2 Testing under Normality Assumption


 Cameron Howard Harrison
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1 1 Hypothesis testing A statistical test is a method of making a decision about one hypothesis (the null hypothesis in comparison with another one (the alternative using a sample of observations of known size. A statistical test is not a proof per se. Accepting the null hypothesis (H 0 doesn t mean it is true, but just that the available observations are not incompatible with this hypothesis, and that there is not enough evidence to favour the alternative hypothesis over the null hypothesis. There are 3 steps: 1. First specify a ull Hypothesis, usually denoted H 0, which describes a model of interest. Usually, we express H 0 as a restricted version of a more general model. 2. Then, construct a test statistic, which is a random variable (because it is a function of other random variables with two features: (a it has a known distribution under the ull Hypothesis (usually, normal or chisquare, t or F. Its distribution is known either because we assume enough about the distribution of the model disturbances to get smallsample distributions, or we assume enough to get asymptotic distributions. (b this known distribution may depend on data, but not on parameters (this is called pivotality: a test statistic is pivotal if it satisfies this condition. 3. Check whether or not the sample value of the test statistic is very far out in its sampling distribution. When we perform a test, we may end up rejecting the null hypothesis even though it is true. In this case we are committing the socalled Type I error. The probability of type I error is the significance level (or size of the test. It is also possible that we fail to reject the null hypothesis even though it is false. In this case we are committing the socalled Type II error. The probability of type II error is the power of the test. 2 Testing under ormality Assumption 2.1 Properties of OLS estimators Suppose Y = Xβ + ε, ε (0, σ 2 I, 1
2 and X is full rank with rank K. Then, (i β = (X X 1 X Y ( β, σ 2 (X X 1 (ii e e σ 2 χ2 ( K (iii s 2 = e e K is an unbiased estimator of σ2 and is independent of β, where e = Y βx. Proof. (i The fact that the disturbances are independent meanzero normals, ε (0, σ 2 I, implies E [X ε] = 0 K and E [εε ] = σ 2 I,so the OLS estimator is still BLUE: ] E [ β = β, V = σ 2 (X X 1. Write out β as a function of ε as follows: β = (X X 1 X Y = β + (X X 1 X ε is a linear combination of a normally distributed vector. Since, for any vector x (µ, Σ, (a + Ax (a + µ, AΣA (See Kennedy s All About Variances Appendix, we have β ( β + 0 K, (X X 1 X σ 2 I X(X X 1, or β ( β, σ 2 (X X 1. (ii The residual vector can be written as e = Y X β = Y P X Y = (I P X Y = M X Y = M X Xβ + M X ε = M X ε where M X is the residual projection matrix that creates the residuals from a regression of something on X: M X = I X(X X 1 X = I P X The matrix M X is a projection matrix that is, M X is symmetric (i.e. M X = M X, idempotent (i.e. M 2 X = M X, and rank(m X =rank(irank(p X = tr(p X = K. 2
3 We can then write e e σ 2 = ε σ M XM ε X σ = ε σ M X Since ε σ (0, I, it follows that e e σ 2 = ε σ M ε X σ χ2 (rank(m X = χ 2 ( K [ e ] e (iii From (ii we have E σ 2 = K, so that E [ [ s 2] e ] e = E = σ 2. K To proof that s 2 and β are independent, it is sufficient to show that the normal random variables e and β are uncorrelated. cov(e, β = cov(m X ε, (X X 1 X y = M X cov(ε, yx(x X 1 = σ 2 M X X(X X 1 = 0 because M X X = Test of equalities We consider 3 types of tests of equalities: single linear, multiple linear, and general nonlinear. Tests of equalities are fully specified when you specify the ull hypothesis: the ull is either true or not true, and you don t care how exactly it isn t true, just that it isn t true Single linear tests: ztest and ttest A single linear test could be written as Rβ r = 0, where R is 1 K and r is a scalar. The discrepancy vector, d, is the sample value of the null hypothesis evaluated at the sample estimate of β, β. Using the terminology above, for a linear hypothesis, d = R β r, Even if the hypothesis is true, we would not expect d to be exactly zero, because β is not exactly equal to β. However, if the hypothesis is true, we would expect d to be close to 0. In contrast if the hypothesis is false, we d have no real prior about where we d see d. 1. The ztest is performed by the zstatistic defined by T z = 1 σ ε σ ( R(X X 1 R 1/2 d (0, 1, It is called a ztest because it follows a standard normal distribution and standard normal variables are often denoted z. 3
