# [BioC] A question regarding the mean of M-values.

marcus marcusb at biotech.kth.se
Wed Apr 27 17:02:24 CEST 2005

```Hello all users.

I have a question regarding the mean calculations of the M-values in LIMMA.

I guess that the fit\$coeff is the mean of the M-values used for the linear
model. The fit\$coeff has the mean value of the data derived from a specific
RNA source (as defined in the design matrix), and the value in fit\$coeff[1]
is the same as mean(MS[1,1:2]) (if I for example had Sample 1 on 2 arrays in
my matrix containing the data.

So...if you take the mean of two values (in the log 2 scale), for example
M = 8 and M = 1, the mean (and hence the fit\$coef ?) will be 4,5.

If you want to look at the foldchange I guess that 2^fit\$coeff is correctly
calculated, so for the example it will be 2^4,5 = 22,6 times upregulated.

But if you look at the values independently, M=8 will give 2^8 = 256 times,
and 2^1 = 2 times upregulation. The mean of these values are (256 + 2) / 2 =
129 times.

I know that the question is a bit naive, but how should one do when you take
the mean of logarithms since the numbers are not related to each other as
normal numbers are. E.g. the number 8 is not twice the size of 4 on a
logarithmic scale, it is 10000 times more (on a log10 scale).

So....how should one do, when I want to take the average of log values?
Shouldn't I calculate the ratios back (not in log2 scale) and calculate the
mean, and transform the data back, If I would like to have an average M
value?

Regards

Marcus

Marcus Gry Björklund

Royal Institute of Technology
AlbaNova University Center
Department of Molecular Biotechnology
106 91 Stockholm, Sweden

Phone (office): +46 8 553 783 44
Fax: + 46 8 553 784 81
Visiting address: Roslagstullsbacken 21, Floor 3
Web: http://www.biotech.kth.se/molbio/microarray/index.html
-----Original Message-----
From: bioconductor-bounces at stat.math.ethz.ch
[mailto:bioconductor-bounces at stat.math.ethz.ch] On Behalf Of Marcus Davy
Sent: Wednesday, April 27, 2005 15:41
To: ramasamy at cancer.org.uk; naomi at stat.psu.edu; saurin_jani at yahoo.com
Cc: bioconductor at stat.math.ethz.ch
Subject: Re: [BioC] LIMMA : design (1, 2, 3, 3 ) , I gotEXCITINGresults,
what could be the logic, since i h

Im another in agreement that you probably need to increase your biological
replication of microarray slides. Any microarray facility has a certain
experimental sensitivity. If the amount of information you have is low then
the actual distribution of the observed proportion of false discoveries
under simulation can become highly variable as pi_0, the proportion of
truely equilalently expressing genes tends to 1 (depending on your cutoff
alpha level), remembering that FDR control is an expectation for any
realization of an experiment. So your list of differentially expressing
genes could contain many (or few) false discoveries but in the long run
should maintain control at ~pi_0*alpha under nonadaptive FDR control.

Saurin, you mention that you have a large list of differentially expressing
genes using a nonadaptive FDR cutoff of alpha=0.01. As an alternative
approach you could analyze some of the chip comparisons of interest using
EBarrays (after checking distribution assumptions), which models the entire
exprSet using hierarchal Gamma Gamma Bernoulli, or Log Normal Normal
Bernoulli models, modelling the null and alternate distributions as a
mixture with unknown mixing proportion p. The advantage of EBarrays is that
it can analyse a single cDNA two channel array or two affymetrix single
channel arrays, providing reasonable estimates of pi_0. The same problem
applies though with a lack of sensitivity without biological replication
(buts it better than fold change at least). You could see *if* EBarrays
predicts a value of pi_0 which is moderately less than 1 for comparisons of
interest.

Marcus

>>> Naomi Altman <naomi at stat.psu.edu> 27/04/2005 4:55:11 a.m. >>>
Significance should be based on biological replication.  If the 2 chips for
group 3 are technical replicates, then the variance estimate for the test
is probably too small.

