[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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