Lecture 07 Hypothesis Testing with Multivariate Regression

Transcription

1 Lecture 07 Hypothesis Testing with Multivariate Regression 23 September 2015 Taylor B. Arnold Yale Statistics STAT 312/612

2 Goals for today 1. Review of assumptions and properties of linear model 2. The multivariate T-test 3. Hypothesis test simulations 4. The multivariate F-test

3 Review from Last Time

4 Classical linear model assumptions I. Linearity Y = Xβ + ϵ II. Strict exogeneity E (ϵ X) = 0 III. No multicollinearity P [rank(x) = p] = 1 IV. Spherical errors V (ϵ X) = σ 2 I n V. Normality ϵ X N (0, σ 2 I n )

5 Finite sample properties Under assumptions I-III: (A) E( β X) = β Under assumptions I-IV: (B) V( β X) = σ 2 (X t X) 1 (C) β is the best linear unbiased estimator (Gauss-Markov) (D) Cov( β, r X) = 0 (E) E(s 2 X) = σ 2 Under assumptions I-V: (F) β achieves the Cramér Rao lower bound

6 The T-test

7 Hypothesis tests Consider testing the hypothesis H 0 : β j = b j. Under assumptions I-V we have the following: β j b j X, H0 N (0, σ 2 ( (X t X) 1 ) jj )

8 Z-test This suggests the following test statistic: z = β j b j σ 2 ( (X t X) 1 jj ) With, z X, H 0 N (0, 1)

9 T-test As in the simple linear linear regression case, we generally need to estimate σ 2 with s 2. This yields the following test statistic: t = β j b j s 2 ( (X t X) 1 jj ) = β j b j S.E.( β j )

10 T-test The test statistic has a T-distribution with (n p) degrees of freedom under the null hypothesis: t X, H 0 t n p This time around, we ll actually prove this.

11 To start, we re-write the test statistic as: t = = = β b σ σ ( 2 (X t X) 1 ) 2 s 2 jj z s 2 /σ 2 z q/(n p) Where q = r t r/σ 2. We need to show that (1) q X χ 2 n p and (2) z q X.

12 Lemma If B is a symmetric idempotent matrix and u N (0, I n ), then u t Bu χ 2 tr(b).

13 Proof: The symmetric matrix B can be written as by its eigen-decompostion; for some orthonormal Q matrix and diagonal matrix Λ: B = Q t ΛQ

14 Notice that as B is idempotent: (Q t ΛQ) = (Q t ΛQ)(Q t ΛQ) = Q t ΛQQ t ΛQ = Q t Λ 2 Q Since Q t = Q 1 : Λ = Λ 2 Therefore all the elements of Λ are 0 or 1.

15 By rotating the columns of Q, we can actually write: ( ) Irank(B) 0 Λ = 0 0 Notice that this also shows that: tr(b) = tr(q t ΛQ) = tr(λqq t ) = tr(λ) = rank(b)

16 Now let v = Qu. We can see that the mean of v is zero: And the variance of v is: Ev = EQu = QEu = 0 Evv t = QE(uu t )Q = QI n Q t = I n

17 Now we calculate the quadtratic form u t Bu by the transform of v. Substitute that u = (Q) 1 v = Q t v: u t Bu = v t QBQ t v = v t Q(Q t ΛQ)Q t v = v t QQ t ΛQQ t v = v t Λv = v t ( Itr(B) = Which completes the lemma. tr(b) v 2 i i=1 χ 2 tr(b) ) v

18 (1) We know that r t r = ϵ t Mϵ, so: q = rt r σ 2 = ϵt σ M ϵ σ Under Assumption V, ϵ σ N (0, I n). Therefore from the lemma, then q χ 2 tr(m) ; last time we showed the tr(m) = n p, which finishes the first part of the proof.

19 (2) The random variables β and r are both linear combinations of ϵ, and are therefore jointly normally distributed. As they are uncorrelated (problem set 2), this implies that they are independent. Finally, this implies that z = f( β) and q = g(r) and themselves independent.

20 So, therefore, we have: t = = β b σ 2 ( (X t X) 1 jj z q/(n p) t n p ) σ 2 s 2

21 Simulation

22 F-Test

23 F-test Now consider the hypothesis test H 0 : Dβ = d for a matrix D with k rows and rank k.

24 We ll form the following test statistic: F = (D β d) t [D(X t X) 1 D t ] 1 (D β d)/k s 2 And prove that is has an F-distribution with k and n p degrees of freedom.

25 We can re-write the test statistics as: Where: F = w/k q/(n p) w = (D β d) t [σ 2 D(X t X) 1 D t ] 1 (D β d) q = r t r/σ 2 We ve already shown that q X χ 2 n p, and can see that q w by the same argument as before. All that is left is to show that w χ 2 k.

26 Let v = D β d. Under the null hypothesis v = D( β β). Therefore: V(v X) = V(D( β β) X) = DV( β β X)D t = σ 2 D(X t X) 1 D t Now, for the multivariate normally distributed v with zero mean, we have: w = v t V(v X) 1 v

27 Now, decompose V(v X) 1 = Σ 1 as Q t ΛQ. Notice that this implies: Σ 1 = Q t Λ 1/2 Λ 1/2 Q QΣ 1 = Λ 1/2 Λ 1/2 Q Λ 1/2 QΣ 1 = Λ 1/2 Q

28 From here, we can write w as the inner product of a suitably defined vector u: With a mean of zero: w = v t Σ 1 v = v t Q t Λ 1/2 Λ 1/2 Qv = u t u E(u X) = E(Λ 1/2 Qv X) = Λ 1/2 QE(v X) = 0

29 And variance of And, finally: E(uu t X) = E(Λ 1/2 Qvv t Q t Λ 1/2 X) = Λ 1/2 QE(vv t X)Q t Λ 1/2 = Λ 1/2 QΣQ t Λ 1/2 = Λ 1/2 QΣ 1 ΣQ t Λ 1/2 = Λ 1/2 Λ 1/2 = I k Which finishes the proof. w = uu t χ 2 k

30 An alternative F-test Now, consider the following estimator: β = arg min y Xb 2 2, s.t. Db = d b We then define the restricted residuals r = y X β.

31 An alternative F-test An alternative expression for the F-test statistic is: F = ( rt r r t r)/k r t r/(n p) Conceptually, it should make sense that this is large whenever the null hypothesis is false.

32 An alternative F-test If we let SSR U be the sum of squared residuals of the unrestricted model (r t r) and SSR R be the sum of squared residuals of the restricted model, then this can be re-written as: F = (SSR R SSR U )/k SSR U /(n p) This is the way it is written in the homework and in the Fumio Hayashi text. It is left for you to prove that this is equivalent to the other F test.

33 Application