How fast would the observed autocorrelation in a sample converge to the true autocorrelation (say, an AR(x) process)? Are there any results like the Central Limit Theorem - which says that the sample mean of a random sample converges to the true mean at a rate of $\frac{1}{\sqrt{n}}$ - or the Berry-Esseen theorem that would apply to the observed autocorrelation?

This question sprung from a minor point around this answer (rate of convergence of the observed autocorrelation vs rate of convergence of the confidence band in an ACF - we do know the latter) and I thought it was sufficiently general to merit its own question.

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    $\begingroup$ The Berry-Esseen theorem says much more than just asserting convergence at rate $1/\sqrt{n}$, this is the job of the Central Limit Theorem. So it is not clear whether you are asking for a CLT for sample auto-correlations, or for the more specific results obtained in B-E theorem. $\endgroup$ – Alecos Papadopoulos Nov 20 '17 at 13:08
  • $\begingroup$ @AlecosPapadopoulos Thanks for correcting me. I am asking about a CLT for sample autocorrelation (if there is one). $\endgroup$ – Candamir Nov 20 '17 at 14:18
  • $\begingroup$ @AlecosPapadopoulos I have now amended the question accordingly, thanks again. $\endgroup$ – Candamir Nov 20 '17 at 16:51

A theorem is $6.7$. in Hall and Heyde (1980) p. 188. For a stationary process, we have

$$n^{1/2}[\hat \rho(j) - \rho(j)] \to_d N(0,v)$$

for autocorrelation of $j$-distance, and in fact, all the autocorrelations converge jointly to a multivariate normal. The authors mention that "results are available also outside the context of stationarity".

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  • $\begingroup$ Thank you! Does this imply a convergence rate of $n^{-1/2}$? Apologies if this is obvious to you. $\endgroup$ – Candamir Nov 20 '17 at 19:43
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    $\begingroup$ @Candamir In the sense that $[\hat \rho(j) - \rho(j)] = O_p(n^{-1/2})$, yes. $\endgroup$ – Alecos Papadopoulos Nov 20 '17 at 20:26
  • $\begingroup$ See also stats.stackexchange.com/questions/207264/… $\endgroup$ – Christoph Hanck Sep 17 '18 at 11:50

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