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What happens if the noise has no zero mean? I mean, if the exercise is something like: $$y(k) = A x + \eta(k)$$

When I have zero mean, I start from: $$y = A x$$ $$\Rightarrow \hat{y} = A \hat{x}$$ Using algebra, I to get to this equation: $$A^H y = A^H A \hat{x}$$

But what happens if it has no zero mean? I have to use the following inner product, I suppose: $$\langle x,y \rangle = \mathbb{E}[(x-\mathbb{E}[x])^H(x-\mathbb{E}[x])]$$ But I can't see how to get to an equation from that.

Thanks!

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    $\begingroup$ $y(k)=Ax-\overline{{\eta}(k)}+(\eta(k)-\overline{{\eta}(k)})$ is such that the parenthesis is a zero-mean noise. $\endgroup$ – Yves Daoust Aug 11 '16 at 14:26
  • $\begingroup$ do you assume to know the mean of $\eta$? if yes Yves comments is the way to go. $\endgroup$ – LJSilver Aug 11 '16 at 15:05
  • $\begingroup$ I assume that the mean of $\eta$ is known, but I don't understand what Yves said. $\endgroup$ – Euler Aug 11 '16 at 15:41
  • $\begingroup$ Just subtract the mean from the samples of $y(k)$ $\endgroup$ – LJSilver Aug 12 '16 at 6:25
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    $\begingroup$ In that way you are making the change of coordinates $z=y-\bar \eta$. In such coordinates the problem is $z(k)=Ax(k)+w(k)$ being $w(k):=\eta(k)-\bar\eta$ zero mean $\endgroup$ – LJSilver Aug 12 '16 at 6:28
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Since this is a linear model if you add noise which isn't centered (Non zero mean noise) your estimation will be good up to a bias term.

The easy way to do so is to remove the bias from $ y $ and solve.

In case you know the mean of the add noise, just remove it from your measurement and you model becomes the classic Linear Least Squares.

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