60 views

### Understanding “we lose degrees of freedom in product deviations”? [duplicate]

I'm having a tricky time understanding the statement: we lose $n−1$ [degrees of freedom] by computing the product deviations [in $cov(x,y)$] I'm not necessarily looking for an overview of ...
56 views

### Sample Covariance [duplicate]

The sample covariance is defined as $\hat{\sigma}_{xy}:=\frac1{n-1} \sum_{i=1}^n (x_i -\bar{x})(y_i-\bar{y})$. What is the intuition for using the correction term $n-1$ instead of $n-2$. Because we ...
36 views

### Degrees of freedom in covariance calculation [duplicate]

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 ...
27 views

### Why does computing product derivations eliminate degrees of freedom? [duplicate]

I'm trying to get an intuition as to why degrees of freedom is n - 1 when calculating sample covariance. This is what I found: 1) Start with 2n degrees of freedom from the bivariate data 2) Lose 2 ...
15 views

### Sample covariance vs population variance of means [duplicate]

Was looking at this link and wondering why for population covariance, the denominator is n, while for sample covariance, denominator is n-1. How does this 1/n(n-1) replace the 1/n^2 in the proof in ...
4k views