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Exercise Sheet 6: Solutions R.G. Pierse 1. (a) Regression yields: Dependent Variable: LC Date: 10/29/02 Time: 18:37 Sample(adjusted): 1950 1985 Included observations: 36 after adjusting endpoints C 0.244716 0.213958 1.143759 0.2618 LC(-1) 0.708128 0.150811 4.695459 0.0001 LY 0.660755 0.067809 9.744325 0.0000 LY(-1) -0.392093 0.124026-3.161394 0.0036 INF -0.164111 0.052693-3.114502 0.0040 INF(-1) 0.015959 0.052421 0.304433 0.7629 R-squared 0.999118 Mean dependent var 10.89442 Adjusted R-squared 0.998971 S.D. dependent var 0.252008 S.E. of regression 0.008085 Akaike info criterion -6.646632 Sum squared resid 0.001961 Schwarz criterion -6.382712 Log likelihood 125.6394 F-statistic 6795.131 Durbin-Watson stat 2.023213 Prob(F-statistic) 0.000000 1

(i) The estimated short run income elasticity is 0.660755. The estimated long run income elasticity is β 3 + β 4 1 β 2 0.660755 0.392093 = 1 0.708128 = 0.920479 Using the EViews Wald coefficient restrictions test facility, we can test the hypothesis that this long-run elasticity is zero: Wald Test on Long Run Elasticity Wald Test: Equation: Untitled Null Hypothesis: (C(3)+C(4))/(1-C(2)) F-statistic 733.3756 Probability 0.000000 Chi-square 733.3756 Probability 0.000000 (ii) The test for serial correlation yields: Breusch-Godfrey Serial Correlation LM Test: F-statistic 0.069339 Probability 0.794163 Obs*R-squared 0.085871 Probability 0.769494 On the basis of the Breusch-Godfrey serial correlation test, we would not reject the null of no autocorrelation. 2

(iii) Wald Test of hypothesis: β 2 + β 3 + β 4 = 1 Wald Test: Equation: Untitled Null Hypothesis: C(2)+C(3)+C(4)=1 F-statistic 1.254592 Probability 0.271565 Chi-square 1.254592 Probability 0.262677 We do not reject H 0 at the 5% significance level. (iv) Subtracting C t 1 from both sides of we get LC t = β 1 + β 2 LC t 1 + β 3 LY t + β 4 LY t 1 +β 5 INF t + β 6 INF t 1 + ε t LC t = β 1 + β 2 LC t 1 LC t 1 + β 3 LY t + β 4 LY t 1 +β 5 INF t + β 6 INF t 1 + ε t, and, adding and subtracting β 2 LY t 1 and β 3 LY t 1, we obtain LC t = β 1 + β 2 (LC t 1 LY t 1 ) LC t 1 + β 3 LY t +(β 2 + β 3 + β 4 )LY t 1 + β 5 INF t + β 6 INF t 1 + ε t. If the restriction β 2 + β 3 + β 4 = 1 holds then we can rewrite this as or β 3 + β 4 1 β 2 = 1 LC t = β 1 + (β 2 1)(LC t 1 LY t 1 ) + β 3 LY t +β 5 INF t + β 6 INF t 1 + ε t which is the same as the restricted model LC t = γ 1 + γ 2 LY t + γ 3 (LY t 1 LC t 1 ) +γ 4 INF t + γ 5 INF t 1 + ε t. with the parameter correspondences: γ 1 = β 1, γ 2 = β 3, γ 3 = 1 β 2, and γ 4 = β 5 and γ 5 = β 6. 3

2. Unrestricted Distributed Lag Dependent Variable: Y Date: 10/29/02 Time: 19:28 Sample(adjusted): 1953:08 1957:12 Included observations: 53 after adjusting endpoints C 92.95484 56.37294 1.648927 0.1063 X 0.057709 0.052537 1.098451 0.2780 X(-1) 0.102538 0.078054 1.313676 0.1958 X(-2) 0.214550 0.077012 2.785941 0.0078 X(-3) 0.184946 0.076605 2.414266 0.0200 X(-4) 0.114580 0.077278 1.482704 0.1453 X(-5) 0.041156 0.076684 0.536696 0.5942 X(-6) 0.133245 0.079779 1.670182 0.1020 X(-7) 0.110570 0.058545 1.888639 0.0655 R-squared 0.989442 Mean dependent var 3229.849 Adjusted R-squared 0.987523 S.D. dependent var 1157.760 S.E. of regression 129.3240 Akaike info criterion 12.71604 Sum squared resid 735886.7 Schwarz criterion 13.05062 Log likelihood -327.9750 F-statistic 515.4453 Durbin-Watson stat 0.425647 Prob(F-statistic) 0.000000 (a) The estimated impact (short-run) multiplier is 0.057709. The estimated long-run multiplier is 0.057709 + 0.102538 + 0.214550 + 0.184946 +0.114580 + 0.041156 + 0.133245 + 0.110570 = 0.95930 Using the EViews Wald coefficient restrictions test facility, we can test the hypothesis that this long-run multiplier is zero: 4

