Exercises Chapter 4 Statistical Hypothesis Testing

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1 Exercises Chapter 4 Statistical Hypothesis Testing Advanced Econometrics - HEC Lausanne Christophe Hurlin University of Orléans December 5, 013 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

2 Exercise 1 Parametric tests and the eyman Pearson lemma Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, 013 / 88

3 Problem We consider two continuous independent random variables U and W normally distributed with 0, σ. The transformed variable X de ned by: X = p U + W has a Rayleigh distribution with a parameter σ : X Rayleigh σ with a pdf f X x; σ de ned by: f X x; σ = x x σ exp σ 8x [0, + [ Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

4 Problem (cont d) Question 1: we consider an i.i.d. sample fx 1, X,.., X g. Derive the MLE estimator of σ Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

5 Solution f X x; σ = x σ exp x σ 8x [0, + [ The log-likelihood of the i.i.d. sample fx 1, x,.., x g is ` σ ; x = ln f X x i ; σ = ln (x i ) ln σ 1 σ xi The ML estimator bσ is de ned as to be: bσ = arg max σ >0 ` σ ; x Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

6 Solution (cont d) bσ = arg max σ >0 FOC (log-likelihood equations) ` σ ; x σ So, the ML estimator of σ is ln (x i ) ln σ 1 σ xi bσ = bσ + 1 bσ 4 xi = 0 bσ = 1 Xi Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

7 Solution (cont d) ` σ ; x σ = σ + 1 σ 4 xi SOC: ` σ ; x σ 4 bσ = 1 bσ 4 bσ 6 xi = bσ bσ 4 bσ 6 = bσ 4 < 0 since x i = bσ. So, we have a maximum. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

8 Problem (cont d) Question : what is the asymptotic distribution of the MLE estimator bσ? Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

9 Solution The average Fisher information matrix associated to the sample is: I σ ` σ ; X! = E σ σ 4! = E σ σ σ 6 Xi = σ σ 4 E σ X i σ Since (X /σ) = (U/σ) + (W /σ) where U/σ and W /σ are two independent standard normal variables, then X /σ χ () with E σ X i σ = Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

10 Solution (cont d) So, we have I σ = = σ σ 4 σ 4 + σ 4 E σ X i σ = σ 4 Since the sample is i.i.d., the average Fisher information matrix is: I σ = 1 I σ = 1 σ 4 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

11 Solution (cont d) The regularity conditions hold, and we have: p bσ σ d! 0, I 1 σ Here p bσ σ d! 0, σ 4 where σ denotes the true value of the parameter. Or equivalently: bσ asy σ, σ4 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

12 Problem (cont d) Question 3: consider the test H 0 : σ = σ 0 H 1 : σ = σ 1 with σ 1 > σ 0. Determine the critical region of the UMP test of size α. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

13 Solution Given the eyman Pearson lemma, the rejection region is given by: ( L σ 0 W = x 1,.., x j ; x ) 1,., x L (σ 1 ; x 1,., x ) < K where K is a constant determined by the level of the test α. So, we have ` σ 0; x ` σ 1; x < ln (K ) () ln (x i ) ln σ 0 1 σ xi ln (x i ) 0 + ln σ σ xi < ln (K ) 1 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

14 Solution (cont d) ln σ 1 ln σ σ 1 1 σ 0 1 xi < ln (K ) 1 () σ 1 with K 1 = ln (K ) log σ 1 1 σ 0 1 xi < K 1 log σ 0. or equivalently: σ 0 σ 1 σ 0 σ 1 1 xi < K 1 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

15 Solution (cont d) σ 0 σ 1 σ 0 σ 1 1 xi < K 1 Since σ 1 > σ 0, we have: 1 xi > A where A = K 1 σ 0 σ 1 / σ 0 σ 1 is a constant determined by α. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

16 Solution (cont d) The rejection region of the UMP test of size α H 0 : σ = σ 0 H 1 : σ = σ 1 with σ 1 > σ 0 is: n o W = x : bσ (x) > A where the critical value A is a constant determined by the size α and bσ (x) is the realisation of the ML estimator bσ (the test statistic): bσ = 1 Xi Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

