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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 added noise, just remove it from your measurement and your model becomes the classic Linear Least Squares.

If you don't know it, you have to estimate it. You can solve the problem either in 2 steps, namely estimate, remove, solve classic Linear Least Squares, or build a model which estimates the noise and solve the least squares at once.

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$$(A^TA)^{-1}A^T(y + noise) = \hat x$$

where $noise$ is vector of the same size as $y$ with all elements equal to unknown constant. That is all. You can get your estimated solution as a function of noise mean position in explicit form.

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