6
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $y_t=\sum_0^\infty \phi^i e_{t-i}- \sum_0^\infty \phi^{i+1} e_{t-i-1}$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.