Let $\{x_t\}_{t\in\mathbb{N}}$ be a zero mean strictly stationary sequence of random variables and $c:\mathbb{N}\to\mathbb{R}$ the (auto)covariance function. If the process follows the AR(1) model $$x_t=\rho x_{t-1}+e_t$$ with $e_t\sim N(0,\sigma^2_e)$ i.i.d. , then I read that the following holds: $$c(0)+2\sum_{r=1}^\infty c(r)=\sigma^2\frac{1+\rho}{1-\rho}$$ My question

In my case, I'm dealing with the expression $c(0)+2\sum_{r=1}^\infty c(kr)$ where $k>1$ is a strictly positive natural number. Does it change the value?

My solution (Update)

\begin{align} \Big[c(0)+2\sum_{r=1}^\infty c(kr)\Big]=\sigma^2\Big(1+2\sum_{r\geq 1}\rho^{k r} \Big )=\sigma^2\frac{1+\rho^k}{1-\rho^k} \end{align} given that $0<\lvert \rho \rvert<1$

  • $\begingroup$ Yes, that is correct. In time series language, you're looking at the ACF of a seasonal AR(1). The quantity you're computing is the long-run variance of a seasonal AR(1), with seasonal lag k. $\endgroup$
    – Michael
    Commented Jun 10, 2020 at 20:59

1 Answer 1


"Then the sequence of partial sums $\{\sum_{r=1}^{a_n-1} c(kr)\}_{n\in\mathbb{N}}$ is a subsequence of $\{\sum_{r=1}^{n-1} c(r)\}_{n\in\mathbb{N}}$"---this statement is not correct.

Your final result is correct. A stationary AR(1) process has autocovariance function $\gamma(r) = \rho^r$ (using more standard notation $\gamma$ instead of $c$) When you $k$-downsampe an AR(1) process (keeping elements at multiples of $k$), the resulting process has autocovariance function $\tilde{\gamma}(r) =\gamma(kr)$. What you computed is the long run variance for the downsampled process---sum of ACF elements $\sum_r \tilde{\gamma}(r)$. Equivalently, it is the Fourier transform of $\tilde{\gamma}$ evaluated at zero frequency.

("Limiting variance" is a bit ambiguous. There is the variance of the stationary distribution $\frac{\sigma^2}{1-\rho^2}$ and there is the long-run variance.)

  • $\begingroup$ Thanks. From this result, for the MA(1) model $x_t=e_t+be_{t-1}$, we can conclude that the long-run variance is simply $\sigma^2(1+2b)$ with $\sigma^2=\sigma^2_e (1+b^2)$. This should hold since the autocovariance for lags greater than one is zero. Right? $\endgroup$ Commented Jun 12, 2020 at 21:06
  • $\begingroup$ Sorry, it's just $\sigma^2$, by hypothesis ($k>1$). $\endgroup$ Commented Jun 12, 2020 at 21:11

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