# 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 ${\displaystyle y_{1},\ldots ,y_{N}}$ the (quantitative) phenotypes (e.g. expression level at a given gene), and ${\displaystyle g_{1},\ldots ,g_{N}}$ 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: ${\displaystyle \forall i\in \{1,\ldots ,N\},\;y_{i}=\mu +\beta _{1}g_{i}+\beta _{2}\mathbf {1} _{g_{i}=1}+\epsilon _{i}{\text{ with }}\epsilon _{i}{\overset {i.i.d}{\sim }}{\mathcal {N}}(0,\tau ^{-1})}$

where ${\displaystyle \beta _{1}}$ is in fact the additive effect of the SNP, noted ${\displaystyle a}$ from now on, and ${\displaystyle \beta _{2}}$ is the dominance effect of the SNP, ${\displaystyle d=ak}$.

Let's now write in matrix notation:

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

which gives the following conditional distribution for the phenotypes:

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

The likelihood of the parameters given the data is therefore:

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

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

• Priors: we use the usual conjugate prior

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

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

${\displaystyle B|\tau \sim {\mathcal {N}}({\vec {0}},\,\tau ^{-1}\Sigma _{B}){\text{ with }}\Sigma _{B}=diag(\sigma _{\mu }^{2},\sigma _{a}^{2},\sigma _{d}^{2})}$

• Joint posterior:

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

• Conditional posterior of B:

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

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

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

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

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

We use the prior and likelihood and keep only the terms in ${\displaystyle B}$:

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

We expand:

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

We factorize some terms:

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

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

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

This now becomes easy to factorizes totally:

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

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

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

• Posterior of ${\displaystyle \tau }$:

Similarly to the equations above:

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

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

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

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 ${\displaystyle B}$!):

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

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

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

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

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

We finally recognize the following Gamma distribution:

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