GaussMarkov Theorem. The GaussMarkov Theorem is given in the following regression model and assumptions:


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1 GaussMarkov Theorem The GaussMarkov Theorem is given in the following regression model and assumptions: The regression model y i = β 1 + β x i + u i, i = 1,, n (1) Assumptions (A) or Assumptions (B): Assumptions (A) Eu i = 0 for all i Var(u i ) = σ for all i (homoscedasticity) Cov(u i, u j ) = 0 for all i j x i is nonstochastic constant The text book uses Assumptions (B) (see p588 of the text): Assumptions (B) E(u i x 1,, x n ) = 0 for all i Var(u i x 1,, x n )) = σ for all i (homoscedasticity) Cov(u i, u j x 1,, x n ) = 0 for all i j If we use Assumptions (B), we need to use the law of iterated expectations in proving the BLUE With Assumptions (B), the BLUE is given conditionally on x 1,, x n Let us use Assumptions (A) The GaussMarkov Theorem is stated in the boxed statement below: 1
2 Gauss Markov Theorem Under Assumptions (A), the OLS estimators, β 1 and β are the Best Linear Unbiased Estimator (BLUE), that is 1 Unbias: E β 1 = β 1 and E β = β Best: β1 and β have the smallest variances among the class of all linear unbiased estimators Real data seldomly satisfy Assumptions (A) or Assumptions (B) Accordingly we should think that the GaussMarkov theorem only holds in the nevernever land However, it is important to understand the GaussMarkov theorem on two grounds: 1 We may treat the world of the GaussMarkov theorem as equivalent to the world of perfect competition in micro economic theory The mathematical exercises are good for your souls We shall prove the GaussMarkov theorem using the simple regression model of equation (1) We can prove the GaussMarkov theorem using the multiple regression model y i = β 1 + β x i + + β k x ik + u i, i = 1,, n () To do so, however, we need to use vector and matrix language (linear algebra) Actually, once you learn linear algebra the proof of GaussMarkov theorem is far more straight forward than the proof for the simple regression model of (1) In the text book the GaussMarkov theorem is discussed on the following pages: You should take a look at these pages
3 Proving the GaussMarkov Theorem The unbiasedness of β 1 and of β are given in the Comments on the Midterm Examination and the answers to Assignment #5 So, we prove here the minimum variance properties There are generally two ways to prove bestness: (i) using linear algebra, and (ii) using calculus We prove bestness using linear algebra first, and we leave the proof using calculus to the Appendix First we prove that β 1 has the smallest variance among all other linear estimators of β 1 Proof that β 1 is best We need to reexpress β 1 first β 1 = ȳ β x = 1 n yi ( (xi x)y i ) x = 1 n y i (x i x) x = n y i ( 1 n (x ) i x) x y i, where = x i n x = w i y i, where w i = 1 n (x i x) x The BLUE only looks at linear estimators of β 1 The linear estimators are defined by n β 1 = a i y i In passing we notice that if a i = w i, for all i = 1,, n then β 1 = β 1 We have to make β 1 unbiased To take expectation of β 1 we first substitute equation (1): y i = β 1 + β x i + u i for y i : β 1 = n a i y i = n a i (β 1 + β x i + u i ) = β 1 ai + β ai x i + a i u i E β 1 = β 1 ai + β ai x i + a i Eu i = β 1 ai + β ai x i, 3
4 since Eu i = 0 for all i We see that E β 1 = β 1 a i = 1 a i x i = 0 ( ) means if and only if We take variance of β 1 : Var( β 1 ) E( β 1 E β 1 ) = E( β 1 β 1 ) since E β 1 = β 1 = E( a i u i ) = (a 1Eu a neu n + a 1 a Eu 1 u + + a n 1 a n Eu n 1 u n ) = σ (a a n) = σ a i, since Eu i = σ and Eu i u j = 0, i j The variance of the OLS estimator, Var( β 1 ) is given by Var( β 1 ) = σ wi We see Var( β 1 ) Var( β 1 ) n a i n wi Since a i is an arbitrary nonstochastic constant we can rewrite a i as a i = w i + d i Earlier we saw that β 1 is unbiased if and only if a i = 1 and a i x i = 1 So, ai = w i + d i = 1 ai x i = w i x i + d i x i = 0 But wi = ( 1 n (x ) i x) x = 1 x wi x i = ( 1 n (x i x) x (xi x) = 1 ) x i = 1 n xi x i n x x = x x = 0 Hence di = 0 and di x i = 0 We square a i and sum with respect to i = 1,, n: a i = (w i + d i ) = w i + d i + w i d i = w i + d i 4