17 Solution (cont d) Given the de nition of the size: α = Pr (Wj H 0 ) = Pr bσ > A H 0 Under the null, for large, we have: bσ asy H 0 σ 0, σ4 0 Then 1 α = Pr bσ σ 0 σ 0 /p < A σ 0 σ 0 /p! H 0 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

18 Solution (cont d) 1 α = Pr bσ σ 0 σ 0 /p < A σ 0 σ 0 /p! H 0 Denote by Φ (.) the cdf of the standard normal distribution: A = σ 0 + σ 0 p Φ 1 (1 α) The rejection region of the UMP test of size α H 0 : σ = σ 0 H 1 : σ = σ 1 with σ 1 > σ 0 is: W = x : bσ (x) > σ 0 + σ 0 p Φ 1 (1 α) Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

19 Problem (cont d) Question 4: consider the test H 0 : σ = H 1 : σ > For a sample of size = 100, we have xi = 470 What is the conclusion of the test for a size of 10%? Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

20 Solution Consider the test H 0 : σ = σ 0 H 1 : σ = σ 1 with σ 1 > σ 0. The rejection region of the UMP test of size α is given by: W = x : bσ (x) > σ 0 + σ 0 p Φ 1 (1 α) This region does not depend on the value of σ 1. So, it corresponds to the rejection region of the one-sided UMP of size α : H 0 : σ = σ 0 H 1 : σ > σ 0 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

21 Solution (cont d) H 0 : σ = H 1 : σ > W = x : bσ (x) > σ 0 + σ 0 p Φ 1 (1 α) A: = 100, α = 10% : W = x : bσ (x) > + 10 Φ 1 (0.9) W = n o x : bσ (x) >.563 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

22 Solution (cont d) W = n o x : bσ (x) >.563 For this sample ( = 100) we have x i bσ (x) = 1 xi = =.35 = 470, and as a consequence For a signi cance level of 10%, we reject the null H 0 : σ =. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, 013 / 88

23 Problem (cont d) Question 5: determine the power of the one-sided UMP test of size α for: H 0 : σ = σ 0 H 1 : σ > σ 0 umerical application: = 100, σ 0 = and α = 10%. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

24 Solution The rejection region of the UMP of size α is: n o W = x : bσ (x) > A with σ 0 + Φ 1 (1 α) σ 0 /p. By of the power, we have: power = Pr (Wj H 1 ) = Pr bσ > σ 0 + σ 0 p Φ 1 (1 α) H 1 Under the alternative hypothesis, for large, we have: bσ asy σ, σ4 σ > σ 0 H 1 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

25 Solution (cont d) Then, the power is equal to:! bσ σ power = 1 Pr σ / p < A σ σ / p H 1 A σ = 1 Φ σ / p Given the de nition of the critical value A = σ 0 + Φ 1 (1 α) σ 0 /p, we have: σ power = 1 Φ 0 σ σ / p + σ 0 σ Φ 1 (1 α) 8σ > σ 0 A: σ 0 =, = 100 and α = 10% σ power = 1 Φ σ /10 + σ Φ 1 (0.9) 8σ > σ 0 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

26 power Solution (cont d) σ Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

27 Solution (cont d) Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

28 Problem (cont d) Question 6: consider the two-sided test H 0 : σ = σ 0 H 1 : σ 6= σ 0 What is the critical region of the test of size α? Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

29 Solution Consider the one-sided tests: Test A: H 0 : σ = σ 0 against H 1 : σ < σ 0 Test B: H 0 : σ = σ 0 against H 1 : σ > σ 0 The non-rejection regions of the UMP one-sided tests of size α/ are: W A = x : bσ (x) > σ 0 + σ 0 p Φ 1 α W B = x : bσ (x) < σ 0 + σ 0 p Φ 1 1 α The non rejection region of the two-sided test corresponds to the intersection of these two regions: W = W A \ W B Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

