# Invertibility after differencing

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?

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.
• 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? – Mobeus Zoom May 31 '20 at 20:20