3
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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
3
$\begingroup$

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".

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.