[Eeglablist] question about variance
Jeng-Ren Duann
duann at sccn.ucsd.edu
Tue Jun 28 10:33:00 PDT 2005
Since the basis spaned by ICs is not orthogonal, so
PVAF(ICi+ICj) is not necessarily equal to PVAF(ICi) + PVAF(ICj). If you
want to know what the PVAF is for multiple ICs, you have to back-project
those ICs and calculate PVAF as a whole.
JR
--
Jeng-Ren Duann, Ph. D.,
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duann at sccn.ucsd.edu
On Thu, 23 Jun 2005, Phillip M.Gilley wrote:
> Howdy,
> When using standard PCA we can easily calculate the percentage of
> variance explained from the eigenvalues for each principal component.
> Is the calculation similar, or even possible for ICA?
>
> For example, the ICA solution should (theoretically) explain 100% of
> the variance in the original data set. However, when looking at the
> percentage of variance accounted for (pvaf) from the different
> functions in EEGLAB, I see several different types of results. Under
> the envtopo() function, there is a specified output for pvaf. This
> function calculates the pvaf for each specified component across all
> channels; and the calculation appears to be made by subtracting the
> variance of the whole data set from the variance of the back projection
> of the component. However, it is often the case that the sum of the
> values exceeds 100%, which I interpret as the variance in the
> distribution of that component irrespective of the contribution of
> other components. Is this a correct interpretation? Is there a pvaf
> function that returns a set of values for each component for the
> original dataset (e.g., for the entire epoch, or for a specified time
> range), such that the sum would equal 100%?
>
> Also, the function eeg_pvaf() returns three sets of variances, which
> don't necessarily seem to match the pvaf values returned from the
> envtopo() function. Are these different calculations altogether? And
> if so, what different variances are these values explaining?
>
> Thanks for any insight or comments.
>
>
> Phillip M. Gilley
> The University of Texas at Dallas
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