2
$\begingroup$

Suppose for the AR(1) model, $$Y_t=\phi_1Y_{t-1}+e_t$$

I want to find the variance $Var(Y_t)$ using lag operator: $$Y_t=(1-\phi_1L)^{-1}e_t$$

My way is simply taking the variance, $$Var(Y_t)=(1-\phi_1L)^{-2}\sigma^2=\sigma^2/(1-\phi_1)^2$$

But obviously, it is not the correct answer, which is $\sigma^2/(1-\phi_1^2)$.

I am new to this topic but the above approach seems logical to me. Can anyone point out the mistake of this method? Thank you.

$\endgroup$
  • $\begingroup$ What exactly is $L$? $\endgroup$ – Winkelried Mar 31 '18 at 14:32
  • 1
    $\begingroup$ @Winkelried - $L$ is the "lag operator"; $Lx_t = x_{t-1}$, for example. $\endgroup$ – jbowman Mar 31 '18 at 14:47
4
$\begingroup$

Assuming $|\phi_1|<1$, you have

$ Y_t = (1-\phi_1 L)^{-1} e_t = \sum_{i=0}^{\infty} (\phi_1L)^i e_t = \sum_{i=0}^{\infty} \phi_1^i e_{t-i}. $

Hence,

$ Var(Y_t) = \sum_{i=0}^{\infty} \phi_1^{2i} \sigma^2 = \frac{\sigma^2}{1-\phi_1^2}. $

$\endgroup$
  • 1
    $\begingroup$ +1, this answer is great, plus it demonstrates the relationship between an AR(1) and an MA($\infty$) process. $\endgroup$ – Lucas Roberts Mar 31 '18 at 15:56

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.