Lecture Notes. Introduction

Transcription

1 5/3/016 Lecture Notes R. Rekaya June 1-10, 016 Introduction Variance components play major role in animal breeding and genetic (estimation of BVs) It has been an active area of research since early 1950 Dr. Henderson proposed 3 methods for VC estimation (not covered in these notes) MIVQUE and MINQUE 1

2 5/3/016 Introduction Maximum Likelihood (ML) Restricted Maximum likelihood (REML) Bayesian Methods Quadratic form Let y~n(μ, V) Introduction E y Qy = tr QV + μ, Qμ var y Qy = tr QV + 4μ, QVQμ p tr V = v ii i=1

3 5/3/016 Introduction n n y Qy = q ij y i y j i=1 j=1 n n E y Qy = E q ij y i y j i=1 j=1 n n = q ij E y y i=1 j=1 n n = q ij [cov y i, y j + E y i E(y j )] i=1 j=1 n n = q ij [v ij + μ i μ j ] i=1 j=1 n n n n = q ij v ij + i=1 j=1 i=1 j=1 n n q ij μ i = q ij v ij + μ Qμ i=1 j=1 = tr QV + μ Qμ μ j Introduction y μ) Q(y μ = y Qy μ Qy y Qμ + μ Qμ E[ y μ) Q(y μ ] = E(y Qy) μ QE(y) E(y )Qμ + μ Qμ = E y Qy μ Qμ μ Qμ + μ Qμ = E(y Qy) μ Qμ and E y Qy = E y μ) Q(y μ + μ Qμ because = tr QV + μ Qμ E y μ) Q(y μ = E tr y μ) Q(y μ = E tr[q y μ)(y μ ] = tr E[Q y μ)(y μ ] = tr QE[ y μ)(y μ ] = tr QV 3

4 5/3/016 Examples y~n(0, I) E y y = n var y y = n y~n(0, Iσ ) E y y = nσ var y y = nσ 4 y~n(μ, Iσ ) E y y = nσ + nμ var y y = nσ 4 + 4nμ σ Example Let Q 1 = I; Q = 1 n 11 ; and Q 3 = Z(Z Z) 1 Z Then y Q 1 y = y y y Q y = 1 n y 11 y y Q 3 y = y (Z Z Z 1 Z )y var y = V = ZZ σ u + Iσ e E y y = tr V + nμ = nσ u + nσ e 4

5 5/3/016 Translation Invariance Let y~n(xb + Zu, Iσ ) If QX = 0 Quadratic form is translation invariant E y Qy var y Qy = tr QV = tr QV Variance components are estimated using translation invariant quadratic forms Maximum Likelihood (ML) ML approach (Hartley and Rao; 1967) Conceptually simple Efficient in the use of information Assumes the fixed effect are known Estimates of fixed effects are used instead Biased estimates of variance components Bias depends on the number of fixed effects and the amount of information in each class 5

6 5/3/016 ML Simple model y = μ + e with e~n(0, Iσ e ) Conditional distribution of the data p y μ, σ e = p(y i μ, σ e ) n i=1 = (πσ e ) 1 exp[ n (y i μ) i=1 σ ] e It is a PDF where y is the random variable Likelihood function ML l(μ, σ e ; y) = (πσ e ) n (y i μ) exp[ ] σ i=1 e In the likelihood function, the random variables are the unknown parameters Thus, the likelihood function is not a PDF when viewed as a function of the unknown parameters n 6

7 5/3/016 ML To find the ML estimates, we need to maximize the likelihood function or its log L μ, σ e ; y = 0.5n ln π + ln σ e n i=1 (y i μ) = 0.5n[ln π + ln σ e + 1 n (y nσ i=1 e i μ) ] One way to find the ML estimates is to the take derivatives σ e ML Derivatives L μ, σ n e ; y i=1 y i nμ = μ σ e L μ, σ e ; y = n SSE (1 σ e σ e nσ ) e Where SSE is the sum of the squares of the errors 7

8 5/3/016 ML Setting the derivate equal to zero leads to: n i=1 y i nμ=0 μ = 1 y n i=1 i nσ e = SSE σ e = SSE n As μ is unknown, SSE is calculated using estimates of the mean SSE = 1 n (y n i=1 i μ) Estimate of the mean is independent of the variance. However, the latter is not. ML estimate of the variance is biased n ML Mixed Model General mixed model y = Xβ + Zu + e With u~n 0, G and e~n 0, R Likelihood function l(β, G, R; y) E y β Var y β = Xβ = ZGZ + R = ZAZ σ u + Iσ e = V = (π) n V 1 exp[ 1 y Xβ V 1 y Xβ ] 8

