Why is there no divsion by degrees of freedom in sample correlation? The same seems to hold for the coefficent of the depandt variable using linear regression.

To me it look like some kind of sample variance and sample covariance but not divided by $\frac{1}{n-1}$

Any ideas?


1 Answer 1


If by "$\textit{no divsion by degrees of freedom in sample correlation}$" you mean that there is no $\frac{1}{n-1}$in the expression:

corr$(x, y)$ = $\frac{\displaystyle \sum_{n=1}^n(x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\displaystyle \sum_{n=1}^n(x_i - \bar{x})^2 \displaystyle \sum_{n=1}^n(y_i - \bar{y})^2}}$

Simply consider the case where you actually add it:

corr$(x, y)$ = $\frac{\displaystyle \sum_{n=1}^n\frac{(x_i - \bar{x})(y_i - \bar{y})}{n-1}}{\sqrt{\displaystyle \sum_{n=1}^n \frac{(x_i - \bar{x})^2}{n-1} \displaystyle \sum_{n=1}^n \frac{(y_i - \bar{y})^2}{n-1}}}$


corr$(x, y)$ = $\frac{\displaystyle \frac{1}{n-1}\sum_{n=1}^n(x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\displaystyle \left(\frac{1}{n-1}\right)^2\sum_{n=1}^n (x_i - \bar{x})^2 \displaystyle \sum_{n=1}^n (y_i - \bar{y})^2}}$


corr$(x, y)$ = $\frac{\displaystyle \frac{1}{n-1}\sum_{n=1}^n(x_i - \bar{x})(y_i - \bar{y})}{\displaystyle \frac{1}{n-1}\sqrt{\sum_{n=1}^n (x_i - \bar{x})^2 \displaystyle \sum_{n=1}^n (y_i - \bar{y})^2}}$

Which is equal to the original expression as $\frac{1}{n-1}$ in the numerator and denominator cancels out.

In case you are wondering why no degrees of freedom correction is made in some expressions you might have found on the internet or textbooks, it is probably because those examples are part of Large Sample Theory. As the number of observations goes to infinity, biased estimators are not that important provided that these estimators are consistent. Just consider this:

$\lim_{n \to \infty} \frac{1}{n} = \lim_{n \to \infty} \frac{1}{n-1}$.

Note that some assumptions like random sampling or covariance stationarity might be needed.


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