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01272014, 10:25 AM  #21 
Member
Location: Mc Gill  Montreal Join Date: Dec 2013
Posts: 37

p value is just a widely used joke. The signification of p value is hard to get and imply assumptions that lot of people don t know.
FDR is just a bigger joke. Your best pvalue will most of the time be multiply by your number of p value. So if you have 10 genes to test giving you 10 pvalues, the best is multiply by 10, the second best by 5, then by 3.3333, then by 2.5, then by 2 etc ..... Here is the code in R Code:
# produce a vector of FDR with an ordered pval vector fdr = function(pval){ size=length(pval) if(size<2) return(pval) #the worst pval is multiply by (size) / (size1) FDR=c( min( 1 , pval[size]*(size)/(size1) )) for( i in 1:(size1)) FDR=c(FDR,min(FDR[i] , pval[sizei]*(size)/(sizei))) # We have to revers the vector to be consistant return(rev(FDR)) } 
01272014, 11:39 AM  #22 
Senior Member
Location: Santa Fe, NM Join Date: Oct 2010
Posts: 250

I don't know for sure. I first noticed the problem doing some metaanalysis hypothesis testing on coverage merging pvalues of bases with each base having a different statistical power. Using Fishers metaanalysis procedure, it became obvious that the underpowered bases were dominating the the test and that Fishers test assumed all published results were adequately powered. It would be nice if that theory were also better. I came up with a heuristic involving weighted sums of log(pvalues) and information entropy as degrees of freedom, which has a certain appeal to it.

01272014, 12:09 PM  #23 
Senior Member
Location: Norway Join Date: Aug 2013
Posts: 266

Interesting, thanks!
Another question, excuse my ignorance.. But look at these codes: > FDR < p.adjust(lrt$table$PValue, method="BH") > sum(FDR < 0.05) Is this the way to choose FDR < 0.1: > FDR < p.adjust(lrt$table$PValue, method="BH") > sum(FDR < 0.1) 
02072014, 08:58 AM  #24  
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Location: Nebraska Join Date: Oct 2011
Posts: 25

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