[Eeglablist] Reliability

Makoto Miyakoshi mmiyakoshi at ucsd.edu
Mon Dec 15 15:54:41 PST 2014


Dear Norbert,

Fiorenzo was with with until the last month.
If I understand his work correctly, what he did was a *bootstrap test*
using datapoints, so it's not simply repeating ICA with all datapoints.

Now this is an interesting point in his work. What is an artifact for ICA?
Is it EMG or EOG? No. As long as ICA can decompose it reliably (i.e. having
a fixed source, not moving around, always there...), it is NOT an artifact.
Rather, something that is turning on and off randomly (which is called
*non-stationary) *is the problem because it violates the ICA's assumption
about the stationarity. If you bootstrap the data, you can test this data
stationarity explicitly... that's what his work means to be.

> If You try to delete a component after the first run, it would be another
after the next run.

Do not reject components for the second ICA run because it is meaningless.
I know this is a very common trap that people would think rejecting
artifactual ICs for the second ICA run should reveal a 'new' component....
but actually if you do this you reduce data rank and most likely you'll see
the exactly the same results for the second run.

Makoto

On Tue, Dec 9, 2014 at 5:46 AM, Norbert Franke <nfranke at uni-bonn.de> wrote:
>
> Dear Collegues,
>
> I have a question about reliability.
>
> Fiorenzo Artoni has sent me his article:
>
>     http://www.sciencedirect.com/science/article/pii/S1053811914007526
>
> When I repeat an ICA run, the result differs.
>
> My question to the developers of EEGLAB is: Fiorenzo
> Artori talked about introducing a RELICA toolbox in EEGLAB. When? Will it
> be a plugin or an additional tool?
>
> And a question to the collegues: What do You do when the results differ
> after simply repeating an ICA run? If You try to delete a component after
> the first run, it would be another after the next run. How can You decide
> which component has to be deleted? How much variation is acceptable? What
> do You do when a 2d image of a component is exactly the same but the
> polarity is inverted?
>
> I am courious to Your experience, it is important for me.
>
> Yours,
>
> Norbert Franke
>
>
>
>
>
>
>
>
>
> nfranke at uni-bonn.de
> Praxis: 0049 (0)228 - 364663
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-- 
Makoto Miyakoshi
Swartz Center for Computational Neuroscience
Institute for Neural Computation, University of California San Diego
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