计量经济学练习

Exercises1. Using a cross-section of n =100 workers, an investigator specified the following equation for the log of earnings:201234log(),i i i i i i w x x e m sex u βββββ=+++++where:log (w i ): refers to the log of earnings,x i : refers to years of experience (the mean of the series equals 15 years), e i : refers to years of education andmsex i : refers to a dummy variable taking the value of 1 for male workers and 0 for female workers.OLS estimation yielded 0ˆ0.305262β=, 1ˆ0.038000β=, 2ˆ0.00087β=-,3ˆ0.035000β=, 4ˆ0.040000β= and R 2= 0.50. The standard errors (SE ) are: SE (0ˆβ)=0.10000, SE (1ˆβ)=0.00580, SE (2ˆβ)=0.00016, SE (3ˆβ)=0.01600 and SE (4ˆβ)=0.01000.a) Interpret the estimated coefficients on experience, education and sex and comment on their statistical significance at the 5% level.b)Describe the implied relationship between earnings and experience. According to the parameter estimates above, after how many years of experience does a worker reach maximum earnings?c)Test the null hypothesis that experience has no effect on earnings when x i =10 years. Assume that the covariance between the estimated coefficients on x i and x i 2 is zero.Brief answer:a) Coefficient on education: An extra year of education raises earnings by 3.5%.Coefficient on experience: We note that the partial derivative12log()2i ii w x x ββ∂=+∂. When x i =15 years (the mean of the series),121212log()2215300.03800030*(0.00087)0.0119i i iw x x ββββββ∂=+=+=+=+-=∂based on the estimates of 1ˆβ and 2ˆβ. Therefore, an extra year of experience above the mean (of 15 years) raises earning by 1.19%.Coefficient on sex: Male workers earn 4% more than female workers. To test the statistical significance of the estimated coefficients, construct the t -ratios :1106.551()t ratio SE ββ--==,2205.437()t ratio SE ββ--==-,3302.187()t ratio SE ββ--==and 4404.000()t ratio SE ββ--==Then, compare their absolute value with the critical valuet n -5,α/2=t 95,0.025=1.984 (approximately). Therefore, we reject the null hypothesis of no effects from education, experience and sex. Testofthejointsignificanceoftheslopecoefficients:01234:0ββββH ====, versus1:H At least one of the '0,i s β≠ for 1,2,3,4i =.Compute the F-test= (R 2/m )/[(1-R 2)/(n -k )], where m is the number of restrictions (m =4) and k =5 is the total number of regressors. In our case, F=(0.5/4)/[(1-0.5)/95]=23.75 which is higher than F 4,95,0.05= 2.45 (at the 5% level of statistical significance). Hence, reject the null hypothesis.b) The implied relationship between earnings and experience is non-linear.Indeed,12log()2i iiw x x ββ∂=+∂ and222log()2i iw x β∂=∂. Using the estimate2ˆ0.00087β=-, we note that 2ˆ22*(0.00087)0.001740β=-=-<. Therefore, the impact of experience on earnings reaches a maximum. This happens at12log()20i i iw x x ββ∂=+=∂, or 122i x ββ=-. Using the estimates 1ˆ0.038000β=and 2ˆ0.00087β=-, this happens at 21.83 years of experience.c) 012:20i x ββH += versus 112:20i x ββH +≠, when x i =10 years. In this case, 12125,/212122*10020~(2*10)(20)n a t ratio t SE SE ββββββββ-+-+-==++, where the levelof statistical significance is α=5%=0.05.Notice that 12(20)SE ββ+= 2121212(20)()20()2*1*20*(,)Var Var Var Cov ββββββ+=++.Using Var (1ˆβ)=(0.0058)2, Var (2ˆβ)=(0.00016)2 and 12(,)Cov ββ =0, we derive12(20)0.00662SE ββ+=and t -ratio =3.111, which (in absolute value) ishigher than 1.984 (the critical value). Hence, we reject the null hypothesis that experience has no effect on earnings when x i =10 years.2. The Eviews file capm.wf1 contains Belgian monthly financial data over the 1988-1996 period. The dataset includes:rpet: return on the “Petrofina” firm, rm: return on the market, r f : risk-free interest rate.a) Construct the excess return variables exretpet=rpet-r f and exretmar=rm-r f .Plot the two variables together. Calculate the mean and standard deviation of each variable.b) Test the validity of the CAPM model by regressing rpet-r f on rm-r f . Showhow the econometric package derives p-values associated with the estimated coefficients.c) Re-estimate the CAPM model allowing for a January effect generated (inEviews) as jandum=@seas(1). [Note: The financial literature suggests that returns might be higher in January. This is typically the