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Earl Austin
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1 Review We have covered so far: Single variant association analysis and effect size estimation GxE interaction and higher order >2 interaction Measurement error in dietary variables (nutritional epidemiology) Today s lecture: set-based association

2 Set-based association analysis i = 1,, n m variants on a certain region Genotype G i = (G i1, G i2,, G im ), g ij = 0, 1, 2 Covariates X i : intercept, age, sex, principal components for population structure. Model: where g( ) is a link function. m g{e(y i )} = X i α + G ij β j j=1 No association for a set means β = (β 1,, β m ) = 0

3 Why? In gene expression studies, a list of differentially expressed genes fails to provide mechanistic insights into the underlying biology to many investigators Pathway analysis extracts meaning by grouping genes into small sets of related genes Function databases are curated to help with this task, e.g., biological processes in which genes are involved in or interact with each other. Analyzing high-throughput molecular measurements at the functional or pathway level is very appealing Reduce complexity Improve explanatory power than a simple list of differentially expressed genes

4 Why? Variants in a defined set (e.g. genes, pathways) may act concordantly on phenotypes. Combining these variants aggregates the signals; as a result, may improve power Power is particularly an issue when variants are rare. The challenge is that not all variants in a set are associated with phenotypes and those who are associated may have either positive or negative effects

5 Outline Burden test Variance component test Mixed effects score test

6 Burden Test If m is large, multivariate test β = 0 is not very powerful Population genetics evolutionary theory suggests that most rare missense alleles are deleterious, and the effect is therefore generally considered one-sided (Kryukov et al., 2007) Collapsing: Suppose β 1 = = β m = η g{e(y i )} = X i α + B iη B i = m j=1 g ij : genetic burden/score. With weight (adaptive burden test) B i = m w j G ij. j=1 Test H 0 : η = 0 (d.f.=1).

7 Weight Threshold-based method { 1 if MAF j t w j (t) = 0 if MAF j > t Burden score B i (t) = m j=1 w j(t)g ij, and the corresponding Z-statistic is denoted by Z (t) Variable threshold (VT) test Z max = max t Z (t) P-value can be calculated by permutation (Price et al 2010) or numerical integration using normal approximation. P(Z max z) = 1 P(Z max < z) = 1 P(Z (t 1 ) < z,, Z (t b ) < z) where {Z (t 1 ),, Z (t b )} follows a multivariate normal distribution MVN(0, Σ).

8 Weight Variant effects can be positive or negative and the strength can be different too. Fit the marginal model for each variant g{e(y i )} = X i α + G ijγ j

9 Weight Adaptive sum test (Han & Pan, Hum Hered 2010) { 1 if γ j < 0 and p-value < α 0 ; w j = 1 otherwise If α 0 = 1, the weight is the sign of γ j, but the corresponding weighted burden test has low power because the null distribution has heavy tails. α 0 is chosen such that only when H 0 likely does not hold, the sign is changed if γ j is negative. The authors suggest α 0 = 0.1, but it is data dependent

10 Weight Estimated regression coefficient (EREC) test (Lin & Tang Am J Hum Genet 2011) Score statistic T EREC = 1 n w j = γ j + c, for c 0 n m {( ( γ j + c)g ij )(y i µ i ( α))} N(0, Σ) i=1 j=1 µ i ( α) is estimated under the null of no association If c = 0, T EREC = 1 n n i=1 {( m j=1 γ jg ij )(y i µ i ( α))} is not asymptotically normal c = 1 for binary traits, c = 2 for standardized quantitative traits Compute p-values using permutation.

11 Burden Tests Burden tests lose a significant amount of power if there are variants with different association directions or a large # of variants are neutral. Adaptive burdent tests have robust power, but they rely on resampling to compute p-values. Computationally intensive, not suitable for genome-wide discovery.

