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

(Difference between revisions)
 Revision as of 20:47, 22 November 2012 (view source) (→Bayesian model of univariate linear regression for QTL detection: add choice of hyperparam + list of future points)← Previous diff Revision as of 22:00, 22 November 2012 (view source) (→Bayesian model of univariate linear regression for QTL detection: add R code)Next diff → Line 213: Line 213: $\mathrm{BF} = \left( \frac{|\Omega|}{\Omega_0} \right)^{\frac{1}{2}} \frac{1}{\sigma_a \sigma_d} \left( \frac{Y^TY - Y^TX\Omega X^TY + \lambda}{Y^TY - \Omega_0 N^2 \bar{Y}^2 + \lambda} \right)^{-\frac{N+\kappa}{2}}$ $\mathrm{BF} = \left( \frac{|\Omega|}{\Omega_0} \right)^{\frac{1}{2}} \frac{1}{\sigma_a \sigma_d} \left( \frac{Y^TY - Y^TX\Omega X^TY + \lambda}{Y^TY - \Omega_0 N^2 \bar{Y}^2 + \lambda} \right)^{-\frac{N+\kappa}{2}}$ - In genetics, effect sizes are usually small, therefore difficult to detect. + When the Bayes factor is large, we say that there is enough evidence in the data to ''support the alternative''. - Hence it is usual to be interested in Bayes factors only when they exceed $10^4$ for instance. + - In such cases, 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. 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''. 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. 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. Line 238: Line 238: - * '''R code''': + * '''Implementation''': the following R function is adapted from Servin & Stephens supplementary text 1. + + + 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 [http://openwetware.org/wiki/User:Timothee_Flutre/Notebook/Postdoc/2011/06/28 here], we can simulate data and check the BF: - to do + + N <- 100 + MAF <- 0.3 + G <- rbinom(n=N, size=2, prob=MAF) + PVE <- 0.4 + tau <- 1 + a <- sqrt((2/5) * (PVE / (tau * MAF * (1-MAF) * (1-PVE)))) + 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) + BF(G, Y, 0.1, 0.1/4) +

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## 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 $y_1,\ldots,y_N$ the (quantitative) phenotypes (e.g. expression levels at a given gene), and $g_1,\ldots,g_N$ 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

$\forall i \in \{1,\ldots,N\}, \; y_i = \mu + \beta_1 g_i + \beta_2 \mathbf{1}_{g_i=1} + \epsilon_i \text{ with } \epsilon_i \overset{i.i.d}{\sim} \mathcal{N}(0,\tau^{-1})$

where β1 is in fact the additive effect of the SNP, noted a from now on, and β2 is the dominance effect of the SNP, d = ak.

Let's now write the model in matrix notation:

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

This gives the following multivariate Normal distribution for the phenotypes:

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

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, 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:

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

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

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

A Gamma distribution for τ:

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

which means:

$\mathsf{P}(\tau) = \frac{\frac{\lambda}{2}^{\kappa/2}}{\Gamma(\frac{\kappa}{2})} \tau^{\frac{\kappa}{2}-1} e^{-\frac{\lambda}{2} \tau}$

And a multivariate Normal distribution for B:

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

which means:

$\mathsf{P}(B | \tau) = \left(\frac{\tau}{2 \pi}\right)^{\frac{3}{2}} |\Sigma_B|^{-\frac{1}{2}} exp \left(-\frac{\tau}{2} B^T \Sigma_B^{-1} B \right)$

• Joint posterior (1):

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

• Conditional posterior of B:

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

Let's neglect the normalization constant for now:

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

Similarly, let's keep only the terms in B for the moment:

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

We expand:

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

We factorize some terms:

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

Importantly, let's define:

$\Omega = (\Sigma_B^{-1} + X^TX)^{-1}$

We can see that ΩT = Ω, which means that Ω is a symmetric matrix. This is particularly useful here because we can use the following equality: Ω − 1ΩT = I.

