User:Timothee Flutre/Notebook/Postdoc/2011/11/10: Difference between revisions
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<math>Y | X, \tau, B \sim \mathcal{N}(XB, \tau^{-1} I_N)</math> | <math>Y | X, \tau, B \sim \mathcal{N}(XB, \tau^{-1} I_N)</math> | ||
Even though we can write the likelihood as a multivariate Normal, I still keep the term "univariate" in the title because the | Even though we can write the likelihood as a multivariate Normal, I still keep the term "univariate" in the title because the regression has a single response, <math>Y</math>. | ||
It is usual to keep the term "multivariate" for the case where there is a matrix of responses (i.e. multiple phenotypes). | |||
The likelihood of the parameters given the data is therefore: | The likelihood of the parameters given the data is therefore: | ||
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* '''Bayes Factor''': to do | * '''Bayes Factor''': one way to answer our goal above ("is there an effect of the genotype on the phenotype?") is to do [http://en.wikipedia.org/wiki/Hypothesis_testing hypothesis testing]. | ||
We want to test the following [http://en.wikipedia.org/wiki/Null_hypothesis null hypothesis]: | |||
<math>H_0: \; a = d = 0</math> | |||
* ''' | In Bayesian modeling, hypothesis testing is performed with a [http://en.wikipedia.org/wiki/Bayes_factor Bayes factor], which in our case can be written as: | ||
<math>BF = \frac{\mathsf{P}(Y | X, a \neq 0, d \neq 0)}{\mathsf{P}(Y | X, a = 0, d = 0)}</math> | |||
We can shorten this into: | |||
<math>BF = \frac{\mathsf{P}(Y | X)}{\mathsf{P}_0(Y)}</math> | |||
Note that, compare to frequentist hypothesis testing which focuses on the null, the Bayes factor requires to explicitly model the data under the alternative. | |||
Let's start with the numerator: | |||
<math>\mathsf{P}(Y | X) = \int \mathsf{P}(\tau) \mathsf{P}(Y | X, \tau) \mathsf{d}\tau</math> | |||
First, let's calculate what is inside the integral: | |||
<math>\mathsf{P}(Y | X, \tau) = \frac{\mathsf{P}(B | \tau) \mathsf{P}(Y | X, \tau, B)}{\mathsf{P}(B | Y, X, \tau)}</math> | |||
Using the formula obtained previously and doing some algebra gives: | |||
<math>\mathsf{P}(Y | X, \tau) = \left( \frac{\tau}{2 \pi} \right)^{\frac{N}{2}} \left( \frac{|\Omega|}{|\Sigma_B|} \right)^{\frac{1}{2}} exp\left( -\frac{\tau}{2} (Y^TY - Y^TX\Omega X^TY) \right)</math> | |||
Now we can integrate out <math>\tau</math> (note the small typo in equation 9 of supplementary text S1 of Servin & Stephens): | |||
<math>\mathsf{P}(Y | X) = (2\pi)^{-\frac{N}{2}} \left( \frac{|\Omega|}{|\Sigma_B|} \right)^{\frac{1}{2}} \frac{\frac{\lambda}{2}^{\frac{\kappa}{2}}}{\Gamma(\frac{\kappa}{2})} \int \tau^{\frac{N+\kappa}{2}-1} exp \left( -\frac{\tau}{2} (Y^TY - Y^TX\Omega X^TY + \lambda) \right)</math> | |||
Inside the integral, we recognize the almost-complete pdf of a Gamma distribution. | |||
As it has to integrate to one, we get: | |||
<math>\mathsf{P}(Y | X) = (2\pi)^{-\frac{N}{2}} \left( \frac{|\Omega|}{|\Sigma_B|} \right)^{\frac{1}{2}} \left( \frac{\lambda}{2} \right)^{\frac{\kappa}{2}} \frac{\Gamma(\frac{N+\kappa}{2})}{\Gamma(\frac{\kappa}{2})} \left( \frac{Y^TY - Y^TX\Omega X^TY + \lambda}{2} \right)^{-\frac{N+\kappa}{2}}</math> | |||
* '''Choosing the hyperparameters''': | |||
invariance properties motivate the use of limits for some "unimportant" hyperparameters | invariance properties motivate the use of limits for some "unimportant" hyperparameters | ||
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* '''R code''': to do | * '''R code''': | ||
to do | |||
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Bayesian model of univariate linear regression for QTL detectionSee Servin & Stephens (PLoS Genetics, 2007).