12 Variance Component Test Model m g{e(y i )} = X i α + G ij β j j=1 Burden tests are derived assuming β 1 = = β m. Variance component test Assume βj F(0, τ 2 ), where F( ) is an arbitrary distribution, and the mean of β j s = 0. H0 : β 1 = = β p = 0 H 0 : τ 2 = 0.

13 Derivation of Variance Component Test Suppose g( ) is linear and Y X, G Normal. That is, m y i = X i α + G ij β j + ε, ε N(0, σ 2 ) j=1 Suppose β j Normal (0, τ 2 ), j = 1,..., m Marginal model: Y n 1 MVN(X n p α, τ 2 G n m G m n + σ 2 I)

14 Derivation of Variance Component Test Log likelihood l = (Y Xα) (τ 2 GG + σ 2 I) 1 (Y Xα) 2 Let V (τ 2 ) = τ 2 GG + σ 2 I Score function 1 2 log τ 2 GG + σ 2 I n 2 log(2π) l τ 2 = (Y Xα) V (τ 2 ) 1 GG V (τ 2 ) 1 (Y Xα) 2 tr(v (τ 2 ) 1 (GG )) 2

15 Derivation of Variance Component Test Score test statistic Q = l τ 2 τ=0 = 1 2 (Y Xα) V 1 GG V 1 (Y Xα) 1 2 tr(v 1 (GG )) = 1 2 (Y Xα) M(Y Xα) 1 2 tr(v 1 2 MV 1 2 ) M = V 1 GG V 1, V = σ 2 I

16 Derivation of Variance Component Test Q is not asymptotically normal Q = 1 2 (Y Xα) M(Y Xα) 1 2 tr(v 1 2 MV 1 2 ) = 1 2Ỹ (V 1 2 MV 1 2 ) Ỹ 1 2 tr(v 1 2 MV 1 2 ) where Ỹ = V 1 2 (Y Xα) N(0, I) Let {λ j, u j, j = 1,..., m} be the eigenvalues and eigenvectors of V 1 2 MV 1 2. Then Q = m λ j ((u j Ỹ )2 1) = j=1 Q is not asymptotically normal Zhang and Lin (2003) show that Ỹ (V 1 2 MV 1 2 ) Ỹ m j=1 m λ j χ 2 1,j j=1 λ j (Z 2 j 1)

17 Variance Component Test The exact probability associated with a mixture of χ 2 distributions is difficult to calculate. Satterthwaite method to approximate the distribution by a scaled χ 2 distribution, κχ 2 ν, where κ and ν are calculated by matching the first and second moments of the two distributions. To adjust for α, replace V 1 by projection matrix P = V 1 V 1 X(X V 1 X) 1 X V 1.

18 General Form of Variance Component Test Linear model y i = X i α + h(g i ) + ε, ε N(0, σ 2 ) h( ) is a centered unknown smooth function H generated by a positive definite kernel function K (, ). K (, ) implicitly specifies a unique function space spanned by a set of orthogonal basis functions {φ j (G), j = 1,..., J} and any h( ) can be represented by linear combination of these basis h(g) = J j=1 ζ jφ j (G)(the primal representation)

19 General Form Equivalently, h( ) can also be represented using K (, ) as h(g i ) = n j=1 ω jk (G i, G j ) (the dual representation) For a multi-dimensional G, it is more convenient to specify h(g) using the dual representation, because explicit basis functions might be complicated to specify, and the number might be high

20 Estimation Penalized likelihood function (Kimeldorf and Wahba, 1970) l = 1 2 n n y i X i α ω j K (G i, G j ) i=1 j= λω K ω where λ is a tuning parameter which controls the tradeoff between goodness of fit and complexity of the model α = { X (I + λ 1 K ) 1 X} 1 X (I + λ 1 K ) 1 y and ω = λ 1 (I + λ 1 K ) 1 (y X α) ĥ = K ω

