THE AITKEN MODEL y = Xβ + , ∼ (0, σ 2 V) Identical to the Gauss-Markov linear model except that Var() = σ 2 V instead of σ 2 I. V is assumed to be a known nonsingular variance matrix. σ 2 is an unknown positive variance parameter. c Copyright 2012 (Iowa State University) Statistics 511 1 / 11 A Transformation of the Model By the Spectral Decomposition Theorem, there exists a nonsingular symmetric matrix V 1/2 such that V 1/2 V 1/2 = V. Using V −1/2 to denote (V 1/2 )−1 , we have V −1/2 y = V −1/2 Xβ + V −1/2 . With z = V −1/2 y, W = V −1/2 X, and δ = V −1/2 , we have z = Wβ + δ, δ ∼ (0, σ 2 I) because Var(δ) = Var(V −1/2 ) = V −1/2 σ 2 VV −1/2 = σ 2 V −1/2 V 1/2 V 1/2 V −1/2 = σ 2 I. c Copyright 2012 (Iowa State University) Statistics 511 2 / 11 Thus, after transformation, we are back to the Gauss-Markov model we are familiar with. We can apply all the results we have established previously to the Gauss-Markov model z = Wβ + δ, δ ∼ (0, σ 2 I). c Copyright 2012 (Iowa State University) Statistics 511 3 / 11 Estimation of E(y) under the Aitken Model Note that E(y) = E(V 1/2 V −1/2 y) = V 1/2 E(V −1/2 y) = V 1/2 E(z). Because the Gauss-Markov model holds for z, we already know that the best estimate of E(z) is ẑ = = = = PW z = W(W 0 W)− W 0 z V −1/2 X((V −1/2 X)0 V −1/2 X)− (V −1/2 X)0 V 1/2 y V −1/2 X(X0 V −1/2 V −1/2 X)− X0 V −1/2 V −1/2 y V −1/2 X(X0 V −1 X)− X0 V −1 y. Thus, to estimate E(y) = V 1/2 E(z), we should use V 1/2 ẑ = V 1/2 V −1/2 X(X0 V −1 X)− X0 V −1 y = X(X0 V −1 X)− X0 V −1 y. c Copyright 2012 (Iowa State University) Statistics 511 4 / 11 Likewise, if Cβ is estimable, we know the BLUE is the ordinary least squares (OLS) estimator. C(W 0 W)− W 0 z = C(X0 V −1/2 V −1/2 X)− X0 V −1/2 V −1/2 y = C(X0 V −1 X)− X0 V −1 y. C(X0 V −1 X)− X0 V −1 y = Cβ̂ V is called a Generalized Least Squares (GLS) estimator. c Copyright 2012 (Iowa State University) Statistics 511 5 / 11 β̂ V = (X0 V −1 X)− X0 V −1 y is a solution to the Aitken Equations: X0 V −1 Xb = X0 V −1 y which follow from the Normal Equations W 0 Wb = W 0 z c Copyright 2012 (Iowa State University) ⇐⇒ ⇐⇒ X0 V −1/2 V −1/2 Xb = X0 V −1/2 V −1/2 y X0 V −1 Xb = X0 V −1 y. Statistics 511 6 / 11 Recall that solving the Normal Equations is equivalent to minimizing (z − Wb)0 (z − Wb) over b ∈ IRp . Note that (z − Wb)0 (z − Wb) = = = = c Copyright 2012 (Iowa State University) (V −1/2 y − V −1/2 Xb)0 (V −1/2 y − V −1/2 Xb) ||V −1/2 y − V −1/2 Xb||2 ||V −1/2 (y − Xb)||2 (y − Xb)0 V −1 (y − Xb). Statistics 511 7 / 11 Thus, β̂ V = (X0 V −1 X)− X0 V −1 y is a solution to this generalized least squares problem. When V is diagonal, the term “Weighted Least Squares” (WLS) is often used instead of GLS. c Copyright 2012 (Iowa State University) Statistics 511 8 / 11 An unbiased estimator of σ 2 is z0 (I − PW )z k(I − PW )zk2 = n − rank(W) n − rank(W) 2 k(I − W(W 0 W)− W 0 )zk = n − rank(W) (I − V −1/2 X(X0 V −1 X)− X0 V −1/2 )V −1/2 y2 = n − rank(V −1/2 X) c Copyright 2012 (Iowa State University) Statistics 511 9 / 11 = = = 2 −1/2 y − V −1/2 X(X0 V −1 X)− X0 V −1 y V n − rank(X) 2 −1/2 (y − X(X0 V −1 X)− X0 V −1 y) V n−r 2 −1/2 (y − Xβ̂ V ) V n−r (y − Xβ̂ V )0 V −1 (y − Xβ̂ V ) = n−r 2 = σ̂V . c Copyright 2012 (Iowa State University) Statistics 511 10 / 11 Inference Under the Normal Theory Aitken Model The Normal Theory Aiken Model: y = Xβ + , ∼ N(0, σ 2 V). Under the Normal Theory Aitken Model, we can back transform to convert known formulas in terms of z and W to formulas in terms of y and X to allow inference about estimable Cβ under the Normal Theory Aitken Model. c Copyright 2012 (Iowa State University) Statistics 511 11 / 11