# prove that coefficient of correlation always lies between -1 to +1

coefficient of correlation = r = $\frac{cov\left(x,y\right)}{{\sigma }_{x}{\sigma }_{y}}$

cov(x,y) = $\frac{\sum \left(x-\overline{x}\right)\left(y-\overline{y}\right)}{n}$

we have to prove $-1\le r\le 1$
let us consider two terms $\left[\frac{x-\overline{x}}{{\sigma }_{x}}\right]$ and $\left[\frac{y-\overline{y}}{{\sigma }_{y}}\right]$

take sum of the square of these two variables
$\sum {\left[\left[\frac{x-\overline{x}}{{\sigma }_{x}}\right]±\left[\frac{y-\overline{y}}{{\sigma }_{y}}\right]\right]}^{2}$
since it is a square term then it is surely a positive number
that is $\sum {\left[\left[\frac{x-\overline{x}}{{\sigma }_{x}}\right]±\left[\frac{y-\overline{y}}{{\sigma }_{y}}\right]\right]}^{2}\ge 0$
opening its square we get

on further simplifying it, we get
$\frac{\sum \left(x-\overline{x}{\right)}^{2}}{{{\sigma }^{2}}_{x}}+\frac{\sum \left(y-\overline{y}{\right)}^{2}}{{{\sigma }^{2}}_{y}}±2\frac{\sum \left(x-\overline{x}\right)\left(y-\overline{y\right)}}{{\sigma }_{x}{\sigma }_{y}}\ge 0$
divide the entire expression by n
$\frac{\sum \left(x-\overline{x}{\right)}^{2}}{n{{\sigma }^{2}}_{x}}+\frac{\sum \left(y-\overline{y}{\right)}^{2}}{n{{\sigma }^{2}}_{y}}±2\frac{\sum \left(x-\overline{x}\right)\left(y-\overline{y\right)}}{n{\sigma }_{x}{\sigma }_{y}}\ge 0$
using the formula of variance of x, variance of y and covariance of x and y, we get
$\frac{{{\sigma }^{2}}_{x}}{{{\sigma }^{2}}_{x}}+\frac{{{\sigma }^{2}}_{y}}{{{\sigma }^{2}}_{y}}±2\frac{cov\left(x,y\right)}{{\sigma }_{x}{\sigma }_{y}}\ge 0$

1+1$±$2r$\ge 0$

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