Square of Covariance is Less Than or Equal to Product of Variances
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Theorem
Let $X$ and $Y$ be random variables.
Let the variances of $X$ and $Y$ exist and be finite.
Then:
- $\paren {\cov {X, Y} }^2 \le \var X \, \var Y$
where $\cov {X, Y}$ denotes the covariance of $X$ and $Y$.
Proof
We have, by the definition of variance, that both:
- $\expect {\paren {X - \expect X}^2}$
and:
- $\expect {\paren {Y - \expect Y}^2}$
exist and are finite.
Therefore:
\(\ds \paren {\cov {X, Y} }^2\) | \(=\) | \(\ds \paren {\expect {\paren {X - \expect X} \paren {Y - \expect Y} } }^2\) | Definition of Covariance | |||||||||||
\(\ds \) | \(\le\) | \(\ds \expect {\paren {X - \expect X}^2} \expect {\paren {Y - \expect Y}^2}\) | Square of Expectation of Product is Less Than or Equal to Product of Expectation of Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \var X \, \var Y\) | Definition of Variance |
$\blacksquare$
Sources
- 2011: Morris H. DeGroot and Mark J. Schervish: Probability and Statistics (4th ed.): $4.6$: Covariance and Correlation: Theorem $4.6.3$