Absolute Value of Pearson Correlation Coefficient is Less Than or Equal to 1
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Theorem
Let $X$ and $Y$ be random variables.
Let the variances of $X$ and $Y$ exist and be finite.
Then:
- $\size {\map \rho {X, Y} } \le 1$
where $\map \rho {X, Y}$ denotes the Pearson correlation coefficient of $X$ and $Y$.
Proof
\(\ds \paren {\map \rho {X, Y} }^2\) | \(=\) | \(\ds \paren {\frac {\map {\operatorname {Cov} } {X, Y} } {\sqrt {\var X \var Y} } }^2\) | Definition of Pearson Correlation Coefficient | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {\map {\operatorname {Cov} } {X, Y} }^2} {\var X \var Y}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds 1\) | Square of Covariance is Less Than or Equal to Product of Variances |
So:
- $\size {\map \rho {X, Y} } \le 1$
$\blacksquare$
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Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): correlation coefficient: 1.
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): correlation coefficient: 1.
- 2011: Morris H. DeGroot and Mark J. Schervish: Probability and Statistics (4th ed.): $4.6$: Covariance and Correlation: Theorem $4.6.3$