30 Solution (cont d) So, the non rejection region of the two-sided test of size α is: W = x : σ 0 + σ 0 p Φ 1 α < bσ (x) < σ 0 + σ 0 p Φ 1 1 α Since, Φ 1 (α/) = Φ 1 (1 α/), this region can be rewritten as: W = x : bσ (x) σ 0 > σ 0 p Φ 1 1 α The rejection region of the two-sided of size α is: W = x : bσ (x) σ 0 < σ 0 p Φ 1 1 α Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

31 Problem (cont d) Question 7: determine the power of the two-sided test of size α for: H 0 : σ = σ 0 H 1 : σ 6= σ 0 umerical application: = 100, σ 0 = and α = 10%. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

32 Solution The non rejection region of the two-sided test of size α is: n o W = x : A < bσ (x) < B A = σ 0 + σ 0 p Φ 1 α By de nition of the power: B = σ 0 + σ 0 p Φ 1 1 power = Pr (Wj H 1 ) = 1 Pr W H 1 α So, we have: power = 1 Pr bσ < B + Pr bσ < A Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

33 Solution (cont d) power = 1 Pr bσ < B + Pr bσ < A Under the alternative bσ asy H 1 σ, σ4 σ 6= σ 0 So, we have power = 1 B σ A σ Φ σ / p + Φ σ / p Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

34 Solution (cont d) We have power = 1 A = σ 0 + σ 0 p Φ 1 α B σ A σ Φ σ / p + Φ σ / p B = σ 0 + σ 0 p Φ 1 1 α So 8σ 6= σ 0, the power function of the two sided test is de ned by: σ power = 1 Φ 0 σ σ / p + σ 0 σ Φ 1 1 σ +Φ 0 σ σ / p + σ 0 σ Φ 1 α α Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

35 Solution (cont d) A: = 100, α = 10% and σ 0 =. 8σ 6= σ power = 1 Φ σ /10 + σ Φ 1 (0.95) σ +Φ σ /10 + σ Φ 1 (0.05) Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

36 power Solution (cont d) σ Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

37 Solution (cont d) Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

38 Problem (cont d) Question 8: show that the two-sided test is unbiased and consistent. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

39 Solution The power function is de ned as to be: P σ σ = 1 Φ 0 σ σ / p + σ 0 σ Φ 1 1 σ +Φ 0 σ σ / p + σ 0 σ Φ 1 α α If σ < σ 0, then: If σ > σ 0, then: lim P σ = 1 Φ (+ ) + Φ (+ ) = = 1! lim! P σ = 1 Φ ( ) + Φ ( ) = = 1 The test is consistent. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

40 Solution (cont d) The power function is de ned as to be: P σ σ = 1 Φ 0 σ σ / p + σ 0 σ Φ 1 1 σ +Φ 0 σ σ / p + σ 0 σ Φ 1 α α This function reaches a minimum when σ tends to σ 0. lim P σ = 1 Φ Φ 1 1 σ!σ 0 The test is unbiased. = 1 = α 1 α + α α + Φ Φ 1 α Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

41 Subsection 4. The trilogy: LRT, Wald, and LM tests Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

42 Problem (Greene, 007, page 531) We consider two random variables Y and X such that the pdf of the conditional distribution Y j X = x is given by f Y jx (yj x; β) = 1 β + x exp y β + x For convenience, let β i = 1 β + x i This exponential density is a restricted form of a more general gamma distribution, f Y jx (yj x; β, ρ) = βρ i Γ (ρ) y ρ 1 i exp ( y i β i ) The restriction is ρ = 1. We want to test the hypothesis H 0 : ρ = 1 versus H 1 : ρ 6= 1 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

43 Reminder: the gamma function The gamma function Γ (p) is de ned as to be: Γ (p) = Z 0 t p 1 exp ( t) dt 8p > 0 The gamma function obeys the recursion So for integer values of p, we have Γ (p) = (p 1) Γ (p 1) 1 Γ = p π Γ (p) = (p 1)! Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

44 Reminder: the gamma function (cont d) The derivatives of the gamma function are k Γ (p) p k = Z 0 (ln (t)) k t p 1 exp ( t) dt The rst two derivatives of ln (Γ (p)) are denoted ln (Γ (p)) p ln (Γ (p)) p = = Γ0 Γ = Ψ (p) ΓΓ " Γ 0 Γ = Ψ 0 (p) where Ψ (p) and Ψ 0 (p) are the digamma and trigamma functions (see polygamma function and function psy in Matlab). Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