4 2. The ttest is performed by using s in place of σ in the above formula when σ 2 is unknown. This corresponds to taking σ s times T z. Tz s = σ s T z σ (0, 1. s So, if we need to figure out the distribution of σ s (0, 1. Recall that: then ( K s2 σ 2 = e e σ 2 χ2 K, s s σ = 2 σ 2 = χ 2 K K, the squareroot of a chisquare divided by its own degrees of freedom. Returning to the expression, we have Tz s = σ s T z σ (0, 1 s (0, 1. χ 2 K K The distribution of a normal divided by a square root of a chisquare divided by its own degrees of freedom is called a Student s t distribution, denoted t K, where K is the number of degrees of freedom in the denominator. The test statistic is then called a t Test Statistic T t = 1 ( R(X X 1 R 1/2 (0, 1 d = t s χ 2 K, K K where t K means t distribution with K degrees of freedom which means a standard normal divided by the square root of a chisquare divided by its own degrees of freedom. 3. The ztest and the ttest are related in practise. The ztest requires knowledge of σ 2, whereas the ttest uses an estimate of it. However, when the sample is very large, the estimate of σ 2 is very close to its true value, so the estimate is almost the same as prior knowledge. This means that the ztest and ttest statistics have nearly the same distribution when the sample is large. 4. Examples: (a An exclusion restriction, e.g., that the second variable does not belong in the model would have R = [ ], r = 0. 4
5 (b A symmetry restriction, e.g., that the second and third variables had identical effects, would have R = [ ], r = 0. (c A value restriction, e.g, that the second variable s coefficient is 1, would have R = [ ], r = Multiple Linear Tests: the finitesample Ftest A multiple linear test could be written as Rβ r = 0, where R is a J K matrix of rank J and r is a J vector. 1. Since T z = 1 σ ( R(X X 1 R 1/2 d (0, IJ, we have 1 σ 2 d ( R(X X 1 R 1 d = T z T z χ 2 J. This provides a test statistic distributed as a χ 2 J, called a Wald Test. This test requires knowledge of σ If we substitute in s 2 for σ 2, then we have 1 s d ( 2 R(X X 1 R 1 σ d 2 χ 2 = s 2 χ2 J J = χ 2 K / K by the same reasoning as for the Single Linear ttest. If we divide the numerator by J, we get a ratio of chisquareds divided by their own degrees of freedom which follows the socalled F distribution, with degrees of freedom given by its numerator and denominator degrees of freedom. This test uses the estimate s 2. The resulting test statistic is then called a Ftest statistic 3. Examples: F = 1 Js 2 d ( R(X X 1 R 1 d = χ 2 J /J χ 2 K / K F J, K. 5
6 (a A set of exclusion restrictions, e.g., that the second and third variables do not belong in the model, would have [ ] R =, [ ] 0 r =. 0 (b A set of symmetry restrictions, that the first, second and third variables all have the same coefficients, would have [ ] R =, [ ] 0 r =. 0 (c Given that we write the restriction as Rβ r = 0 for both single and multiple linear hypotheses, you can think of the single hypothesis as a case of the multiple hypothesis. 3 Testing without the ormality Assumption Without the ormality assumption, we can still get the approximate distributions of test statistics in large samples. In this case, the law of large numbers, the central limit theorem and the Slutsky s lemma can be invoked. 3.1 Properties of OLS estimators Suppose Y = Xβ + ε, E [X ε] = 0 K, E [εε ] = σ 2 I, Let β = (X X 1 X Y and e = Y βx be the OLS estimator and residual. Then, the following properties follow from the law of large numbers, the central limit theorem and the Slutsky s lemma: (i β = approx ( β, σ 2 (X X 1 e e (ii s 2 = K is a consistent estimator of σ2. Proof. (i β = β + (X X 1 X ε, so ( X 1 X X ε β =. 6
