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Independent Implies Uncorrelated

It can be shown that independent zero-mean random numbers are also uncorrelated, since, referring to (C.26),

$\displaystyle E\{\overline{v(n)}v(n+m)\} = \left\{\begin{array}{ll} E\{\left\vert v(n)\right\vert^2\} = \sigma_v^2, & m=0 \\ [5pt] E\{\overline{v(n)}\}\cdot E\{v(n+m)\}=0, & m\neq 0 \\ \end{array} \right. \isdef \sigma_v^2 \delta(m)$ (C.28)

For Gaussian distributed random numbers, being uncorrelated also implies independence [201]. For related discussion illustrations, see §6.3.

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