9 5/3/016 ML-Mixed model Maximization L β, G, R; y =.5n ln π.5 ln V Partial derivatives L β, V; y β 1 y Xβ V 1 y Xβ = X V 1 (y Xβ) L β, V; y σ u Recall ML-mixed model =.5tr V 1 ZAZ +. 5( y Xβ V 1 ZAZ V 1 y Xβ =.5tr V 1 V u +. 5( y Xβ V 1 V u V 1 y Xβ ln V x V 1 x V = tr(v 1 x ) V σ u = ZAZ = V 1 V x V 1 9

10 5/3/016 ML-mixed model Further, L β, V; y σ u =.5tr V 1 V u +. 5 y Xβ V 1 V u V 1 y Xβ +. 5 β β V 1 V u V 1 β β Similarly, L β, V; y σ e =.5tr V y Xβ V 1 V 1 y Xβ Because V σ =I e +. 5 β β V 1 V 1 β β ML-mixed model ML of fixed effects β = (X V 1 X) 1 X V 1 y Setting β = β and the partial derivatives respect to the variance components equal to zero leads to: tr V 1 V u = y Xβ V 1 V u V 1 y Xβ tr V 1 = ( y Xβ V 1 V 1 y Xβ 10

11 5/3/016 ML-mixed model Let P = V 1 V 1 X(X V 1 X) 1 X V 1 Py = V 1 y V 1 X X V 1 X 1 X V 1 y = V 1 y Xβ Thus, tr V 1 V u = y PV u Py tr V 1 ZAZ = y P(ZAZ )Py tr V 1 = y PPy Note that P is function of V ML-mixed model With V = ZAZ σ u + Iσ e Special case: one observation per animal (Z=I) V = Aσ u + Iσ e Several random effects V = k i=1 Z i G i Z i σ i + Iσ e where k is the number of random effects and p u i G i, σ i ~N(0, G i σ i ) for i=(1,, k) 11

12 5/3/016 ML-mixed model Problems Estimates of fixed effects are function of variance components The matrix V is a function of the unknown variance components Solutions for variance components involve non-linear equations due the inverse for V Estimates of variance components are biased (remember we replaced β by β ) Iterative approaches are needed to solve the system of equations Restricted Maximum Likelihood (REML) ML estimates of variance components are biased unless the fixed effect are known without errors Fixed effects are seldom known and estimating them will not solve the problem Can we estimate variance components after removing the fixed effects from the model? REML does just that! We will visit this again from a Bayesian perspective 1

13 5/3/016 REML Basic idea: Remove the fixed effects from the model! This could be accomplished by some linear transformation of the observed data REML Mixed model y = Xβ + Zu + e With u~n 0, G and e~n 0, R Imagine there is a transformation matrix K such that: KX = 0 13

14 5/3/016 REML Multiplying both side of the mixed model with K y = Ky = K Xβ + Zu + e = KXβ + KZu + Ke = KZu + Ke The likelihood function using the transformed data X n r l(g, R; y) = (π) KVK 1 exp[ 1 Ky KVK 1 Ky ] Keep in mind: ML REML Y KY X KX=0 Z KZ V KVK n n-r(x) r(x)=rank of X REML Log-likelihood function L G, R; y =.5(n r X ) ln π.5 ln KVK 1 Ky (KVK ) 1 Ky Using the following results (Searle, 1979) ln KVK = ln V + ln X V 1 X P = K (KVK ) 1 K 14

15 5/3/016 REML y K KVK 1 Ky = y Py = y Xβ V 1 y Xβ The log likelihood can be written as: L G, R; y =.5 ln V 1 ln X V 1 X.5 y Xβ V 1 y Xβ REML Derivative of log-likelihood L G, R; y σ i For i=e, u V =.5tr V 1 σ i.5tr Remember: X V 1 X 1 V X V 1 σ V 1 X i +. 5 y Xβ V V 1 σ V 1 y Xβ i =.5tr V 1. V 1 X X V 1 X 1 X V 1 V σ i +. 5y P V σ i Py Py = V 1 y Xβ 15

16 5/3/016 REML Remember: V σ = ZAZ and V u σ =I e Then L G, R; y σ e =.5tr V 1 V 1 X X V 1 X 1 X V y PPy =.5tr P +. 5y PPy L G, R; y σ u =.5tr V 1 V 1 X X V 1 X 1 X V 1 ZAZ +. 5y P(ZAZ )Py =.5tr PZAZ +. 5y PZAZ Py REML Setting the derivatives to zero leads to: tr P = y PPy for σ e tr PZAZ = y PZAZ Py for σ u Note that P is a function of V = ZAZ σ u + Iσ e 16

