[BioC] How does Limma calculate the log2 fold change?

[Gordon Smyth]

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