# 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:

- $\size {\map \rho {X, Y} } \le 1$

$\blacksquare$

This article is complete as far as it goes, but it could do with expansion.In particular: There's a proof available here that uses Cauchy's Inequality, but that means pulling apart the Definition of Pearson Correlation Coefficient and covariance and so onYou can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding this information.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Expand}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Sources

- 2011: Morris H. DeGroot and Mark J. Schervish:
*Probability and Statistics*(4th ed.): $4.6$: Covariance and Correlation: Theorem $4.6.3$