On 12-Jun-11 01:36:00, Thomas Lumley wrote: > On Sun, Jun 12, 2011 at 12:44 PM, Tiago Pereira > <tiago.pere...@mbe.bio.br> wrote: > >> The test I am working on has an asymptotic 0.5*chi2(1)+0.5*chi2(2) >> distribution, where numbers inside parenthesis stand for the >> degrees of freedom. Is is possible to compute quickly in R >> the cumulative distribution of that distribution? > > There appear to be pchibar() functions in both the ibdreg and ic.infer > packages that should do want you want. Simulation is also fairly > efficient. > > -thomas > > -- > Thomas Lumley > Professor of Biostatistics > University of Auckland
Is there anything wrong with the following argument: In Tiago's notation, let X have the mixed chi2 distribution: X ~ chi2(1) with probability p X ~ chi2(2) with probability q = (1-p) Then the cumulative distribution of X is Prob(X <= x) = p*Prob(chi2(1) <= x) + q*prob(chi2(2) <= x) so the CDF is p*pchisq(x,1) + (1-p)*pchisq(x,2) Inverting this to find the value of x for a given value P of Prob(X <= x) = P is another matter, but should be reliably solvable by using R's uniroot() function, with lower and upper endpoints taken from whichever of qchisq(P,1), qchisq(P,2) is smallest, and whichever is largest. Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <ted.hard...@wlandres.net> Fax-to-email: +44 (0)870 094 0861 Date: 12-Jun-11 Time: 09:37:46 ------------------------------ XFMail ------------------------------ ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.