21 Connection with Linear Mixed Models The same estimators can be re-written as [ X V 1 X X V 1 ] [ ] α V 1 X V 1 + (τk ) 1 = h [ X V 1 y V 1 y ] where τ = λ 1 σ 2 and V = σ 2 I Estimators α and ĥ are best linear unbiased predictors under the linear mixed model y = X α + h + ε where h is a n 1 vector of random effects with distribution N(0, τk ) and ε N(0, V )

22 General Form of Variance Component Test Testing H 0 : h = 0 is equivalent to testing the variance component τ as H 0 : τ = 0 versus H 1 : τ > 0 The REML under the linear mixed model is l = 1 2 log V (τ 2 ) 1 2 X V 1 (τ 2 )X Score statistic for H 0 : τ 2 = 0 is 1 2 (y X α) V (τ 2 ) 1 (y X α) Q = (Y X α) K (Y X α) tr(kp), which follows a mixture of χ 2 1 distribution.

23 Kernel Kernel function K (, ) measures similarity for pairs of subjects Linear kernel: K (G i, G k ) = m j=1 G ijg kj Something about K (, ) Ability to incorporate high-dimension and different types of features (e.g., SNPs, expression, environmental factors) K (, ) is a symmetric semipositive definite matrix Eigenvalues are interpreted as % of the variation explained by the corresponding eigenvectors, but a negative eigenvalue implying negative variance is not sensible. No guarantee that the optimization algorithms that work for positive semidefinite kernels will work when there are negative eigenvalues Mathematical foundation moves from real numbers to complex numbers

24 Some Kernels Some kernels K (G i, G k ) = m j=1 G ijg kj =< G i, G k > K (G i, G k ) = 1 2m m j=1 IBS(G ij, G kj ), where IBS is identity-by-state K (G i, G k ) = (< G i, G k >) p : polynomial kernel, p > 0 Modeling higher-order interaction m (< G i, G k >) 2 = ( G ij G kj ) 2 = j=1 m m (G ij G ij )(G kj G kj ) j=1 j =1 K (Gi, G k ) = exp( G i G j 2 /σ 2 ): Gaussian kernel Schaid DJ. (2010) Genomic similarity and kernel methods I: advancements by building on mathematical and statistical foundations. Hum Hered 70: Schaid DJ. (2010) Genomic similarity and kernel methods II: methods for genomic information. Hum Hered 70:

25 Choice of Kernels An advantage of the kernel method is its expressive power to capture domain knowledge in a general manner. Generally difficult to construct a good kernel for a specific problem Basic operations to create new kernels from existing kernels: multiplying by a positive scalar adding kernels multiplying kernels (element-wise).

26 Generalized Linear Model Observations for the linear model apply to the generalized linear model Penalized log-likelihood function n n l = y i (X i α + ω j K (G i, G j )) i=1 j=1 n log{1 + exp(x i α + ω j K (G i, G j ))} 1 2 λω K ω j=1 The logistic kernel machine estimator [ X DX X ] [ ] D α DX D + (τk ) 1 = h [ X Dỹ Dỹ ] where τ = λ 1 σ 2, D = diag{e(y i )(1 E(y i ))}, and ỹ = Xα + K ω + var(y) 1 (y µ)

27 Generalized Linear Model The same estimators can be obtained from maximizing the penalized quasilikelihood from a logistic mixed model logite(y i ) = X i α + h i where h = (h 1,..., h n ) is a n 1 vector of random effects following h N(0, τk ) with τ = 1/λ The score statistic for τ is Q = (y X α) K (y X α), which follows a mixture of χ 2 distributions

28 Exponential Family Suppose y i follows a distribution in the exponential family with density p(y i ; θ i, φ) = exp{ y iθ i a(θ i ) φ + c(y i, φ)}, where θ i = X i α + h(g i) is the canonical parameter, a( ) and c( ) are known functions, φ is a dispersion parameter µ i = E(y i ) = a (θ i ) and var(y i ) = φa (θ i ) Gaussian: φ = σ 2, a(θ i ) = θ 2 i /2, and a (θ i ) = θ i Logistic: φ = 1, a(θ i ) = log(1 + exp(θ i )), a (θ i ) = exp(θ i ) 1+exp(θ i ) Other distributions: log-normal, Poisson, etc.