45 Problem (cont d) Question 1: consider an i.i.d. sample fx i, Y i g and write its log-likelihood under H 1 (unconstrained model) and under H 0 (constrained model). Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

46 Solution Under H 1, with θ = (β : ρ) >, we have f Yi jx i (y i j x i ; θ) = βρ i Γ (ρ) y ρ 1 i exp ( y i β i ) with β i = 1 β + x i ` (yj x; θ) = ln f Yi jx i (y i j x i ; θ) The log-likelihood under H 1 (unconstrained model) is: ` (yj x; θ) = ρ ln (β i ) ln (Γ (ρ)) + (ρ 1) ln (y i ) y i β i Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

47 Solution (cont d) Under H 0 : ρ = 1, we have f Yi jx i (y i j x i ; β) = β i exp ( y i β i ) with β i = 1 β + x i ` (yj x; β) = ln f Yi jx i (y i j x i ; β) The log-likelihood under H 0 (constrained model) is: ` (yj x; β) = ln (β i ) y i β i Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

48 Problem (cont d) Question : write the gradient vectors and the Hessian matrices associated to the unconstrained log-likelihood (under H 1 ) and to the constrained log-likelihood (under H 0 ). Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

49 Solution Under H 1 : ` (yj x; θ) = ρ ln (β i ) ln (Γ (ρ)) + (ρ 1) ln (y i ) y i β i Remarks: β i β = (1/ (β + x i )) = β 1 (β + x i ) = β i ln (β i ) β = ( ln (β + x i )) β ln (Γ (ρ)) ρ = = Ψ (ρ) 1 β + x i = β i Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

50 Solution (cont d) ` (yj x; θ) = ρ ln (β i ) ln (Γ (ρ)) + (ρ 1) ln (y i ) y i β i The gradient vector under H 1 is: g (yj x; θ) = ` (yj x; θ) θ 0 ` ( y jx;θ) β ` ( y jx;θ) ρ 1 A with ` (yj x; θ) ρ ` (yj x; θ) β = = ρ β i + ln (β i ) Ψ (ρ) + y i β i ln (y i ) Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

51 Solution (cont d) ` (yj x; θ) β = ρ β i + y i β i So, we have: ` (yj x; θ) β = ρ ` (yj x; θ) β ρ = β i β i y i β 3 i Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

52 Solution (cont d) So, we have ` (yj x; θ) ρ = ln (β i ) Ψ (ρ) + ` (yj x; θ) ρ = Ψ 0 (ρ) ` (yj x; θ) ρ β = β i ln (y i ) Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

53 Solution (cont d) The Hessian matrix associated to the log-likelihood under H 1 is: 0 H (yj x; θ) = ` (yj x; θ) B θ θ > ` ( y jx;θ) β ` ( y jx;θ) ρ β ` ( y jx;θ) β ρ ` ( y jx;θ) ρ 1 C A with H (yj x; θ) = ρ β i y i β 3 i β i β i Ψ 0 (ρ)! Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

54 Solution (cont d) Under H 0 : ρ = 1, the gradient (scalar) is The Hessian (scalar) is: g (yj x; β) = ` (yj x; β) β = β i + y i β i H (yj x; β) = ` (yj x; β) β = β i y i β 3 i Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

55 Problem (cont d) Question 3: write the average Fisher information matrices under H 1 and under H 0. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

56 Solution Under H 1 (unconstrained model), the Hessian (stochastic) is! H i (Y i j x i ; θ) = ρβ i Y i β 3 i β i β i Ψ 0 (ρ) The average Fisher information matrice can be de ned (one of the three de nitions) as: I (θ) = E X E θ ( H i (Y i j x i ; θ)) ρβ i + E θ (Y i ) β 3 i β i I (θ) = E X β i Ψ 0 (ρ) since β i = 1/ (β + X i ) depends on the random variable X i.! Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

