Chapter 4: Statistical Hypothesis Testing


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1 Chapter 4: Statistical Hypothesis Testing Christophe Hurlin November 20, 2015 Christophe Hurlin () Advanced Econometrics  Master ESA November 20, / 225
2 Section 1 Introduction Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
3 1. Introduction The outline of this chapter is the following: Section 2. Statistical hypothesis testing Section 3. Tests in the multiple linear regression model Subsection 3.1. The Student test Subsection 3.2. The Fisher test Section 4. MLE and Inference Subsection 4.1. The Likelihood Ratio (LR) test Subsection 4.2.The Wald test Subsection 4.3. The Lagrange Multiplier (LM) test Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
4 1. Introduction References Amemiya T. (1985), Advanced Econometrics. Harvard University Press. Greene W. (2007), Econometric Analysis, sixth edition, Pearson  Prentice Hil (recommended) Ruud P., (2000) An introduction to Classical Econometric Theory, Oxford University Press. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
5 1. Introduction Notations: In this chapter, I will (try to...) follow some conventions of notation. f Y (y) F Y (y) Pr () y Y probability density or mass function cumulative distribution function probability vector matrix Be careful: in this chapter, I don t distinguish between a random vector (matrix) and a vector (matrix) of deterministic elements (except in section 2). For more appropriate notations, see: Abadir and Magnus (2002), Notation in econometrics: a proposal for a standard, Econometrics Journal. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
6 Section 2 Statistical hypothesis testing Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
7 2. Statistical hypothesis testing Objectives The objective of this section is to de ne the following concepts: 1 Null and alternative hypotheses 2 Onesided and twosided tests 3 Rejection region, test statistic and critical value 4 Size, power and power function 5 Uniformly most powerful (UMP) test 6 Neyman Pearson lemma 7 Consistent test and unbiased test 8 pvalue Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
8 2. Statistical hypothesis testing Introduction 1 A statistical hypothesis test is a method of making decisions or a rule of decision (as concerned a statement about a population parameter) using the data of sample. 2 Statistical hypothesis tests de ne a procedure that controls ( xes) the probability of incorrectly deciding that a default position (null hypothesis) is incorrect based on how likely it would be for a set of observations to occur if the null hypothesis were true. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
9 2. Statistical hypothesis testing Introduction (cont d) In general we distinguish two types of tests: 1 The parametric tests assume that the data have come from a type of probability distribution and makes inferences about the parameters of the distribution 2 The nonparametric tests refer to tests that do not assume the data or population have any characteristic structure or parameters. In this course, we only consider the parametric tests. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
10 2. Statistical hypothesis testing Introduction (cont d) A statistical test is based on three elements: 1 A null hypothesis and an alternative hypothesis 2 A rejection region based on a test statistic and a critical value 3 A type I error and a type II error Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
11 2. Statistical hypothesis testing Introduction (cont d) A statistical test is based on three elements: 1 A null hypothesis and an alternative hypothesis 2 A rejection region based on a test statistic and a critical value 3 A type I error and a type II error Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
12 2. Statistical hypothesis testing De nition (Hypothesis) A hypothesis is a statement about a population parameter. The formal testing procedure involves a statement of the hypothesis, usually in terms of a null or maintained hypothesis and an alternative, conventionally denoted H 0 and H 1, respectively. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
13 2. Statistical hypothesis testing Introduction 1 The null hypothesis refers to a general or default position: that there is no relationship between two measured phenomena or that a potential medical treatment has no e ect. 2 The costs associated to the violation of the null must be higher than the cost of a violation of the alternative. Example (Choice of the null hypothesis) In a credit scoring problem, in general we have: H 0 : the client is not risky(acceptance of the loan) versus H 1 : the client is risky (refusal of the loan). Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
14 2. Statistical hypothesis testing De nition (Simple and composite hypotheses) A simple hypothesis speci es the population distribution completely. A composite hypothesis does not specify the population distribution completely. Example (Simple and composite hypotheses) If X t (θ), H 0 : θ = θ 0 is a simple hypothesis. H 1 : θ > θ 0, H 1 : θ < θ 0, and H 1 : θ 6= θ 0 are composite hypotheses. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
15 2. Statistical hypothesis testing De nition (Onesided test) A onesided test has the general form: H 0 : θ = θ 0 or H 0 : θ = θ 0 H 1 : θ < θ 0 H 1 : θ > θ 0 Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
