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Mary Putt
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40
@mary-putt-356
Last seen 10.3 years ago
Hi,
I have a question about the output of se.exprs. I normalized a data
set
using expresso and the model from Irizarry et al (2003) Biostatistics
paper.
rma.dta<-expresso(raw.data, normalize.method="quantiles.robust",
bkground.correct="rma", summary.method="medianpolish",
pmcorrect.method="pmonly")
I can extract the expression values using
exprs.rma<-exprs(rma.dta)
and could compute a sample standard deviation from
se.rma<-sqrt(apply(exprs.rma, 1, var)). The square of this estimates
the
variance of the individual expression values.
I can also extract the model-based estimate of the standard deviation
from
se.model.rma<-se.exprs(rma.dta)
I'm not sure whether se.model.rma (squared) estimates the variance of
the individual espression values. For further clarification and
statistical notation pls see the attached pdf file.
This question was motivated by my observation that the sample standard
deviation is substantially larger than the model-based standard
deviation in a dataset that I am working with.
Many thanks in advance, Mary Puttse.
exprs.pdf
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