29 Summary Burden tests are more powerful when a large number of variants are causal and all causal variants are harmful or protective. Variance component test is more powerful when a small number of variants are causal, or mixed effects exist. Both scenarios can happen across the genome and the underlying biology is unknown in advance.

30 Combined Test SKAT (SNP-set/Sequence Kernel Association Test): variance component test Combine the SKAT variance component and burden test statistics (Lee et al. 2012) Q ρ = (1 ρ)q SKAT + ρq burden, 0 ρ 1 ρ = 0: SKAT ρ = 1: Burden Instead of assuming {β j } are iid from F(0, τ 2 ), assume β 1. β m F 1 ρ... ρ 0, τ ρ... 1

31 SKAT-O Q ρ = (1 ρ)q SKAT + ρq burden, which is asymptotically equivalent to (1 ρ)κ + a(ρ)η 0, where κ follows a mixture of χ 2 1 and η 0 χ 2 1. Use the smallest P-value from different ρs: T = inf 0 ρ 1 P ρ In practice, evaluate Q ρ on a set of pre-selected grid points, 0 = ρ 1 < < ρ B = 1 T = min ρ {ρ1,,ρ B }P ρ

32 Summary Have robust power under a wide range of models Q SKAT and Q burden are not independent. The underlying model for SKAT-O is not natural.

33 Mixed Effects Model Model m g{e(y i )} = X i α + G ij β j (1) j=1 Burden: β1 = = β m SKAT: βj F (0, τ 2 ) independently SKAT-O: βj F(0, τ 2 ) with pairwise correlation ρ Hierarchical model of β β j = w j η + δ j (2) wj : known features for the jth variant (e.g., w j = 1 for all j s) δj F (0, τ 2 )

34 Mixed Effects Model Plug (2) into (1) m g{e(y i )} = X i α + ( m w j G ij )η + G ij δ j j=1 j=1 Some examples: If w = 0, δj = β j, the model becomes m g{e(y i )} = X i α + G ij β j, β j F(0, τ 2 ) j=1 If w = 1 and δj = 0, the model becomes m g{e(y i )} = X i α + ( G ij )η j=1

35 Mixed Effects Model Some examples: wj = (w j1, w j2 ) where m w j1 = { 1 for j = 1,, m 1 if jth variant is a missense w j2 = 0 otherwise j=1 w jg ij = ( m j=1 G ij, m j=1 w j2g ij ) η1 : average effect of m variants η 2 : effect of missense variants relative to the average δj : residual variant specific effect F(0, τ 2 )

36 Mixed Effects Model-based Test Mixed effects model m g{e(y i )} = X i α + ( m w j G ij )η + G ij δ j j=1 j=1 Null hypothesis is H 0 : η = 0 and τ 2 = 0 η: fixed effects; τ 2 : variance component The score test statistic for τ 2 and η is S η = (Y X α) (GW )(GW ) (Y X α), and S τ 2 = ( GG Y X α) ( ) Y X α, where α is MLE of α under H 0. However, S τ 2 and S η are not independent.

37 Independence of score test statistics We made a minor (but important) modification S τ 2 = ( GG Y X α GW η) ( ) Y X α GW η, where ( α T, π T ) are obtained under τ 2 = 0. We can show that S τ 2 and S η are independent. G } E{(GW ) (Y X α)((y X α GW η) = σ 2 E{(GW ) (I P1)(I P2)G} = 0, where P 1 is the projection onto X and P 2 is the projection onto (X, (GW )).

