Difference between revisions of "User:Timothee Flutre/Notebook/Postdoc/2011/11/10"
(Autocreate 2011/11/10 Entry for User:Timothee_Flutre/Notebook/Postdoc) 
m (→Bayesian model of univariate linear regression for QTL detection) 

(28 intermediate revisions by the same user not shown)  
Line 6:  Line 6:  
 colspan="2"   colspan="2"  
<! ##### DO NOT edit above this line unless you know what you are doing. ##### >  <! ##### DO NOT edit above this line unless you know what you are doing. ##### >  
−  ==  +  ==Bayesian model of univariate linear regression for QTL detection== 
−  
+  
+  ''This page aims at helping people like me, interested in quantitative genetics, to get a better understanding of some Bayesian models, most importantly the impact of the modeling assumptions as well as the underlying maths. It starts with a simple model, and gradually increases the scope to relax assumptions. See references to scientific articles at the end.''  
+  
+  
+  * '''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 levels at a given gene), and <math>g_1,\ldots,g_N</math> the genotypes at a given SNP (encoded 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''': we start by writing the usual linear regression for one individual  
+  
+  <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 the model in matrix notation:  
+  
+  <math>Y = X B + E \text{ where } B = [ \mu \; a \; d ]^T</math>  
+  
+  This gives the following [http://en.wikipedia.org/wiki/Multivariate_normal_distribution multivariate Normal distribution] for the phenotypes:  
+  
+  <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 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:  
+  
+  <math>\mathcal{L}(\tau, B) = \mathsf{P}(Y  X, \tau, B)</math>  
+  
+  <math>\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>  
+  
+  
+  * '''Priors''': we use the usual [http://en.wikipedia.org/wiki/Conjugate_prior conjugate prior]  
+  
+  <math>\mathsf{P}(\tau, B) = \mathsf{P}(\tau) \mathsf{P}(B  \tau)</math>  
+  
+  A [http://en.wikipedia.org/wiki/Gamma_distribution Gamma distribution] for <math>\tau</math>:  
+  
+  <math>\tau \sim \Gamma(\kappa/2, \, \lambda/2)</math>  
+  
+  which means:  
+  
+  <math>\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>B</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>  
+  
+  which means:  
+  
+  <math>\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>  
+  
+  
+  * '''Joint posterior (1)''':  
+  
+  <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) = \frac{\mathsf{P}(B, Y  X, \tau)}{\mathsf{P}(Y  X, \tau)}</math>  
+  
+  Let's neglect the normalization constant for now:  
+  
+  <math>\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>B</math> for the moment:  
+  
+  <math>\mathsf{P}(B  Y, X, \tau) \propto exp(B^T \Sigma_B^{1} B) exp((YXB)^T(YXB))</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>  
+  
+  Importantly, 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 [http://en.wikipedia.org/wiki/Symmetric_matrix 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 [http://en.wikipedia.org/wiki/Kernel_%28statistics%29 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^{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>\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^{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>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 Y  Y^T X \Omega X^T Y) \right]</math>  
+  
+  We finally recognize a Gamma distribution, allowing us to write the posterior as:  
+  
+  <math>\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>  
+  
+  
+  * '''Joint posterior (2)''': sometimes it is said that the joint posterior follows a Normal Inverse Gamma distribution:  
+  
+  <math>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>\lambda^\ast = Y^T Y  Y^T X \Omega X^T Y + \lambda</math>  
+  
+  
+  * '''Marginal posterior of B''': we can now integrate out <math>\tau</math>:  
+  
+  <math>\mathsf{P}(B  Y, X) = \int \mathsf{P}(\tau) \mathsf{P}(B  Y, X, \tau) \mathsf{d}\tau</math>  
+  
+  <math>\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 [http://en.wikipedia.org/wiki/Gamma_function#Integration_problems integrate the Gamma function]:  
+  
+  <math>\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 [http://en.wikipedia.org/wiki/Multivariate_tdistribution multivariate Student's tdistribution]:  
+  
+  <math>\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>B  Y, X \sim \mathcal{S}_{N+\kappa}(\Omega X^TY, \; (Y^T Y  Y^T X \Omega X^T Y + \lambda) \Omega)</math>  
+  
+  
+  * '''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>\mathrm{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>\mathrm{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.  
+  This makes a big difference when interpreting the results (see below).  
+  
+  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 almostcomplete 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>  
+  
+  We can use this expression also under the null.  
+  In this case, as we need neither <math>a</math> nor <math>d</math>, <math>B</math> is simply <math>\mu</math>, <math>\Sigma_B</math> is <math>\sigma_{\mu}^2</math> and <math>X</math> is a vector of 1's.  
+  We can also defines <math>\Omega_0 = ((\sigma_{\mu}^2)^{1} + N)^{1}</math>.  
