Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am CS major, please be patient if my question is not well-stated.

The dataset is quantitative mass spectrometry (MS) data. By labeling proteins of two different samples A and B, we get the relative abundance of 100 to thousands of proteins in A/B. Alongside with this ratio, we can estimate its variance based on the signal intensities.

Wanted: A list of proteins significantly different from the set of all protein ratios.

Most proteins remain unchanged between A and B. The population of log-ratios distributes around 1. The histogram shows a bell shape with fat tails. Two-term Gaussian mixture model has been found to provide a good fit to experimental noise. I suppose it would work good for this data - think of experimental and biological noise.

How to test for significantly different ratios on such a two-term Gaussian mixture model?

Thanks for your responses!

share|cite|improve this question
You might want to cross-post this to the stats site. – J. M. Nov 5 '10 at 14:46
Thanks for the hint. I cross-posted on stats:… – Florian Bw Nov 5 '10 at 15:13
Definitely this is more of a question for the stats site IMO. What error rate are you trying to control exactly? This sounds like the sort of problem where people typically try to control FDR, in which case I'm not sure how you implement it for a mixture model. – guy Oct 5 '11 at 3:29

There is robust detection to test such outliers:=significantly different ratios. It is quite easy here because Gaussian mixture model is a sub model of epsilon contamination model and there exist least favourable densities to perform the likelihood ratio test.

To find such densities and accordingly the likelihood ratio test, you need to solve two non-linear equations and find two constants. Using these coefficients you get the least favourable densities and accordingly the likelihood ratio test.

The best thing here is that robust detection will provide you no loss of performance under outliers. Your performance can never degrade due to the boundedness property of the set of densities you consider around the nominal density.

It is also known that the likelihood ratio test based on least favourable densities is censored version of the original likelihood ratio test. I suggest to clip the likelihood ratio test and check the performance if you dont wanna deal with solving non linear equations.

share|cite|improve this answer

When you say the data fits one distribution or another, what are you plotting?

One hypothesis would be that the two samples are drawn from the same distribution. Then you would expect A/B to be the same over all proteins. If the distribution is the frequency of A/B for all the different proteins you observe you can test the hypothesis if you have a model of the errors in A and B.

How you model the data should be dependent upon how you will use the model. It is not clear to me what the model is for. You can certainly fit any curve through the data.

share|cite|improve this answer
Thanks for pointing it out, I rewrote the text. I hope it is clearer now. I meant of course that the pdf of the distribution fits the density of the data as visualized by a histogram. – Florian Bw Nov 5 '10 at 15:05

If I understand properly, you have a histogram of A/B values for various proteins. You don't need any model to find the most extreme, just find the 2.5 percentile and 97.5 percentile by counting. The proteins outside these bounds are the 5% most extreme.

share|cite|improve this answer
Thanks for your answer. I improved my formulation: I think a two-term Gaussian mixture model is appropriate and like to select - with a p-value - ratios which are significantly different. – Florian Bw Nov 8 '10 at 11:38
I think it is better to upvote instead of thanking. – Seyhmus Güngören Aug 15 '12 at 7:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.