How does Limma calculate the log2 fold change?
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@groot-philip-de-1307
Last seen 8.2 years ago

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

limma • 1.8k views
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@gordon-smyth
Last seen 9 hours ago
WEHI, Melbourne, Australia

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

(This answer was originally posted 12 May 2006.)