In theory, statistical tests need only 2 replicate in a single condition,
as the null distribution accounts for the number of replicates.   However,
for this theory to hold, the normality of the samples must be pretty
good.  When the data are exactly normally distributed (and the assumptions
for limma for the distribution of variance hold) then the FDR values should
be pretty good, but the FNR will be poor (as you have no power).

However, I don't think anyone believes that microarray data are normally
distributed.  So, I would not really trust these results, even if you have
a biological replicate.  Of course the 2-fold rule is even worse, so really
you should do more biological replication.

--Naomi

At 09:51 PM 4/26/2005, Saurin Jani wrote:
>
>Yes, you are right. I have 4 samples :
>
>Group1 = Growth Effect for Day 1 : 1 Affy GeneChip.
>Group2 = Growth Effect for Day 2 : 1 Affy GeneChip.
>Group3 = Growth Effect for Day 3 : 2 Affy GeneChips.
>
>so, my design matrix is:
>design <- model.matrix(~ -1+factor(c(1,2,3,3)));
>
>LIMMA did not give any error or waring even it has 1
>sample per group...! ( I thought similar thing,  since
>it needs technical replicates per group to make a
>decision). The results are very interesting. I have
>many genes for 0.01 FDR, which is very good.
>
>Somehow,I don't understand the logic. Do you think is
>this a valid design? Or You think I should go by Fold
>Change Logic. Please, let me know.
>
>Thank you very much,
>Saurin
>
>
>
>
>
>--- Adaikalavan Ramasamy <ramasamy at cancer.org.uk>
>wrote:
> > PLEASE correct me if I am wrong.
> >
> > You have a total of 4 samples that could be
> > classified into one of 3
> > groups ? How do you plan on distinguishing
> > biological from technical
> > variation ? Shouldn't limma come with some sort of
> > warning or error if
> > there are only one sample per group ?
> >
> >
> >
> >
> > On Tue, 2005-04-26 at 10:01 -0700, Saurin Jani
> > wrote:
> > > Hi BioC,
> > >
> > > I have 3 groups but I have only 2 replicates for
> > last
> > > group. so, group 1 and 2 has only one Affy CEL
> > file. I
> > > Did..LIMMA as below and I got some Exciting
> > results:
> > >
> > > #----------------------------------
> > > design <- model.matrix(~ -1+factor(c(1,2,3,3)));
> > > colnames(design) <-  c("g1","g2","g3");
> > > fit <- lmFit(myRMA,design);
> > >
> > > contrast.matrix <-
> > > makeContrasts(g1-g2,g1-g3,g2-g3,levels = design);
> > >
> > > fit2 <- contrasts.fit(fit,contrast.matrix);
> > > fit2 <- eBayes(fit2);
> > >
> > > results <-
> > >
> > > myGenes <- geneNames(myRMA);
> > > i <- apply(results,c(1,2),all);
> > >
> > > a <- i[,1];
> > > b <- i[,2];
> > > c <- i[,3];
> > > tempgenes1 <- myGenes[a];
> > > tempgenes2 <- myGenes[b];
> > > tempgenes3 <- myGenes[c];
> > >
> > > tempall <- c(tempgenes1,tempgenes2,tempgenes3);
> > > myDEGenes <- tempall;
> > >
> > > esetSub2X <- MatrixRMA[myDEGenes,];
> > > esetSub2 <- new("exprSet",exprs = esetSub2X);
> > > pData(esetSub2) <- pData(myRMA);
> > > heatmap(esetSub2X);
> > > #----------------------------------
> > >
> > > I got EXCITING results, what could be the
> > logic,since
> > > i have 2 replicates for 3rd group only ?
> > >
> > > Could anyone point me out ?
> > >
> > > I highly appreciate your help , Thank you in
> > >
> > > Thank you,
> > > Saurin
> > >
> > > _______________________________________________
> > > Bioconductor mailing list
> > > Bioconductor at stat.math.ethz.ch
> > > https://stat.ethz.ch/mailman/listinfo/bioconductor
> > >
> >
> >
>
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Naomi S. Altman                                814-865-3791 (voice)
Associate Professor
Bioinformatics Consulting Center
Dept. of Statistics                              814-863-7114 (fax)
Penn State University                         814-865-1348 (Statistics)
University Park, PA 16802-2111

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