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Hi Bernadet, Yes, this is valid, and is the canonical way of doing the correction. It was proposed by Tom Nichols some time ago, and if you search the archives you'll find the steps there. I can't recall if Tom's original email to Lisa (the 2nd author of Licata et al, 2013) went to the list, but there is for sure an email in the archives (this one from me) in which the exact steps are outlined. However, this was in 2012. Now the method has been implemented in PALM <http://fsl.fmrib.ox.ac.uk/fsl/fslwiki/PALM>, which does the correction on the fly, without the need for custom scripts or to save files to the disk. You'd just enter each IC as one modality (with its own "-i"), and use the option "-corrmod". The remaining of the syntax is very similar to randomise. The paper that describes the method in complete generality is currently under review. All the best, Anderson On 23 September 2015 at 11:18, Bernadet Klaassens < [log in to unmask]> wrote: > Dear FSL experts, > > I use dual regression to investigate connectivity with 10 functional > networks. The tfce output of stage 3 gives me the p-map, corrected for > multiple comparisons across voxels, but not across components. Therefore, I > might use a Bonferroni correction and look at p-values < 0.005 instead of > 0.05. > > However, I recently read the following article where a different approach > is suggested (page 214): > > http://www.sciencedirect.com/science/article/pii/S1053811912012414 > > (Licata et al. (2013). NeuroImage) > > It basically comes down to creating one new null distribution of the > maxima across components, by using the 10 vectors (1 per component) of > maximum tfce values that result from Randomise. A new threshold can then be > obtained from this new distribution to examine the 10 raw tfce maps at p< > 0.05 (corrected for multiple comparisons across voxels and components), by > looking only at tfce values in the top 5% of this distribution. > > I was wondering whether this correction method is accepted and look > forward to your opinion. > > Best, > > Bernadet >