How to estimate the auto-correlation function value r(k)=E[X(n)X(n-k)]

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If X(n) is an available time series, I want to estimate the auto-correlation function value of r(k). The definition of r(k) is r(k)=E[X(n)X(n-k)]. Should I use the command "xcorr(x)"? But I will obtain a sequence, how could I get a value? Is this way right? I define y(n)=x(n-k), and use the xcorr(x,y)? But still I need a value and how to do the time delay process y(n)=x(n-k)?

Accepted Answer

Nirmal Gunaseelan
Nirmal Gunaseelan on 12 Jul 2011
XCORR will return the autocorrelation values and the associated lags at which these values were calculated. The examples section has one:
ww = randn(1000,1);
[c_ww,lags] = xcorr(ww,10,'coeff');
stem(lags,c_ww)
What do you mean by just one value representing ACF? The sequence of values represent correlation between values at different lag indices (time) and a single value does not make sense.
  4 Comments
Jiazeng Shan
Jiazeng Shan on 12 Jul 2011
Yes, that is what I need. I want the ACF value when lag are 0, 1 and so on. If the definition r(1)=E[X(n)X(n-1)],should I use xcorr(ww,10,'biased') instead of xcorr(ww,10,'coeff')? How to determine the options? Thanks.
Nirmal Gunaseelan
Nirmal Gunaseelan on 15 Jul 2011
Option depends on what you want - the document clearly gives the various options and default (none) should work for most cases.

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