+  In the end, this gives:  
+  
+  <math>\mathsf{P}_0(Y) = (2\pi)^{\frac{N}{2}} \frac{\Omega_0^{\frac{1}{2}}}{\sigma_{\mu}} \left( \frac{\lambda}{2} \right)^{\frac{\kappa}{2}} \frac{\Gamma(\frac{N+\kappa}{2})}{\Gamma(\frac{\kappa}{2})} \left( \frac{Y^TY  \Omega_0 N^2 \bar{Y}^2 + \lambda}{2} \right)^{\frac{N+\kappa}{2}}</math>  
+  
+  We can therefore write the Bayes factor:  
+  
+  <math>\mathrm{BF} = \left( \frac{\Omega}{\Omega_0} \right)^{\frac{1}{2}} \frac{1}{\sigma_a \sigma_d} \left( \frac{Y^TY  Y^TX\Omega X^TY + \lambda}{Y^TY  \Omega_0 N^2 \bar{Y}^2 + \lambda} \right)^{\frac{N+\kappa}{2}}</math>  
+  
+  When the Bayes factor is large, we say that there is enough evidence in the data to ''support the alternative''.  
+  Indeed, the Bayesian testing procedure corresponds to measuring support for the specific alternative hypothesis compared to the null hypothesis.  
+  Importantly, note that, for a frequentist testing procedure, we would say that there is enough evidence in the data to ''reject the null''.  
+  However we wouldn't say anything about the alternative as we don't model it.  
+  
+  The threshold to say that a Bayes factor is large depends on the field. It is possible to use the Bayes factor as a test statistic when doing permutation testing, and then control the false discovery rate. This can give an idea of a reasonable threshold.  
+  
+  
+  * '''Hyperparameters''': the model has 5 hyperparameters, <math>\{\kappa, \, \lambda, \, \sigma_{\mu}, \, \sigma_a, \, \sigma_d\}</math>. How should we choose them?  
+  Such a question is never easy to answer. But note that all hyperparameters are not that important, especially in typical quantitative genetics applications. For instance, we are mostly interested in those that determine the magnitude of the effects, <math>\sigma_a</math> and <math>\sigma_d</math>, so let's deal with the others first.  
+  
+  As explained in Servin & Stephens, the posteriors for <math>\tau</math> and <math>B</math> change appropriately with shifts (<math>y+c</math>) and scaling (<math>y \times c</math>) in the phenotype when taking their limits.  
+  This also gives us a new Bayes factor, the one used in practice (see Guan & Stephens, 2008):  
+  
+  <math>\mathrm{lim}_{\sigma_{\mu} \rightarrow \infty \; ; \; \lambda \rightarrow 0 \; ; \; \kappa \rightarrow 0 } \; \mathrm{BF} = \left( \frac{N}{\Sigma_B^{1} + X^TX} \right)^{\frac{1}{2}} \frac{1}{\sigma_a \sigma_d} \left( \frac{Y^TY  Y^TX (\Sigma_B^{1} + X^TX)^{1} X^TY}{Y^TY  N \bar{Y}^2} \right)^{\frac{N}{2}}</math>  
+  
+  Now, for the important hyperparameters, <math>\sigma_a</math> and <math>\sigma_d</math>, it is usual to specify a grid of values, i.e. <math>M</math> pairs <math>(\sigma_a, \sigma_d)</math>. For instance, Guan & Stephens used the following grid:  
+  
+  <math>M=4 \; ; \; \sigma_a \in \{0.05, 0.1, 0.2, 0.4\} \; ; \; \sigma_d = \frac{\sigma_a}{4}</math>  
+  
+  Then, we can average the Bayes factors obtained over the grid using, as a first approximation, equal weights:  
+  
+  <math>\mathrm{BF} = \sum_{m \, \in \, \text{grid}} \frac{1}{M} \, \mathrm{BF}(\sigma_a^{(m)}, \sigma_d^{(m)})</math>  
+  
+  In eQTL studies, the weights can be estimated from the data using a hierarchical model (see below), by pooling all genes together as in Veyrieras ''et al'' (PLoS Genetics, 2010).  
+  
+  
+  * '''Implementation''': the following R function is adapted from Servin & Stephens supplementary text 1.  