7 By the central limit theorem, X ε = ( 1 where X i is the i th row of X. (ii s 2 = ( e e K = ε M X ε K = ε ε K = K ε ε By the law of large numbers, = 1 1 i=1 X P iε i E [X ε] = 0 K, as. 3.2 Wald Tests Linear hypothesis i=1 X iε i approx ( 0, σ 2 (X X, ε X(X X 1 X ε ( ε ( ε ε X X X i=1 ε2 i 1 X ε (1 (2 (3 P E [ ε 2] = σ 2, and X ε = Consider a multiple linear hypothesis with a linear model and possibly nonnormal (but finite variance ε: Y = Xβ + ε, E [X ε] = 0 K, E [εε ] = σ 2 I, H 0 : Rβ r = 0. Here, ε may be nonnormal (for example uniform as long as ε is finitevariance. Since β approx (β, σ 2 (X X 1 the Wald vector defined by ( T W v = 1 σ R(X X 1 R 1/2 d is asympotically approximately a vector of standard normals (0 J, I J. Hence its inner product, called a Wald Test, is asymptotically approximately a chisquare: T W = T W vt W v = 1 σ 2 d ( R(X X 1 R 1 d approx χ2 J. Moreover, because s 2 is asymptotically equal to σ 2 the approximation is not affected by replacing σ 2 with s 2 when σ 2 is unknown. Thus, we have that T s W = T s W vt s W v = 1 s 2 d ( R(X X 1 R 1 d approx χ2 J. The Wald Statistic approximately follows the chisquare distribution as the sample size gets really large, even if one uses s 2. 7
8 3.2.2 onlinear hypothesis A multiple nonlinear test could be written as c(β = 0, where c is a J vector function of β. Consider the model in which we have a set of J nonlinear restrictions c(β = 0 that we wish to test: Y = Xβ + ε, E [X ε] = 0 K, E [εε ] = σ 2 I, H 0 : c(β = 0. The discrepancy vector d gives the distance between the sample value of the hypothesis and its hypothesized value of 0: d = c. Since we have not assumed normality of the ε, all we have for the finitesample distribution of β is a mean and variance: ] E [ β = β, [ β] V = σ 2 (X X 1. Application of the deltamethod allows us to calcuate an approximate asymptotic distribution of c :By firstorder Taylor approximation, we have c c(β + β c (β β = β c (β β where β c (β is the matrix of derivatives of the vectorfunction c(β with respect to the rowvector β (each row of β c (β gives the derivatives of an element of c(β with respect to β. Since β β approx (0, σ2 (X X 1, then c approx ( 0 J, σ 2 β c (β (X X 1 ( β c (β. Since β goes to β asymptotically, we can replace β c (β with β c c approx (0 J, σ 2 β c ( (X X 1 β c. : 8
9 ow, we use this information to create the Wald Statistic. Premultiplying the sample value of the hypothesis by the minusonehalf matrix of its variance gives the Wald Vector distributed as a vector of standard normals: T W v = 1 σ ( β c ( 1/2 (X X 1 β c c approx (0 J, I J. Finally, we take the inner product of this to create the Wald Statistic T W = 1 ( σ 2 c β c ( 1 (X X 1 β c c approx χ2 J. Since this is an approximate asymptotic result, it also works with s instead of σ: T W = 1 ( s 2 c β c ( 1 (X X 1 β c c approx χ2 J. 4 Testing the exogeneity of regressors: the Hausman test Suppose we have the model Y = Xβ + ε, E [εε ] = σ 2 I, This model can be estimated by Instrumental variables using 2SLS method. However, if the regressors X are exogeneous, then 2SLS estimator is less efficient (i.e. larger variance. Therefore, it is important to test for exogeneity first, in order to avoid using an IV estimator that is: (i More computationally intensive (two stages is more difficult than one and (ii less efficient. An exogeneity test for the regresssors X could be written as H 0 : E [X ε] = 0 K, Instrumental variable estimation requires to find an instrument Z such that E (Z X 0, and E (Z ε = 0 where rank(z=j > K. Let β OLS be the ordinary least squares estimator and β 2SLS be twostage least squares instrumental variable estimation of β: β OLS = (X X 1 X Y, β2sls = (X P Z X 1 X P Z Y 9
