Basic Statistics and Data Analysis

Lecture notes, MCQS of Statistics

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. 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 r2 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 Correlation Coefficient values lies between -1 and +1.

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The Author

Muhammad Imdadullah

Student and Instructor of Statistics and business mathematics. Currently Ph.D. Scholar (Statistics), Bahauddin Zakariya University Multan. Like Applied Statistics and Mathematics and Statistical Computing. Statistical and Mathematical software used are: 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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