+  
+  <nowiki>  
+  BF < function(G=NULL, Y=NULL, sigma.a=NULL, sigma.d=NULL, get.log10=TRUE){  
+  stopifnot(! is.null(G), ! is.null(Y), ! is.null(sigma.a), ! is.null(sigma.d))  
+  subset < complete.cases(Y) & complete.cases(G)  
+  Y < Y[subset]  
+  G < G[subset]  
+  stopifnot(length(Y) == length(G))  
+  N < length(G)  
+  X < cbind(rep(1,N), G, G == 1)  
+  inv.Sigma.B < diag(c(0, 1/sigma.a^2, 1/sigma.d^2))  
+  inv.Omega < inv.Sigma.B + t(X) %*% X  
+  inv.Omega0 < N  
+  tY.Y < t(Y) %*% Y  
+  log10.BF < as.numeric(0.5 * log10(inv.Omega0)   
+  0.5 * log10(det(inv.Omega))   
+  log10(sigma.a)  log10(sigma.d)   
+  (N/2) * (log10(tY.Y  t(Y) %*% X %*% solve(inv.Omega)  
+  %*% t(X) %*% cbind(Y))   
+  log10(tY.Y  N*mean(Y)^2)))  
+  if(get.log10)  
+  return(log10.BF)  
+  else  
+  return(10^log10.BF)  
+  }  
+  </nowiki>  
+  
+  In the same vein as what is explained [http://openwetware.org/wiki/User:Timothee_Flutre/Notebook/Postdoc/2011/06/28 here], we can simulate data under different scenarios and check the BFs:  
+  
+  <nowiki>  
+  N < 300 # play with it  
+  PVE < 0.1 # play with it  
+  grid < c(0.05, 0.1, 0.2, 0.4, 0.8, 1.6, 3.2)  
+  MAF < 0.3  
+  G < rbinom(n=N, size=2, prob=MAF)  
+  tau < 1  
+  a < sqrt((2/5) * (PVE / (tau * MAF * (1MAF) * (1PVE))))  
+  d < a / 2  
+  mu < rnorm(n=1, mean=0, sd=10)  
+  Y < mu + a * G + d * (G == 1) + rnorm(n=N, mean=0, sd=tau)  
+  for(m in 1:length(grid))  
+  print(BF(G, Y, grid[m], grid[m]/4))  
+  </nowiki>  
+  
+  
+  * '''Binary phenotype''': using a similar notation, we model casecontrol studies with a [http://en.wikipedia.org/wiki/Logistic_regression logistic regression] where the probability to be a case is <math>\mathsf{P}(y_i = 1) = p_i</math>.  
+  
+  There are many equivalent ways to write the likelihood, the usual one being:  
+  
+  <math>y_i  p_i \; \overset{i.i.d}{\sim} \; Bernoulli(p_i)</math> with the [http://en.wikipedia.org/wiki/Logodds logodds] (logit function) being <math>\mathrm{ln} \frac{p_i}{1  p_i} = \mu + a \, g_i + d \, \mathbf{1}_{g_i=1}</math>  
+  
+  Let's use <math>X_i^T=[1 \; g_i \; \mathbf{1}_{g_i=1}]</math> to denote the <math>i</math>th row of the design matrix <math>X</math>. We can also keep the same definition as above for <math>B=[\mu \; a \; d]^T</math>. Thus we have:  
+  
+  <math>p_i = \frac{e^{X_i^TB}}{1 + e^{X_i^TB}}</math>  
+  
+  As the <math>y_i</math>'s can only take <math>0</math> and <math>1</math> as values, the likelihood can be written as:  
+  
+  <math>\mathcal{L}(B) = \mathsf{P}(Y  X, B) = \prod_{i=1}^N p_i^{y_i} (1p_i)^{1y_i}</math>  
+  
+  We still use the same prior as above for <math>B</math> (but there is no <math>\tau</math> anymore), so that:  
+  
+  <math>B  \Sigma_B \sim \mathcal{N}_3(0, \Sigma_B)</math>  
+  
+  where <math>\Sigma_B</math> is a 3 x 3 matrix with <math>[\sigma_\mu^2 \; \sigma_a^2 \; \sigma_d^2]</math> on the diagonal and 0 elsewhere.  
+  
+  As above, the Bayes factor is used to compare the two models:  
+  
+  <math>\mathrm{BF} = \frac{\mathsf{P}(Y  X, M1)}{\mathsf{P}(Y  X, M0)} = \frac{\mathsf{P}(Y  X, a \neq 0, d \neq 0)}{\mathsf{P}(Y  X, a=0, d=0)} = \frac{\int \mathsf{P}(B) \mathsf{P}(Y  X, B) \mathrm{d}B}{\int \mathsf{P}(\mu) \mathsf{P}(Y  X, \mu) \mathrm{d}\mu}</math>  
+  
+  The interesting point here is that there is no way to analytically calculate these integrals (marginal likelihoods). Therefore, we will use [http://en.wikipedia.org/wiki/Laplace_approximation Laplace's method] to approximate them, as in Guan & Stephens (2008).  
