Variance of M as a function of A in MA plots
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@andrew-harrison-302
Last seen 9.7 years ago
Hi, I've generated several MA/RI plots after using RMA and noticed that the standard deviation in M (fold change) is related to A (average intensity) - to derive the SD in M, only sample fold changes over a limited range in A (a similar calculation to the Quackenbush sliding Z-score). When using RMA, the standard deviation in M is proportional to A, except for the largest intensities. However, when using expresso with all the options set to MAS, the standard deviation is independent of A. I believe the variance in M should be independent of A, in which case this suggests MAS is the way to go. Has anybody else seen this effect, or can suggest a remedy, or more likely a gap in my logic? Best wishes and thanks in advance, Harry -- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~ Dr Andrew Harrison Tel: 44 (0) 207 679 3890 Biomolecular Structure and Modelling Unit Fax: 44 (0) 207 679 7193 Biochemistry and Molecular Biology Dept. University College London Gower Street Email: harry@biochem.ucl.ac.uk London, WC1E 6BT, UK http://www.biochem.ucl.ac.uk/~harry ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~
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Susan Holmes ▴ 120
@susan-holmes-158
Last seen 9.7 years ago
Dear Andrew, This is very often the case, I have found that Wolfgang Huber's vsn package does a very good job of taking out the variance's dependence on the intensity especially in the lower ranges. Best Susan Holmes Statistics On Thu, 22 May 2003, Andrew Harrison wrote: > Hi, > > I've generated several MA/RI plots after using RMA and > noticed that the standard deviation in M (fold change) is related > to A (average intensity) - to derive the SD in M, only sample fold > changes over a limited range in A (a similar calculation to the > Quackenbush sliding Z-score). > > When using RMA, the standard deviation in M is proportional > to A, except for the largest intensities. However, when using > expresso with all the options set to MAS, the standard deviation is > independent of A. I believe the variance in M should be independent > of A, in which case this suggests MAS is the way to go. > > Has anybody else seen this effect, or can suggest a remedy, > or more likely a gap in my logic? > > Best wishes and thanks in advance, > Harry > > -- > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~ > Dr Andrew Harrison Tel: 44 (0) 207 679 3890 > Biomolecular Structure and Modelling Unit Fax: 44 (0) 207 679 7193 > Biochemistry and Molecular Biology Dept. > University College London > Gower Street Email: harry@biochem.ucl.ac.uk > London, WC1E 6BT, UK http://www.biochem.ucl.ac.uk/~harry > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~ > > _______________________________________________ > Bioconductor mailing list > Bioconductor@stat.math.ethz.ch > https://www.stat.math.ethz.ch/mailman/listinfo/bioconductor >
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@rafael-a-irizarry-205
Last seen 9.7 years ago
ive oberved exactly the oposite. SDs for RMA,dCHip and MAS 5.0 depend on A. but MAS 5.0 is by way more dependent. look at http://www.biostat.jhsph.edu/~ririzarr/sd.html for some pictures. keep in mind that RMA gives back log2 values. expresso with mas gives back linear scale values. On Thu, 22 May 2003, Andrew Harrison wrote: > Hi, > > I've generated several MA/RI plots after using RMA and > noticed that the standard deviation in M (fold change) is related > to A (average intensity) - to derive the SD in M, only sample fold > changes over a limited range in A (a similar calculation to the > Quackenbush sliding Z-score). > > When using RMA, the standard deviation in M is proportional > to A, except for the largest intensities. However, when using > expresso with all the options set to MAS, the standard deviation is > independent of A. I believe the variance in M should be independent > of A, in which case this suggests MAS is the way to go. > > Has anybody else seen this effect, or can suggest a remedy, > or more likely a gap in my logic? > > Best wishes and thanks in advance, > Harry > >
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