Correlation Coeficient values lies between +1 and -1?

We know that the ratio of the explained variation to the total variation is called the coefficient of determination which is the square of the correlation coefficient. This ratio is non-negative, therefore denoted by $r^2$, thus

r^2&=\frac{\text{Explained Variation}}{\text{Total Variation}}\\
&=\frac{\sum (\hat{Y}-\overline{Y})^2}{\sum (Y-\overline{Y})^2}

It can be seen that if the total variation is all explained, the ratio $r^2$ (Coefficient of Determination) is one and if the total variation is all unexplained then the explained variation and the ratio $r^2$ is zero.

The square root of the coefficient of determination is called the correlation coefficient, given by

r&=\sqrt{ \frac{\text{Explained Variation}}{\text{Total Variation}} }\\
&=\pm \sqrt{\frac{\sum (\hat{Y}-\overline{Y})^2}{\sum (Y-\overline{Y})^2}}


\[\sum (\hat{Y}-\overline{Y})^2=\sum(Y-\overline{Y})^2-\sum (Y-\hat{Y})^2\]


r&=\sqrt{ \frac{\sum(Y-\overline{Y})^2-\sum (Y-\hat{Y})^2} {\sum(Y-\overline{Y})^2} }\\
&=\sqrt{1-\frac{\sum (Y-\hat{Y})^2}{\sum(Y-\overline{Y})^2}}\\
&=\sqrt{1-\frac{\text{Unexplained Variation}}{\text{Total Variation}}}=\sqrt{1-\frac{S_{y.x}^2}{s_y^2}}

where $s_{y.x}^2=\frac{1}{n} \sum (Y-\hat{Y})^2$ and $s_y^2=\frac{1}{n} \sum (Y-\overline{Y})^2$

\Rightarrow r^2&=1-\frac{s_{y.x}^2}{s_y^2}\\
\Rightarrow s_{y.x}^2&=s_y^2(1-r^2)

Since variances are non-negative

\[\frac{s_{y.x}^2}{s_y^2}=1-r^2 \geq 0\]

Solving for inequality we have

1-r^2 & \geq 0\\
\Rightarrow r^2 \leq 1\, \text{or}\, |r| &\leq 1\\
\Rightarrow & -1 \leq r\leq 1

Alternative Proof

Since $\rho(X,Y)=\rho(X^*,Y^*)$ where $X^*=\frac{X-\mu_X}{\sigma_X}$ and $Y^*=\frac{Y-Y^*}{\sigma_Y}$

and as covariance is bi-linear and X* ,Y* have zero mean and variance 1, therefore

&=\frac{Cov(X,Y)}{\sigma_X \sigma_Y}=\rho(X,Y)

We also know that the variance of any random variable is ≥0, it could be zero i.e .(Var(X)=0) if and only if X is a constant (almost surely), therefore

\[V(X^* \pm Y^*)=V(X^*)+V(Y^*)\pm2Cov(X^*,Y^*)\]

As Var(X*)=1 and Var(Y*)=1, the above equation would be negative if $Cov(X^*,Y^*)$ is either greater than 1 or less than -1. Hence \[1\geq \rho(X,Y)=\rho(X^*,Y^*)\geq -1\].

If $\rho(X,Y )=Cov(X^*,Y^*)=1$ then $Var(X^*- Y ^*)=0$ making X* =Y* almost surely. Similarly, if $\rho(X,Y )=Cov(X^*,Y^*)=-1$ then X*=−Y* almost surely. In either case, Y would be a linear function of X almost surely.

For proof with Cauchy-Schwarz Inequality please follow the link

We can see that the Correlation Coefficient values lie between -1 and +1.

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