User:Timothee Flutre/Notebook/Postdoc/2011/11/10: Difference between revisions

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<math>Y | X, B, \tau \sim \mathcal{N}(XB, \tau^{-1} I_N)</math>
<math>Y | X, B, \tau \sim \mathcal{N}(XB, \tau^{-1} I_N)</math>


The likelihood of the parameters given the data is therefore:


* '''Priors''': conjugate
<math>\mathcal{L}(\tau, B) = \mathsf{P}(Y | X, \tau, B)</math>
 
<math>\mathcal{L}(\tau, B) = \left(\frac{\tau}{2 \pi}\right)^{n/2} exp \left( -\frac{\tau}{2} (Y - XB)^T (Y - XB) \right)</math>
 
 
* '''Priors''': we use the usual conjugate prior
 
<math>\mathsf{P}(\tau, B) = \mathsf{P}(\tau) \mathsf{P}(B | \tau)</math>


<math>\tau \sim \Gamma(\kappa/2, \, \lambda/2)</math>
<math>\tau \sim \Gamma(\kappa/2, \, \lambda/2)</math>

Revision as of 10:57, 21 November 2012

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Bayesian model of univariate linear regression for QTL detection

See Servin & Stephens (PLoS Genetics, 2007).


  • Data: let's assume that we obtained data from N individuals. We note [math]\displaystyle{ y_1,\ldots,y_N }[/math] the (quantitative) phenotypes (e.g. expression level at a given gene), and [math]\displaystyle{ g_1,\ldots,g_N }[/math] the genotypes at a given SNP (as allele dose, 0, 1 or 2).


  • Goal: we want to assess the evidence in the data for an effect of the genotype on the phenotype.


  • Assumptions: the relationship between genotype and phenotype is linear; the individuals are not genetically related; there is no hidden confounding factors in the phenotypes.


  • Likelihood: [math]\displaystyle{ \forall i \in \{1,\ldots,N\}, \; y_i = \mu + \beta_1 g_i + \beta_2 \mathbf{1}_{g_i=1} + \epsilon_i \text{ with } \epsilon_i \overset{i.i.d}{\sim} \mathcal{N}(0,\tau^{-1}) }[/math]

where [math]\displaystyle{ \beta_1 }[/math] is in fact the additive effect of the SNP, noted [math]\displaystyle{ a }[/math] from now on, and [math]\displaystyle{ \beta_2 }[/math] is the dominance effect of the SNP, [math]\displaystyle{ d = a k }[/math].

Let's now write in matrix notation:

[math]\displaystyle{ Y = X B + E \text{ where } B = [ \mu \; a \; d ]^T }[/math]

which gives the following conditional distribution for the phenotypes:

[math]\displaystyle{ Y | X, B, \tau \sim \mathcal{N}(XB, \tau^{-1} I_N) }[/math]

The likelihood of the parameters given the data is therefore:

[math]\displaystyle{ \mathcal{L}(\tau, B) = \mathsf{P}(Y | X, \tau, B) }[/math]

[math]\displaystyle{ \mathcal{L}(\tau, B) = \left(\frac{\tau}{2 \pi}\right)^{n/2} exp \left( -\frac{\tau}{2} (Y - XB)^T (Y - XB) \right) }[/math]


  • Priors: we use the usual conjugate prior

[math]\displaystyle{ \mathsf{P}(\tau, B) = \mathsf{P}(\tau) \mathsf{P}(B | \tau) }[/math]

[math]\displaystyle{ \tau \sim \Gamma(\kappa/2, \, \lambda/2) }[/math]

[math]\displaystyle{ B | \tau \sim \mathcal{N}(\vec{0}, \, \tau^{-1} \Sigma_B) \text{ with } \Sigma_B = diag(\sigma_{\mu}^2, \sigma_a^2, \sigma_d^2) }[/math]


  • Joint posterior:

[math]\displaystyle{ \mathsf{P}(\tau, B | Y, X) = \mathsf{P}(\tau | Y, X) \mathsf{P}(B | Y, X, \tau) }[/math]


  • Conditional posterior of B:

[math]\displaystyle{ \mathsf{P}(B | Y, X, \tau) = \mathsf{P}(B, Y | X, \tau) }[/math]

[math]\displaystyle{ \mathsf{P}(B | Y, X, \tau) = \frac{\mathsf{P}(B, Y | X, \tau)}{\mathsf{P}(Y | X, \tau)} }[/math]

[math]\displaystyle{ \mathsf{P}(B | Y, X, \tau) = \frac{\mathsf{P}(B | \tau) \mathsf{P}(Y | X, B, \tau)}{\int \mathsf{P}(B | \tau) \mathsf{P}(Y | X, \tau, B) \mathsf{d}B} }[/math]

Here and in the following, we neglect all constants (e.g. normalization constant, [math]\displaystyle{ Y^TY }[/math], etc):

[math]\displaystyle{ \mathsf{P}(B | Y, X, \tau) \propto \mathsf{P}(B | \tau) \mathsf{P}(Y | X, \tau, B) }[/math]

We use the prior and likelihood and keep only the terms in [math]\displaystyle{ B }[/math]:

[math]\displaystyle{ \mathsf{P}(B | Y, X, \tau) \propto exp(B^T \Sigma_B^{-1} B) exp((Y-XB)^T(Y-XB)) }[/math]

We expand:

[math]\displaystyle{ \mathsf{P}(B | Y, X, \tau) \propto exp(B^T \Sigma_B^{-1} B - Y^TXB -B^TX^TY + B^TX^TXB) }[/math]

We factorize some terms:

[math]\displaystyle{ \mathsf{P}(B | Y, X, \tau) \propto exp(B^T (\Sigma_B^{-1} + X^TX) B - Y^TXB -B^TX^TY) }[/math]

Let's define [math]\displaystyle{ \Omega = (\Sigma_B^{-1} + X^TX)^{-1} }[/math]. We can see that [math]\displaystyle{ \Omega^T=\Omega }[/math], which means that [math]\displaystyle{ \Omega }[/math] is a symmetric matrix. This is particularly useful here because we can use the following equality: [math]\displaystyle{ \Omega^{-1}\Omega^T=I }[/math].

[math]\displaystyle{ \mathsf{P}(B | Y, X, \tau) \propto exp(B^T \Omega^{-1} B - (X^TY)^T\Omega^{-1}\Omega^TB -B^T\Omega^{-1}\Omega^TX^TY) }[/math]

This now becomes easy to factorizes totally:

[math]\displaystyle{ \mathsf{P}(B | Y, X, \tau) \propto exp((B^T - \Omega X^TY)^T\Omega^{-1}(B - \Omega X^TY)) }[/math]

We recognize the kernel of a Normal distribution, allowing us to write the conditional posterior as:

[math]\displaystyle{ B | Y, X, \tau \sim \mathcal{N}(\Omega X^TY, \tau^{-1} \Omega) }[/math]


  • Posterior of [math]\displaystyle{ \tau }[/math]:

Similarly to the equations above:

[math]\displaystyle{ \mathsf{P}(\tau | Y, X) \propto \mathsf{P}(\tau) \mathsf{P}(Y | X, \tau) }[/math]

But now, to handle the second term, we need to integrate over [math]\displaystyle{ B }[/math], thus effectively taking into account the uncertainty in [math]\displaystyle{ B }[/math]:

[math]\displaystyle{ \mathsf{P}(\tau | Y, X) \propto \mathsf{P}(\tau) \int \mathsf{P}(B | \tau) \mathsf{P}(Y | X, \tau, B) \mathsf{d}B }[/math]


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