16 2. Statistical hypothesis testing De nition (Twosided test) A twosided test has the general form: H 0 : θ = θ 0 H 1 : θ 6= θ 0 Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
17 2. Statistical hypothesis testing Introduction (cont d) A statistical test is based on three elements: 1 A null hypothesis and an alternative hypothesis 2 A rejection region based on a test statistic and a critical value 3 A type I error and a type II error Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
18 2. Statistical hypothesis testing De nition (Rejection region) The rejection region is the set of values of the test statistic (or equivalently the set of samples) for which the null hypothesis is rejected. The rejection region is denoted W. For example, a standard rejection region W is of the form: or equivalently W = fx : T (x) > cg W = fx 1,.., x N : T (x 1,.., x N ) > cg where x denotes a sample fx 1,.., x N g, T (x) the realisation of a test statistic and c the critical value. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
19 2. Statistical hypothesis testing Remarks 1 A (hypothesis) test is thus a rule that speci es: 1 For which sample values the decision is made to fail to reject H0 as true; 2 For which sample values the decision is made to reject H0. 3 Never say "Accept H1", "fail to reject H1" etc.. 2 The complement of the rejection region is the nonrejection region. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
20 2. Statistical hypothesis testing Remark The rejection region is de ned as to be: W = fx : T (x) {z } test statistic 7 c {z} g critical value T (x) is the realisation of the statistic (random variable): T (X ) = T (X 1,.., X N ) The test statistic T (X ) has an exact or an asymptotic distribution D under the null H 0. T (X ) H0 D or T (X ) d! H0 D Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
21 2. Statistical hypothesis testing Introduction (cont d) A statistical test is based on three elements: 1 A null hypothesis and an alternative hypothesis 2 A rejection region based on a test statistic and a critical value 3 A type I error and a type II error Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
22 2. Statistical hypothesis testing Decision Fail to reject H 0 Reject H 0 Truth H 0 Correct decision Type I error H 1 Type II error Correct decision Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
23 2. Statistical hypothesis testing De nition (Size) The probability of a type I error is the (nominal) size of the test. This is conventionally denoted α and is also called the signi cance level. α = Pr (Wj H 0 ) Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
24 2. Statistical hypothesis testing Remark For a simple null hypothesis: α = Pr (Wj H 0 ) For a composite null hypothesis: α = sup Pr (Wj H 0 ) θ 0 2H 0 A test is said to have level if its size is less than or equal to α. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
25 2. Statistical hypothesis testing De nition (Power) The power of a test is the probability that it will correctly lead to rejection of a false null hypothesis: power = Pr (Wj H 1 ) = 1 β where β denotes the probability of type II error, i.e. β = Pr W denotes the nonrejection region. W H 1 and Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
26 2. Statistical hypothesis testing Example (Test on the mean) Consider a sequence X 1,.., X N of i.i.d. continuous random variables with X i N m, σ 2 where σ 2 is known. We want to test H 0 : m = m 0 H 1 : m = m 1 with m 1 < m 0. An econometrician propose the following rule of decision: W = fx : x N < cg where X N = N 1 N i=1 X i denotes the sample mean and c is a constant (critical value). Question: calculate the size and the power of this test. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
27 2. Statistical hypothesis testing Solution The rejection region is W= fx : x N < cg. Under the null H 0 : m = m 0 : X N H0 N So, the size of the test is equal to: m 0, σ2 N α = Pr (Wj H 0 ) = Pr X N < c H 0 X = N m 0 Pr σ/ p N < c m 0 σ/ p N H 0 c m0 = Φ σ/ p N Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
28 2. Statistical hypothesis testing Solution (cont d) The rejection region is W= fx : x N < cg. Under the alternative H 1 : m = m 1 : X N N m 1, σ2 H1 N So, the power of the test is equal to: power = Pr (Wj H 1 ) X = N m 1 Pr σ/ p N < c m 1 σ/ p N H 1 c m1 = Φ σ/ p N Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
29 2. Statistical hypothesis testing Solution (cont d) In conclusion: c m0 α = Φ σ/ p N c m1 β = 1 power = 1 Φ σ/ p N We have a system of two equations with three parameters: α, β (or power) and the critical value c. 1 There is a tradeo between the probabilities of the errors of type I and II, i.e. α and β : if c decreases, α decreases but β increases. 2 A solution is to impose a size α and determine the critical value and the power. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
30 2. Statistical hypothesis testing Solution (cont d) In order to illustrate the tradeo between α and β given the critical value c, take an example with σ 2 = 1 and N = 100: H 0 : m = m 0 = 1.2 H 1 : m = m 1 = 1 X N N m 0, σ2 X N N m 1, σ2 H0 N H1 N We have W = fx : x N < cg c m0 α = Pr (Wj H 0 ) = Φ σ/ p = Φ (10 (c 1.2)) N β = Pr W c m1 H 1 = 1 Φ σ/ p = 1 Φ (10 (c 1)) N Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
