Dear Philip, limma does all analysis on the log-scale, which appears to be your second method. This is appropriate because the expression values are more nearly normally distributed and homoscedastic on the log-scale, hence taking means on the log-scale is statistically more powerful and less influenced by individual arrays. Best wishes Gordon >Date: Fri, 12 May 2006 16:16:08 +0200 >From: "Groot, Philip de" <philip.degroot at="" wur.nl=""> >Subject: [BioC] How does Limma calculate the log2 fold change? >To: <bioconductor at="" stat.math.ethz.ch=""> > >Hello, > > The topic of this email looks easy enough, but I cannot find in the >Limma documentation how the (log2) fold changes are actually calculated. > >As you probably now, Limma uses log2-scaled input (originating from e.g. >RMA). So, if you want to calculate a log2 fold change, it is possible to >keep this log2-transformation into account or to discard it. What I mean >with this is that the mean of logged values is lower than the mean of >the unlogged values. > >Take for example the series: 2, 3, and 4 > >log2(mean(c(2^2, 2^3, 2^4))) > >[1] 3.222392 > >mean(c(2,3,4)) > >[1] 3 > > >To my opinion, the top method, unlog, calculate the mean, and >log2-transform again, is the proper way and I do expect that Limma does >it the same way. My question is: is this true? > > >Regards, >Dr. Philip de Groot >Wageningen University