case for small firms and for firms whose price has already declined during the year. Hence, an investor can make money by buying in December and selling in January. The most likely cause of the year-end effect is tax selling; for ordinary investors, the relevant consideration is whether to realise enough of their losses to getsome benefit on their taxes. Capital losses are fully deductible against gains, and in the US, up to $3,000 of losses can also be deducted from ordinary income. This last can save an investor in the 28 percent tax bracket $840 on his tax return in April.]Brief answer to part a):From the above graph, the two excess return variables move very close to each other. This seems to suggest a beta coefficient which is close to 1. To get a summary of statistics for each variable, open the two variables as a group. Then click on the “Descriptive Stats” option:E XRE TPE T E XRE TMARMean -0.003764 0.001300Median -0.005510 -0.002904Maximum 0.165594 0.220758Minimum -0.231116 -0.116153Std. Dev. 0.058685 0.047400Skew ness -0.079797 0.876733Kurtosis 5.096003 7.160984Jarque-Bera 18.04303 83.25273P robability 0.000121 0.000000The excess return on Petrofina has a lower mean return (and a higher volatility) than the excess return on the market portfolio.Brief answer to part b): Check the icon “part_b”.Dependent Variable: EXRETPETMethod: Least SquaresDate: 05/20/04 Time: 10:07Sample: 1988:01 1996:02Included observations: 98Variable Coefficient Std. Error t-Statistic Prob.C -0.004991 0.003863 -1.291955 0.1995R-squared 0.580082 Mean dependent var -0.003764Adjusted R-squared 0.575708 S.D. dependent var 0.058685S.E. of regression 0.038226 A kaike info criterion -3.670404Sum squared resid 0.140278 Schwarz criterion-3.617649Log likelihood 181.8498 F-statistic 132.6162Durbin-Watson stat 1.959643 Prob(F-statistic) 0.000000p-value equals the probability of getting a value of the test at least as extreme as the one observed. Hence,p-value = P(abs(t) > 1.29) = P(t> 1.29) + P(t< -1.29) = 2P(t> 1.29) (due to symmetry). This equals (roughly) 2*0.10 = 0.1995 (roughly), where 0.10 comes from the statistical tables related to t n-2 = t98-2 = t96 = 1.29. This is a two-tailed test. A one-tailed test implies a p-value = 0.1995 / 2.Brief answer to part c):Check the icon “part_c”, where the coefficient on the January dummy variable is equal to -0.002519 with a t-ratio of -0.185159.3. The Eviews file assets.wf1 contains quarterly financial data on:r1: excess return on portfolio 1,rm: excess return on the market,r f: risk-free interest rate,jandum: January dummy,rsize: size effects,rbm: book-to-market equity,rmom: one-year momentum.a)Estimate the CAPM model, that is, regress r1 on a constant term and rm.b)Estimate the CAPM model augmented by the January dummy variable.c)Estimate the five factor model, that is, regress r1 on a constant term, rm, rsize,rbm, rmom and the January dummy.d)Using the model you estimated in part c), test jointly that rsize, rbm and rmomhave no statistical effect on r1. Do this by using the …WALD test‟ option in Eviews.Brief answers to all parts: Check the icons: “part_a”, “part_b”, “part_c” and “part_d_table”.Brief answer to part d):Model with all regressors:Dependent Variable: R1Method: Least SquaresDate: 05/18/00 Time: 16:14Sample: 1963:07 1993:10Included observations: 364Variable Coefficient Std. Error t-Statistic Prob.C 0.004244 0.001323 3.207312 0.0015RM 1.070355 0.030284 35.34394 0.0000RSIZE 1.264048 0.045677 27.67385 0.0000RBM 0.636377 0.051550 12.34487 0.0000RMOM -0.237296 0.037799 -6.277853 0.0000JANDUM 0.018590 0.004915 3.782270 0.0002R-squared 0.907951 Mean dependent var 0.015091Adjusted R-squared 0.906666 S.D. dependent var 0.073042S.E. of regression 0.022315 A kaike info criterion -4.750778Sum squared resid 0.178268 Schwarz criterion-4.686539Log likelihood 870.6416 F-statistic 706.2486Wald Test:Chi-s quare 1073.943 3 0.0000Null Hypothesis Summary:Normalized Restriction (= 0) Value Std. Err.C(3) 1.264048 0.045677C(4) 0.636377 0.051550C(5) -0.237296 0.037799Restrictions are linear in coefficients.Manual derivation of the WALD test:The unrestricted five factor model has RSS U=0.178268. The restricted model in part b) has RSS R=0.713043. Hence, to test