31 2. Statistical hypothesis testing Density under H0 Density under H α=pr(w[h0)=5% β=1 Pr(W H1)=36.12% Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
32 2. Statistical hypothesis testing Click me! Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
33 2. Statistical hypothesis testing Fact (Critical value) The (nominal) size α is xed by the analyst and the critical value is deduced from α. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
34 2. Statistical hypothesis testing Example (Test on the mean) Consider a sequence X 1,.., X N of i.i.d. continuous random variables with X i N m, σ 2, N = 100 and σ 2 = 1. We want to test H 0 : m = 1.2 H 1 : m = 1 An econometrician propose the following rule of decision: W = fx : x N < cg where X N = N 1 N i=1 X i denotes the sample mean and c is a constant (critical value). Questions: (1) what is the critical value of the test of size α = 5%? (2) what is the power of the test? Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
35 2. Statistical hypothesis testing Solution We know that: c m0 α = Pr (Wj H 0 ) = Φ σ/ p N So, the critical value that corresponds to a signi cance level of α is: c = m 0 + σ p N Φ 1 (α) NA: if m 0 = 1.2, m 1 = 1, N = 100, σ 2 = 1 and α = 5%, then the rejection region is W = fx : x N < g Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
36 2. Statistical hypothesis testing Solution (cont d) W = x : x N < m 0 + p σ Φ 1 (α) N The power of the test is: c m1 power = Pr (Wj H 1 ) = Φ σ/ p N Given the critical value, we have: m0 m 1 power = Φ σ/ p N + Φ 1 (α) NA: if m 0 = 1.2, m 1 = 1, N = 100, σ 2 = 1 and α = 5%: power = Φ 1/ p Φ 1 (0.05) = Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
37 2. Statistical hypothesis testing Example (Test on the mean) Consider a sequence X 1,.., X N of i.i.d. continuous random variables with X i N m, σ 2 with σ 2 = 1 and N = 100. We want to test H 0 : m = 1.2 H 1 : m = 1 The rejection region for a signi cance level α = 5% is: W = fx : x N < g where X N = N 1 N i=1 X i denotes the sample mean. Question: if the realisation of the sample mean is equal to 1.13, what is the conclusion of the test? Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
38 2. Statistical hypothesis testing Solution (cont d) For a nominal size α = 5%, the rejection region is: W = fx : x N < g If we observe x N = 1.13 This realisation does not belong to the rejection region: x N /2 W For a level of 5%, we do not reject the null hypothesis H 0 : m = 1.2. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
39 2. Statistical hypothesis testing De nition (Power function) In general, the alternative hypothesis is composite. In this case, the power is a function of the value of the parameter under the alternative. power = P (θ) 8θ 2 H 1 Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
40 2. Statistical hypothesis testing Example (Test on the mean) Consider a sequence X 1,.., X N of i.i.d. continuous random variables with X i N m, σ 2 where σ 2 is known. We want to test H 0 : m = m 0 H 1 : m < m 0 Consider the following rule of decision: W = x : x N < m 0 + p σ Φ 1 (α) N Questions: What is the power function of the test? Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
41 2. Statistical hypothesis testing Solution As in the previous case, we have: m0 m power = P (m) = Φ σ/ p N + Φ 1 (α) with m < m 0 NA: if m 0 = 1.2, N = 100, σ 2 = 1 and α = 5%. 1.2 m P (m) = Φ with m < m 0 1/10 Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
42 2. Statistical hypothesis testing Power function P (m) Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
43 2. Statistical hypothesis testing Example (Power function) Consider a test H 0 : θ = θ 0 versus H 1 : θ 6= θ 0, the power function has this general form: Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
44 2. Statistical hypothesis testing De nition (Most powerful test) A test (denoted A) is uniformly most powerful (UMP) if it has greater power than any other test of the same size for all admissible values of the parameter. α A = α B = α for any test B of size α. β A β B Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
45 2. Statistical hypothesis testing UMP tests How to derive the rejection region of the UMP test of size α? => the Neyman Pearson lemma Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
46 2. Statistical hypothesis testing Lemma (Neyman Pearson) Consider a hypothesis test between two point hypotheses H 0 : θ = θ 0 and H 1 : θ = θ 1. The uniformly most powerful (UMP) test has a rejection region de ned by: W = xj L N (θ 0 ; x) L N (θ 1 ; x) < K where L N (θ 0 ; x) denotes the likelihood of the sample x and K is a constant determined by the size α such that: LN (θ 0 ; X ) Pr L N (θ 1 ; X ) < K H 0 = α Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
47 2. Statistical hypothesis testing Example (Test on the mean) Consider a sequence X 1,.., X N of i.i.d. continuous random variables with X i N m, σ 2 where σ 2 is known. We want to test H 0 : m = m 0 H 1 : m = m 1 with m 1 > m 0. Question: What is the rejection region of the UMP test of size α? Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
48 2. Statistical hypothesis testing Solution Since X 1,.., X N are N.i.d. m, σ 2, the likelihood of the sample fx 1,.., x N g is de ned as to be (cf. chapter 2):! N 1 L N (θ; x) = σ N (2π) N /2 exp 1 2σ 2 (x i m) 2 i=1 Given the Neyman Pearson lemma the rejection region of the UMP test of size α is given by: L N (θ 0 ; x) L N (θ 1 ; x) < K where K is a constant determined by the size α. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
