} $$

This process is invertible in the case where the roots of the equation $$1-\theta'_1x-\theta'_2x^2-\theta'_3x^3$$ are greater than 1 in modulus and where $$\begin{aligned} \theta_1'&= \theta_1+1\\ \theta_2'&=\theta_2-\theta_1 \\ \theta'_3&=-\theta_2 \end{aligned}$$ A general solutions can be found using the cubic formula.

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    $\begingroup$ Thanks. But isn't it the case, looking at it like this, that (given you can always(?) factorise out $(1-x)$) you should always get a unit root and therefore the process will always be non-invertible? In other words. won't the roots simply be those of $1-\theta_1x-\theta_2x^2$ (from the original $Y_t$) with the additional root $x=1$ that guarantees non-invertibility? $\endgroup$ – Mobeus Zoom May 31 '20 at 20:20
  • $\begingroup$ Good point! That should give you your answer. $\endgroup$ – Ryan Volpi May 31 '20 at 20:34
  • $\begingroup$ It just seems strange that differencing the process would make it non-invertible. I thought the whole point of differencing was to make non-stationary processes stationary (and similar for invertibility)? $\endgroup$ – Mobeus Zoom May 31 '20 at 20:55
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    $\begingroup$ Differencing is typically employed for purposes of stationarity. I don't believe it has thing to do with invertibility. A non invertible MA process can almost always be represented by an equivalent invertible MA process. See stats.stackexchange.com/a/394659/240024 It's therefore not an issue necessarily of the process or the data, but the representation. $\endgroup$ – Ryan Volpi May 31 '20 at 21:12
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    $\begingroup$ Invertibility is more a property of the model, not the time series. Invertible models are preferable for logical and mathematical reasons as mentioned in Forecasting Principles and Practice section 8.3. So, it makes sense to only use invertible models since you lose hardly any degree of generality by restricting yourself to the set of invertible models. Nonstationarity is a property of the timeseries (eg. the variance explodes). You cannot rewrite the model coefficients to prevent the variance from exploding without meaningfully changing the behavior of the model. $\endgroup$ – Ryan Volpi Jun 2 '20 at 22:24

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