+  
+  Starting with the numerator:  
+  
+  <math>\mathsf{P}(YX,M1) = \int \exp \left[ N \left( \frac{1}{N} \mathrm{ln} \, \mathsf{P}(B) + \frac{1}{N} \mathrm{ln} \, \mathsf{P}(Y  X, B) \right) \right] \mathsf{d}B</math>  
+  
+  <math>\mathsf{P}(YX,M1) = \int \exp \left\{ N \left[ \frac{1}{N} \left( \mathrm{ln} \left( (2 \pi)^{\frac{3}{2}} \, \frac{1}{\sigma_\mu \sigma_a \sigma_d} \, \exp\left( \frac{1}{2} (\frac{\mu^2}{\sigma_\mu^2} + \frac{a^2}{\sigma_a^2} + \frac{d^2}{\sigma_d^2}) \right) \right) \right) + \frac{1}{N} \left( \sum_{i=1}^N \left( y_i \, \mathrm{ln} (p_i) + (1y_i) \, \mathrm{ln} (1p_i) \right) \right) \right] \right\} \mathsf{d}B</math>  
+  
+  Let's use <math>f</math> to denote the function inside the exponential:  
+  
+  <math>\mathsf{P}(YX,M1) = \int \exp \left( N \; f(B) \right) \mathsf{d}B</math>  
+  
+  The function <math>f</math> is defined by:  
+  
+  <math>f: \mathbb{R}^3 \rightarrow \mathbb{R}</math>  
+  
+  <math>f(B) = \frac{1}{N} \left( \frac{3}{2} \mathrm{ln}(2 \pi)  \frac{1}{2} \mathrm{ln}(\Sigma_B)  \frac{1}{2}(B^T \Sigma_B^{1} B) \right) + \frac{1}{N} \sum_{i=1}^N \left( y_i \, X_i^T B  \mathrm{ln}(1 + e^{X_i^TB}) \right)</math>  
+  
+  This function will then be used to approximate the integral, like this:  
+  
+  <math>\mathsf{P}(YX,M1) \approx N^{3/2} (2 \pi)^{3/2} H(B^\star)^{1/2} e^{N f(B^\star)}</math>  
+  
+  where <math>H</math> is the [http://en.wikipedia.org/wiki/Hessian_matrix Hessian] of <math>f</math> and <math>B^\star = [\mu^\star a^\star d^\star]^T</math> is the point at which <math>f</math> is maximized.  
+  
+  We therefore need to find <math>B^\star</math>. As it maximizes <math>f</math>, we need to calculate the first derivatives of <math>f</math>. Let's do this the univariate way:  
+  
+  <math>\frac{\partial f}{\partial \beta} =  \frac{\beta}{N \, \sigma_\beta^2} + \frac{1}{N} \sum_{i=1}^N \left(\frac{y_i}{p_i}  \frac{1y_i}{1p_i} \right) \frac{\partial p_i}{\partial \beta}</math>  
+  
+  where <math>\beta</math> is <math>\mu</math>, <math>a</math> or <math>d</math>.  
+  
+  A simple form for the first derivatives of <math>p_i</math> also exists when writing <math>p_i = e^{X_i^tB} (1 + e^{X_i^tB})^{1}</math>:  
+  
+  <math>\frac{\partial p_i}{\partial \beta} = \left[ e^{X_i^tB} (1 + e^{X_i^tB})^{1} + e^{X_i^tB} \left( e^{X_i^tB} (1 + e^{X_i^tB})^{2} \right) \right] \frac{\partial X_i^TB}{\partial \beta}</math>  
+  
+  <math>\frac{\partial p_i}{\partial \beta} = \left[ \frac{e^{X_i^tB} (1 + e^{X_i^tB})  (e^{X_i^tB})^2}{(1 + e^{X_i^tB})^2} \right] \frac{\partial X_i^TB}{\partial \beta}</math>  
+  
+  <math>\frac{\partial p_i}{\partial \beta} = \left[ p_i (1  p_i) \right] \frac{\partial X_i^TB}{\partial \beta}</math>  
+  
+  where <math>\frac{\partial X_i^TB}{\partial \beta}</math> is equal to <math>1, \, g_i, \, \mathbf{1}_{g_i=1}</math> when <math>\beta</math> corresponds respectively to <math>\mu, \, a, \, d</math>.  
+  
+  This simplifies the first derivatives of <math>f</math> into:  
+  
+  <math>\frac{\partial f}{\partial \beta} =  \frac{\beta}{N \, \sigma_\beta^2} + \frac{1}{N} \sum_{i=1}^N (y_i  p_i ) \frac{\partial X_i^TB}{\partial \beta}</math>  
+  
+  When setting <math>\frac{\partial f}{\partial \beta}(\beta^\star) = 0</math>, we observe that <math>\beta^\star</math> is present not only alone but also inside the sum, in the <math>p_i</math>'s: indeed <math>p_i</math> is a nonlinear function of <math>B</math>. This means that an iterative procedure is required, typically [http://en.wikipedia.org/wiki/Newton_method_in_optimization Newton's method].  
+  
+  To use it, we need the second derivatives of <math>f</math>:  
+  
+  <math>\frac{\partial^2 f}{\partial \beta^2} =  \frac{1}{N \, \sigma_\beta^2} + \frac{1}{N} \sum_{i=1}^N \left[ (p_i(1p_i)\frac{\partial X_i^TB}{\partial \beta}) + (y_ip_i)\frac{\partial^2 X_i^TB}{\partial \beta^2} \right]</math>  
+  
+  The second derivatives of <math>X_i^TB</math> are all equal to 0:  
+  
+  <math>\frac{\partial^2 f}{\partial \beta^2} =  \frac{1}{N \, \sigma_\beta^2}  \frac{1}{N} \sum_{i=1}^N p_i(1p_i)\frac{\partial X_i^TB}{\partial \beta}</math>  
+  
+  Note that the second derivatives of <math>f</math> are strictly negative. Therefore, <math>f</math> is globally convex, which means that it has a unique global maximum, at <math>B^\star</math>. As a consequence, we have the right to use Laplace's method to approximate the integral around its maximum.  