49 2. Statistical hypothesis testing Solution (cont d) 1 exp σ N (2π) N /2 1 σ N (2π) N /2 exp 1 2σ 2 N i=1 (x i 1 2σ 2 N i=1 (x i m 0 ) 2 m 1 ) 2 < K This expression can rewritten as: 1 exp 2σ 2 N i=1 (x i m 1 ) 2 N i=1 (x i m 0 ) 2 < K () N i=1 (x i m 1 ) 2 N i=1 (x i m 0 ) 2 < K 1 where K 1 = 2σ 2 ln (K ) is a constant. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
50 2. Statistical hypothesis testing Solution (cont d) N i=1 (x i m 1 ) 2 N i=1 (x i m 0 ) 2 < K 1 () 2 (m 0 m 1 ) N i=1 x i + N m 2 1 m 2 0 < K1 () (m 0 m 1 ) N i=1 x i < K 2 where K 2 = K 1 N m 2 1 m 2 0 /2 is a constant. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
51 2. Statistical hypothesis testing Solution (cont d) Since m 1 > m 0, we have where K 3 = K 2 / (N (m 0 (m 0 m 1 ) N i=1 x i < K 2 1 N N i=1 x i > K 3 m 1 )) is a constant. The rejection region of the UMP test for H 0 : m = m 0 against H 0 : m = m 1 with m 1 > m 0 has the general form: where A is a constant. W = fx : x N > Ag Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
52 2. Statistical hypothesis testing Solution (cont d) W = fx : x N > Ag Determine the critical value A from the nominal size: α = Pr (Wj H 0 ) = Pr (x N > Aj H 0 ) X = N m 0 1 Pr σ/ p N < A m 0 σ/ p N H 0 A m0 = 1 Φ σ/ p N Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
53 2. Statistical hypothesis testing Solution (cont d) So, we have α = 1 A = m 0 + A m0 Φ σ/ p N σ p N Φ 1 (1 α) The rejection region of the UMP test of size α for H 0 : m = m 0 against H 0 : m = m 1 with m 1 > m 0 is: W = x : x N > m 0 + p σ Φ 1 (1 α) N Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
54 2. Statistical hypothesis testing Fact (UMP onesided test) For a onesided test H 0 : θ = θ 0 against H 1 : θ > θ 0 (or H 1 : θ < θ 1 ) the rejection region W of the UMP test is equivalent to the rejection region obtained for the test H 0 : θ = θ 0 against H 1 : θ = θ 1 with for θ 1 > θ 0 (or θ 1 < θ 0 ) if this region does not depend on the value of θ 1. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
55 2. Statistical hypothesis testing Example (Test on the mean) Consider a sequence X 1,.., X N of i.i.d. continuous random variables with X i N m, σ 2 where σ 2 is known. We want to test H 0 : m = m 0 H 1 : m > m 0 Question: What is the rejection region of the UMP test of size α? Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
56 2. Statistical hypothesis testing Solution Consider the test: H 0 : m = m 0 H 1 : m = m 1 with m 1 > m 0. The rejection region of the UMP test of size α is: W = x : x N > m 0 + p σ Φ 1 (1 α) N W does not depend on m 1. It is also the rejection region of the UMP onesided test for H 0 : m = m 0 H 1 : m > m 0 Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
57 2. Statistical hypothesis testing Fact (Twosided test) For a twosided test H 0 : θ = θ 0 against H 1 : θ 6= θ 0 the non rejection region W of the test of size α is the intersection of the non rejection regions of the corresponding onesided UMP tests of size α/2 Test A: H 0 : θ = θ 0 against H 1 : θ > θ 0 Test B: H 0 : θ = θ 0 against H 1 : θ < θ 0 So, we have: W = W A \ W B Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
58 2. Statistical hypothesis testing Example (Test on the mean) Consider a sequence X 1,.., X N of i.i.d. continuous random variables with X i N m, σ 2 where σ 2 is known. We want to test H 0 : m = m 0 H 1 : m 6= m 0 Question: What is the rejection region of the test of size α? Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
59 2. Statistical hypothesis testing Solution Consider the onesided tests: Test A: H 0 : m = m 0 against H 1 : m < m 0 Test B: H 0 : m = m 0 against H 1 : m > m 0 The rejection regions of the UMP test of size α/2 are: W A = x : x N < m 0 + p σ Φ 1 (α/2) N W B = x : x N > m 0 + p σ Φ 1 (1 N α/2) Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
60 2. Statistical hypothesis testing Solution (cont d) The nonrejection regions of the UMP test of size α/2 are: W A = x : x N m 0 + p σ Φ 1 (α/2) N W B = x : x N m 0 + p σ Φ 1 (1 N α/2) The non rejection region of the twosided test corresponds to the intersection of these two regions: W = W A \ W B Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
61 2. Statistical hypothesis testing Solution (cont d) So, non rejection region of the twosided test of size α is: W = x : m 0 + p σ Φ 1 (α/2) x N m 0 + p σ Φ 1 (1 N N α/2) Since, Φ 1 (α/2) = Φ 1 (1 α/2), this region can be rewritten as: W = x : jx N m 0 j p σ Φ 1 (1 α/2) N Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
62 2. Statistical hypothesis testing Solution (cont d) W = x : jx N m 0 j σ p N Φ 1 (1 α/2) Finally, the rejection region of the twosided test of size α is: W = x : jx N m 0 j > p σ Φ 1 (1 α/2) N Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
63 2. Statistical hypothesis testing Solution (cont d) W = x : jx N m 0 j > σ p N Φ 1 (1 α/2) NA: if m 0 = 1.2, N = 100, σ 2 = 1 and α = 5%: W = x : jx N 1.2j > 1 10 Φ 1 (0.975) W = fx : jx N 1.2j > g If the realisation of jx N 1.2j is larger than , we reject the null H 0 : m = 1.2 for a signi cance level of 5%. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
64 2. Statistical hypothesis testing De nition (Unbiased Test) A test is unbiased if its power P (θ) is greater than or equal to its size α for all values of the parameter θ. P (θ) α 8θ 2 H 1 By construction, we have P (θ 0 ) = Pr (Wj Hj 0 ) = α. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
