Difference between revisions of "User:Timothee Flutre/Notebook/Postdoc/2011/11/10"
(→Bayesian model of univariate linear regression for QTL detection: add multivar eq for logistic reg) 
(→Bayesian model of univariate linear regression for QTL detection: add info about confounders in phenotype) 

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−  * '''  +  * '''Confounders in phenotype''': it is well known in molecular biology that any experiment involving several assays (e.g. measuring gene expression levels with a DNA microarray) suffers from "unknown confounders", the most (in)famous being the socalled "batch effects". 
−  to do  +  For instance, samples from individual 1 and 2 are correlated with each other because they were treated another day than all the other samples. Such a correlations has nothing to do with the genotype at a given SNP (in most cases). However, the core model, <math>y_i = \mu + \beta g_i + \epsilon_i</math> assumes that the errors are uncorrelated between individuals: <math>\epsilon_i \overset{\mathrm{i.i.d}}{\sim} \mathcal{N}(0,\tau^{1})</math>. If this is not the case, i.e. if the <math>y_i</math>'s are correlated but this correlation has nothing to do with the <math>g_i</math>'s, then more variance in the errors won't be accounted for, and we'll loose power when trying to detect weak, yet nonzero <math>\beta</math>. 
+  
+  An intuitive way of removing these confounders is to realize that we can use all gene expression levels to try to identify them. Indeed, batch effects are very likely to affect all genes in a sample (though possibly at different magnitudes). As the effect of the confounders are, as a first approximation, typically much bigger than the effect of a SNP genotype, we can try to learn the confounders using all gene expression levels, and only them. So let's put all of them into a <math>G \times N</math> matrix <math>Y_1</math> with genes in rows and individuals in columns.  
+  
+  For the moment, the data are expressed in the [http://en.wikipedia.org/wiki/Standard_basis standard basis], i.e. the basis of the observations. But some confounders are present in these data, they contribute with noise and redundancy and hence dilute the signal. The idea is, first, to identify a new basis which will correspond to a "mix" of the original samples (e.g. one component of this "mix" may correspond to the day at which the samples were processed), and second, to remove these components from the data in order to only keep the signal.  
+  
+  to be continued  
+  
+  see also factor analysis, see Stegle ''et al'' (PLoS Computational Biology, 2010)  
−  * '''  +  * '''Confounders in genotype''': mainly pop structure and genetic relatedness, linear mixed model (LMM), see Zhou & Stephens (Nature Genetics, 2012) 
to do  to do  
−  * '''Discrete phenotype''': count data  +  * '''Discrete phenotype''': count data (e.g. from RNAseq), Poissonlike likelihood, generalized linear model (GLM), see Sun (Biometrics, 2012) 
to do  to do  
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−  * '''Nonindependent genes''': enrichment in known pathways, learn "modules"  +  * '''Nonindependent genes''': enrichment in known pathways, learn "modules", distributions on networks 
to do  to do 
Revision as of 17:46, 27 February 2013
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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.
[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 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]
[math]\mathsf{P}(\tau, B) = \mathsf{P}(\tau) \mathsf{P}(B  \tau)[/math] A 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]
[math]\mathsf{P}(\tau, B  Y, X) = \mathsf{P}(\tau  Y, X) \mathsf{P}(B  Y, X, \tau)[/math]
[math]\mathsf{P}(B  Y, X, \tau) = \frac{\mathsf{P}(B, Y  X, \tau)}{\mathsf{P}(Y  X, \tau)}[/math] 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 symmetric matrix. This is particularly useful here because we can use the following equality: [math]\Omega^{1}\Omega^T=I[/math]. [math]\mathsf{P}(B  Y, X, \tau) \propto exp(B^T \Omega^{1} B  (X^TY)^T\Omega^{1}\Omega^TB B^T\Omega^{1}\Omega^TX^TY)[/math] This now becomes easy to factorizes totally: [math]\mathsf{P}(B  Y, X, \tau) \propto exp((B^T  \Omega X^TY)^T\Omega^{1}(B  \Omega X^TY))[/math] We recognize the kernel of a Normal distribution, allowing us to write the conditional posterior as: [math]B  Y, X, \tau \sim \mathcal{N}(\Omega X^TY, \tau^{1} \Omega)[/math]
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]
[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]
[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 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 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]
We want to test the following null hypothesis: [math]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]\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.
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).
