User:Hussein Alasadi/Notebook/stephens/2013/10/03

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(Notes from Meeting)
(Notes from Meeting)
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Thus <math>  \bar{\mu} = \vec{\mu_2} + \Sigma_{21} \Sigma_{11}^{-1} (x_1 - \mu_1), \bar{\Sigma} = \Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12} </math>
Thus <math>  \bar{\mu} = \vec{\mu_2} + \Sigma_{21} \Sigma_{11}^{-1} (x_1 - \mu_1), \bar{\Sigma} = \Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12} </math>
*'''Likelihood for frequency a the test SNP t given all data'''
<math> L(f_{i,k,t}^{true}) = P(\prod_{j \not= t} f_{i,k,j} | y_j, M) </math>

Revision as of 21:51, 16 October 2013

analyzing pooled sequenced data with selection Main project page
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Notes from Meeting

Consider a single lineage for now.

Xj = frequency of "1" allele at SNP j in the pool (i.e. the true frequency of the 1 allele in the pool)

  • Data:

 (n_j^0, n_j^1) = number of "0", "1" alleles at SNP j ( n_j = n_j^0 + n_j^1 )

  • Normal approximation

 n_j^1 ~ Bin(n_j, X_j) \approx N(n_jX_j, n_jX_j(1-X_j)) Normal approximation to binomial

 \frac{n_j^1}{n_j} \approx N(X_j, \frac{X_j(1-X_j)}{n_j}) The variance of this distribution results from error due to binomial sampling.

To simplify, we just plug in \hat{X_j} = \frac{n_j^1}{n_j} for Xj

 \implies \frac{n_j^1}{n_j} | X_j \approx N(X_j, \frac{\hat{X_j}(1-\hat{X_j})}{n_j})

  • notation

fi,k,j = frequency of reference allele in group i, replicate and SNP j.

 \vec{f_{i,k}} =  vector of frequencies

Without loss of generality, we assume that the putative selected site is site j = 1

  • Model

We assume a prior on our vector of frequencies based on our panel of SNPs (M) of dimension 2mxp

 \vec{f_{i,k}} ~ MVN(μ,Σ)

 \mu = (1-\theta)f^{panel} + \frac{\theta}{2} 1

 \Sigma = (1-\theta)^2 S + \frac{\theta}{2}(1 - \frac{\theta}{2})I

where  S_{i,j} = \sum_{i,j}^{panel} if i = j or  e^{-\frac{\rho_{i,j}}{2m} \sum_{i,j}^{panel}} if i not equal to j

 \theta = \frac{(\sum_{i=1}^{2m-1} \frac{1}{i})^{-1}}{2m + (\sum_{i=1}^{2m-1} \frac{1}{i})^{-1}}

  • at selected site

 log \frac{f_{i,k,1}}{1-f_{i,k,1}} = \mu + \beta g_i + \epsilon_{i,k}

  • conditional distribution

(fi,k,2,....,fi,k,p) | fi,k,1,M ~  MVN(\bar{\mu}, \bar{\Sigma}) The conditional distribution is easily obtained when we use a result derived here.

let X2 = (fi,k,2,....,fi,k,p) and X1 = fi,k,1

X2 | X1,M ~  N(\vec{\mu_2} + \Sigma_{21} \Sigma_{11}^{-1} (x_1 - \mu_1), \Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12})

Thus   \bar{\mu} = \vec{\mu_2} + \Sigma_{21} \Sigma_{11}^{-1} (x_1 - \mu_1), \bar{\Sigma} = \Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12}

  • Likelihood for frequency a the test SNP t given all data

 L(f_{i,k,t}^{true}) = P(\prod_{j \not= t} f_{i,k,j} | y_j, M)

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