65 2. Statistical hypothesis testing De nition (Consistent Test) A test is consistent if its power goes to one as the sample size grows to in nity. lim N! P (θ) = 1 8θ 2 H 1 Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
66 2. Statistical hypothesis testing Example (Test on the mean) Consider a sequence X 1,.., X N of i.i.d. continuous random variables with X i N m, σ 2 where σ 2 is known. We want to test H 0 : m = m 0 H 1 : m < m 0 The rejection region of the UMP test of size α is W = x : x N < m 0 + p σ Φ 1 (α) N Question: show that this test is (1) unbiased and (2) consistent. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
67 2. Statistical hypothesis testing Solution W = x : x N < m 0 + p σ Φ 1 (α) N The power function of the test is de ned as to be: P (m) = Pr (Wj H 1 ) = Pr X N < m 0 + p σ Φ 1 (α) N m < m 0 m0 m = Φ σ/ p N + Φ 1 (α) Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
68 2. Statistical hypothesis testing Solution m0 m P (m) = Φ σ/ p N + Φ 1 (α) The test is consistent since: 8m < m 0 lim P (m) = 1 N! The test is unbiased since P (m) α 8m < m 0 lim m!m 0 P (m) = Φ Φ 1 (α) = α Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
69 2. Statistical hypothesis testing Solution The decision Reject H0 or fail to reject H0 is not so informative! Indeed, there is some arbitrariness to the choice of α (level). Another strategy is to ask, for every α, whether the test rejects at that level. Another alternative is to use the socalled pvalue the smallest level of signi cance at which H 0 would be rejected given the value of the teststatistic. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
70 2. Statistical hypothesis testing De nition (pvalue) Suppose that for every α 2 [0, 1], one has a size α test with rejection region W α. Then, the pvalue is de ned to be: pvalue = inf fα : T (y) 2 W α g The pvalue is the smallest level at which one can reject H 0. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
71 2. Statistical hypothesis testing The pvalue is a measure of evidence against H 0 : pvalue evidence < 0.01 Very strong evidence against H Strong evidence against H Weak evidence against H 0 > 0.10 Little or no evidence against H 0 Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
72 2. Statistical hypothesis testing Remarks 1 A large pvalue does not mean strong evidence in favor of H0. 2 A large pvalue can occur for two reasons: 1 H0 is true; 2 H0 is false but the test has low power. 3 The pvalue is not the probability that the null hypothesis is true! Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
73 2. Statistical hypothesis testing For a nominal size of 5%, we reject the null H 0 : β SP500 = 0. For a nominal size of 5%, we fail to reject the null H 0 : β C = 0. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
74 2. Statistical hypothesis testing Summary Hypothesis testing is de ned by the following general procedure Step 1: State the relevant null and alternative hypotheses (misstating the hypotheses muddies the rest of the procedure!); Step 2: Consider the statistical assumptions being made about the sample in doing the test (independence, distributions, etc.) incorrect assumptions mean that the test is invalid! Step 3: Choose the appropriate test (exact or asymptotic tests) and thus state the relevant test statistic (say, T). Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
75 2. Statistical hypothesis testing Summary (cont d) Step 4: Derive the distribution of the test statistic under the null hypothesis (sometimes it is wellknown, sometimes it is more tedious!) for example, the Student tdistribution or the Fisher distribution. Step 5: Determine the critical value (and thus the critical region). Step 6: Compute (using the observations!) the observed value of the test statistic T, say t obs. Step 7: Decide to either fail to reject the null hypothesis or reject in favor of the alternative assumption the decision rule is to reject the null hypothesis H 0 if the observed value of the test statistic, t obs is in the critical region, and to fail to reject the null hypothesis otherwise Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
76 2. Statistical hypothesis testing Key concepts 1 Null and alternative hypotheses 2 Simple and composite hypotheses 3 Onesided and twosided tests 4 Rejection region, test statistic and critical value 5 Type I and type II errors 6 Size, power and power function 7 Uniformly most powerful (UMP) test 8 Neyman Pearson lemma 9 Consistent test and unbiased test 10 pvalue Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
77 Section 3 Tests in the multiple linear regression model Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
78 3. Tests in the multiple linear regression model Objectives In the context of the multiple linear regression model (cf. chapter 3), the objective of this section is to present : 1 the Student test 2 the tstatistic and the zstatistic 3 the Fisher test 4 the global Ftest 5 To distinguish the case with normality assumption and the case without any assumption on the distribution of the error term (semiparametric speci cation) Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