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

This now becomes easy to factorizes totally:

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

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

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

• Posterior of τ:

Similarly to the equations above:

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

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

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

Again, we use the priors and likelihoods specified above (but everything inside the integral is kept inside it, even if it doesn't depend on B!):

$\mathsf{P}(\tau | Y, X) \propto \tau^{\frac{\kappa}{2} - 1} e^{-\frac{\lambda}{2} \tau} \int \tau^{3/2} \tau^{N/2} exp(-\frac{\tau}{2} B^T \Sigma_B^{-1} B) exp(-\frac{\tau}{2} (Y - XB)^T (Y - XB)) \mathsf{d}B$

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

$\mathsf{P}(\tau | Y, X) \propto \tau^{\frac{N+\kappa}{2} - 1} e^{-\frac{\lambda}{2} \tau} \int \tau^{3/2} exp \left[-\frac{\tau}{2} \left( (B - \Omega X^T Y)^T \Omega^{-1} (B - \Omega X^T Y) - Y^T X \Omega X^T Y + Y^T Y \right) \right] \mathsf{d}B$

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

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

We finally recognize a Gamma distribution, allowing us to write the posterior as:

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

• Joint posterior (2): sometimes it is said that the joint posterior follows a Normal Inverse Gamma distribution:

$B, \tau | Y, X \sim \mathcal{N}IG(\Omega X^TY, \; \tau^{-1}\Omega, \; \frac{N+\kappa}{2}, \; \frac{\lambda^\ast}{2})$

where $\lambda^\ast = Y^T Y - Y^T X \Omega X^T Y + \lambda$

• Marginal posterior of B: we can now integrate out τ:

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

$\mathsf{P}(B | Y, X) = \frac{\frac{\lambda^\ast}{2}^{\frac{N+\kappa}{2}}}{(2\pi)^\frac{3}{2} |\Omega|^{\frac{1}{2}} \Gamma(\frac{N+\kappa}{2})} \int \tau^{\frac{N+\kappa+3}{2}-1} exp \left[-\tau \left( \frac{\lambda^\ast}{2} + (B - \Omega X^TY)^T \Omega^{-1} (B - \Omega X^TY) \right) \right] \mathsf{d}\tau$

Here we recognize the formula to integrate the Gamma function:

$\mathsf{P}(B | Y, X) = \frac{\frac{\lambda^\ast}{2}^{\frac{N+\kappa}{2}} \Gamma(\frac{N+\kappa+3}{2})}{(2\pi)^\frac{3}{2} |\Omega|^{\frac{1}{2}} \Gamma(\frac{N+\kappa}{2})} \left( \frac{\lambda^\ast}{2} + (B - \Omega X^TY)^T \Omega^{-1} (B - \Omega X^TY) \right)^{-\frac{N+\kappa+3}{2}}$

And we now recognize a multivariate Student's t-distribution:

$\mathsf{P}(B | Y, X) = \frac{\Gamma(\frac{N+\kappa+3}{2})}{\Gamma(\frac{N+\kappa}{2}) \pi^\frac{3}{2} |\lambda^\ast \Omega|^{\frac{1}{2}} } \left( 1 + \frac{(B - \Omega X^TY)^T \Omega^{-1} (B - \Omega X^TY)}{\lambda^\ast} \right)^{-\frac{N+\kappa+3}{2}}$

We hence can write:

$B | Y, X \sim \mathcal{S}_{N+\kappa}(\Omega X^TY, \; (Y^T Y - Y^T X \Omega X^T Y + \lambda) \Omega)$

• Bayes Factor: one way to answer our goal above ("is there an effect of the genotype on the phenotype?") is to do hypothesis testing.

We want to test the following null hypothesis:

$H_0: \; a = d = 0$

In Bayesian modeling, hypothesis testing is performed with a Bayes factor, which in our case can be written as:

$\mathrm{BF} = \frac{\mathsf{P}(Y | X, a \neq 0, d \neq 0)}{\mathsf{P}(Y | X, a = 0, d = 0)}$

We can shorten this into:

$\mathrm{BF} = \frac{\mathsf{P}(Y | X)}{\mathsf{P}_0(Y)}$

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).

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

First, let's calculate what is inside the integral:

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

Using the formula obtained previously and doing some algebra gives:

$\mathsf{P}(Y | X, \tau) = \left( \frac{\tau}{2 \pi} \right)^{\frac{N}{2}} \left( \frac{|\Omega|}{|\Sigma_B|} \right)^{\frac{1}{2}} exp\left( -\frac{\tau}{2} (Y^TY - Y^TX\Omega X^TY) \right)$

Now we can integrate out τ (note the small typo in equation 9 of supplementary text S1 of Servin & Stephens):

$\mathsf{P}(Y | X) = (2\pi)^{-\frac{N}{2}} \left( \frac{|\Omega|}{|\Sigma_B|} \right)^{\frac{1}{2}} \frac{\frac{\lambda}{2}^{\frac{\kappa}{2}}}{\Gamma(\frac{\kappa}{2})} \int \tau^{\frac{N+\kappa}{2}-1} exp \left( -\frac{\tau}{2} (Y^TY - Y^TX\Omega X^TY + \lambda) \right)$

