User:Hussein Alasadi/Notebook/stephens/2013/10/03
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*'''Likelihood for frequency a the test SNP t given all data'''  *'''Likelihood for frequency a the test SNP t given all data'''  
  <math> L(f_{i,k,t}^{true}) = P(  +  
+  let <math>f_{obs} = \prod_{j \not= t} f_{i,k,j} </math>  
+  
+  <math> L(f_{i,k,t}^{true}) = P(f_{obs}  f_{i,k,t}^{true}, M) = \frac{P(_j^{true}  M, f_{obs} P(f^{obs}M)}{P(f_{i,k,t}^{true}  M}</math>  
Revision as of 21:03, 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
let