38 Combining independent statistics MiST (Mixed effects Score Test) P-value combination Fisher s combination: reject H0 at significance level α if 2 log(p τ 2) 2 log(p η ) χ 2 4,α Tippitt s combination: reject H 0 at significance level α if min(p τ 2, P η ) 1 (1 α) 1/2 Other combinations, e.g., linear combination S = ρs η + (1 ρ)s τ 2 Jianping Sun, Yingye Zheng, and Li Hsu (2013). A Unified Mixed-Effects Model for Rare-Variant Association in Sequencing Studies. Genetic Epidemiology, 37:

39 Power Comparison m=10 variants, n=1000 subjects, α = 0.01 Burden SKAT SKAT-O MiST F MiST T β j = c/{p j (1 p j )} 1/2, j = 1,..., β 3 = 1.5c, β 4 = 1.5c, β 5 = c, β 6 = c; β 1 = β 4 = β 7 = c β 1 = c, β 4 = 0.5c, β 7 = 0.25c

40 Dallas Heart Study Dallas Heart Study (Victor et al. 2004). n=3409 subjects, 3 genes (ANGPTL3, ANGPTL4 and ANGPTL5) were sequencied. We analyzed these genes in association with log(triglyceride). ANGPTL3 ANGPTL4 ANGPTL5 Burden SKAT SKAT-O EREC MiST F MiST T MiST F (Z ) MiST T (Z ) The component p-values of ANGPTL5: p π = 5x10 6 and p τ 2 = Furthermore, p = for nonsense variants and p=0.24 for frame shift variants.

41 Summary MiST (Mixed effects Score Test) is based on hierarchical models for a set of variants The model includes the usual appealing features for regression models such as adjusting for confounders and being able to accommodate different types of outcomes by using appropriate link functions. It models the variant effects as a function of (known) variant characteristics to leverage information across loci while still allowing for individual variant effects.

42 Combining K studies We have discussed for single variant analysis: Pooling the data from K studies. Since all score statistics are derived from regression models, it is easy to account for the differences between studies by adjusting for study and/or study covariates Pooling the data ensures consistency in data QC and model fitting Pooling can be logistically difficult and time consuming Sometimes protection of human subjects prohibit sharing the data Meta-analysis of combining summary statistics from K studies is still a viable alternative

43 Revisit Score Statistics Weighted burden test n m U burden = w j G ij (y i X i α) = i=1 m j=1 n w j j=1 i=1 G ij (y i X i α) }{{} U j : Score of single variant model U j = n i=1 G ij(y i X i α) is a score function of a single variant model. y i = X i α + G ij β j + ε i, ε N(0, σ 2 )

44 Variance Component test Q SKAT is a weighted sum of squared score statistics of the single SNP marginal model. Q SKAT = (Y X α) GG m n(y X α) n 1 m n = { G ij (Y i X i α)} 2 = j=1 m j=1 i=1 U 2 j

45 Key Elements A vector of single variant score statistics, U = (U 1,..., U m ) with covariance V = cov(u) Burden score statistic U burden = W U, var(u burden ) = W VW Variance component score statistic Q SKAT = U U, which follows a mixture of chi 2 distribution with weights being the eigenvalues of V

46 Fixed effects model For k = 1,, K, let U k and V k denote the score statistics and covariance for the kth study. Score statistic over K studies is Burden test U = k U k V = k=1 k k=1 V k U burden = W U var(u burden ) = W VW U burden var(u burden) 1 U burden χ 2 p

47 Fixed effects model Variance component test Q SKAT = U U m λ j χ 2 1,j j=1 where λ j is the jth eigenvalue of V = K k=1 V k Combination of burden and score statistics Q ρ = (1 ρ)q SKAT + ρq burden where ρ is adaptively chosen and the p-value can be obtained by one-dimensional numerical integration

48 Fixed effects model Summary of single variant score statistic may not enough for MiST score statistics S τ 2 = S η = (Y X α) (GW )(GW ) (Y X α), ( GG Y X α GW η) ( ) Y X α GW η, where ( α T, η T ) are obtained under τ 2 = 0.