79 3. Tests in the multiple linear regression model Be careful: in this section, I don t distinguish between a random vector (matrix) and a vector (matrix) of deterministic elements. For more appropriate notations, see: Abadir and Magnus (2002), Notation in econometrics: a proposal for a standard, Econometrics Journal. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
80 3. Tests in the multiple linear regression model Model Consider the (population) multiple linear regression model: where (cf. chapter 3): y = Xβ + ε y is a N 1 vector of observations y i for i = 1,.., N X is a N K matrix of K explicative variables x ik for k = 1,., K and i = 1,.., N ε is a N 1 vector of error terms ε i. β = (β 1..β K ) > is a K 1 vector of parameters Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
81 3. Tests in the multiple linear regression model Assumptions Fact (Assumptions) We assume that the multiple linear regression model satisfy the assumptions A1A5 (cf. chapter 3) We distinguish two cases: 1 Case 1: assumption A6 (Normality) holds and ε N 0, σ 2 I N 2 Case 2: the distribution of ε is unknown (semiparametric speci cation) and ε?? Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
82 3. Tests in the multiple linear regression model Parametric tests The β k are unknown features of the population, but: 1 One can formulate a hypothesis about their value; 2 One can construct a test statistic with a known nite sample distribution (case 1) or an asymptotic distribution (case 2); 3 One can take a decision meaning reject H0 if the value of the test statistic is too unlikely. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
83 3. Tests in the multiple linear regression model Three tests of interest: H 0 : β k = a k or H 0 : β k = a k H 1 : β k < a k H 1 : β k > a k H 0 : β k = a k H 1 : β k 6= a k where a k = 0 or a k 6= 0. H 0 : Rβ = q H 1 : Rβ 6= q Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
84 3. Tests in the multiple linear regression model For that, we introduce two types of test 1 The Student test or ttest 2 The Fisher test of Ftest Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
85 Subsection 3.1 The Student test Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
86 3.1. The Student test Case 1: Normality assumption A6 Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
87 3.1. The Student test Assumption 6 (normality): the disturbances are normally distributed. εj X N 0 N 1, σ 2 I N Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
88 3.1. The Student test Reminder (cf. chapter 3) Fact (Linear regression model) Under the assumption A6 (normality), the estimators β b and bσ 2 have a nite sample distribution given by: bβ 1 N β,σ 2 X > X bσ 2 σ 2 (N K ) χ2 (N K ) Moreover, b β and bσ 2 are independent. This result holds whether or not the matrix X is considered as random. In this last case, the distribution of b β is conditional to X. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
89 3.1. The Student test Remarks 1 Any linear combination of β b is also normally distributed: Aβ b 1 N Aβ,σ 2 A X > X A > 2 Any subset of b β has a joint normal distribution. bβ k N β k, σ 2 m kk where m kk is k th diagonal element of X > X 1. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
90 3.1. The Student test Reminder If X and Y are two independent random variables such that then the variable Z de ned as to be X N (0, 1) Y χ 2 (θ) Z = X p Y /θ has a Student s tdistribution with θ degrees of freedom Z t (θ) Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
91 3.1. The Student test Student test statistic Consider a test with the null: Under the null H 0 : H 0 : β k = a k bβ k a k σ p N (0, 1) m kk H0 bσ 2 σ 2 (N K ) H 0 χ 2 (N K ) and these two variables are independent... Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
92 3.1. The Student test Student test statistic (cont d) So, under the null H 0 we have: bβ k a k σ p N (0, 1) m kk H0 bσ 2 σ 2 (N K ) H 0 χ 2 (N K ) bβ k a k σ p m q kk bσ 2 (N K ) σ 2 (N K ) = b β k a k bσ p m kk H0 t (N K ) Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
93 3.1. The Student test De nition (Student tstatistic) Under the null H 0 : β k = a k, the Student teststatistic or tstatistic is de ned to be: a k T k = b β k t (N K ) bse b β H0 k where N is the number of observations, K is the number of explanatory variables (including the constant term), t (N K ) is the Student tdistribution with N K degrees of freedom and bse b β k = bσ p m kk with m kk is k th diagonal element of X > X 1. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
94 3.1. The Student test Remarks 1 Under the assumption A6 (normality) and under the null H 0 : β k = a k, the Student teststatistic has an exact ( nite sample) distribution. T k t (N K ) H0 2 The term bse b β k denotes the estimator of the standard error of the OLS estimator bβ k and it corresponds to the square root of the k th diagonal element of bv bβ (cf. chapter 3): bv 1 bβ = bσ 2 X > X Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
95 3.1. The Student test Consider the onesided test: k (y) < Aj T k t (N H0 H 0 : β k = a k H 1 : β k < a k The rejection region is de ned as to be: W = fy : T k (y) < Ag where A is a constant determined by the nominal size α. α = Pr (Wj H 0 ) = Pr T K ) Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
96 3.1. The Student test α = Pr T k (y) < Aj T k t (N K ) = F N K (A) H0 where F N K (.) denotes the cdf of the Student s tdistribution with N K degrees of freedom. Denote c α the αquantile of this distribution:. A = FN 1 K (α) = c α The rejection region of the test of size α is de ned as to be: W = fy : T k (y) < c α g Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
