Check out Steffener et al. NeuroImage 49:2113-2122 (2010).
They only use 2 basis functions; although reading the paper may give you
helpful references or ideas.
Also discussed is using area under the curve (AUC); if there is a large
difference in HDR shape (their example of elderly vs. young), the AUC
results will be biased.
Jan Bernard Marsman wrote:
> Dear SPM users,
>
>
> We have a question about how to do second level analysis if multiple
> basis functions are used at first level.
> We have modeled conditions at the first level using the standard hrf,
> derivative and dispersed basis function provided by SPM.
> We calculated the area under the curve. (a negative number will
> indicate that most of the curve is below 0) and taken this value to a
> second level analysis.
> We would like to know if this solution has been used before (see
> below), and if there are drawbacks and/or issues we should be aware of
> when we want to interpret the results.
>
> Some background information:
> Of course we have been searching the list and literature. We have
> found several methods to combine information of multiple basis
> function at first level in a second level analysis.
> There is the option to only take the (contrast of) beta belonging to
> the first basis function (hrf). However Calhoun et al (2004) has shown
> that this can give biased results.
> In this paper, a solution is suggested. But the article uses different
> basis function. Applying this solution to the basis functions in SPM
> is not straightforward. Two solutions can be found
> in http://mindhive.mit.edu/node/1320. On the mailing list code for the
> latter solution has been provided by Dr. McLaren
> (https://www.jiscmail.ac.uk/cgi-bin/wa.exe?A2=ind0908&L=SPM&P=R26741
> <https://www.jiscmail.ac.uk/cgi-bin/wa.exe?A2=ind0908&L=SPM&P=R26741>).
> In principle the amplitude is calculated of the reconstructed hrf, and
> beta 1 is corrected (boosted) such that the amplitude of the first
> basis function will be equal.
>
> %%% comment taken from the original script (which only takes the
> temporal derivative into account)...
> % sc = max(b1*xi+b2*xi)/max(b1*xi); OR min(b1*xi+b2*xi)/min(b1*xi) -- when
> % b1 is negative.
> %
> % Then your 'boosted' beta is:
> %
> % boosted_b1 = b1*sc;
> %%% end of citation
>
> Next only the corrected beta (boosted_b1) is taken to second level.
> However we have noticed that this solution is unstable (see formula)
> if beta 1 is near 0. -- The solution in the provided code to this
> instability is by allowing a maximal scaling of 1.29 --
> Therefore, we opted for a different solution. Instead of taking the
> (equivalent of the) amplitude of the reconstructed signal to second
> level, we looked at the area under the curve.
>
> We are looking forward to your comments/suggestions,
>
> Yours,
>
>
> Remco Renken & Jan-Bernard Marsman
>
> Neuroimaging Center
> University Medical Center Groningen
> The Netherlands
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