Mean subtraction before covariance computation

[ap...@nu.edu.kz]

Hello,

I wanted to ask about the subtraction of the average from the LFP data. I think it is common to rectify EEG data but not LFP (what I have seen from the papers). In your lectures about the General Eigendecomposition you said it is important to subtract the mean before computation of the covariance matrix, so my question is: Should I compute the mean from the rectified signal or original one?

I am asking because there are two different ways to do it:

1) I can just subtract the mean from the original data. In this case it will be close to 0, because the signal is not rectified. Not big difference from the original signal.

LFP = LFP - mean(LFP)

2) I can take absolute of the signal and then compute the mean then subtract the mean from the rectified signal. After this, revert negative values back.

mask = LFP<0

mean = abs(LFP).mean()

LFP_norm = abs(LFP) - mean

LFP_norm[mask] =  LFP_norm[mask] * (-1)

I attached plots generated from two different approaches.    

The same applies to the baseline subtraction, which way is the proper way of doing this?

Thanks,

[Mike X Cohen]

Hello. I don't think it's common in EEG to rectify the signal. Mostly that's done in EMG, where the total energy is the primary dependent variable. I'm not sure what your goal is, but I would not normally recommend rectifying an LFP signal. For one thing, it changes the frequency characteristics. To see why that's the case, you can plot the power spectrum of a simulated sine wave at 10 Hz, and then plot the power spectrum of the same sine wave but rectified.

As for subtracting the mean before computing the covariance matrix, it depends on what you want to do with that matrix. For an eigendecomposition (PCA or GED), you definitely should subtract the mean, otherwise the first component will reflect the mean offset, not the variance in the data.

As for how to subtract the mean, your first suggestion was correct -- subtract the mean from the signal as the final step before computing its covariance.

Mike

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Mike X Cohen, PhD
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[ap...@nu.edu.kz]

Thanks for a quick reply, I was confused because mean computed from the original signal is close to 0, so it does not really change the signal. 

[Mike X Cohen]

If the mean was close to zero, then the signal was already almost demeaned. This is probably due to a high-pass filter. 

Mike