xcov sample covariance / dividing by N?
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Dear matlab community,
I looked into using xcov to calculate auto-covariances of a Tx1 time series vector. Using xcov in the first instance, it seemed as if this function would not devide by T as opposed to what the general formula for sample autocovariance would suggest:
?
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/452313/image.png)
Can someone confirm whether I need to devide by T to get a vector of sample auto-covariances? If I do not devide by T, the covariances seem to explode, as more and more terms are added. It is also not clear from the documentation of xcov.
Best
Marcel
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回答 (1 件)
Pratyush Roy
2020 年 12 月 29 日
Hi,
xcov in MATLAB computes raw covariances with no normalization. Hence after obtaining the results, one can divide by T to obtain the results suggested by the formula in the question.
Hope this helps!
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