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.
 A: 
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.