[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 the model in matrix notation: [math]\displaystyle{ Y = X B + E \text{ where } B = [ \mu \; a \; d ]^T }[/math] This gives the following multivariate Normal distribution for the phenotypes: [math]\displaystyle{ Y | X, \tau, B \sim \mathcal{N}(XB, \tau^{-1} I_N) }[/math] Even though we can write the likelihood as a multivariate Normal, I still keep the term "univariate" in the title because the regression has a single response, [math]\displaystyle{ Y }[/math]. It is usual to keep the term "multivariate" for the case where there is a matrix of responses (i.e. multiple phenotypes). 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)^{\frac{N}{2}} exp \left( -\frac{\tau}{2} (Y - XB)^T (Y - XB) \right) }[/math]
[math]\displaystyle{ \mathsf{P}(\tau, B) = \mathsf{P}(\tau) \mathsf{P}(B | \tau) }[/math] A Gamma distribution for [math]\displaystyle{ \tau }[/math]: [math]\displaystyle{ \tau \sim \Gamma(\kappa/2, \, \lambda/2) }[/math] which means: [math]\displaystyle{ \mathsf{P}(\tau) = \frac{\frac{\lambda}{2}^{\kappa/2}}{\Gamma(\frac{\kappa}{2})} \tau^{\frac{\kappa}{2}-1} e^{-\frac{\lambda}{2} \tau} }[/math] And a multivariate Normal distribution for [math]\displaystyle{ B }[/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] which means: [math]\displaystyle{ \mathsf{P}(B | \tau) = \left(\frac{\tau}{2 \pi}\right)^{\frac{3}{2}} |\Sigma_B|^{-\frac{1}{2}} exp \left(-\frac{\tau}{2} B^T \Sigma_B^{-1} B \right) }[/math]
[math]\displaystyle{ \mathsf{P}(\tau, B | Y, X) = \mathsf{P}(\tau | Y, X) \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] Let's neglect the normalization constant for now: [math]\displaystyle{ \mathsf{P}(B | Y, X, \tau) \propto \mathsf{P}(B | \tau) \mathsf{P}(Y | X, \tau, B) }[/math] Similarly, let's keep only the terms in [math]\displaystyle{ B }[/math] for the moment: [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] Importantly, 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]
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] 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]\displaystyle{ B }[/math]!): [math]\displaystyle{ \mathsf{P}(\tau | Y, X) \propto \tau^{\frac{\kappa}{2} - 1} e^{-\frac{\lambda}{2} \tau} \int \tau^{3/2} \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]\displaystyle{ \tau }[/math], we know that we expect a Gamma distribution for the posterior. Therefore, we can take [math]\displaystyle{ \tau^{N/2} }[/math] out of the integral and start guessing what looks like a Gamma distribution. We also factorize inside the exponential: [math]\displaystyle{ \mathsf{P}(\tau | Y, X) \propto \tau^{\frac{N+\kappa}{2} - 1} e^{-\frac{\lambda}{2} \tau} \int \tau^{3/2} 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]\displaystyle{ B }[/math]. This allows us to use the fact that the pdf of the Normal distribution integrates to one: [math]\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 Y - Y^T X \Omega X^T Y) \right] }[/math] We finally recognize a Gamma distribution, allowing us to write the posterior as: [math]\displaystyle{ \tau | Y, X \sim \Gamma \left( \frac{N+\kappa}{2}, \; \frac{1}{2} (Y^T Y - Y^T X \Omega X^T Y + \lambda) \right) }[/math]
[math]\displaystyle{ B, \tau | Y, X \sim \mathcal{N}IG(\Omega X^TY, \; \tau^{-1}\Omega, \; \frac{N+\kappa}{2}, \; \frac{\lambda^\ast}{2}) }[/math] where [math]\displaystyle{ \lambda^\ast = Y^T Y - Y^T X \Omega X^T Y + \lambda }[/math]
