Question

If $r$ is a coefficient of correlation of two variables $x$ and $y$, then prove that $r=\frac{{\sigma }_x^2+{\sigma }_y^2-{\sigma }_{x-y}^2}{2{\sigma }_x{\sigma }_y}$.
where ${\sigma }_x^2,{\sigma }_y^2$ and ${\sigma }_{x-y}^2$ are the variance of $x,y$ and $x-y$ respectively.

Answer 1

Coefficient of correlation of $x$ and $y$ is $r$ and ${\sigma }_x^2,{\sigma }_y^2$ and ${\sigma }_{x-y}^2$ are varience of x, y and $x-y$
${\sigma }_{x-y}^2=\frac{1}{n}\left[\right(x-y)-(\overline{x}-\overline{y}){]}^2$
${\sigma }_{x-y}^2=\frac{1}{n}\left[\right(x-y)-(\overline{x})+(\overline{y}){]}^2$
${\sigma }_{x-y}^2=\frac{1}{n}\left[\right(x-\overline{x})-(y-\overline{y}){]}^2$
${\sigma }_{x-y}^2=\frac{1}{n}\left[\right(x-\overline{x}{)}^2+(y-\overline{y}{)}^2-2(x-\overline{x}\left)\right(y-\overline{y}\left)\right]$
${\sigma }_{x-y}^2=\frac{1}{n}\left[\right(x-\overline{x}{)}^2]+\frac{1}{n}\left[\right(y-\overline{y}{)}^2]-2\frac{1}{n}(x-\overline{x})(y-\overline{y})$
${\sigma }_{(x-y)}^2={\sigma }_x^2+{\sigma }_y^2-2r{\sigma }_x{\sigma }_y$
$2r{\sigma }_x{\sigma }_y={\sigma }_x^2+{\sigma }_y^2-{\sigma }_{(x-y)}^2$
$\therefore r=\frac{{\sigma }_x^2+{\sigma }_y^2-{\sigma }_{(x-y)}^2}{2{\sigma }_x{\sigma }_y}$ We know that $r=\frac{\sum (x-\overline{x})(y-\overline{y})}{n{\sigma }_x{\sigma }_y}$.