49 Random Effects Model For k = 1,, K, β k = (β k1,, β km ) is the effect of m variants for the kth study. Random effects model β k = β 0 + ξ k where β 0 = (β 01,, β 0m ) represents the average effect among the studies, ξ k is a set of random effects representing the deviation of the kth study from the average effect ξ k N(0, Σ)

50 Heterogeneity Assume Σ = σ 2 B, where B is a pre-specified matrix to constrain the potential many parameters in Σ. A choice of B is b1 2 b 1 b 2 r b 1 b m r. B = b 2 b 1 r b m b 1 r bm 2 (b 1, b m ) controls the relative degrees of heterogeneity for the m variates (e.g., MAF), and r specifies the correlation of heterogeneity. Choice of B has no effect on the type I error but may affect the power.

51 New Random Effects Burden Test The null hypothesis H 0 : β 0 = 0, σ 2 = 0 For k = 1,..., K, β k N(β 0, Ω k = V 1 k + σ 2 B). The log-likelihood function is l = 1 2 K ( β k β 0 ) Ω 1 k ( β k β 0 ) 1 2 k=1 K log Ω k Let β k V 1 k U k, the random effects (RE) test for fixed effects where U σ = 1 2 U RE burden = U V 1 U + U2 σ V σ k=1 K k=1 U k BU k 1 2 tr(vb), V σ = 1 2 tr( K k=1 V kbv k B) For burden test, replace U by W U, and V by W VW.

52 New Random Effects Variance Component Test β 0 N(0, τ 2 W ), where W is a pre-specified matrix, e.g., w 2 1 w 1 w 2 ρ W =. w 2 w 1 ρ... w 2 d where (w 1,, w d ) controls the relative magnitude of the d average genetic effects, and ρ indicates the correlation. The null hypothesis H 0 : τ 2 = 0, σ 2 = 0 Let β = ( β 1,..., β K ), then ( ) β MVN 0, τ 2 (J K W ) + σ 2 (I K B) + diag(v 1 1,, V 1 K ) where denotes Kronecker product { β k V 1 k U k }, the score statistic is a function of U k, V k, k = 1,..., K.

53 Summary Pooled- vs meta-analysis For meta-analysis rare variant association tests can be constructed from multivariate summary statistics, i.e., the score vector U and information matrix V Fixed vs random effects model

54 Set-based gene-environment interaction m variants, G i = (G i1,, G im ) E i : environmental covariate X i : covariates Gene-environment interaction (GxE) model g{e(y i )} =X i α + E iβ E + m m G ij βj G + j=1 j=1 No interaction means β = (β GE 1,, β GE m ) = 0 (E i G ij )β GE j

55 Hierarchical model for β GE Model the interaction effect β GE j = w j η + δ j wj : a vector of known features δ j F (0, τ 2 ) The interaction effect term m j=1 (E i G ij )β GE j = ( m E i G ij w j )η + j=1 m = E i ( G ij w j )η + i=1 No interaction means H 0 : η = 0, τ 2 = 0 m E i G ij δ j j=1 m (E i G ij )δ j j=1

56 Challenges Main effects {β G 1,, βg m} may not be estimated reliably if m is large or variants are rare. Assume the main effects {β G j } are random effects such that β G j F(0, ν 2 ) Need to derive score statistics for the mixed GxE effects (η, τ 2 ) in the presence of another random effects β G j.