+  
+  finding the maximums: iterative procedure, update equations or generic solver > to do  
+  
+  implementation: in R > to do  
+  
+  finding the effect sizes and their std error: to do  
+  
+  
+  * '''Link between Bayes factor and Pvalue''': see Wakefield (2008)  
+  
+  to do  
+  
+  
+  * '''Hierarchical model''': pooling genes, learn weights for grid and genomic annotations, see Veyrieras ''et al'' (PLoS Genetics, 2010)  
+  
+  to do  
+  
+  
+  * '''Multiple SNPs with LD''': joint analysis of multiple SNPs, handle correlation between them, see Guan & Stephens (Annals of Applied Statistics, 2011) for MCMC, see Carbonetto & Stephens (Bayesian Analysis, 2012) for Variational Bayes  
+  
+  to do  
+  
+  
+  * '''Confounding factors in phenotype''': factor analysis, see Stegle ''et al'' (PLoS Computational Biology, 2010)  
+  
+  to do  
+  
+  
+  * '''Genetic relatedness''': linear mixed model, see Zhou & Stephens (Nature Genetics, 2012)  
+  
+  to do  
+  
+  
+  * '''Discrete phenotype''': count data as from RNAseq, Poissonlike likelihood, see Sun (Biometrics, 2012)  
+  
+  to do  
+  
+  
+  * '''Multiple phenotypes''': matrixvariate distributions, tensors  
+  
+  to do  
+  
+  
+  * '''Nonindependent genes''': enrichment in known pathways, learn "modules"  
+  
+  to do  
+  
+  
+  * '''References''':  
+  ** Servin & Stephens (PLoS Genetics, 2007)  
+  ** Guan & Stephens (PLoS Genetics, 2008)  
+  ** Stephens & Balding (Nature Reviews Genetics, 2009)  
<! ##### DO NOT edit below this line unless you know what you are doing. ##### >  <! ##### DO NOT edit below this line unless you know what you are doing. ##### > 
Revision as of 14:41, 3 February 2013
Project name  <html><img src="/images/9/94/Report.png" border="0" /></html> Main project page <html><img src="/images/c/c3/Resultset_previous.png" border="0" /></html>Previous entry<html> </html>Next entry<html><img src="/images/5/5c/Resultset_next.png" border="0" /></html> 
Bayesian model of univariate linear regression for QTL detectionThis page aims at helping people like me, interested in quantitative genetics, to get a better understanding of some Bayesian models, most importantly the impact of the modeling assumptions as well as the underlying maths. It starts with a simple model, and gradually increases the scope to relax assumptions. See references to scientific articles at the end.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \beta_1} is in fact the additive effect of the SNP, noted Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle a} from now on, and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \beta_2} is the dominance effect of the SNP, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle d = a k} . Let's now write the model in matrix notation: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle Y = X B + E \text{ where } B = [ \mu \; a \; d ]^T} This gives the following multivariate Normal distribution for the phenotypes: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle Y  X, \tau, B \sim \mathcal{N}(XB, \tau^{1} I_N)} 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, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle Y} . 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: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathcal{L}(\tau, B) = \mathsf{P}(Y  X, \tau, B)} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathsf{P}(\tau, B) = \mathsf{P}(\tau) \mathsf{P}(B  \tau)} A Gamma distribution for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \tau} : Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \tau \sim \Gamma(\kappa/2, \, \lambda/2)} which means: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathsf{P}(\tau) = \frac{\frac{\lambda}{2}^{\kappa/2}}{\Gamma(\frac{\kappa}{2})} \tau^{\frac{\kappa}{2}1} e^{\frac{\lambda}{2} \tau}} And a multivariate Normal distribution for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B} : Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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)} which means: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathsf{P}(\tau, B  Y, X) = \mathsf{P}(\tau  Y, X) \mathsf{P}(B  Y, X, \tau)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathsf{P}(B  Y, X, \tau) = \frac{\mathsf{P}(B, Y  X, \tau)}{\mathsf{P}(Y  X, \tau)}} Let's neglect the normalization constant for now: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathsf{P}(B  Y, X, \tau) \propto \mathsf{P}(B  \tau) \mathsf{P}(Y  X, \tau, B)} Similarly, let's keep only the terms in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B} for the moment: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathsf{P}(B  Y, X, \tau) \propto exp(B^T \Sigma_B^{1} B) exp((YXB)^T(YXB))} We expand: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \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: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathsf{P}(B  Y, X, \tau) \propto exp(B^T (\Sigma_B^{1} + X^TX) B  Y^TXB B^TX^TY)} Importantly, let's define: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \Omega = (\Sigma_B^{1} + X^TX)^{1}} We can see that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \Omega^T=\Omega} , which means that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \Omega} is a symmetric matrix. This is particularly useful here because we can use the following equality: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \Omega^{1}\Omega^T=I} . Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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)} This now becomes easy to factorizes totally: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \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: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B  Y, X, \tau \sim \mathcal{N}(\Omega X^TY, \tau^{1} \Omega)}
