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In linear prediction we can say that in case of optimum linear predictor the error with be orthogonal to data. And when we derive minimum mean square error for $\underline{y} = \mathbf{a}\underline{x} + \underline{b}$ case we find following relation

$\xi = \sigma^2_y (1 - \rho_{xy} )^2$

Where $\rho_{xy} $ is correlation coefficient and if we talk about cross correlation $r_{xy}(k,l)$ of zero mean and uncorrelated random variables x and y, then how does this conclude that if $r_{xy}(k,l)$ will be then the signals will be orthogonal?

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  • $\begingroup$ What is the relationship between $\rho_{xy}$ and $r_{xy}(k,l)$ ? I'm assuming $\rho$ is the correlation coefficient and $r$ is the cross correlation, but it would be good to spell this out. $\endgroup$ – Peter K. May 6 '16 at 13:12
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    $\begingroup$ I apologize for ambiguity in the question, I edited it. I hope it is not missing anything now. $\endgroup$ – Tab May 6 '16 at 13:22
  • $\begingroup$ No need to apologise! Thanks for the edit. It's much clearer now. $\endgroup$ – Peter K. May 6 '16 at 13:23
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I suppose you mean the cross-correlation at lag zero. Well take an Hilbert space $H$ (i.e. a metric space in which you can define a scalar product $\langle\cdot ,\cdot\rangle$). Then $x,y\in H$ are orthogonal if $\langle x,y\rangle=0$, by definition.
If your Hilbert Space is $L_2(\mathbb{R})$ (the space of real square integrable functions) then the scalar product of $f,g\in L_2(\mathbb{R})$ is defined as
$$ \langle f, g\rangle = \int_{-\infty}^\infty \overline{f(t)}g(t) dt $$ If the signals are assumed to be ergodic, which is actually a costumary assumption, then the cross-correlation is simply a sliding scalar product, you have $$ \text{corr}(f,g)(\tau) = \int_{-\infty}^\infty\overline{f(t)}g(t+\tau) dt $$ hence $$ \text{corr}(f,g)(0) = \langle f ,g \rangle $$ and uncorrelation means orthogonality.
The same applies for discrete time signals in $l_2$

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    $\begingroup$ Correlation is, more generally, $E[\overline{f(t)}g(t+\tau)]$. The definition you're using can only be used if $f$ and $g$ are ergodic. $\endgroup$ – Peter K. May 6 '16 at 12:32
  • $\begingroup$ yes, thanks for pointing it out! I assumed implicitly that ergodicity was an assumption on the signals $\endgroup$ – LJSilver May 6 '16 at 12:39
  • $\begingroup$ Most people do. I gave this a + 1, but thought we should mention the ergodicity piece. $\endgroup$ – Peter K. May 6 '16 at 12:40
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    $\begingroup$ I added a piece saying that this holds under ergodicity assumption $\endgroup$ – LJSilver May 7 '16 at 10:35

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