[math]\displaystyle{ \mathsf{P}(B | Y, X) = \int \mathsf{P}(\tau) \mathsf{P}(B | Y, X, \tau) \mathsf{d}\tau }[/math] [math]\displaystyle{ \mathsf{P}(B | Y, X) = \frac{\frac{\lambda^\ast}{2}^{\frac{N+\kappa}{2}}}{(2\pi)^\frac{3}{2} |\Omega|^{\frac{1}{2}} \Gamma(\frac{N+\kappa}{2})} \int \tau^{\frac{N+\kappa+3}{2}-1} exp \left[-\tau \left( \frac{\lambda^\ast}{2} + (B - \Omega X^TY)^T \Omega^{-1} (B - \Omega X^TY) \right) \right] \mathsf{d}\tau }[/math] Here we recognize the formula to integrate the Gamma function: [math]\displaystyle{ \mathsf{P}(B | Y, X) = \frac{\frac{\lambda^\ast}{2}^{\frac{N+\kappa}{2}} \Gamma(\frac{N+\kappa+3}{2})}{(2\pi)^\frac{3}{2} |\Omega|^{\frac{1}{2}} \Gamma(\frac{N+\kappa}{2})} \left( \frac{\lambda^\ast}{2} + (B - \Omega X^TY)^T \Omega^{-1} (B - \Omega X^TY) \right)^{-\frac{N+\kappa+3}{2}} }[/math] And we now recognize a multivariate Student's t-distribution: [math]\displaystyle{ \mathsf{P}(B | Y, X) = \frac{\Gamma(\frac{N+\kappa+3}{2})}{\Gamma(\frac{N+\kappa}{2}) \pi^\frac{3}{2} |\lambda^\ast \Omega|^{\frac{1}{2}} } \left( 1 + \frac{(B - \Omega X^TY)^T \Omega^{-1} (B - \Omega X^TY)}{\lambda^\ast} \right)^{-\frac{N+\kappa+3}{2}} }[/math] We hence can write: [math]\displaystyle{ B | Y, X \sim \mathcal{S}_{N+\kappa}(\Omega X^TY, \; (Y^T Y - Y^T X \Omega X^T Y + \lambda) \Omega) }[/math]
We want to test the following null hypothesis: [math]\displaystyle{ H_0: \; a = d = 0 }[/math] In Bayesian modeling, hypothesis testing is performed with a Bayes factor, which in our case can be written as: [math]\displaystyle{ BF = \frac{\mathsf{P}(Y | X, a \neq 0, d \neq 0)}{\mathsf{P}(Y | X, a = 0, d = 0)} }[/math] We can shorten this into: [math]\displaystyle{ BF = \frac{\mathsf{P}(Y | X)}{\mathsf{P}_0(Y)} }[/math] Note that, compare to frequentist hypothesis testing which focuses on the null, the Bayes factor requires to explicitly model the data under the alternative. Let's start with the numerator: [math]\displaystyle{ \mathsf{P}(Y | X) = \int \mathsf{P}(\tau) \mathsf{P}(Y | X, \tau) \mathsf{d}\tau }[/math] First, let's calculate what is inside the integral: [math]\displaystyle{ \mathsf{P}(Y | X, \tau) = \frac{\mathsf{P}(B | \tau) \mathsf{P}(Y | X, \tau, B)}{\mathsf{P}(B | Y, X, \tau)} }[/math] Using the formula obtained previously and doing some algebra gives: [math]\displaystyle{ \mathsf{P}(Y | X, \tau) = \left( \frac{\tau}{2 \pi} \right)^{\frac{N}{2}} \left( \frac{|\Omega|}{|\Sigma_B|} \right)^{\frac{1}{2}} exp\left( -\frac{\tau}{2} (Y^TY - Y^TX\Omega X^TY) \right) }[/math] Now we can integrate out [math]\displaystyle{ \tau }[/math] (note the small typo in equation 9 of supplementary text S1 of Servin & Stephens): [math]\displaystyle{ \mathsf{P}(Y | X) = (2\pi)^{-\frac{N}{2}} \left( \frac{|\Omega|}{|\Sigma_B|} \right)^{\frac{1}{2}} \frac{\frac{\lambda}{2}^{\frac{\kappa}{2}}}{\Gamma(\frac{\kappa}{2})} \int \tau^{\frac{N+\kappa}{2}-1} exp \left( -\frac{\tau}{2} (Y^TY - Y^TX\Omega X^TY + \lambda) \right) }[/math] Inside the integral, we recognize the almost-complete pdf of a Gamma distribution. As it has to integrate to one, we get: [math]\displaystyle{ \mathsf{P}(Y | X) = (2\pi)^{-\frac{N}{2}} \left( \frac{|\Omega|}{|\Sigma_B|} \right)^{\frac{1}{2}} \left( \frac{\lambda}{2} \right)^{\frac{\kappa}{2}} \frac{\Gamma(\frac{N+\kappa}{2})}{\Gamma(\frac{\kappa}{2})} \left( \frac{Y^TY - Y^TX\Omega X^TY + \lambda}{2} \right)^{-\frac{N+\kappa}{2}} }[/math]
invariance properties motivate the use of limits for some "unimportant" hyperparameters average BF over grid
to do | |
