ANOVA Variance Component Estimation. Copyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

Transcription

1 ANOVA Variance Component Estimation Copyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

2 We now consider the ANOVA approach to variance component estimation. The ANOVA approach is based on the method of moments. Copyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

3 Suppose y = Xβ + Zu + e, where u and e are independent random vectors satisfying E(u) = 0 and E(e) = 0. opyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

4 Furthermore, suppose n q Z can be partitioned as Z = [ Z 1 n q 1,..., Z m n q m ] and u can be partitioned correspondingly as u 1 u m u =. so that Zu = m Z j u j. j=1 opyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

5 Suppose u 1 u m Var(u) = Var. = diag(σ2 11 q 1,..., σm1 2 q m ) = σ1 2 I 0 q 1 q σm 2 I q m q m Copyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

6 Then m+1 Var(y) = σj 2 Z j Z j, j=1 where Z m+1 n n I and σm+1 2 = σ2 e. By Lemma 4.1, E(y Ay) = tr(avar(y)) + [E(y)] AE(y) m+1 = σj 2 tr(az j Z j) + β X AXβ j=1 m+1 = σj 2 tr(z jaz j ) + β X AXβ. j=1 opyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

7 Now suppose we choose a set of matrices A 1,..., A m+1 X A i X = 0 i = 1,..., m + 1. Then m+1 E(y A i y) = σj 2 tr(z ja i Z j ) i = 1,..., m + 1. j=1 opyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

8 If we use the method of moments of moments, we replace E(y A i y) with its observed value y A i y to obtain the equations y A i y = m+1 j=1 ˆσ 2 j tr(z ja i Z j ) i = 1,..., m + 1. opyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

9 We can write these equations in matrix form as tr(z 1A 1 Z 1 )... tr(z m+1a 1 Z m+1 ) tr(z 1A 2 Z 1 )... tr(z m+1a 2 Z m+1 )... tr(z 1A m+1 Z 1 )... tr(z m+1a m+1 Z m+1 ) ˆσ 1 2 ˆσ 2 2. ˆσ 2 m+1 y A 1 y = y A 2 y.. Solving for ˆσ 1 2, σ2 2,..., ˆσ2 m+1 gives the ANOVA estimates. y A m+1 y opyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

10 The matrices A 1,..., A m+1 are usually chosen so that y A 1 y,..., y A m+1 y correspond to sums of squares from an ANOVA table. opyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

11 Many strategies for choosing A 1,..., A m+1 have been proposed. The book Variance Components by Searle, Casella, and McCulloch contains an extensive discussion of several strategies. opyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

12 Let M denote the matrix whose (i, j) th element is tr(z ja i Z j ). Let s = [y A 1 y, y A 2 y,..., y A m+1 y]. opyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

13 If M is nonsingular, then the vector of ANOVA variance component estimates is ˆσ 2 ˆσ 2 1. ˆσ 2 m+1 = M 1 s. Copyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

14 Recall M and s were chosen so that Mσ 2 = E(s), where σ 2 = σ σ 2 m+1 Thus, E(ˆσ 2 ) = E(M 1 s) = M 1 E(s) = M 1 Mσ 2 = σ 2. the ANOVA estimator of σ 2 is unbiased. Copyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

15 Find ANOVA-based estimates of σ 2 p and σ 2 e for the seedling dry weight example. opyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

16 Recall X =, Z = Copyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

17 Because there is only one variance component besides the error variance, we have m = 1, Z 1 = Z, and Z 2 = I. We need to identify matrices A 1 and A 2 such that X A 1 X = 0 and X A 2 X = 0. Of course, we also require the quadratic forms y A 1 y and y A 2 y to contain information about σ 2 p and σ 2 e. To find appropriate A 1 and A 2, start by writing down an ANOVA table with columns labeled Source, Matrix, and DF. opyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

18 Which of the matrices in the ANOVA table satisfy X AX = 0? Copyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

19 To apply the ANOVA variance component estimation method, we need to set up the equations [ ] [ tr(z 1A 1 Z 1 ) tr(z 2A 1 Z 2 ) tr(z 1A 2 Z 1 ) tr(z 2A 2 Z 2 ) [ ] [ tr(z (P Z P X )Z) tr(p Z P X ) tr(z (I P Z )Z) tr(i P Z ) ˆσ 2 p ˆσ 2 e ˆσ 2 p ˆσ 2 e ] [ ] y A 1 y = y A 2 y ] [ ] y (P Z P X )y =. y (I P Z )y Copyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

20 Let n ij = number of seedlings in the j th pot for genotype i. Thus, we have n 11 = 3, n 12 = 2, n 21 = 3, n 22 = 2. Write y A 1 y = y (P Z P X )y and y A 2 y = y (I P Z )y using summation notation. opyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

21 Find tr(z (P Z P X )Z), tr(z (I P Z )Z), tr(p Z P X ), and tr(i P Z ). opyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

22 Find expressions for the ANOVA estimators of ˆσ 2 p and ˆσ 2 e. Copyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

23 The ANOVA estimates of the variance components are sometimes equal to the REML estimates. This equality occurs for certain types of balanced designs and when the ANOVA estimates are positive. Copyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32

24 For the seedling dry weight example, the REML and ANOVA estimates agree when the latter are positive. However, if last pot contained a 3 seedlings instead of 2 (for example), the REML and ANOVA estimates would differ. Copyright c 2012 Dan Nettleton (Iowa State University) Statistics / 32