I'm trying to understand the root cancellation (or sometimes pole zero cancellation of an ARMA(1,1) model as an example). I know, root cancellation occurs if the AR parameter is equal to the MA parameter multiplied by $-1$. I also understand that this is explained with the help of the lag operator.

If $a$ denotes the AR parameter and $b$ denotes the MA parameter, we can write

$$y_t = ay_{t-1} + be_{t-1} + e_t$$


$$y_t = \frac{1-aL}{1+bL}e_t.$$

If $a = -b$, we get

$$y_t = e_t.$$

What I do not understand is why the lag operator L is the same for the AR part and the MA part. $L$ is defined as

$$Ly_t = y_{t-1}.$$

Does this imply

$$Le_t = e_{t-1}$$



This is simply what the lag operator does - it operates on the time series by shifting the index one period back - it is defined as $Lx_t\equiv x_{t-1}$. The lag operator behaves very much like a regular operator. For instance, $$L^kx_t=L^{k-1}Lx_{t}=L^{k-1}x_{t-1}=x_{t-k}$$ More generally, $$ (aL^k+bL^m)x_t=ax_{t-k}+bx_{t-m} $$ You will thus get different results if the lag operators are additionally multiplied by different coefficients.

Zero or negative powers are also legitimate, $L^0x_t=x_t$ and $L^{-j}x_t=x_{t+j}$. Some polynomials in $L$ carry a special name. The difference operator $\Delta\equiv1-L$ is a prominent example.

Powers of functions of $L$ work perfectly analogously. E.g., the double difference operator $\Delta^2=\Delta\Delta=(1-L)^2=1-2L+L^2$.


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.