Question: limma: parameter s0^2 for prior of residual variances
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gravatar for Homer
12 days ago by
Homer0
Homer0 wrote:

The limma User's Guide says on p. 62 that the parameter (s_0)^2 is the mean of the inverse chi-squared prior for the true residual variances (sigma_g)^2. However, if I get

Smyth, G. K. (2004). Linear Models and Empirical Bayes Methods for Assessing Differential Expression in Microarray Experiments. Statistical Applications in Genetics and Molecular Biology, 3(1), 1–25. https://doi.org/10.2202/1544-6115.1027

correctly (more precisely: section 3 (p. 6 bottom)), the prior for the (sigma_g)^2 is rather a scaled inverse chi-squared distribution with scaling parameter (s_0)^2. In that case, the mean is d_0 * (s_0)^2 / (d_0 - 2) (see Wikipedia). So my question is: Am I right that the correct mean for the prior of the (sigma_g)^2 is d_0 * (s_0)^2 / (d_0 - 2) and not (s_0)^2 ?

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ADD COMMENTlink modified 11 days ago by Gordon Smyth38k • written 12 days ago by Homer0
Answer: limma: parameter s0^2 for prior of residual variances
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gravatar for Gordon Smyth
11 days ago by
Gordon Smyth38k
Walter and Eliza Hall Institute of Medical Research, Melbourne, Australia
Gordon Smyth38k wrote:

Please see the errata to my 2004 paper in SAGMB, which is on the 3rd page of the pdf from the journal website. Or see the reprint incorporating corrections that is available from my website http://www.statsci.org/smyth/pubs/2003to2014.html

The exponential family canonical parameter here is 1/sigma_g^2, often called the precision parameter, rather than sigma^2_g itself. The expected value of the prior distribution for 1/sigma_g^2 is 1/(s^2_0). The Bayesian calculations are done for the precision rather than for the variance.

The prior distribution for sigma^2_g, on the other hand, does not necessarily have a finite mean or variance and is not very useful.

ADD COMMENTlink modified 4 days ago • written 11 days ago by Gordon Smyth38k

Thank you for your answer that I understand in the way that the limma User's Guide is wrong. Your answer doesn't say that my formula for the prior mean of the (sigma_g)^2 is wrong and the corrected version of the publication still contains the scaled inverse chi-squared distribution as the prior for the (sigma_g)^2 (see p.8 at the top). So I guess that my formula d_0 * (s_0)^2 / (d_0 - 2) is correct. But correct me if I am wrong.

ADD REPLYlink modified 11 days ago • written 11 days ago by Homer0

The published paper is exactly correct and is already as clear as I can make it. I have nothing more to add.

It is not my responsibility to comment on all the probability calculations you make. The prior mean of sigma^2 is not relevant to limma, for the reasons I already explained.

ADD REPLYlink modified 7 days ago • written 10 days ago by Gordon Smyth38k
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