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Independent Implies Uncorrelated

It can be shown that independent zero-mean random numbers are also uncorrelated, since, referring to (D.3),

$\displaystyle E\{\overline{v(n)}v(n+m)\} =
\left\{\begin{array}{ll}
E\{\left\v...
...ot E\{v(n+m)\}=0, & m\neq 0 \\
\end{array}\right. \isdef \sigma_v^2 \delta(m)
$

For Gaussian random numbers, being uncorrelated also implies independence [185]. For further discussion and illustrations, see §5.3 and the examples of the following sections.


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written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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