xcov sample covariance / dividing by N?
4 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
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
0 commentaires
Réponses (1)
Pratyush Roy
le 29 Déc 2020
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!
0 commentaires
Voir également
Catégories
En savoir plus sur Logical dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!