57 Solution ρβ i + E θ (Y i ) β 3 i β i I (θ) = E X β i Ψ 0 (ρ) Consider the score of the unit i. By de nition, we have: E θ (s i (Y i j x i ; θ)) = ρβ i + Y i β i ln (β i ) Ψ (ρ) + ln (Y i )!! = 0 1 So, we have E θ (Y i ) = ρ β i where E θ denotes the expectation with respect to the conditional distribution of Y given X = x. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

58 Solution (cont d) ρβ i + E θ (Y i ) β 3 i β i I (θ) = E X β i Ψ 0 (ρ)! E θ (Y i ) = ρ β i Under H 1 (unconstrained model), the average Fisher information is de ned as to be:! ρβ i β i I (θ) = E X β i Ψ 0 (ρ) Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

59 Solution (cont d) Under H 0 (constrained model), we have: H i (Y i j x i ; β) = β i Y i β 3 i E β (s i (Y i j x i ; β)) = E β β i + Y i β i = 0 The average Fisher information number is de ned by: I (β) = E X E β ( H i (Y i j x i ; β)) = E X β i + E β (Y i ) β 3 i = E X β i The average Fisher information number is equal to: I (β) = E X β i Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

60 Problem (cont d) Question 4: denote bθ H1 the ML estimator of θ = (β : ρ) > obtained under H 1 and bθ H0 = bβ H0 the ML estimator of β obtained under H 0 : ρ = 1. Determine the asymptotic distribution and the asymptotic variance covariance matrix of bθ H1 and bθ H0. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

61 Solution The regularity conditions hold. Under H 1 (unconstrainded model) we have: p b θ H1 θ 1 d! 0, I 1 (θ 1 ) where θ 1 denotes the true value of the parameters (under H 1 ), or equivalently: asy bθ H1 θ 1, 1 H 1 I 1 (θ 1 ) with ρβ i β i I (θ 1 ) = E X β i Ψ 0 (ρ)! Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

62 Solution (cont d) The regularity conditions hold. Under H 0 (constrainded model) we have: p b θ H0 θ 0 d! 0, I 1 (θ 0 ) where θ 0 = β 0 denotes the true value of the parameter (under H 0 ), or equivalently: asy bθ H0 θ 0, 1 H 0 I 1 (θ 0 ) with I (θ 0 ) = E X β i Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

63 Problem (cont d) Question 5: Propose three alternative estimators of the average Fisher information matrices under H 1 and under H 0. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

64 Solution Three alternative estimators of the average Fisher information matrix I (θ) can be used: bi A b θ = 1 bi i b θ bi B b θ = 1 bi c b θ = 1 `i (θ; y i j x i ) θ b θ >! `i (θ; y i j x i ) θ `i (θ; y i j x i ) θ θ > b θ bθ Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

65 Solution (cont d) First estimator: actual Fisher information matrix Under H 1 : Under H 0 : bi A b θ = 1 0 bi A b θ = bρ b β i b β i bi A b θ = 1 bi i b θ b β i Ψ 0 (bρ) bi i b θ = 1 b β i where bβ i = 1/ b β + x i and where the estimators bβ and bρ are obtained under H 1 (unconstrained model) or H 0 (constrained model) given the case. 1 A Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

66 Solution (cont d) Second estimator: BHHH estimator bi B b θ = 1 `i (θ; y i j x i ) θ b θ >! `i (θ; y i j x i ) θ bθ Under H 1 : Under H 0 : bi B b θ 0 = 1 bρbβ i + y i bβ 1 A ln b β i Ψ (bρ) + ln (y i ) bρbβ i + y i bβ i ln b β i Ψ (bρ) + ln (y i ) bi B b θ = 1 b β i + y i bβ i Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

67 Solution (cont d) Third estimator: Hessian bi C b θ = 1 `i (θ; y i j x i ) θ θ > b θ Under H 1 : 0 bi C b θ = bρ bβ i + y i bβ 3 i bβ i bβ i Ψ 0 (bρ) 1 A Under H 0 : bi C b θ = 1 b β i + y i bβ 3 i Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