Inside the integral, we recognize the almost-complete pdf of a Gamma distribution. As it has to integrate to one, we get:

$\mathsf{P}(Y | X) = (2\pi)^{-\frac{N}{2}} \left( \frac{|\Omega|}{|\Sigma_B|} \right)^{\frac{1}{2}} \left( \frac{\lambda}{2} \right)^{\frac{\kappa}{2}} \frac{\Gamma(\frac{N+\kappa}{2})}{\Gamma(\frac{\kappa}{2})} \left( \frac{Y^TY - Y^TX\Omega X^TY + \lambda}{2} \right)^{-\frac{N+\kappa}{2}}$

We can use this expression also under the null. In this case, as we need neither a nor d, B is simply μ, ΣB is $\sigma_{\mu}^2$ and X is a vector of 1's. We can also defines $\Omega_0 = ((\sigma_{\mu}^2)^{-1} + N)^{-1}$. In the end, this gives:

$\mathsf{P}_0(Y) = (2\pi)^{-\frac{N}{2}} \frac{|\Omega_0|^{\frac{1}{2}}}{\sigma_{\mu}} \left( \frac{\lambda}{2} \right)^{\frac{\kappa}{2}} \frac{\Gamma(\frac{N+\kappa}{2})}{\Gamma(\frac{\kappa}{2})} \left( \frac{Y^TY - \Omega_0 N^2 \bar{Y}^2 + \lambda}{2} \right)^{-\frac{N+\kappa}{2}}$

We can therefore write the Bayes factor:

$\mathrm{BF} = \left( \frac{|\Omega|}{\Omega_0} \right)^{\frac{1}{2}} \frac{1}{\sigma_a \sigma_d} \left( \frac{Y^TY - Y^TX\Omega X^TY + \lambda}{Y^TY - \Omega_0 N^2 \bar{Y}^2 + \lambda} \right)^{-\frac{N+\kappa}{2}}$

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.

• Hyperparameters: the model has 5 hyperparameters, $\{\kappa, \, \lambda, \, \sigma_{\mu}, \, \sigma_a, \, \sigma_d\}$. 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, σa and σd, so let's deal with the others first.

As explained in Servin & Stephens, the posteriors for τ and B change appropriately with shifts (y + c) and scaling ($y \times c$) in the phenotype when taking their limits. This also gives us a new Bayes factor, the one used in practice (see Guan & Stephens, 2008):

$\mathrm{lim}_{\sigma_{\mu} \rightarrow \infty \; ; \; \lambda \rightarrow 0 \; ; \; \kappa \rightarrow 0 } \; \mathrm{BF} = \left( \frac{N}{|\Sigma_B^{-1} + X^TX|} \right)^{\frac{1}{2}} \frac{1}{\sigma_a \sigma_d} \left( \frac{Y^TY - Y^TX (\Sigma_B^{-1} + X^TX)^{-1} X^TY}{Y^TY - N \bar{Y}^2} \right)^{-\frac{N}{2}}$

Now, for the important hyperparameters, σa and σd, it is usual to specify a grid of values, i.e. M pairs ad). For instance, Guan & Stephens used the following grid:

$M=4 \; ; \; \sigma_a \in \{0.05, 0.1, 0.2, 0.4\} \; ; \; \sigma_d = \frac{\sigma_a}{4}$

Then, we can average the Bayes factors obtained over the grid using, as a first approximation, equal weights:

$\mathrm{BF} = \sum_{m \, \in \, \text{grid}} \frac{1}{M} \, \mathrm{BF}(\sigma_a^{(m)}, \sigma_d^{(m)})$

• Implementation: the following R function is adapted from Servin & Stephens supplementary text 1.
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 and check the BF:

N <- 100
MAF <- 0.3
G <- rbinom(n=N, size=2, prob=MAF)
PVE <- 0.4
tau <- 1
a <- sqrt((2/5) * (PVE / (tau * MAF * (1-MAF) * (1-PVE))))
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)
BF(G, Y, 0.1, 0.1/4)



• Binary phenotype: case-control like in GWAS, logistic regression, see Guan & Stephens (2008) for Laplace approximation

to do

• Link between Bayes factor and P-value: see Wakeley (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

to do

• Discrete phenotype: count data as from RNA-seq, Poisson-like likelihood

to do

• Multiple phenotypes: matrix-variate distributions, tensors

to do

• Non-independent 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)