User:Timothee Flutre/Notebook/Postdoc/2011/11/10

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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 $y_1,\ldots,y_N$ the (quantitative) phenotypes (e.g. expression level at a given gene), and $g_1,\ldots,g_N$ 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: $\forall i \in \{1,\ldots,N\}, \; y_i = \mu + \beta_1 g_i + \beta_2 \mathbf{1}_{g_i=1} + \epsilon_i \text{ with } \epsilon_i \overset{i.i.d}{\sim} \mathcal{N}(0,\tau^{-1})$

where $β 1$ is in fact the additive effect of the SNP, noted $a$ from now on, and $β 2$ is the dominance effect of the SNP, $d = a k$ .

Let's now write in matrix notation:

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

which gives the following conditional distribution for the phenotypes:

$Y | X, B, \tau \sim \mathcal{N}(XB, \tau^{-1} I_N)$

The likelihood of the parameters given the data is therefore:

$\mathcal{L}(\tau, B) = \mathsf{P}(Y | X, \tau, B)$

$\mathcal{L}(\tau, B) = \left(\frac{\tau}{2 \pi}\right)^{N/2} exp \left( -\frac{\tau}{2} (Y - XB)^T (Y - XB) \right)$

Priors: we use the usual conjugate prior

$\mathsf{P}(\tau, B) = \mathsf{P}(\tau) \mathsf{P}(B | \tau)$

$\tau \sim \Gamma(\kappa/2, \, \lambda/2)$

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

Joint posterior:

$\mathsf{P}(\tau, B | Y, X) = \mathsf{P}(\tau | Y, X) \mathsf{P}(B | Y, X, \tau)$

Conditional posterior of B:

$\mathsf{P}(B | Y, X, \tau) = \mathsf{P}(B, Y | X, \tau)$

$\mathsf{P}(B | Y, X, \tau) = \frac{\mathsf{P}(B, Y | X, \tau)}{\mathsf{P}(Y | X, \tau)}$

$\mathsf{P}(B | Y, X, \tau) = \frac{\mathsf{P}(B | \tau) \mathsf{P}(Y | X, B, \tau)}{\int \mathsf{P}(B | \tau) \mathsf{P}(Y | X, \tau, B) \mathsf{d}B}$

Here and in the following, we neglect all constants (e.g. normalization constant, $Y T Y$ , etc):

$\mathsf{P}(B | Y, X, \tau) \propto \mathsf{P}(B | \tau) \mathsf{P}(Y | X, \tau, B)$

We use the prior and likelihood and keep only the terms in $B$ :

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

We expand:

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

We factorize some terms:

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

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

$\mathsf{P}(B | Y, X, \tau) \propto exp(B^T \Omega^{-1} B - (X^TY)^T\Omega^{-1}\Omega^TB -B^T\Omega^{-1}\Omega^TX^TY)$

This now becomes easy to factorizes totally:

$\mathsf{P}(B | Y, X, \tau) \propto exp((B^T - \Omega X^TY)^T\Omega^{-1}(B - \Omega X^TY))$

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

$B | Y, X, \tau \sim \mathcal{N}(\Omega X^TY, \tau^{-1} \Omega)$

Posterior of $τ$ :

Similarly to the equations above:

$\mathsf{P}(\tau | Y, X) \propto \mathsf{P}(\tau) \mathsf{P}(Y | X, \tau)$

But now, to handle the second term, we need to integrate over $B$ , thus effectively taking into account the uncertainty in $B$ :

$\mathsf{P}(\tau | Y, X) \propto \mathsf{P}(\tau) \int \mathsf{P}(B | \tau) \mathsf{P}(Y | X, \tau, B) \mathsf{d}B$

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 $B$ !):

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

As we used a conjugate prior for $τ$ , we know that we expect a Gamma distribution for the posterior. Therefore, we can take $τ N / 2$ out of the integral and start guessing what looks like a Gamma distribution. We also factorize inside the exponential:

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

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

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

We finally recognize the following Gamma distribution:

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

@@ Line 37: / Line 37: @@
 <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>
@@ Line 101: / Line 101: @@
 <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>
 <!-- ##### DO NOT edit below this line unless you know what you are doing. ##### -->

User:Timothee Flutre/Notebook/Postdoc/2011/11/10

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Revision as of 18:22, 21 November 2012

Bayesian model of univariate linear regression for QTL detection

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