5 since the cross product term is zero: wi d i = ( 1 n (x ) i x) x d i = 1 n di 1 ( d i x i x d i ) = 0 Hence a i = w i + d i w i, and this concludes the proof Proof that β is best where β = (xi x)y i We shall use the fact that vi = 0 = ( ) xi x y i = v i y i v i = x i x and The variance of β, Var( β ), is given by Let β be a linear estimator of β : v i = Var( β ) = σ v i β = b i y i We need to find the conditions that make β unbiased Taking expectation we have E β = E b i (β 1 + β x i + u i ) = β 1 bi + β bi x i and thus E( β ) = β b i = 0, bi x i = 1 The variance of β, Var( β ), is Var( β ) = σ b i Let b i = v i + c i 5
6 then bi = v i + c i = c i = 0 bi x i = v i x i + c i x i = c i x i = 0 since vi x i = 1 x (xi x)x i = i n x = 1 So the variance of β becomes Var( β ) = σ b i = σ, (v i + c i ) = σ ( v i + c i + v i c i ) σ v i + σ c i Var( β ) since vi c i = 1 (xi x)c i = 1 ( x i c i x c i ) = 0 Appendix: Proving Bestness using calculus Another way to prove that the OLS estimators, β 1 and β, are best is to use calculus to find the minimum variance Since variance is a quadratic function, it is twice differentiable and thus we may use calculus to find the minimum Proving that β 1 is best The variance of a linear unbiased estimator is given by σ a i with two linear constraints ai = 1 and ai x i = 0 Hence we may form the following minimization problem subject to the linear constraints: min a 1,,a n σ a i subject to ai = 1 ai x i = 0 6
7 We form the Lagrangian Λ = σ a i λ 1 ( a i 1) λ ai x i The first order conditions are a 1 = σ a 1 λ 1 x 1 λ = 0 (1) a = σ a λ 1 x λ = 0 () a n = σ a n λ 1 x n λ = 0 (n) λ 1 = a i + 1 = 0 (n+1) λ = a i x i = 0 (n+) Adding the left hand and right hand sides of equations (1) (n) we have Since a i = 1 σ a i n λ 1 λ xi = 0 σ n λ 1 λ n x = 0 (*) Multiplying the left hand and right hand sides of equations (1) (n) by x 1, x, x n respectively and adding up we have Since a i x i = 0 we have σ a i x i λ x i λ x i = 0 n x λ 1 λ x i = 0 (**) Equations (*) and (**) form a linear equation system in λ 1 and in λ : n λ 1 + λ n x = σ Solving for λ 1 and for λ we have From equations (1) (n)we have n xλ 1 + λ x i = 0 λ 1 = σ n x i, and λ = σ x σ a i = λ 1 + λ x i, i = 1,, n 7
8 Substituting for λ 1 and for λ we obtain σ a i = σ x n s i σ xx i i = 1,, n xx or x a i = i x x i = + n x x x i n n since x i = + n x, w i is for the OLS estimator of β 1, β1 The second order conditions are = 1 n (x i x) x = w i, i = 1,, n a 1 = σ,, a n = σ, λ 1 = 0, λ = 0, and the crossderivatives = 0,, ; = 0 a 1 a a n a n 1 λ 1 a i = 1, i = 1,, n λ a i = x i, i = 1,, n Hence the bordered Hessian becomes a 1 a 1 a H = a n a 1 λ 1 a 1 λ a 1 a n a λ 1 a λ a a 1 a n a n λ 1 a n λ a n a 1 λ 1 a n λ 1 λ 1 λ λ 1 a 1 λ a n λ λ 1 λ λ 8
9 This becomes σ x 1 0 σ x H = σ 1 x n x 1 x x 3 x n 0 0 and it can be proved that H is negative definite, and hence the solutions yield the minimum variance a 1 = w 1, a = w,, a n = w n Proving that β is best The constrained minimization problem becomes The first order conditions are min σ b i b 1,,b n bi = 0 subject to bi x i = 1 b 1 = σ b 1 λ 1 x 1 λ = 0 (1) b = σ b λ 1 x λ = 0 () b n = σ b n λ 1 x n λ = 0 (n) λ 1 = b i = 0 (n+1) λ = b i x i + 1 = 0 (n+) We proceed just in the same way as we did before and obtain λ 1 + x λ = 0 n x λ 1 + ( x i )λ = σ 9
10 Solving for λ 1 and for λ we obtain λ 1 = σ x λ = σ Substituting for λ 1 and for λ we obtain Hence σ b i = λ 1 + λ x i = σ x ( ) = σ xi x + σ x i b i = x i x = v i, i = 1,, n The second order conditions are obtained in a similar way and the bordered Hessian is negative definite 10
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