68 Problem (cont d) Question 6: Consider the dataset provided by Greene (007) in the le Chapter4_Exercise.xls. Write a Matlab code (1) to estimate the parameters of model under H 1 ( unconstrained model) by MLE, and () to compute three alternative estimates of the asymptotic variance covariance matrix. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

69 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

70 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

71 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

72 Remarks 1 Asymptotically the three estimators of the asymptotic variance covariance matrix are equivalent. But, this exercise con rms that these estimators can give very di erent results for small samples 3 The striking di erence of the BHHH estimator is typical of its erratic performance in small samples Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

73 Problem (cont d) Question 7: Write a Matlab code (1) to estimate the parameters of model under H 0 ( constrained model) by MLE, and () to compute three alternative estimates of the asymptotic variance covariance matrix. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

74 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

75 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

76 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

77 Problem (cont d) Question 8: test the hypothesis H 0 : ρ = 1 versus H 1 : ρ 6= 1 with a likelihood ratio (LR) test for a signi cance level of 5%. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

78 Solution The likelihood ratio (LR) test-statistic is de ned by: LR = b θ H0 ; yj x b θ H1 ; yj x ` In this sample, we have a realisation equal to The critical region is LR (y) = ( ) = ` W = y : LR (y) > χ 0.95 (1) = where χ 0.95 (1) is the critical value of the chi-squared distribution with p = 1 degrees of freedom. Conclusion: for a signi cance level of 5%, we reject the null H 0 : ρ = 1. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

79 Problem (cont d) Question 9: test the hypothesis H 0 : ρ = 1 versus H 1 : ρ 6= 1 with a Wald test for a signi cance level of 5%. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

80 Solution The null hypothesis H 0 : ρ = 1 can be expressed as with c (θ) = ρ Wald = c Here, we have Then, we get H 0 : c (θ) = 0 1. The Wald test-statistic is de ned by: > b c θ H1 b θ > θ H1 V b asy b c > 1 θ H1 b θ > θ H1 c b θ H1 Wald (y) = c b θ > θ H1 = 0 1 bρ H1 1 V 1 asy bρ H1 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

81 Solution (cont d) Wald (y) = bρ H1 1 V 1 asy bρ H1 Given the estimator chosen for the asymptotic variance, we get: Wald A (y) = Wald B (y) = Wald C (y) = The critical region is ( ) ( ) ( ) = = = W = y : Wald (y) > χ 0.95 (1) = Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

82 Solution (cont d) Conclusion: 1 For a signi cance level of 5%, the Wald test-statistic based on the estimators A and C (actual Fisher matrix and Hessian) of the asymptotic variance covariance matrix lead to reject the null H 0 : ρ = 1. For a signi cance level of 5%, the Wald test-statistic based on the estimators B (BHHH estimator) of the asymptotic variance covariance matrix fails to reject the null H 0 : ρ = 1. 3 In most of software, the Hessian (estimator C) is preferred and the Wald test-statistics are computed with this estimator. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

83 Problem (cont d) Question 10: test the hypothesis H 0 : ρ = 1 versus H 1 : ρ 6= 1 with a Lagrange Multiplier test for a signi cance level of 5%. Write a Matlab code to compute the three possibles values of the LM test-statistic. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

84 Solution The Lagrange multiplier test is based on the restricted estimators. The LM test-statistic is de ned by: or equivalently LM = s b θ H0 ; y i j x i > b I b θ H0 1 s b θ H0 ; y i j x i > 1 LM = s b θ H0 ; y i j x i bi b θ H0 s b θ H0 ; y i j x i Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

85 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

86 Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

87 Solution (cont d) So, given the estimator chosen for bi b θ H0, we have: LM A (y) = Lm B (y) = The critical region is LM C (y) = W = y : LM (y) > χ 0.95 (1) = Conclusion: for a signi cance level of 5%, we reject the null H 0 : ρ = 1, whatever the choice of the estimator. Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

88 End of Exercices - Chapter 4 Christophe Hurlin (University of Orléans) Christophe Hurlin (University of Orléans) Advanced Econometrics - HEC Lausanne December 5, / 88

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