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

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(Bayesian model of univariate linear regression for QTL detection: add lik and joint prior)
(Bayesian model of univariate linear regression for QTL detection: finish posterior tau)
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<math>\mathcal{L}(\tau, B) = \mathsf{P}(Y | X, \tau, B)</math>
 
<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>
+
<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>
  
  
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<math>\mathsf{P}(\tau | Y, X) \propto \mathsf{P}(\tau) \int \mathsf{P}(B | \tau) \mathsf{P}(Y | X, \tau, B) \mathsf{d}B</math>
 
<math>\mathsf{P}(\tau | Y, X) \propto \mathsf{P}(\tau) \int \mathsf{P}(B | \tau) \mathsf{P}(Y | X, \tau, B) \mathsf{d}B</math>
 +
 +
Again, we use the priors and likelihoods specified above (but everything inside the integral is kept inside it, even if it doesn't depend on <math>B</math>!):
 +
 +
<math>\mathsf{P}(\tau | Y, X) \propto \tau^{\frac{\kappa}{2} - 1} e^{-\frac{\lambda}{2} \tau} \int \tau^{1/2} \tau^{N/2} exp(-\frac{\tau}{2} B^T \Sigma_B^{-1} B) exp(-\frac{\tau}{2} (Y - XB)^T (Y - XB)) \mathsf{d}B</math>
 +
 +
As we used a conjugate prior for <math>\tau</math>, we know that we expect a Gamma distribution for the posterior.
 +
Therefore, we can take <math>\tau^{N/2}</math> out of the integral and start guessing what looks like a Gamma distribution.
 +
We also factorize inside the exponential:
 +
 +
<math>\mathsf{P}(\tau | Y, X) \propto \tau^{\frac{N+\kappa}{2} - 1} e^{-\frac{\lambda}{2} \tau} \int \tau^{1/2} exp \left[-\frac{\tau}{2} \left( (B - \Omega X^T Y)^T \Omega^{-1} (B - \Omega X^T Y) - Y^T X \Omega X^T Y + Y^T Y \right) \right] \mathsf{d}B</math>
 +
 +
We recognize the conditional posterior of <math>B</math>.
 +
This allows us to use the fact that the pdf of the Normal distribution integrates to one:
 +
 +
<math>\mathsf{P}(\tau | Y, X) \propto \tau^{\frac{N+\kappa}{2} - 1} e^{-\frac{\lambda}{2} \tau} exp\left[-\frac{\tau}{2} (Y^T X \Omega X^T Y + Y^T Y) \right]</math>
 +
 +
We finally recognize the following Gamma distribution:
 +
 +
<math>\tau | Y, X \sim \Gamma \left( \frac{N+\kappa}{2}, \; \frac{1}{2} (Y^T X \Omega X^T Y + Y^T Y + \lambda) \right)</math>
  
 
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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]y_1,\ldots,y_N[/math] the (quantitative) phenotypes (e.g. expression level at a given gene), and [math]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]\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]\beta_1[/math] is in fact the additive effect of the SNP, noted [math]a[/math] from now on, and [math]\beta_2[/math] is the dominance effect of the SNP, [math]d = a k[/math].

Let's now write in matrix notation:

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

which gives the following conditional distribution for the phenotypes:

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

The likelihood of the parameters given the data is therefore:

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


  • Conditional posterior of B:

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

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

[math]\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]Y^TY[/math], etc):

[math]\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]B[/math]:

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

We expand:

[math]\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]\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]\Omega = (\Sigma_B^{-1} + X^TX)^{-1}[/math]. We can see that [math]\Omega^T=\Omega[/math], which means that [math]\Omega[/math] is a symmetric matrix. This is particularly useful here because we can use the following equality: [math]\Omega^{-1}\Omega^T=I[/math].

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


  • Posterior of [math]\tau[/math]:

Similarly to the equations above:

[math]\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]B[/math], thus effectively taking into account the uncertainty in [math]B[/math]:

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

Again, we use the priors and likelihoods specified above (but everything inside the integral is kept inside it, even if it doesn't depend on [math]B[/math]!):

[math]\mathsf{P}(\tau | Y, X) \propto \tau^{\frac{\kappa}{2} - 1} e^{-\frac{\lambda}{2} \tau} \int \tau^{1/2} \tau^{N/2} exp(-\frac{\tau}{2} B^T \Sigma_B^{-1} B) exp(-\frac{\tau}{2} (Y - XB)^T (Y - XB)) \mathsf{d}B[/math]

As we used a conjugate prior for [math]\tau[/math], we know that we expect a Gamma distribution for the posterior. Therefore, we can take [math]\tau^{N/2}[/math] out of the integral and start guessing what looks like a Gamma distribution. We also factorize inside the exponential:

[math]\mathsf{P}(\tau | Y, X) \propto \tau^{\frac{N+\kappa}{2} - 1} e^{-\frac{\lambda}{2} \tau} \int \tau^{1/2} exp \left[-\frac{\tau}{2} \left( (B - \Omega X^T Y)^T \Omega^{-1} (B - \Omega X^T Y) - Y^T X \Omega X^T Y + Y^T Y \right) \right] \mathsf{d}B[/math]

We recognize the conditional posterior of [math]B[/math]. This allows us to use the fact that the pdf of the Normal distribution integrates to one:

[math]\mathsf{P}(\tau | Y, X) \propto \tau^{\frac{N+\kappa}{2} - 1} e^{-\frac{\lambda}{2} \tau} exp\left[-\frac{\tau}{2} (Y^T X \Omega X^T Y + Y^T Y) \right][/math]

We finally recognize the following Gamma distribution:

[math]\tau | Y, X \sim \Gamma \left( \frac{N+\kappa}{2}, \; \frac{1}{2} (Y^T X \Omega X^T Y + Y^T Y + \lambda) \right)[/math]