97 3.1. The Student test De nition (Onesided Student test) The critical region of the Student test is that H 0 : β k = a k is rejected in favor of H 1 : β k < a k at the α (say, 5%) signi cance level if: W = fy : T k (y) < c α g where c α is the α (say, 5%) critical value of a Student tdistribution with N K degrees of freedom and T k (y) is the realisation of the Student teststatistic. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
98 3.1. The Student test Example (Onesided test) Consider the CAPM model (cf. chapter 1) and the following results (Eviews). We want to test the beta of MSFT as H 0 : β MSFT = 1 against H 1 : β MSFT < 1 Question: give a conclusion for a nominal size of 5%. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
99 3.1. The Student test Solution Step 1: compute the tstatistic T MSFT (y) = b β MSFT 1 = = bse b β MSFT Step 2: Determine the rejection region for a nominal size α = 5%. T MSFT H0 t (20 2) W = fy : T k (y) < g Conclusion: for a signi cance level of 5%, we fail to reject the null H 0 : β MSFT = 1 against H 1 : β MSFT < 1 Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
100 3.1. The Student test Solution (cont d) Density of Ts under H α=5% DF = 18 Critical value = Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
101 3.1. The Student test Consider the onesided test k (y) > Aj T k t (N H0 H 0 : β k = a k H 1 : β k > a k The rejection region is de ned as to be: W = fy : T k (y) > Ag where A is a constant determined by the nominal size α. α = Pr (Wj H 0 ) = Pr T K ) Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
102 3.1. The Student test α = 1 Pr T k (y) < Aj T k t (N K ) H0 or equivalently 1 α = F N K (A) where F N K (.) denotes the cdf of the Student s tdistribution with N K degrees of freedom. Denote c 1 α the 1 α quantile of this distribution: A = F 1 N K (1 α) = c 1 α The rejection region of the test of size α is de ned as to be: W = fy : T k (y) > c 1 α g Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
103 3.1. The Student test De nition (Onesided Student test) The critical region of the Student test is that H 0 : β k = a k is rejected in favor of H 1 : β k > a k at the α (say, 5%) signi cance level if: W = fy : T k (y) > c 1 α g where c 1 α is the 1 α (say, 95%) critical value of a Student tdistribution with N K degrees of freedom and T k (y) is the realisation of the Student teststatistic. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
104 3.1. The Student test Example (Onesided test) Consider the CAPM model (cf. chapter 1) and the following results (Eviews). We want to test the beta of MSFT as H 0 : β MSFT = 1 against H 1 : β MSFT > 1 Question: give a conclusion for a nominal size of 5%. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
105 3.1. The Student test Solution Step 1: compute the tstatistic T MSFT (y) = b β MSFT 1 = = bse b β MSFT Step 2: Determine the rejection region for a nominal size α = 5%. T MSFT H0 t (20 2) W = fy : T k (y) > g Conclusion: for a signi cance level of 5%, we reject the null H 0 : β MSFT = 1 against H 1 : β MSFT > 1 Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
106 3.1. The Student test Solution (cont d) Density of Ts under H DF = 18 Critical value = α=5% Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
107 3.1. The Student test Consider the twosided test H 0 : β k = a k H 1 : β k 6= a k The nonrejection region is de ned as the intersection of the two nonrejection regions of the onesided test of level α/2: W = W A \ W B H 0 : β k = a k against H 1 : β k < a k W A = fy : T k (y) > c α/2 g H 0 : β k = a k against H 1 : β k > a k W B = fy : T k (y) < c 1 α/2 g Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
108 3.1. The Student test W = fy : c α/2 < T k (y) < c 1 α/2 g Since the Student s tdistribution is symmetric, c α/2 = c 1 α/2 W = fy : c 1 α/2 < T k (y) < c 1 α/2 g The rejection region is then de ned as to be: W = fy : jt k (y)j > c 1 α/2 g Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
109 3.1. The Student test De nition (Twosided Student test) The critical region of the Student test is that H 0 : β k = a k is rejected in favor of H 1 : β k 6= a k at the α (say, 5%) signi cance level if: W = fy : jt k (y)j > c 1 α/2 g where c 1 α/2 is the 1 α/2 (say, 97.5%) critical value of a Student tdistribution with N K degrees of freedom and T k (y) is the realisation of the Student teststatistic. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
110 3.1. The Student test Example (Onesided test) Consider the CAPM model (cf. chapter 1) and the following results (Eviews). We want to test the beta of MSFT as H 0 : β MSFT = 1 against H 1 : β MSFT 6= 1 Question: give a conclusion for a nominal size of 5%. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
111 3.1. The Student test Solution Step 1: compute the tstatistic T MSFT (y) = b β MSFT 1 = = bse b β MSFT Step 2: Determine the rejection region for a nominal size α = 5%. T MSFT H0 t (20 2) W = fy : jt k (y)j > g Conclusion: for a signi cance level of 5%, we reject the null H 0 : β MSFT = 1 against H 1 : β MSFT 6= 1 Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
112 3.1. The Student test Density of Ts under H α=2.5% DF = 18 Critical value = α=2.5% Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
