Alternative formula for sample covariance

Question

Is this an equivalent formula for the sample covariance?

$$\frac{1}{n-1}\left(\sum_{i=1}^nx_iy_i -n\overline{x}\overline{y}\right)$$

Thanks

Answer 1

It is an equivalent formula to $\frac{1}{n-1}\sum_{i=1}^{n} (x_i-\overline x)(y_i-\overline y)$

Firstly you can multiply out the brackets.

$\sum_{i=1}^{n} (x_i-\overline x)(y_i-\overline y)=\sum_{i=1}^{n}x_iy_i-\sum_{i=1}^{n}x_i\overline y-\sum_{i=1}^{n}\overline x y_i+ \overline x \ \overline y\sum_{i=1}^{n} 1$

$\overline x$ and $\overline y$ are both constants. Thus they can be put in front of the sigma signs.

$=\sum_{i=1}^{n}x_iy_i-\overline y\overline x-\overline x\sum_{i=1}^{n} y_i+\overline x \ \overline y\sum_{i=1}^{n} 1$

• $\overline x=\frac{1}{n}\sum_{i=1}^n x_i\Rightarrow n\cdot \overline x=\sum_{i=1}^n x_i$.

• Similar for $y_i$

$=\sum_{i=1}^{n}x_iy_i-n\cdot \overline y \ \overline x-n \cdot \overline x \ \overline y+\overline x \ \overline y\sum_{i=1}^{n} 1$

• $\sum_{i=1}^n 1=n$

$=\sum_{i=1}^{n}x_iy_i-n\cdot \overline y \ \overline x\underbrace{-n \cdot \overline x \ \overline y+n\cdot \overline x \ \overline y}_{=0}$

Finally we get

$\sum_{i=1}^{n} (x_i-\overline x)(y_i-\overline y)=\sum_{i=1}^{n}x_iy_i-n\cdot \overline y \ \overline x$