Wald Test on Long Run Multiplier Wald Test: Equation: Untitled Null Hypothesis: C(2)+C(3)+C(4)+C(5)+C(6)+C(7)+C(8)+C(9)=0 F-statistic 3209.189 Probability 0.000000 Chi-square 3209.189 Probability 0.000000 (b) Form the Almon variables: Z0 = X + X( 1) + X( 2) + X( 3) + X( 4) +X( 5) + X( 6) + X( 7) Z1 = X( 1) + 2 X( 2) + 3 X( 3) + 4 X( 4) +5 X( 5) + 6 X( 6) + 7 X( 7) Z2 = X( 1) + 4 X( 2) + 9 X( 3) + 16 X( 4) +25 X( 5) + 36 X( 6) + 49 X( 7) Z3 = X( 1) + 8 X( 2) + 27 X( 3) + 64 X( 4) +125 X( 5) + 6 36 X( 6) + 7 49 X( 7) Z4 = X( 1) + 16 X( 2) + 81 X( 3) + 256 X( 4) +625 X( 5) + 36 36 X( 6) + 49 49 X( 7) 5

Fourth Degree Almon Lag Polynomial Dependent Variable: Y Date: 10/29/02 Time: 19:45 Sample(adjusted): 1953:08 1957:12 Included observations: 53 after adjusting endpoints C 91.15685 55.03268 1.656413 0.1043 Z0 0.036566 0.045387 0.805649 0.4245 Z1 0.162404 0.172128 0.943511 0.3502 Z2-0.051813 0.115507-0.448572 0.6558 Z3 0.003884 0.025901 0.149944 0.8815 Z4 7.05E-05 0.001845 0.038189 0.9697 R-squared 0.989220 Mean dependent var 3229.849 Adjusted R-squared 0.988073 S.D. dependent var 1157.760 S.E. of regression 126.4396 Akaike info criterion 12.62368 Sum squared resid 751387.9 Schwarz criterion 12.84673 Log likelihood -328.5275 F-statistic 862.5750 Durbin-Watson stat 0.490582 Prob(F-statistic) 0.000000 We can see from the p-value that the coefficient on the fourth order term, Z4, is not significant so we drop it and estimate the third degree polynomial Almon lag. 6

Third Degree Almon Lag Polynomial Dependent Variable: Y Date: 10/29/02 Time: 19:49 Sample(adjusted): 1953:08 1957:12 Included observations: 53 after adjusting endpoints C 91.31976 54.29340 1.681968 0.0991 Z0 0.035553 0.036435 0.975776 0.3341 Z1 0.168311 0.074723 2.252456 0.0289 Z2-0.056093 0.027670-2.027187 0.0482 Z3 0.004867 0.002645 1.840519 0.0719 R-squared 0.989220 Mean dependent var 3229.849 Adjusted R-squared 0.988321 S.D. dependent var 1157.760 S.E. of regression 125.1175 Akaike info criterion 12.58597 Sum squared resid 751411.2 Schwarz criterion 12.77185 Log likelihood -328.5283 F-statistic 1101.125 Durbin-Watson stat 0.489410 Prob(F-statistic) 0.000000 We can see from the p-value that the coefficient on the third order term, Z3, is not significant so we drop it and estimate the second degree polynomial Almon lag. 7

Second Degree Almon Lag Polynomial Dependent Variable: Y Date: 10/29/02 Time: 19:56 Sample(adjusted): 1953:08 1957:12 Included observations: 53 after adjusting endpoints C 98.01741 55.47535 1.766864 0.0835 Z0 0.092330 0.019855 4.650281 0.0000 Z1 0.034612 0.017935 1.929833 0.0594 Z2-0.005375 0.002569-2.092453 0.0416 R-squared 0.988459 Mean dependent var 3229.849 Adjusted R-squared 0.987752 S.D. dependent var 1157.760 S.E. of regression 128.1294 Akaike info criterion 12.61643 Sum squared resid 804440.6 Schwarz criterion 12.76513 Log likelihood -330.3354 F-statistic 1398.878 Durbin-Watson stat 0.493917 Prob(F-statistic) 0.000000 This time the p-value of the coefficient on the highest order term, Z2, shows that it is significant and so we stop searching and choose a second degree polynomial. (c) One way to obtain an estimate of the long-run (equilibrium) multiplier is to note that in equilibrium so that X t = X t 1 = X t 1 = = X t 7 Y t = c + 7 α 0 X t + (1 + 2 + 3 + + 7) α 1 X t +(1 + 4 + 9 + + 49) α 2 X t. The long-run multiplier is therefore given by 7 α 0 + (1 + 2 + 3 + + 7) α 1 + (1 + 4 + 9 + + 49) α 2 = 0.86298. 8

A Wald test for the hypothesis that the long-run multiplier is zero is given by: Wald Test on Long Run Multiplier Wald Test: Equation: Untitled Null Hypothesis: 7*C(2)+7*4*C(3)+(1+4+9+16+25+36+49)*C(4)=0 F-statistic 1742.687 Probability 0.000000 Chi-square 1742.687 Probability 0.000000 9

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