ARMA models and invertibility

I'm reading the book Time Series Models by Franses et al. It says that if we have an $ARMA(1,1)$ model with $\phi=1$ and $\theta=-1$ we have $y_t=\epsilon_t$. So, this means that in the equation $(1-L)y_t=(1-L)\epsilon_t$ we can cancel out $(1-L)$ in both sides. I'm wondering if this is a licit operation since having $\theta=-1$ means that the $MA$ model is not invertible.

• if L=1, then we have like this $0*x=0*y$ this does not mean that $x=y$ – dato datuashvili Feb 6 '15 at 5:59
• It is illuminating to write the process as $y_t = y_{t-1} + \epsilon_t - \epsilon_{t-1}$ and do repetitive substitutions of the lagged $y_t$; e.g. $y_{t-1} = y_{t-2} + \epsilon_{t-1} - \epsilon_{t-2}$ and hence $y_t = y_{t-2} + \epsilon_{t-1} - \epsilon_{t-2} + \epsilon_t - \epsilon_{t-1} = y_{t-2} - \epsilon_{t-2} + \epsilon_t$. Then substitute the model for $y_{t-2}$ and so on. You will see that $y_t = y_0 - \epsilon_0 + \epsilon_t$, for some initial values $y_0$ and $\epsilon_0$. – javlacalle Feb 6 '15 at 7:49

I'm wondering if this is a licit operation

The operation is licit. In fact, you can multiply both sides with $(1-L)^{-1},$ which results in

$$y_t=\epsilon_t$$

and then, you can multiply it with $(1-\phi L), \;\phi \in \mathbb{R}$ and get a more general representation of this process:

$$(1-\phi L)y_t=(1-\phi L)\epsilon_t.$$

This process is not identifiable (i.e., a case of parameter indeterminacy) and $y_t$ is in fact a white noise process. In ARMA literature, it is usually ruled out by the assumption that the AR and MA operators have no common factors or equivalently, $y_t$ is not white noise. (Lutkepohl (2005), page 449-450)

To explain another aspect of this question, please note that for a process such as

$$a(z) = 1-z,$$

there is an invertible process, such

$$a^{-1}(z) = \frac{1}{1-z} = 1+z+z^2+z^3+\ldots,$$

with which we have

$$a(z)a^{-1}(z) = 1.$$

However, in time-series analysis, these process are not very attractive. If an increase in the power of $z$ is interpreted as the passing of time, the effect of further values of $z$ does not change (does not decrease) over time.

Hope it helps.