variance of an autoregressive process

Question

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 $0<\lvert \rho \rvert<1$

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.)