tag:blogger.com,1999:blog-8499895524521663926.post7305287101835669194..comments2024-11-06T04:50:42.713-05:00Comments on Phylogenetic Tools for Comparative Biology: The difference between different methods for fitting the Mk model in R: It's all about the priorLiam Revellhttp://www.blogger.com/profile/04314686830842384151noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-8499895524521663926.post-1065129456202382462020-11-07T07:02:48.511-05:002020-11-07T07:02:48.511-05:00Dear Liam,
First, thank you very much for your qu...Dear Liam,<br /><br />First, thank you very much for your quick and detailed reply to my previous question a couple of days ago.<br /><br />For my tree I do ancestral character estimation, using "ace" in ape package and then I do a G test to check which is the best model to use. However, if I use "fitDiscrete" in geiger and test which is the best model, there are differences. If using "ace" ARD seems to be the best model in my particular case, and with "fitDiscrete" - SYM. I would simply use "ace" and plot it, but I also need fitDiscrete to check for phylogenetical signal (Pagel's lambda). Can I present the "ace" plots in my paper and use "fitDiscrete" results to check for phylogenetical signal? I prefer "ace", because it gives me a good information about the root and all knots, and I can also add pie graphs to the tree. Do you know if I can I do the same with "fitDiscrete"? <br /><br />Cheers,<br />KostadinAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-77580377356464570172018-05-22T17:05:54.275-04:002018-05-22T17:05:54.275-04:00Why is the likelihood returned by ace (-85.35101) ...Why is the likelihood returned by ace (-85.35101) not the same as the likelihood returned by the geiger likelihood function (-86.44962), when evaluated at the same rate value (0.8865379)? The examples in the help page for fitDiscrete give a clue: one can get the same likelihood as fitMk (-86.44962) like this:<br /><br />fit(0.8865379, root="given", root.p=c(1/3,3))<br /><br />or one can get the same likelihood as ace (-85.35101) like this, using un-normalized root probabilities (1,1,1 instead of .33,.33,.33):<br /><br />fit(0.8865379, root="given", root.p=c(1,3))<br /><br />So it looks like the likelihood output by ace has an extra term log(3) = 1.098612 = -85.35101 - (- 86.44962). That's log(3) for 3 states. Weird. It's the same constant term for all choices of rate parameters, so the rates estimated by ace are correct. Any likelihood ratio would still be correct, too. But still, I find this strange.<br /><br />Thanks Liam for your post, and thanks Luke for the help in the manual page of fitDiscrete!Cecile Anehttps://www.blogger.com/profile/08451803342899072267noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-63221158083361226802018-03-21T23:41:03.059-04:002018-03-21T23:41:03.059-04:00The replica watches uk
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Another option out their is the ...Great post Liam,<br /><br />Another option out their is the way that diversitree handles this by default. Where the likelihoods at the root are weighted in proportion to how well they explain the observed tip data. Rich does a nice job of explaining the approach in the first appendix of Fitzjohn, Maddison and Otto 2009. Also in diversitree you can choose just to return the multiple probabilites calculated at the root rather than summing over them. This can be nice at least for exploration so you can see how or to what extent the choice of root prior is going to effect your result<br /><br />cheersHeathhttps://www.blogger.com/profile/02076735319410745326noreply@blogger.com