113 3.1. The Student test Rejection regions H 0 H 1 Rejection region β k = a k β k > a k W = fy : T k (y) > c 1 α g β k = a k β k < a k W = fy : T k (y) < c α g β k = a k β k 6= a k W = fy : jt k (y)j > c 1 α/2 g where c β denotes the βquantile (critical value) of the Student tdistribution with N K degrees of freedom. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
114 3.1. The Student test De nition (Pvalues) The pvalues of Student tests are equal to: Twosided test: pvalue = 2 F N K ( jt k (y)j) Right tailed test: pvalue = 1 F N K (T k (y)) Left tailed test: pvalue = F N K ( T k (y)) where T k (y) is the realisation of the Student teststatistic and F N K (.) the cdf of the Student s tdistribution with N K degrees of freedom. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
115 3.1. The Student test Example (Onesided test) Consider the previous CAPM model. We want to test: H 0 : c = 0 against H 1 : c 6= 0 H 0 : β MSFT = 0 against H 1 : β MSFT 6= 0 Question: nd the corresponding pvalues. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
116 3.1. The Student test Solution Since we consider twosided tests with N = 20 and K = 2: pvalue c = 2 F 18 ( jt c (y)j) = 2 F 18 ( ) = pvalue c = 2 F 18 ( jt MSFT (y)j) = 2 F 18 ( ) = 5.7e 006 Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
117 3.1. The Student test Fact (Student test with large sample) For a large sample size N T k t (N K ) N (0, 1) H0 Then, the rejection region for a Student twosided test becomes W = y : jt k (y)j > Φ 1 (1 α/2) where Φ (.) denotes the cdf of the standard normal distribution. For α = 5%, Φ 1 (0.975) = 1.96, so we have: W = fy : jt k (y)j > 1.96g Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
118 3.1. The Student test Case 2: Semiparametric model Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
119 3.1. The Student test Assumption 6 (normality): the distribution of the disturbances is unknown, but satisfy (assumptions A1A5): E ( εj X) = 0 N 1 V ( εj X) = σ 2 I N Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
120 3.1. The Student test Problem 1 The exact ( nite sample) distribution of bβ k and bσ 2 are unknown. 2 As a consequence the nite sample distribution of T k (y) is also unknown. 3 But, we can use the asymptotic properties of the OLS estimators (cf. chapter 3). In particular, we have: p N bβ β d! N 0, σ 2 Q 1 where Q = p lim 1 N X> X = E X x i x > i Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
121 3.1. The Student test De nition (Zstatistic) Under the null H 0 : β k = a k, if the assumptions A1A5 hold (cf. chapter 3), the zstatistic de ned by Z k = b β k a k bse asy b β k d! H0 N (0, 1) where bse asy b β k = bσ p m kk denotes the estimator of the asymptotic standard error of the estimator bβ k and m kk is k th diagonal element of X > X 1. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
122 3.1. The Student test Rejection regions The rejection regions have the same form as for the ttest (except for the distribution) H 0 H 1 Rejection region β k = a k β k > a k W = y : Z k (y) > Φ 1 (1 α) β k = a k β k < a k W = y : Z k (y) < Φ 1 (α) β k = a k β k 6= a k W = y : jz k (y)j > Φ 1 (1 α/2) where Φ (.) denotes the cdf of the standard normal distribution. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
123 3.1. The Student test De nition (Pvalues) The pvalues of the Ztests are equal to: Twosided test: pvalue = 2 Φ ( jz k (y)j) right tailed test: pvalue = 1 left tailed test: pvalue = Φ ( Φ (Z k (y)) Z k (y)) where Z k (y) is the realisation of the Zstatistic and Φ (.) the cdf of the standard normal distribution. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
124 3.1. The Student test Summary Normality Assumption Non Assumption Teststatistic tstatistic zstatistic De nition T k = b β k a k bσ p m kk Z k = b β k a k bσ p m kk Exact distribution T k H0 t (N K ) Asymptotic distribution Z K d!h0 N (0, 1) Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
125 3.1. The Student test Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
126 Subsection 3.2 The Fisher test Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
127 3.2. The Fisher test Consider the twosided test associated to p linear constraints on the parameters β k : H 0 : Rβ = q H 1 : Rβ 6= q where R is a p K matrix and q is a p 1 vector. Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
128 3.2. The Fisher test Example (Linear constraints) If K = 4 and if we want to test H 0 : β 1 + β 2 = 0 and β 2 we have p = 2 linear constraints with: 3β 3 = 4, then R β (24) (4,1) = q (21) β 1 β 2 β 3 β 4 1 C A = 0 4 Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
129 3.2. The Fisher test Example (Linear constraints) If K = 4 and if we want to test H 0 : β 2 = β 3 = β 4 = 0, then we have p = 3 linear constraints with: R β (34) (4,1) A = q (31) β 1 β 2 β 3 β 4 1 C A = A Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
130 3.1. The Student test Case 1: Normality assumption A6 Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
131 3.2. The Fisher test De nition (Fisher teststatistic) Under assumptions A1A6 (cf. chapter 3), the Fisher teststatistic is de ned as to be: F = 1 Rβ p b > 1 1 q bσ 2 R X > X R > Rβ b q where b β denotes the OLS estimator. Under the null H 0 : Rβ = q, the F statistic has a Fisher exact ( nite sample) distribution F H0 F (p,n K ) Christophe Hurlin (University of Orléans) Advanced Econometrics  Master ESA November 20, / 225
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