Econometrics II - Problem Set 2

Transcription

1 Deadline for solutions: Econometrics II - Problem Set Problem 1 The senior class in a particular high school had thirty boys. Twelve boys lived on farms and the other eighteen lived in town. A test was devised to see if farm boys in general were more physically fit than town boys. Each boy in the class was given a physical fitness test in which a low score indicates a poor physical condition. The scores of the farm boys (X i ) and the town boys (Y i ) are as follows. X i Y i One is now interested in testing H : Farm boys do not tend to be more fit, physically, than town boys against H 1 : Farm boys tend to be more fit than town boys. a) Apply the Wilcoxon-Mann-Whitney-Test and interpret your result. a) Apply the Kolomogorov-Smirnov two-sample test and interpret your result. a) X i Rank Y i Rank Y i Rank We know that E(W N ) = 1 31 = 18 and V ar(w N ) = = 37. With 1 i=1 r(x i) = we get as Test-Statistik =.4. As this value is smaller than z.95 = 1.4, H is accepted. 1

2 b) H : F X (z) F Y (z) H 1 : F X (z) < F Y (z) The testing problem requires to apply the one-sided Kolmogorov-Smirnov test. z obs F X,m (z) F Y,n (z) F Y,n (z) F X,m (z) 1. 1/ / /1 / /1 3/ /1 3/ /1 5/ /1 5/ /1 / /1 / /1 7/ /1 8/ /1 8/ /1 9/ /1 1/ /1 11/ /1 11/ /1 1/ /1 1/ /1 13/ /1 14/ /1 15/ /1 15/ /1 15/ /1 1/ / / As max z F Y,n (z) F X,m (z) =.111 < D 1;18,.5 =.455, H is accepted. Problem a) Generally consider the Wilcoxon-Mann-Whitney-Test. Show that under H the expected value and the variance of the test statistics are indeed given by E(W N ) = m(n+1) and V ar(w N ) = mn(n+1) 1. Hint: i=1 i = N(N+1), i=1 i = N(N+1)(N+1)

3 i=1 i3 = 1 4 N N N b) In addition to its home country A, a company sells products in two other countries B and C. In a market research study consumer satisfaction is measured for random samples of n = 1 customers drawn separately for each country. An econometrician now uses different nonparametric tests in order to check whether there are differences in the distribution of consumer satisfaction between the home country A and the countries B as well as C. Different tests now lead to the following p-values: Test Country A against B Country A against C Kolmogorov-Smirnov..1 Wilcoxon-Mann-Whitney.31.9 van der Waerden.5.8 Siegel-Tukey.1.4 Which of the above tests provides a significant result? Which conclusions can be drawn from the different tests? a) First of all note that under H we have P (V i = 1) = m N. So we get that E(V i ) = 1P (V i = 1) + P (V i = ) = m N and V ar(v i ) = E(V i ) E(V i) = E(V i ) E(V i ) = m N m N = nm N. For i j, Cov(V i, V j ) = E(V i V j ) E(V i )E(V j ). Next consider the term m N1 N E(V i V j ) = P (V i = 1, V j = 1) = P (V i = 1 V j = 1)P (V j = 1) = m1 and hence Cov(V i, V j ) = m1 m N1 N m N = mn N (N1). For the expected value we get E( i=1 i V i) = i=1 ie(v i) = m N(N+1) N = m(n+1) 3

4 The Variance is given by V ar( i V i ) = i=1 i=1 N1 i V ar(v i ) + i=1 j=i+1 i j Cov(V i, V j ) = N1 mn(n + 1) N(N + 1)(N + 1) mn 1 N (N 1) i=1 j=i+1 i j = mn(n + 1) N(N + 1)(N + 1) mn 1 N (N 1) j1 j i = j= i=1 mn(n + 1) 1 mn(n + 1) N(N + 1)(N + 1) mn 1 N (N 1) N(N + 1)(N + 1) mn N (N 1) j= j(j 1) j = ( 1 4 N N N ) N(N + 1)(N + 1) = where we used i=1 i = N(N+1), i=1 i = N(N+1)(N+1) i=1 i3 = 1 4 N N N... = mn(n + 1), 1 b) As we face a multiple testing problem here, we use the Bonferroni adjustment. Note that.5/8 =.3. For the tests Country A against B, K-S and Siegel-Tukey deliver significant results after the adjustment. The Kolmogorov-Smirnov test is consistent against all alternatives, and we can thus generally conclude that there is a significant difference between the two distributions. Since the tests of Wilcoxon-Mann-Whitney and van der Waerden are far from being significant, there does not seem to exist a difference in the locations of the center of the distributions. Since the Siegel-Tukey test yields a significant result, there seems to exist a difference in dispersion. For the test Country A against C, no test delivers significant results after Bonferroni adjustment. But this is a quite conservative procedure, and the small p-values of the Wilcoxon-Mann-Whitney and the van der Waerden tests indicate a possible difference in medians. In view of the multiple testing problem, this results can, however, not be considered as significant. 4