Similarly to the equations above: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B} , thus effectively taking into account the uncertainty in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B} : Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B} !): Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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} As we used a conjugate prior for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \tau} , we know that we expect a Gamma distribution for the posterior. Therefore, we can take Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \tau^{N/2}} out of the integral and start guessing what looks like a Gamma distribution. We also factorize inside the exponential: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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} We recognize the conditional posterior of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B} . This allows us to use the fact that the pdf of the Normal distribution integrates to one: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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]} We finally recognize a Gamma distribution, allowing us to write the posterior as: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B, \tau  Y, X \sim \mathcal{N}IG(\Omega X^TY, \; \tau^{1}\Omega, \; \frac{N+\kappa}{2}, \; \frac{\lambda^\ast}{2})} where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \lambda^\ast = Y^T Y  Y^T X \Omega X^T Y + \lambda}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathsf{P}(B  Y, X) = \int \mathsf{P}(\tau) \mathsf{P}(B  Y, X, \tau) \mathsf{d}\tau} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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} Here we recognize the formula to integrate the Gamma function: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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}}} And we now recognize a multivariate Student's tdistribution: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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}}} We hence can write: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B  Y, X \sim \mathcal{S}_{N+\kappa}(\Omega X^TY, \; (Y^T Y  Y^T X \Omega X^T Y + \lambda) \Omega)}
We want to test the following null hypothesis: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle H_0: \; a = d = 0} In Bayesian modeling, hypothesis testing is performed with a Bayes factor, which in our case can be written as: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathrm{BF} = \frac{\mathsf{P}(Y  X, a \neq 0, d \neq 0)}{\mathsf{P}(Y  X, a = 0, d = 0)}} We can shorten this into: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathrm{BF} = \frac{\mathsf{P}(Y  X)}{\mathsf{P}_0(Y)}} Note that, compare to frequentist hypothesis testing which focuses on the null, the Bayes factor requires to explicitly model the data under the alternative. This makes a big difference when interpreting the results (see below). Let's start with the numerator: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathsf{P}(Y  X) = \int \mathsf{P}(\tau) \mathsf{P}(Y  X, \tau) \mathsf{d}\tau} First, let's calculate what is inside the integral: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathsf{P}(Y  X, \tau) = \frac{\mathsf{P}(B  \tau) \mathsf{P}(Y  X, \tau, B)}{\mathsf{P}(B  Y, X, \tau)}} Using the formula obtained previously and doing some algebra gives: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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)} Now we can integrate out Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \tau} (note the small typo in equation 9 of supplementary text S1 of Servin & Stephens): Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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)} Inside the integral, we recognize the almostcomplete pdf of a Gamma distribution. As it has to integrate to one, we get: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\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}}} We can use this expression also under the null. In this case, as we need neither Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle a} nor Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle d} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B} is simply Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mu} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \Sigma_B} is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \sigma_{\mu}^2} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle X} is a vector of 1's. We can also defines Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \Omega_0 = ((\sigma_{\mu}^2)^{1} + N)^{1}} . In the end, this gives: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathsf{P}_0(Y) = (2\pi)^{\frac{N}{2}} \frac{\Omega_0^{\frac{1}{2}}}{\sigma_{\mu}} \left( \frac{\lambda}{2} \right)^{\frac{\kappa}{2}} \frac{\Gamma(\frac{N+\kappa}{2})}{\Gamma(\frac{\kappa}{2})} \left( \frac{Y^TY  \Omega_0 N^2 \bar{Y}^2 + \lambda}{2} \right)^{\frac{N+\kappa}{2}}} We can therefore write the Bayes factor: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathrm{BF} = \left( \frac{\Omega}{\Omega_0} \right)^{\frac{1}{2}} \frac{1}{\sigma_a \sigma_d} \left( \frac{Y^TY  Y^TX\Omega X^TY + \lambda}{Y^TY  \Omega_0 N^2 \bar{Y}^2 + \lambda} \right)^{\frac{N+\kappa}{2}}} When the Bayes factor is large, we say that there is enough evidence in the data to support the alternative. Indeed, the Bayesian testing procedure corresponds to measuring support for the specific alternative hypothesis compared to the null hypothesis. Importantly, note that, for a frequentist testing procedure, we would say that there is enough evidence in the data to reject the null. However we wouldn't say anything about the alternative as we don't model it. The threshold to say that a Bayes factor is large depends on the field. It is possible to use the Bayes factor as a test statistic when doing permutation testing, and then control the false discovery rate. This can give an idea of a reasonable threshold.
