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

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

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

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

and

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

therefore

\begin{align*}
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}}
\end{align*}

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

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

Since variances are non-negative

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

Solving for inequality we have

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

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

\begin{align*}
\rho(X^*,Y^*)&=Cov(X^*,Y^*)=Cov\{\frac{X-\mu_X}{\sigma_X},\frac{Y-\mu_Y}{\sigma_Y}\}\\
&=\frac{Cov(X-\mu_X,Y-\mu_Y)}{\sigma_X\sigma_Y}\\
&=\frac{Cov(X,Y)}{\sigma_X \sigma_Y}=\rho(X,Y)
\end{align*}

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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Muhammad Imdad Ullah

Currently working as Assistant Professor of Statistics in Ghazi University, Dera Ghazi Khan. Completed my Ph.D. in Statistics from the Department of Statistics, Bahauddin Zakariya University, Multan, Pakistan. l like Applied Statistics, Mathematics, and Statistical Computing. Statistical and Mathematical software used is SAS, STATA, GRETL, EVIEWS, R, SPSS, VBA in MS-Excel. Like to use type-setting LaTeX for composing Articles, thesis, etc.

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