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Consider an ARMA process expressed in lag operator notation, $$ \Phi(L)x_t=\theta(L)\varepsilon_t. $$ Let $\text{Var}(\varepsilon_t)=\sigma^2_{\varepsilon}$.

Question: Can the long-run variance of $x_t$ be lower than the error variance? If so, could you provide an example?

A related thread with an answer that may help answer my question is "Long-run variance of ARMA(p,q)".
A related new question is "Can unconditional variance of an ARMA process be lower than its error variance?".

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  • $\begingroup$ The second link is the same as the first one $\endgroup$ – gunes Feb 2 at 9:08
  • $\begingroup$ @gunes, good catch! Now fixed. $\endgroup$ – Richard Hardy Feb 2 at 9:09
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Let's denote the long-run variance as $\operatorname{var}_L(y_t)$ to prevent confusion. From the formula provided in the linked post, a proper choice of coefficients can lead to smaller long-run variance. Let our process be defined as $$y_t=e_t-0.6e_{t-1}$$ then, the long-run variance will be $$\operatorname{var}_L(y_t)=\sigma^2 \left(\frac{1-0.6}{1}\right)^2=0.16\sigma^2<\sigma^2$$ Note that the missing definiton of coefficients in that post is ($b_0=1$) $$y_t=\sum_{i=1}^p a_py_{t-p}+\sum_{i=0}^q b_qe_{t-q}$$

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  • $\begingroup$ I believe the formula there is wrong. Your example yields $\text{Var}(x_t)=1.36\sigma^2_\varepsilon>\sigma^2_\varepsilon$. Here is a quick simulation to show that: n=1e6; set.seed(1); e=rnorm(n); x=arima.sim(model=list(ma1=-0.6),n=n,innov=e); var(e); var(x). $\endgroup$ – Richard Hardy Feb 2 at 6:05
  • $\begingroup$ @RichardHardy But, long-term variance is not $var(x_t)=\gamma(0)$, where $\gamma(t)$ is the auto covariance function. We also have $\gamma(1)=\gamma(-1)=-0.6\sigma^2$ and higher order covariances $0$, which yields $$\operatorname{var}_L(y_t)=\sum_{k=-\infty}^\infty \gamma(k) = 0.16\sigma^2$$ $\endgroup$ – gunes Feb 2 at 8:38
  • $\begingroup$ Ahh, I forgot what the term long-run variance actually means. I meant to ask about the unconditional variance of $x$ instead, which I actually did by writing $\text{Var}(x_t)$. I will edit the question to clarify. Or better yet, I will post a new question so that your answer gets the credit it deserves. ... And I did: here is my new question. $\endgroup$ – Richard Hardy Feb 2 at 9:04

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