17 5/3/016 Algorithms Derivative based methods Newton-Raphson Let θ = σ e, σ u, NR is an iterative procedure that requires the second order derivatives θ k+1 = θ k H k 1 L θ; y θ θ k Where k is the iteration number and H is the Hessian matrix H = L θ; y θ θ REML Observed information matrix or Fisher information matrix L( θ, y) I H θ θ' If θ is a scalar L(, y) I Expected Fisher matrix L(, y) E ( I) E The expectation being taken over possible values of y for a fixed value of 17

18 5/3/016 Algorithms Derivative based methods Thus, in our case L θ; y σ θ k =.5tr P k +. 5y Pk P k y e L θ; y σ θ k =.5tr P k ZAZ +. 5y P k ZAZ P k y u H k ij = L θ; y σ u σ e =.5tr P k V u P k V e y P k V u P k V e P k y With V u = ZAZ and V e = I Algorithms Derivative based methods Fisher s Scoring Same as NR, except the Hessian matrix is replaced by its expectation Expectation is taking over the data Thus, it is free of the data θ k+1 = θ k + F k 1 L G, R; y θ With F k ij =.5tr P k V i P k V j There have been several variations of these two algorithms (AIREML). θ k 18

19 5/3/016 Recall: V = ZGZ + R Algorithms Derivative Free methods = R I + R 1 ZGZ = R I + R 1 ZGZ because AB = A B = R G 1 + R 1 ZZ G = R G G 1 + Z R 1 Z Algorithms Derivative based methods Let C = X R 1 X X R 1 Z Z R 1 X Z R 1 Z + G 1 C = (X R 1 X) G 1 + Z (R 1 R 1 X X R 1 X 1 X R 1 )Z = (Z R 1 Z + G 1 ) X (R 1 R 1 Z Z R 1 Z + G 1 1 Z R 1 )X Proof: A C B D = A D CA 1 B = D A BD 1 C 19

20 5/3/016 Algorithms Derivative based methods Using previous results ln V = ln R + ln G + ln G 1 + R 1 ZR 1 Z ln X V 1 X = ln C ln G 1 + Z R 1 Z Then L G, R; y =.5 ln R.5 ln G.5 ln C.5 y Xβ V 1 y Xβ =.5 ln R.5 ln G.5 ln C.5y Py Recall Algorithms Derivative based methods ln R = ln Iσ e = n ln(σ e ) ln G = q ln σ u ln C = ln X R 1 X + ln G 1 + Z WZ Where W= (R 1 R 1 X X R 1 X 1 X R 1 ) ln X R 1 X = ln X Xσ e = ln σ e r X X X = r x ln σ e + ln X X G 1 + Z WZ = σ e (G 1 σ e + Z MZ) 0

21 5/3/016 Algorithms Derivative based methods Finally, L G, R; y =.5 n r x q lnσ e.5qlnσ u.5ln C.5y Py where C = X X X Z Z X Z Z + G 1 σ e Expectation/Maximization (EM) Algorithm Basic idea The observed data is incomplete If the missing data is observed, it facilitate greatly the implementation What is the complete data? What is the missing data? If the random effects are observed, estimating variance components will be easy 1

22 5/3/016 EM Algorithm Thus, if u and e are observed σ u = E(u A 1 u) q σ e = E(e e) n Unfortunately, u and e are not observed! EM Algorithm However, E u y = σ u Z V 1 y Xβ = σ u Z Py var u y = σ u I σ 4 u Z V 1 y and, E u A 1 u y = qσ u + σ 4 u [y P ZAZ Py tr(pzaz ) E e e y = nσ e + σ e 4 [y PPy tr P ] Both expectations are function of the variances we are trying to estimates!

23 5/3/016 EM Algorithm To overcome this problem, Dempster et al (1977) suggested the following iterative approach 1. Expectation (E ) step: Using starting values for variances, calculate the expectation of the quadratic forms. Maximization (M) step: expectations from step 1 will be used to obtain VC 3. Repeat steps 1 and until convergence Set k=0; (σ e ) k, (σ u ) k E-Step EM-Algorithm E u A 1 u y k = q(σ u ) k + (σ u 4 ) k [y P ZAZ Py tr(pzaz ) E e e y k = n(σ e ) k + (σ e 4 ) k [y PPy tr P ] M-Step σ (k+1) u = E(u A 1 u) k q σ (k+1) e = E(e e) k n 3