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I read that for any stationary process the autocovariance function approaches 0 as the lag goes to infinity, but I can't find any proof of this concept. Is this really a general statement not depending on the kind of process? Thanks!

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The autocovariance $\gamma(h):=\Bbb E[X_t X_{t+h}]-[\Bbb E[X_0]]^2$ admits a spectral representation $$ \gamma(h)=\int_{\Bbb R} e^{2\pi ihz}G(dz),\qquad h\in\Bbb R, $$ where $G$ is a positive measure on $\Bbb R$, with total mass $\gamma(0)$. If the measure $G$ admits a density with respect to Lebesgue measure, then $\lim_{|h|\to \infty}\gamma(h) =0$ by the Riemann-Lebesgue lemma. The absolute continuity of $G$ is is known hold if and only if the process $X$ admits a moving-average representation.

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  • $\begingroup$ If $\lim_{|h|\to\infty}\gamma(h)=0$, then $G((-\infty,x])$ is continuous in $x$. $\endgroup$ – JGWang Jun 20 '17 at 3:21
  • $\begingroup$ This too is true. $\endgroup$ – John Dawkins Jun 20 '17 at 19:11
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I don't think this is true in general. If $X_t$ is stationary, so that $\mu=\mu_t=\mathbb{E}[X_t]$, $\sigma^2=\sigma^2_t=\mathbb{V}[X_t]$, and $$ C_{X,X}(t,s)=\mathbb{E}[(X_t-\mu_t)(X_s-\mu_s)] = \mathbb{E}[X_tX_{t+\tau}]-\mu^2 = \gamma_X(\tau) $$ where $\tau=s-t$ is the lag and $\gamma_X(\tau)=\mathbb{E}[X_0X_\tau]-\mu^2$.

I believe you can say $|\gamma_X(r)|\leq \gamma_X(0)$ for $r>0$ using the Cauchy-Schwartz inequality $ |\text{cov}(X,Y)|^2\leq \mathbb{V}[X]\mathbb{V}[Y] $, so that: $$ |\gamma_X(t)|=\sqrt{ |\text{cov}(X_0,X_t)|^2 }\leq \sqrt{ \mathbb{V}[X_0]\mathbb{V}[X_t] }=\sqrt{ \sigma^2\sigma^2 } = \sigma^2 = \gamma_X(0) $$ using the fact that $\gamma_X(0)=\mathbb{E}[X_tX_t]-\mu^2=\mathbb{V}[X_t]$.

Note that, for ARMA series (pg. 84), you do get $\lim_{\tau\rightarrow \infty}\gamma_X(\tau)=0$.

Furthermore, in the case that one has $\lim_{\tau\rightarrow \infty}\gamma_X(\tau)=0$, the standard estimator for the stationary sample mean becomes consistent i.e. $$ \lim_{n\rightarrow\infty}\mathbb{E}\left[\left(\mu - \frac{1}{n} \sum_{i=1}^n X_i\right)^2 \right]\rightarrow 0 $$

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