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LS solution vs. pre-averaging

by divB
Tags: preaveraging, solution
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Jan22-13, 09:10 PM
P: 86

I have a system of equations [itex]\mathbf{y} = \mathbf{A}\mathbf{c}[/itex] where the entries in [itex]\mathbf{c}[/itex] are small (say, K=10 elements) and the number equations (i.e., elements in [itex]\mathbf{y}[/itex]) is huge (say, N=10000 elements).

I want to solve now for [itex]\mathbf{c}[/itex]; this can be done using LS with the Pseudo inverse:

[tex]\mathbf{c} = \mathbf{A}^{\dagger} \mathbf{y}[/tex]

However, the vector [itex]\mathbf{y}[/itex] is now heavily corrupted by noise (just assume iid Gaussian).

I could calculate the mean over M consecutive elements in [itex]\mathbf{y}[/itex] and rows in [itex]\mathbf{A}[/itex] in order to average over the noise. The system would be collapsed to a smaller system with N/M entries which would be solved via LS.

Now I ask the question: Is this better than directly using LS with the full system?

I doubt because that's the sense of LS. However, I was not able to "proof" this analytically.

Any help?
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Jan22-13, 09:35 PM
P: 4,573
Hey divB.

Can you use the properties of a psuedo-inverse to show that this holds? (Recall that a pseudo-inverse has the property that C*C'*C = C)

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