Variance of AR(1) process using lag operator

Question

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.

Answer 1

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}. $