10 where P Z = Z(Z Z 1 Z If the regressors X are endogeneous, then the OLS estimates should differ from the endogeneitycorrected 2SLS estimates (as long as the instruments are exogenous. An exogeneity test can therefore be based on the difference between β 2SLS and β OLS. The test of this hypothesis is called a Hausman Test. By the central limit theorem, we have, under H 0, β OLS β approx ( 0, σ 2 (X X 1, β2sls β approx ( 0, σ 2 (X P Z X 1 Hence, β 2SLS β OLS approx ( 0, σ 2 [ (X P Z X 1 (X X 1] ow, we use this information to create a Wald Statistic. Premultiplying the difference of the two estimators by the minusonehalf matrix of its variance gives the Wald Vector distributed as a vector of standard normals: T w = 1 σ [ (X P Z X 1 (X X 1] 1/2 2SLS β OLS approx (0, I K Hence its inner product is asymptotically approximately a chisquare: T wt w = 1 σ 2 2SLS β OLS [(X P Z X 1 (X X 1] 1 2SLS β OLS Since this is an approximate asymptotic result, it also works with s 2 instead of σ 2. So the HausmanWald test statistic for the exogeneity of regressors is H w = 1 s 2 2SLS β OLS [(X P Z X 1 (X X 1] 1 2SLS β OLS ote: we are assuming here that the matrix V = (X P Z X 1 (X X 1 is positive definite. The more general approach is to take a generalized inverse (instead of the inverse in the formula of H w. In that case the statistic H w has an asymptotic approximate chisquared distribution with degree of freedom equal to the rank of the matrix V. 5 Testing the validity of instruments: Overidentification test This is a test that will tell you if the instruments Z are uncorrelated with the error term ε, an essential condition for the validity of instrumental variables. When this condition is not satisfied, the IV estimator could be inconsistent. approx χ2 (K approx χ2 (K 10
11 The degree of overidenfication of an overidentified linear regression model is defined to be J K, where J=rank(Z is the number of instruments and K=rank(X the number of regressors. The model we wish to test is Y = Xβ + ε, E [εε ] = σ 2 I, H 0 : E [Z ε] = 0 Denote e = Y X β 2SLS the residual of the 2SLS estimation of the model. The test statistic is based on the vector 1 σ P Ze = [ P Z P Z X(X P Z X 1 X ] ε P Z σ which is the projection of the residual vector e over the vector space of instruments Z. Orthogonality between e and Z therefore implies that P Z e is equal to 0. An appropriate test statistic can therefore be based on the squared eulidean distance between P Z e and 0, that is, (P Z e P Z e. The rank of P Z is J and the rank of P Z X(X P Z X 1 X P Z is K, so the rank of the (idempotent, symmetric matrix P Z P Z X(X P Z X 1 X P Z is J K. Since ε σ is of mean 0 and variance matrix I, 1 σ 2 e P Z e = ( 1 σ P Ze ( 1 σ P Ze is the sum of squares of J K things that have mean 0 and variance 1. By the central limit theorem, each of these things is approximately asymptotically normal (0, 1. Hence 1 σ 2 e P Z e is approximately asymptotically chisquare with J K degrees of freedom. Since this is an approximate asymptotic result, it also works with s 2 instead of σ 2, where σ 2 = e e = is a consistent estimator of σ2. So the test statistic for the validity of instruments is given by Q = 1 σ 2 e P Z e approx χ2 (J K The same test statistic can be obtained by considering the synthetic regression e = Zγ + u The estimate of γ is γ = (Z Z 1 Z e and the predicted value of e from the synthetic regression is ê = Z γ = Z(Z Z 1 Z e. The sum of squares of this 11
12 value (the explained sum of squares is by definition ESS = ê ê = e Z(Z Z 1 Z Z(Z Z 1 Z e = e Z(Z Z 1 Z e = e P Z e, while the total sum of squares is T SS = e e. By definition, the R 2 of this synthetic model is R 2 = ESS T SS = e P Z e e e. So Q = e P Z e σ 2 = e P Z e e e = R 2 where R 2 is from the second stage regression of the predicted errors e on all exogeneous factors in the model. 6 Testing heteroskedasticity: the White test The model we wish to test is Y = Xβ + ε, E [X ε] = 0 K, H 0 : E [εε ] = σ 2 I, The alternative hypothesis of the White test is that the error variance is affected by any of the regressors and their squares (and possibly their cross product. It therefore tests whether or not any heteroskedasticity present causes the variance matrix of the OLS estimator to differ from its usual formula. The steps to build the test are as follows: 1. Run the OLS regression and take the vector of residuals e = Y X β 2. Run the synthetic regression of the residuals over the regressors and their squares e = Xγ + X 2 θ + u where X 2 is the K 1 matrix of the squares of the regressors (excluding the column of constants. ow, take the coefficient of determination of this synthetic regression, R The White test statistic is defined by W = R 2 and is approximately asymptotically distributed as χ 2 (2K. 12
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