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Hi @AvoDopple, I'm not completely sure I understand what you're doing, but this sounds a lot like a simple result of statistical power. In other words, I don't find the pattern surprising. I.e. more taxa leads to more power in the AU test. The more taxa in tree, the more different two trees (A and B in your scenario) can be. So the more of the bootstrap trees I'd expect to see rejected. Having said that, I'm not 100% sure this is right, and I think it might depend on how you are getting trees A and B in the first place. I.e. what are the alignments and where do they come from? Rob |
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I have been running IQtree2's AU test on a series of analyses. Each analysis has 2 alignments and best trees and sets of bootstraps from each of those alignments. Each analysis has its own set of taxa, the number of which can vary from a handful to over a hundred. These are non-overlapping taxa sets, if that matters. Just the same method being applied to many datasets.
My issue/observation is that all of the small trees have much higher rates of cross-hits in the AU results. I.e., far more trees are accepted ("+"). The larger trees have far fewer (more "-") and the only 4 with zero in both directions are among the 6 largest.
To illustrate with a semi-arbitrary breakpoint, analyses with less than 14 taxa (n=33) average about 55% ("+") in the comparison bootstrap sets. And those with 14+ taxa (n=21) are at 17%. A logarithmic best-fit between these percentages and taxa count returns an R-squared of 0.6141.
I was unaware that the AU test cared much about the size of the trees. I looked back to the Shimodaira 2002 paper and could find no mention of changing the taxa number. Is this a known phenomenon? If so, is there a good paper to read to understand why this is happening? Does this colour how one should read AU test results? Did I jsut get really
I would appreciate any advice or explanation anyone has to help me wrap my head around this behaviour.
Thank you for your time.
** Just in case it matters, the files I am using follow the formula of: -s Alignment_A -z A'sBestTree_Plus_AlignmentB'sBootstraps -n 0 -zb 10000 -zw -au.
And then I also run it with Alignemnt_B and B's best tree + A's Bootstraps
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