# Long-run variance of ARMA(p,q)

Assume you have $$A(L)y_t = B(L)e_t$$ and $$e_t$$ is a zero mean white noise with variance $$\sigma^2$$. Why is the long-run variance of $$y_t$$ equal to $$\sigma^2\left(\frac{B(1)}{A(1)}\right)^2$$?

I know that the long-run variance is the infinite sum of all autocovariances of $$y_t$$ and that it can also be written as: $$\gamma(0) + 2\sum_{j=1}^{\infty}\gamma(j)$$ where $$\gamma(0)$$ is variance of $$y_t$$ and $$\gamma(j)$$ is j-th autocovariance $$Cov(y_t,y_{t-j})$$. But I struggle to reach this form $$\sigma^2\left(\frac{B(1)}{A(1)}\right)^2$$.

By definition, the long-run variance is the infinite sum of all autocovariances of $$y_t$$:

$$LRV(y_t) = \sum_{k=-\infty}^{\infty} \gamma_{j}$$

Rewrite $$y_t$$ in its Wold representation: $$y_t = A^{-1}(L)B(L)e_t = \Psi(L) e_t$$,

Then

$$\sum_{k=-\infty}^{\infty} \gamma_{j} = \sum_{k=-\infty}^{\infty} COV(\sum_{i=0}^{\infty}\psi_i e_{t-i},\sum_{j=0}^{\infty}\psi_j e_{t+k-j}) =$$

$$\sum_{k=-\infty}^{\infty} \sum_{i=0}^{\infty} \sum_{j=0}^{\infty} \psi_i \psi_j COV(e_{t-i},e_{t+k-j}) =$$

$$\sigma ^2 \sum_{k=-\infty}^{\infty} \sum_{i=0}^{\infty} \sum_{j=0}^{\infty} \psi_i \psi_j I(t-i = t + k - j) =$$

$$\sigma ^2 \sum_{k=-\infty}^{\infty} \sum_{i=0}^{\infty} \psi_i \psi_{k+i} =$$

$$\sigma ^2 \sum_{i=0}^{\infty} \sum_{k=-i}^{\infty} \psi_i \psi_{k+i} =$$

$$\sigma^2 (\sum_{i=0}^{\infty} \psi_i)^2 =$$

$$\sigma^2 \psi(1)^2 =$$

$$\sigma^2 \frac{1+b_1 + ... + b_q}{1 - a_1 - ... - a_p} =$$

$$\sigma^2 (\frac{B(1)}{A(1)})^2$$

• Thank you for your answer, it is obviously helpful for the LRV of general linear processes, too. Could you elaborate on what you did in step 4 to 5? To be more precise, why does the starting index of the sum change when interchanging the sums? Sep 23, 2019 at 10:35
• If we take the unconditional variance for $AR(1)$ then it is given by $\gamma(0) = \sigma^2 / (1 - a_1^2)$ which doesn't seem to agree with this formula. Is this LVR the same as the unconditional variance of a stationary process or is it something else? Thank you Nov 15, 2020 at 15:03
• Indeed. What are $b_1,\dots,b_q$ and $a_1,\dots,a_p$? Feb 1, 2021 at 15:02