User:Hussein Alasadi/Notebook/stephens/2013/10/03
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We assume a prior on our vector of frequencies based on our panel of SNPs <math> (M) </math> of dimension <math> 2mxp </math>  We assume a prior on our vector of frequencies based on our panel of SNPs <math> (M) </math> of dimension <math> 2mxp </math>  
  <math> \vec{f_{i,k}} </math> ~ <math> MVN(\mu, \  +  <math> \vec{f_{i,k}} </math> ~ <math> MVN(\mu, \Sigma) </math> 
<math> \mu = (1\theta)f^{panel} + \frac{\theta}{2} 1 </math>  <math> \mu = (1\theta)f^{panel} + \frac{\theta}{2} 1 </math>  
  <math> \  +  <math> \Sigma = (1\theta)^2 S + \frac{\theta}{2}(1  \frac{\theta}{2})I </math> 
where <math> S_{i,j} = \sum_{i,j}^{panel}</math> if i = j or <math> e^{\frac{\rho_{i,j}}{2m} \sum_{i,j}^{panel}} </math> if i not equal to j  where <math> S_{i,j} = \sum_{i,j}^{panel}</math> if i = j or <math> e^{\frac{\rho_{i,j}}{2m} \sum_{i,j}^{panel}} </math> if i not equal to j 
Revision as of 20:22, 16 October 2013
analyzing pooled sequenced data with selection  Main project page Next entry 
Notes from MeetingConsider a single lineage for now. X_{j} = frequency of "1" allele at SNP j in the pool (i.e. the true frequency of the 1 allele in the pool)
= number of "0", "1" alleles at SNP j ()
~ Normal approximation to binomial The variance of this distribution results from error due to binomial sampling. To simplify, we just plug in for X_{j}
f_{i,k,j} = frequency of reference allele in group i, replicate and SNP j. vector of frequencies Without loss of generality, we assume that the putative selected site is site j = 1
We assume a prior on our vector of frequencies based on our panel of SNPs (M) of dimension 2mxp ~ MVN(μ,Σ)
where if i = j or if i not equal to j
(f_{i,k,2},....,f_{i,k,p})  f_{i,k,1},M ~ The conditional distribution is easily obtained when we use a result derived here. let X_{2} = (f_{i,k,2},....,f_{i,k,p}) and X_{1} = f_{i,k,1} X_{2}  X_{1},M ~ Thus
