Independent Implies Uncorrelated

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

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

For Gaussian random numbers, being uncorrelated also implies independence [177]. For further discussion and illustrations, see §5.3 and the examples of the following sections.