When calculating the sample covariance, why do we divide by $n-1$ instead of $n-2$? Don't we lose two degrees of freedom since we need to calculate two sample means? For example, when estimating the variance for a Bayes classifier, we divide by $n-K$ where $K$ is the number of classes since we use $K$ sample means in the calculation.

Could someone please explain this in terms of degrees of freedom?

**UPDATE**

The other answers on this site don't quite make sense to me. So for clarification, I would like to extend this question such that I believe it is sufficiently different to not be marked as a duplicate.

The definition of degrees of freedom as per my understanding is $df = n-p$, where $p$ is the number of parameters calculated en route to the current calculation. I guess the issue boils down to what a "parameter" is considered to be. Clearly in the case of covariance, the individual sample means are not each treated as a parameter, since $df=n-1$. Which leads me to hypthothesize the crude definition that $df=n-p$, where **$p$ is the number of sequentially dependent parameters**.

For example, in the case of covariance, both sample means can be calculated independently of each other. But in calculating residual standard error for simple regression, $df=n-2$ since the intercept term can only be calculated knowing the estimated gradient - or it is sequentially dependent on the estimated gradient.

Is this a valid definition? If so, does that mean that:

$f(X,Y)=\sum_{i=0}^n(X_{i}-\overline{X})^2+\sum_{i=0}^n(Y_{i}-\overline{Y})^2$ has $df=n-1$ also?

If not - with reference to both covariance and the above equation - why so?