[FieldTrip] question about automatic artefact rejection

[De Meyer, Kris]

Thank you for your response, Eelke.

I guess it makes sense but it does make the selection of a threshold more arbitrary.

If your underlying distribution is normal, then a z-score of 3, 4 or 5 has a clear meaning in terms of outlier detection.

For a non-Gaussian distribution, that interpretation no longer holds, and rejection becomes less automated and more open to subjective interpretation.

Isn't it better to z-transform the Hilbert-envelope values with the mean/std or the median/mad of the original filtered EEG distribution rather than the mean/std or median/mad of the hilbert values distribution? That way, the meaning of the threshold magnitude would remain valid. Or is that what is happening in the code already?

Best,

Kris



________________________________
From: fieldtrip <fieldtrip-bounces at science.ru.nl> on behalf of Eelke Spaak <e.spaak at donders.ru.nl>
Sent: 18 February 2020 10:45
To: FieldTrip discussion list <fieldtrip at science.ru.nl>
Subject: Re: [FieldTrip] question about automatic artefact rejection

Dear Kris,

Since there is no statistical test involved, I see no problems with
computing a z-score here. Think of it more as bringing the data into a
sensible range and allowing yourself to forget about details like
measurement unit etc. You could also not do the z-transform, and
simply specify a threshold based on raw amplitude, this would be
equivalent (but might make it potentially more difficult to come up
with a good threshold).

Best,
Eelke

On Tue, 18 Feb 2020 at 09:47, De Meyer, Kris <kris.de_meyer at kcl.ac.uk> wrote:
>
> Dear Fieldtrippers,
>
> In ft_artifact_muscle, an EEG signal is first filtered - usually with a bandpass filter with cutoff frequencies of [110,140]. The documentation then says that the hilbert envelope is calculated, after which the envelope values are z-transformed before they are thresholded.
>
> However, is this sound? Whereas the actual filtered EEG values (the result of the bandpass filter) are usually normally distributed, the hilbert envelope values are far from.
>
> What is the rationale for calculating the z-transform of a non-Gaussian distribution? Doesn't that introduce problems?
>
> Best,
>
> Kris
>
>
>
>
>
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