the m = 3 restrictions, calculateF = [(RSS R- RSS U)/m] / [RSS U/(n-k)] = 357.98 > F(3,358) = 2.60 (at 5%). Hence, reject the null.4. The Eviews file capm.wf1 contains Belgian monthly financial data over the 1988-1996 period. The dataset includes:rpet: return on the “Petrofina” firm,rm: return on the market,r f: risk-free interest rate.a) Construct the excess return variables rpet-r f and rm-r f. Estimate the CAPM equation of rpet-r f on rm-r f over the whole sample period. Test for autororrelation of order 12, heteroskedasticity, and normality. Test for parameter instability by splitting the sample in 1992:02. Test for functional form misspecification using Ramsey‟s RESET test. What do you infer from these tests?b) The relationship between Petrofina‟s return and the market return might be nonlinear. For instance, it might be the case that Petrofina‟s return increases more when the market does very well. This might be the case when investors see investment opportunities in Petrofina‟s stock in periods where the market rallies. Test this by including as an additional regressor (rm-r f)2. A positive and statistical significant coefficient on (rm-r f)2would then confirm the nonlinear and positive relationship between Petrofina‟s return and the market return.Answer:a) Autocorrelation (serial correlation) test of order 12: The model does not fail the test.Breusch-Godfrey Serial Correlation LM Test:F-statistic 0.690664 Prob. F(12,84) 0.7560Heteroskedasticity test: The model fails the test at 5% (but not at 1%) since the p-value is between 0.01 and 0.05.Heteroskedasticity Test: WhiteF-statistic 3.448676 Prob. F(2,95) 0.0358Normality test: The model fails badly the test as the p-value is equal to 0.Parameter stability test: In Eviews, choose: Stability tests, then Chow Forecast test and enter 1992:02. The model does not fail the test as the p-value=0.1125.Chow Forecast Test: Forecast from 1992M02 to 1996M02F-statistic 1.424959 Prob. F(49,47) 0.1125Ramsey‟s RESET test: The test does not reject the null hypothesis of correct functional form at 5% as the p-value is equal to 0.0542 (but rejects at 10%).Ramsey RESET Test:Value df ProbabilityF-statistic 3.798497 (1, 95) 0.0542Unrestricted Test Equation:Dependent Variable: EXRETPETMethod: Least SquaresDate: 11/11/11 Time: 15:07Sample: 1988M01 1996M02Included observations: 98Variable Coefficient Std. Error t-Statistic Prob.C -0.001835 0.004138 -0.443593 0.6583EXRETMAR0.990881 0.084377 11.74345 0.0000FITTED^2 -1.615607 0.828953 -1.948973 0.0542Therefore, it appears the model is well-specified as it passes all tests except normality. Notice also the model has an adjusted R2=0.575. This means that57.5% of the variation in Petrofina‟s return is explained by variations in themarket return in excess of the risk-free interest rate.b) Check the icon “part_b_exercise_4”. The relationship is not linear at 5% as EXRETMAR^2 has a p-value of 0.054. However it is statistically significant at 10% (as the p-value is less than 10%). This result is in accordance with Ramsey‟s RESET test. The negative sign on EXRETMAR^2 implies that the impact of the market return on Petrofina‟s return increases at a decreasing rate. Indeed, from: EXRETPET=1*EXRETMAR-1.43*EXRETMAR^2, we get that the maximum impact of the market on Petrofina‟s return occurs by setting the derivative d(EXRETPET)/d(EXRETMAR) equal to zero and solving for EXRETMAR. The derivative is equal to d(EXRETPET)/d(EXRETMAR)=1-2*1.43*EXRETMAR. This is equal to zero when EXRETMAR=1/(2*1.43) or EXRETMAR =0.35, that is, when the market return exceeds the risk-free interest rate by 35 basis points.Dependent Variable: EXRETPETMethod: Least SquaresSample: 1988M01 1996M02Included observations: 98Variable Coefficient Std. Error t-Statistic Prob.C -0.001876 0.004130 -0.454201 0.6507EXRETMAR 1.006087 0.086972 11.56796 0.0000EXRETMAR^2-1.436547 0.737079 -1.948973 0.0542R-squared 0.596227 Mean dependent var -0.003764Adjusted R-squared 0.587726 S.D. dependent var 0.058685S.E. of regression 0.037681 Akaike info criterion -3.689201Sum squared resid 0.134885 Schwarz criterion -3.610069Log likelihood 183.7708 Hannan-Quinn criter. -3.657194F-statistic 70.14031 Durbin-Watson stat 1.829615Prob(F-statistic) 0.000000。