5 Problem 3 Let (Y 1, X 1 ),..., (Y n, X n ) be an i.i.d. sample of random variables. Assume that Y i = β 1 + β X i + ɛ i, i = 1,..., n, where β 1, β are real-valued parameters, and where ɛ 1, ɛ,... are independent, zero mean error terms with the following property: the conditional distribution of ɛ i given X i is a normal distribution with mean zero and variance σ X i. a) Show that this model can be equivalently rewritten in the form Y i = b 1 (Z i ) + b (Z i )X i, where Z i are i.i.d. U(, 1)-distributed random variables. Identify the structure of the functions b 1 ( ) and b ( ) and establish a relation between Z i and ɛ i. b) Show that E[Y X = x] = 1 Q Y X=x(τ)dτ. a) As stated in the lecture, assume in addition X i > i. We know that ε i X i N(, σ Xi ) or εi N(, 1). As was shown in the lecture it holds true that Φ ( εi ) U(, 1), where Φ( ) is the distribution function of the standard normal distribution. As Z i U(, 1) set ( ) Z i = Φ εi With this in hand we can write ε i = Φ 1 (Z i ) Y i = b 1 (Z i ) + b (Z i )X i, with b 1 (Z i ) = β 1 and b (Z i ) = β + Φ 1 (Z i ) σ. Remark: If we allow X i, we get that The model can then be written in the form ε i σ X i N(, 1) and ε i = Φ 1 (Z i ) σ X i. Y i = b 1 (Z i ) + b (Z i )X i + b 3 (Z i ) X i, where b 1 (Z i ) = β 1, b (Z i ) = β and b 3 (Z i ) = Φ 1 (Z i ) σ. 5

6 b) We know that P ( ε i z τ ) = τ. So Q Y X=x (τ) = β 1 + β x + σ x z τ = E(Y X = x) + σ x z τ, and thus 1 Q Y X=x (τ) = E(Y X = x) + σ x 1 z τ dτ = It therefore only remains to prove that 1 z τ dτ =. If Φ denotes the standard normal distribution function we have z τ = Φ 1 (τ), and φ(x) = Φ (x), where φ is the density of the standard normal distribution. Substitution then leads to 1 z τ dτ = 1 Φ 1 (τ)dτ = φ(x)φ 1 (Φ(x))dx = φ(x)x dx =. Note that this result generalizes to any continuous error distribution with zero mean. Problem 4 Let the discrete variable X have a distribution function F (u) and assume that F (u) has only a finite number of discontinuities occurring at u 1 <... < u m with associated probabilities p 1,..., p m. Show that the Kolomogorov-Smirnov one sample test is conservative in this case. Hint: You may use a continuous random variable Y with distribution G(u) that assigns probability p i to the interval (u i1, u i ], i=1,...,m where u < u 1. In addition construct the random sample x 1,..., x n from a random sample y 1,..., y n from G(u). Let Y be a continuous random variable with distribution G(u) that assigns probability p i to the interval (u i1, u i ], i=1,...,m, where u < u 1. Therefore F (u i ) = G(u i ), i = 1,..., m Any random sample x 1,..., x n from F (u) can be thought of as having been drawn from a random sample y 1,..., y n from G(u) by setting x k = u i if u i1 < y k u i, k=1,...,n, i=1,...,m. If F n (u) is the empirical distribution of the x-sample and G n (u) that of the y- sample, Then it follows that F n (u i ) = G n (u i ), i = 1,..., m. D n = max ui F n (u i ) F (u i ) sup u G n (u) G(u) = D n so that P (D n d) P (D n d),

7 see Noether, Gottfried E. "Note on the Kolmogorov statistic in the discrete case." Metrika 7.1 (193):