User:Hussein Alasadi/Notebook/stephens/2013/10/03
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let <math> X_2 = (f_{i,k,2}, .... , f_{i,k,p}) </math> and <math> X_1 = f_{i,k,1} </math>  let <math> X_2 = (f_{i,k,2}, .... , f_{i,k,p}) </math> and <math> X_1 = f_{i,k,1} </math>  
  <math> X_2  X_1, M </math> ~ <math> N(\vec{\mu_2} + \  +  <math> X_2  X_1, M </math> ~ <math> N(\vec{\mu_2} + \sum_{21} \sum_{11}^{1} (x_1  \mu_1), \sum_{22}  \sum_{21}\sum_{11}^{1}\sum_{12}) </math> 
Revision as of 21:14, 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 ~
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 ~