Such a question is never easy to answer. But note that all hyperparameters are not that important, especially in typical quantitative genetics applications. For instance, we are mostly interested in those that determine the magnitude of the effects, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \sigma_a} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \sigma_d} , so let's deal with the others first. As explained in Servin & Stephens, the posteriors for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \tau} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B} change appropriately with shifts (Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle y+c} ) and scaling (Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle y \times c} ) in the phenotype when taking their limits. This also gives us a new Bayes factor, the one used in practice (see Guan & Stephens, 2008): Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathrm{lim}_{\sigma_{\mu} \rightarrow \infty \; ; \; \lambda \rightarrow 0 \; ; \; \kappa \rightarrow 0 } \; \mathrm{BF} = \left( \frac{N}{\Sigma_B^{1} + X^TX} \right)^{\frac{1}{2}} \frac{1}{\sigma_a \sigma_d} \left( \frac{Y^TY  Y^TX (\Sigma_B^{1} + X^TX)^{1} X^TY}{Y^TY  N \bar{Y}^2} \right)^{\frac{N}{2}}} Now, for the important hyperparameters, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \sigma_a} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \sigma_d} , it is usual to specify a grid of values, i.e. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle M} pairs Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle (\sigma_a, \sigma_d)} . For instance, Guan & Stephens used the following grid: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle M=4 \; ; \; \sigma_a \in \{0.05, 0.1, 0.2, 0.4\} \; ; \; \sigma_d = \frac{\sigma_a}{4}} Then, we can average the Bayes factors obtained over the grid using, as a first approximation, equal weights: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathrm{BF} = \sum_{m \, \in \, \text{grid}} \frac{1}{M} \, \mathrm{BF}(\sigma_a^{(m)}, \sigma_d^{(m)})} In eQTL studies, the weights can be estimated from the data using a hierarchical model (see below), by pooling all genes together as in Veyrieras et al (PLoS Genetics, 2010).
BF < function(G=NULL, Y=NULL, sigma.a=NULL, sigma.d=NULL, get.log10=TRUE){ stopifnot(! is.null(G), ! is.null(Y), ! is.null(sigma.a), ! is.null(sigma.d)) subset < complete.cases(Y) & complete.cases(G) Y < Y[subset] G < G[subset] stopifnot(length(Y) == length(G)) N < length(G) X < cbind(rep(1,N), G, G == 1) inv.Sigma.B < diag(c(0, 1/sigma.a^2, 1/sigma.d^2)) inv.Omega < inv.Sigma.B + t(X) %*% X inv.Omega0 < N tY.Y < t(Y) %*% Y log10.BF < as.numeric(0.5 * log10(inv.Omega0)  0.5 * log10(det(inv.Omega))  log10(sigma.a)  log10(sigma.d)  (N/2) * (log10(tY.Y  t(Y) %*% X %*% solve(inv.Omega) %*% t(X) %*% cbind(Y))  log10(tY.Y  N*mean(Y)^2))) if(get.log10) return(log10.BF) else return(10^log10.BF) } In the same vein as what is explained here, we can simulate data under different scenarios and check the BFs: N < 300 # play with it PVE < 0.1 # play with it grid < c(0.05, 0.1, 0.2, 0.4, 0.8, 1.6, 3.2) MAF < 0.3 G < rbinom(n=N, size=2, prob=MAF) tau < 1 a < sqrt((2/5) * (PVE / (tau * MAF * (1MAF) * (1PVE)))) d < a / 2 mu < rnorm(n=1, mean=0, sd=10) Y < mu + a * G + d * (G == 1) + rnorm(n=N, mean=0, sd=tau) for(m in 1:length(grid)) print(BF(G, Y, grid[m], grid[m]/4))
There are many equivalent ways to write the likelihood, the usual one being: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle y_i  p_i \; \overset{i.i.d}{\sim} \; Bernoulli(p_i)} with the logodds (logit function) being Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathrm{ln} \frac{p_i}{1  p_i} = \mu + a \, g_i + d \, \mathbf{1}_{g_i=1}} Let's use Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle X_i^T=[1 \; g_i \; \mathbf{1}_{g_i=1}]} to denote the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle i} th row of the design matrix Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle X} . We can also keep the same definition as above for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B=[\mu \; a \; d]^T} . Thus we have: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle p_i = \frac{e^{X_i^TB}}{1 + e^{X_i^TB}}} As the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle y_i} 's can only take Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle 0} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle 1} as values, the likelihood can be written as: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathcal{L}(B) = \mathsf{P}(Y  X, B) = \prod_{i=1}^N p_i^{y_i} (1p_i)^{1y_i}} We still use the same prior as above for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B} (but there is no Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \tau} anymore), so that: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B  \Sigma_B \sim \mathcal{N}_3(0, \Sigma_B)} where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \Sigma_B} is a 3 x 3 matrix with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle [\sigma_\mu^2 \; \sigma_a^2 \; \sigma_d^2]} on the diagonal and 0 elsewhere. As above, the Bayes factor is used to compare the two models: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathrm{BF} = \frac{\mathsf{P}(Y  X, M1)}{\mathsf{P}(Y  X, M0)} = \frac{\mathsf{P}(Y  X, a \neq 0, d \neq 0)}{\mathsf{P}(Y  X, a=0, d=0)} = \frac{\int \mathsf{P}(B) \mathsf{P}(Y  X, B) \mathrm{d}B}{\int \mathsf{P}(\mu) \mathsf{P}(Y  X, \mu) \mathrm{d}\mu}} The interesting point here is that there is no way to analytically calculate these integrals (marginal likelihoods). Therefore, we will use Laplace's method to approximate them, as in Guan & Stephens (2008). Starting with the numerator: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathsf{P}(YX,M1) = \int \exp \left[ N \left( \frac{1}{N} \mathrm{ln} \, \mathsf{P}(B) + \frac{1}{N} \mathrm{ln} \, \mathsf{P}(Y  X, B) \right) \right] \mathsf{d}B} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathsf{P}(YX,M1) = \int \exp \left\{ N \left[ \frac{1}{N} \left( \mathrm{ln} \left( (2 \pi)^{\frac{3}{2}} \, \frac{1}{\sigma_\mu \sigma_a \sigma_d} \, \exp\left( \frac{1}{2} (\frac{\mu^2}{\sigma_\mu^2} + \frac{a^2}{\sigma_a^2} + \frac{d^2}{\sigma_d^2}) \right) \right) \right) + \frac{1}{N} \left( \sum_{i=1}^N \left( y_i \, \mathrm{ln} (p_i) + (1y_i) \, \mathrm{ln} (1p_i) \right) \right) \right] \right\} \mathsf{d}B} Let's use Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle f} to denote the function inside the exponential: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathsf{P}(YX,M1) = \int \exp \left( N \; f(B) \right) \mathsf{d}B} The function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle f} is defined by: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle f: \mathbb{R}^3 \rightarrow \mathbb{R}} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle f(B) = \frac{1}{N} \left( \frac{3}{2} \mathrm{ln}(2 \pi)  \frac{1}{2} \mathrm{ln}(\Sigma_B)  \frac{1}{2}(B^T \Sigma_B^{1} B) \right) + \frac{1}{N} \sum_{i=1}^N \left( y_i \, X_i^T B  \mathrm{ln}(1 + e^{X_i^TB}) \right)} This function will then be used to approximate the integral, like this: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathsf{P}(YX,M1) \approx N^{3/2} (2 \pi)^{3/2} H(B^\star)^{1/2} e^{N f(B^\star)}} where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle H} is the Hessian of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle f} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B^\star = [\mu^\star a^\star d^\star]^T} is the point at which Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle f} is maximized. We therefore need to find Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B^\star} . As it maximizes Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle f} , we need to calculate the first derivatives of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle f} . Let's do this the univariate way: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{\partial f}{\partial \beta} =  \frac{\beta}{N \, \sigma_\beta^2} + \frac{1}{N} \sum_{i=1}^N \left(\frac{y_i}{p_i}  \frac{1y_i}{1p_i} \right) \frac{\partial p_i}{\partial \beta}} where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \beta} is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mu} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle a} or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle d} . A simple form for the first derivatives of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle p_i} also exists when writing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle p_i = e^{X_i^tB} (1 + e^{X_i^tB})^{1}} : Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{\partial p_i}{\partial \beta} = \left[ e^{X_i^tB} (1 + e^{X_i^tB})^{1} + e^{X_i^tB} \left( e^{X_i^tB} (1 + e^{X_i^tB})^{2} \right) \right] \frac{\partial X_i^TB}{\partial \beta}} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{\partial p_i}{\partial \beta} = \left[ \frac{e^{X_i^tB} (1 + e^{X_i^tB})  (e^{X_i^tB})^2}{(1 + e^{X_i^tB})^2} \right] \frac{\partial X_i^TB}{\partial \beta}} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{\partial p_i}{\partial \beta} = \left[ p_i (1  p_i) \right] \frac{\partial X_i^TB}{\partial \beta}} where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{\partial X_i^TB}{\partial \beta}} is equal to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle 1, \, g_i, \, \mathbf{1}_{g_i=1}} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \beta} corresponds respectively to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mu, \, a, \, d} . This simplifies the first derivatives of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle f} into: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{\partial f}{\partial \beta} =  \frac{\beta}{N \, \sigma_\beta^2} + \frac{1}{N} \sum_{i=1}^N (y_i  p_i ) \frac{\partial X_i^TB}{\partial \beta}} When setting Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{\partial f}{\partial \beta}(\beta^\star) = 0} , we observe that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \beta^\star} is present not only alone but also inside the sum, in the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle p_i} 's: indeed Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle p_i} is a nonlinear function of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B} . This means that an iterative procedure is required, typically Newton's method. To use it, we need the second derivatives of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle f} : Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{\partial^2 f}{\partial \beta^2} =  \frac{1}{N \, \sigma_\beta^2} + \frac{1}{N} \sum_{i=1}^N \left[ (p_i(1p_i)\frac{\partial X_i^TB}{\partial \beta}) + (y_ip_i)\frac{\partial^2 X_i^TB}{\partial \beta^2} \right]} The second derivatives of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle X_i^TB} are all equal to 0: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{\partial^2 f}{\partial \beta^2} =  \frac{1}{N \, \sigma_\beta^2}  \frac{1}{N} \sum_{i=1}^N p_i(1p_i)\frac{\partial X_i^TB}{\partial \beta}} Note that the second derivatives of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle f} are strictly negative. Therefore, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle f} is globally convex, which means that it has a unique global maximum, at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle B^\star} . As a consequence, we have the right to use Laplace's method to approximate the integral around its maximum. finding the maximums: iterative procedure, update equations or generic solver > to do implementation: in R > to do finding the effect sizes and their std error: to do
to do
to do
to do
to do
to do
to do
to do
to do
