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Two options for the sample covariance between X and Y:

1)(with Bessel's) COV(X,Y) = $1/(n-1)$ * $\Sigma$ $(Xi - mean(X))$*$(Yi - mean(Y))$

2)(without) COV(X,Y) = $1/n$ * $\Sigma$ $(Xi - mean(X))$*$(Yi - mean(Y))$

So the denominator is either n or n-1 (Bessel's correction). Same thing happens with the sample variance of X:

1) (with) VAR(X) = $1/(n-1)$ * $\Sigma$ $(Xi - mean(X))^2$

2) (without) VAR(X) = $1/n$ * $\Sigma$ $(Xi - mean(X))^2$

Now, I know that, regarding the sample variance, the first option is the unbiased one, whereas the second one is biased (but perhaps with lowest prediction error?). Moreover, the second one derives from some "Analysis of Variance" results (how exactly?). Does the same reasoning apply to the sample covariance as well?

Thank you in advance!

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As with sample variance, sample covariance using $n-1$ rather than $n$ in the denominator produces an unbiased estimate, presuming the true population means of $X$ and $Y$ are not known. See https://en.wikipedia.org/wiki/Sample_mean_and_covariance#Unbiasedness

Also see Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1?

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