57 Estimation β G can be estimated by maximum posterior approach (or best linear unbiased prediction, in the linear mixed effects model), but the computation is intensive under a generalized linear model due to m-dimensional integration with no closed form. βg j minimizes ridge regression n β ridge = argmin (y i X i α E iβ E G i β G ) 2 + λ where λ = σ 2 /ν 2 i=1 m j=1 β 2 j

58 Some nice properties about ridge Knight and Fu (2000) states that if λ = o( n) then β λ is a n consistent estimator of β 0 Score statistics for the fixed effects under H 0 : η = 0, τ 2 = 0 u η = (D µ) ( m E( G j w j ) ) ( m V E( G j w j ) ) (D µ) j=1 where µ = Ê(D G, E) under η = 0, τ 2 = 0 Score statistic for the variance component under H 0 : τ 2 = 0 u τ 2 j=1 = (D û)(ge)(ge) (D µ) where û = Ê(D G, E) under τ 2 = 0

59 Combination of score statistics P-value based, Z η = 2 log P η and Z τ 2 = 2 log P τ 2 T f = Z η + Z τ 2 χ 2 4 Grid-search based optimal linear combination T o = max (ρu η + (1 ρ)u τ 2) ρ [0,1] where ρ is restricted on a set of pre-specified grid points {0 = ρ 0, ρ 1,..., ρ d = 1} Adaptive-weighted linear combination T a = Z 2 η + Z 2 τ 2 Give more weight to either burden or variance component if the evidence comes mainly from one Su YR, Di C and Hsu L (2015). A unified powerful set-based test for sequencing data analysis of GxE interactions. Submitted.

60 Power comparison m = 25 variants T o T a T f Burden Var Comp H a : 30% variants β = c H a : Half β = c, other half β = c H a : All β = c

61 Weight Choices of weight Functioncal characteristics (e.g., missense, nonsense) Screening statistics, Mj and C j are the Z statistics from marginal association screening and correlation of G and E screening { Mj if M w j = j > C j otherwise C j Since the screening statistics are independent of GxE test, no need to use permutation to calculate the p-values Jiao S, Hsu L, et al. (2013, 2015)

62 Summary Set-based association testing Mixed effects model that accounts for both burden genetic risk score and variance component Meta-analysis GxE interaction between a set of variants and environmental factor

63 Recommended readings Liu D et al. (2007). Semiparametric regression of multidimensional genetic pathway data: Least-Squares Kernel Machines and Linear Mixed Models. Biometrics 63: Liu D et al. (2008). Estimation and testing for the effect of a genetic pathway on a disease outcome using logistic kernel machine regression via logistic mixed models. BMC Bioinformatics 9:292. Schaid DJ (2010) Genomic similarity and kernel methods I: advancements by building on mathematical and statistical foundations. Hum Hered 70: Schaid DJ (2010) Genomic similarity and kernel methods II: methods for genomic information. Hum Hered 70: Zhang and Lin (2003). Hypothesis testing in semiparametric additive mixed models. Biostatistics 4:57 74.

64 Recommended readings Lee S, Lin X and Wu M (2012). Optimal tests for rare variant effects in sequencing association studies. Biostatistics 13: Lin DY and Tang Z (2011). A general framework for detecting disease associations with rare variants in sequencing studies. AJHG 89: Jianping Sun, Yingye Zheng, and Li Hsu (2013). A Unified Mixed-Effects Model for Rare-Variant Association in Sequencing Studies. Genetic Epidemiology, 37: Jiao S,..., Hsu L (2015) Powerful Set-Based Gene-Environment Interaction Testing Framework for Complex Diseases. Genetic Epidemiology, DOI: /gepi Tang Z and Lin DY (2015). Meta-analysis for discovering rare-variant associations: Statistical methods and software programs. AJHG 97:35 53 Wu et al. (2011). Rare-variant association testing for sequencing data with the sequence kernel association test. AJHG 89:

Lecture 9: Kernel (Variance Component) Tests and Omnibus Tests for Rare Variants. Summer Institute in Statistical Genetics 2017

Lecture 9: Kernel (Variance Component) Tests and Omnibus Tests for Rare Variants Timothy Thornton and Michael Wu Summer Institute in Statistical Genetics 2017 1 / 46 Lecture Overview 1. Variance Component