# Variance of AR(1) process using lag operator

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.

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

Assuming $|\phi_1|<1$, you have
$Y_t = (1-\phi_1 L)^{-1} e_t = \sum_{i=0}^{\infty} (\phi_1L)^i e_t = \sum_{i=0}^{\infty} \phi_1^i e_{t-i}.$
$Var(Y_t) = \sum_{i=0}^{\infty} \phi_1^{2i} \sigma^2 = \frac{\sigma^2}{1-\phi_1^2}.$
• +1, this answer is great, plus it demonstrates the relationship between an AR(1) and an MA($\infty$) process. – Lucas Roberts Mar 31 '18 at 15:56