Here is just a lil "trivia" statistic question for the forum:-) I apologize for my clumsy annotation but i hope I get the question through anyhow:-) For ratios I take it its normal procedure to calculate the average as the geometric mean. That is easiest done with log transformed values, giving something like <r> = mean(Ratio=a/b)=2^ (<log2(a)> - <log2(b)> + - SD) Its the SD thats giving me a little headache as i go back to normal (un-transformed, i use ** for annotating normal) values. SD in log space is not symmetric in normal space, so SD** != 2^(SD)? or? To illustrate my clumsy annotation, if : < log2(R) > + - SD(log2(R)) = 4+-1 in log space it becomes (2^4- 2^3) and 2^4+2^5 ~ 16-8 and 16 + 32. so SD** is not symmetric. I found 2 suggestions in the litterature that doesnt seem to account for this asymmetry. One was giving the standard deviation of the geometric mean to be SD**=2^(SD) just as i reasoned was inappropriate? Another suggestion I found was for propagating errors for exponential transformation : X = e^A, SD(X)/X = SD(A) So should i do SD**(X) = mean(X) * SD(A) --- X ~ Ratio and A ~ log (R) ???? again i dont see how this solved the asymmetric SD from the log space??? Maybe i missed something basic with log-normal distributions, in any case any help will be highly appreciated:-) I have the feeling its rather trivial but I would really like to know how to put (assymetric?) error bars on my (normal scale) ratios correctly. This goes for both Affymterix summary ratios and RT-PCR ratios. What's the correct procedure??? cheers:-) jesper ryge Karolinska Institutet Dep. of Neuroscience

Hi Jesper, the asymmetry only becomes appreciable when SD(X)/X = SD(A) (btw this is called the coefficient of variation) becomes big; say, much bigger than 0.1 (=10%). You can see this from the power series of the exponential function. e^x = 1 + x + O(x^2), and the quadratic and higher order terms that cause the asymmetry become non-negligible when |x|>>0. Hope this helps Wolfgang > Here is just a lil "trivia" statistic question for the forum:-) I > apologize for my clumsy annotation but i hope I get the question > through anyhow:-) > > > For ratios I take it its normal procedure to calculate the average as > the geometric mean. That is easiest done with log transformed values, > giving something like > > <r> = mean(Ratio=a/b)=2^ (<log2(a)> - <log2(b)> + - SD) > > Its the SD thats giving me a little headache as i go back to normal > (un-transformed, i use ** for annotating normal) values. SD in log > space is not symmetric in normal space, so SD** != 2^(SD)? or? > > To illustrate my clumsy annotation, if : < log2(R) > + - SD(log2(R)) > = 4+-1 in log space it becomes (2^4- 2^3) and 2^4+2^5 ~ 16-8 and 16 > + 32. so SD** is not symmetric. > > > I found 2 suggestions in the litterature that doesnt seem to account > for this asymmetry. One was giving the standard deviation of the > geometric mean to be SD**=2^(SD) just as i reasoned was inappropriate? > > Another suggestion I found was for propagating errors for exponential > transformation : > > X = e^A, SD(X)/X = SD(A) > > So should i do SD**(X) = mean(X) * SD(A) --- X ~ Ratio and A ~ log > (R) ???? again i dont see how this solved the asymmetric SD from the > log space??? > > Maybe i missed something basic with log-normal distributions, in any > case any help will be highly appreciated:-) I have the feeling its > rather trivial but I would really like to know how to put > (assymetric?) error bars on my (normal scale) ratios correctly. This > goes for both Affymterix summary ratios and RT-PCR ratios. What's > the correct procedure??? > > > cheers:-) > jesper ryge > Karolinska Institutet > Dep. of Neuroscience > > _______________________________________________ > Bioconductor mailing list > Bioconductor at stat.math.ethz.ch > https://stat.ethz.ch/mailman/listinfo/bioconductor > Search the archives: http://news.gmane.org/gmane.science.biology.informatics.conductor -- ------------------------------------------------------------------ Wolfgang Huber EBI/EMBL Cambridge UK http://www.ebi.ac.uk/huber