# Least Squares with Non Zero Mean Noise

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!

• $y(k)=Ax-\overline{{\eta}(k)}+(\eta(k)-\overline{{\eta}(k)})$ is such that the parenthesis is a zero-mean noise. Aug 11 '16 at 14:26
• do you assume to know the mean of $\eta$? if yes Yves comments is the way to go. Aug 11 '16 at 15:05
• I assume that the mean of $\eta$ is known, but I don't understand what Yves said. Aug 11 '16 at 15:41
• Just subtract the mean from the samples of $y(k)$ Aug 12 '16 at 6:25
• 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 Aug 12 '16 at 6:28

The easy way to do so is to remove the bias from $$y$$ and solve.
$$(A^TA)^{-1}A^T(y + noise) = \hat x$$
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.