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: [math]y_i  p_i \; \overset{i.i.d}{\sim} \; \mathrm{Binomial}(1,p_i)[/math] with the 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 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] 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 \mathrm{ln}(p_i) + (1y_i) \mathrm{ln}(1  p_i) \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 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 two things: [math]H[/math] and [math]B^\star[/math]. Note that for both we need to calculate the first derivatives of [math]f[/math]. As [math]f[/math] is multidimensional (it takes values in [math]\mathbb{R}^3[/math]), we need to calculate its gradient. In the following, some formulas from matrix calculus are sometimes required. In such cases, I will use the numerator layout. [math]\nabla f = \frac{\partial f}{\partial B} = \left( \frac{\partial f}{\partial \mu} \; \frac{\partial f}{\partial a} \; \frac{\partial f}{\partial d} \right)[/math] [math]\nabla f =  \frac{1}{2N} \frac{\partial B^T\Sigma_B^{1}B}{\partial B} + \frac{1}{N} \sum_i \left( y_i \frac{\partial \mathrm{ln}(p_i)}{\partial B} + (1y_i) \frac{\partial \mathrm{ln}(1p_i)}{\partial B} \right)[/math] [math]\nabla f =  \frac{1}{N} B^T\Sigma_B^{1} + \frac{1}{N} \sum_i \left( \frac{y_i}{p_i}  \frac{1y_i}{1p_i} \right) \frac{\partial p_i}{\partial B}[/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 B} = \frac{\partial e^{X_i^TB}}{\partial B} (1 + e^{X_i^TB})^{1} + e^{X_i^TB} \frac{\partial (1 + e^{X_i^TB})^{1}}{\partial B}[/math] [math]\frac{\partial p_i}{\partial B} = e^{X_i^TB} \frac{\partial X_i^TB}{\partial B} (1 + e^{X_i^TB})^{1}  e^{X_i^TB} (1 + e^{X_i^TB})^{2} \frac{\partial (1 + e^{X_i^TB})}{\partial B}[/math] [math]\frac{\partial p_i}{\partial B} = p_i X_i^T  p_i (1 + e^{X_i^TB})^{1} e^{X_i^TB} \frac{\partial X_i^TB}{\partial B}[/math] [math]\frac{\partial p_i}{\partial B} = p_i (1  p_i) X_i^T[/math] This simplifies the gradient of [math]f[/math] into: [math]\nabla f =  \frac{1}{N} B^T\Sigma_B^{1} + \frac{1}{N} \sum_i (y_i  p_i) X_i^T[/math] To find [math]B^\star[/math], we set [math]\nabla f(B^\star) = 0[/math]. However, in this equation, [math]B^\star[/math] is present not only alone but also in the [math]p_i[/math]'s. As [math]p_i[/math] is a nonlinear function of [math]B[/math], the equation can't be solved directly but an iterative procedure is required, typically a conjugate gradient method (as in Guan & Stephens) or Newton's method. The former only requires [math]f[/math] and [math]\nabla f[/math] while the latter also requires [math]H[/math]. Remember that, in any case, we need [math]H[/math] for the Laplace approximation, so let's calculate it: [math]H =  \frac{1}{N} \Sigma_B^{1}  \frac{1}{N} \sum_i \frac{\partial p_i}{\partial B} X_i^T[/math] [math]H =  \frac{1}{N} \Sigma_B^{1}  \frac{1}{N} \sum_i X_i^T p_i (1p_i) X_i[/math] [math]H =  \frac{1}{N} (\Sigma_B^{1} + X^T W X)[/math] where [math]W[/math] is the N x N matrix with [math]p_i(1p_i)[/math] on the diagonal. Note that all 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 of [math]f[/math] around its maximum. implementation in R > to do finding the effect sizes and their std error: to do
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For instance, samples from individual 1 and 2 are correlated with each other because they were treated another day than all the other samples. Such a correlations has nothing to do with the genotype at a given SNP (in most cases). However, the core model, [math]y_i = \mu + \beta g_i + \epsilon_i[/math] assumes that the errors are uncorrelated between individuals: [math]\epsilon_i \overset{\mathrm{i.i.d}}{\sim} \mathcal{N}(0,\tau^{1})[/math]. If this is not the case, i.e. if the [math]y_i[/math]'s are correlated but this correlation has nothing to do with the [math]g_i[/math]'s, then more variance in the errors won't be accounted for, and we'll loose power when trying to detect weak, yet nonzero [math]\beta[/math]. An intuitive way of removing these confounders is to realize that we can use all gene expression levels to try to identify them. Indeed, batch effects are very likely to affect all genes in a sample (though possibly at different magnitudes). As the effect of the confounders are, as a first approximation, typically much bigger than the effect of a SNP genotype, we can try to learn the confounders using all gene expression levels, and only them. So let's put all of them into a [math]G \times N[/math] matrix [math]Y_1[/math] with genes in rows and individuals in columns. For the moment, the data are expressed in the standard basis, i.e. the basis of the observations. But some confounders are present in these data, they contribute with noise and redundancy and hence dilute the signal. The idea is, first, to identify a new basis which will correspond to a "mix" of the original samples (e.g. one component of this "mix" may correspond to the day at which the samples were processed), and second, to remove these components from the data in order to only keep the signal. to be continued see also factor analysis, see Stegle et al (PLoS Computational Biology, 2010)
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