I have a basic question about lag operators, but can't seem to find the answer online anywhere. If I have an equation like so:

$ (1-L)P = XY(1-L) $

can I divide out the (1-L) from each side and just be left with

$ P=XY $

It seems like I should be able to, but then again it also feels weird treating an operator like any other variable.


First of all it is important to notice that $L$ is an operator that works on the following random variable. Then a short answer is No you cannot. Let me give you an example,

Assume you have a time series of the form of $y_t- y_{t-1}=e_t - e_{t-1}$. then using backward shift operator we have, $(1-L)y_t=(1-L)e_t$. Itis obvious that we cannot cancel $1-L$ from right and left side of the equation (and get $y_t=e_t$.

No we consider the case where the left hand side is a stationary process. Let the process $(1-\phi L)y_t=(1-\phi L)e_t$ where $|\phi|>1$. Then using power series we have, $y_t=\sum_0^\infty \phi^i L^i (1-\phi L)e_t$ that is $y_t=\sum_0^\infty \phi^i e_{t-i}- \sum_0^\infty \phi^{i+1} e_{t-i-1}$.

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    $\begingroup$ $y_t=\sum_0^\infty \phi^i e_{t-i}- \sum_0^\infty \phi^{i+1} e_{t-i-1} = e_t + \sum_1^\infty \phi^i e_{t-i}- \sum_0^\infty \phi^{i+1} e_{t-i-1} = e_t + \sum_1^\infty \phi^i e_{t-i} - \sum_1^\infty \phi^{i} e_{t-i} = e_t$ $\endgroup$ – The Laconic Feb 17 '18 at 2:44

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