A Complete Specification of the Asymptotic Variance - Assignment Example

 A Complete Specification of the Asymptotic Variance 1) Continuation of question 1, Assignment 1. In answering question 1 of Assignment 1, you showed that the maximum likelihood estimator of the parameter λ of the Poisson distribution is the sample mean, . Using of Theorem 14.1 of Greene (2012), find the asymptotic distribution of this estimator. (Your answer should include a complete specification of the asymptotic variance of the estimator.) SOLUTION Fitting a Poisson distribution (misspecified case) The variance of the MLE  can be approximated by Using the formulas for the first and second derivative, we find that    And  Where we used that Hence the asymptotic variance of  becomes 2) For the Chow test, the unrestricted model is a) Show that the OLS estimator of this model is SOLUTION For the unrestricted model Where  denotes the column vector of  residuals. That is,  and the sum of the squared residuals is Applying the LS principle The LS principle is to choose  to minimize the residual sum of squares  Now to find the value of  which minimizes the sum of squared residuals we differentiate Equating to zero yields Thus  But . Hence shown as required. b) Show that the sum of squared errors of the unrestricted model is Where  is the vector of residuals from the regression of  on  and  is the vector of residuals from the regression of  on . SOLUTION As earlier mentioned for the unrestricted model Where  denotes the column vector of  residuals. That is,  and the sum of the squared residuals is 3. Continuation of question 4, Assignment 1. The le slid2010HKontD.dta is a Stata data le that that contains data retrieved from the 2010 Survey of Labour and Income Dynamics (SLID). The le contains 6,808 observations for individuals living in Ontario. a) As you did for Assignment 1, construct the variable lhwage, the natural log of hwage. In addition, create the variable exp2, the square of exp. SOLUTION Done in the STATA file b) Estimate two human capital earnings models, one for men and one for women. Use lhwage as the dependent variable and include exp, exp2, educ, notgrad, hsgrad somepse, uni1, uni2 and marr as independent variables. Discuss the goodness of the of the two equations. SOLUTION Three statistics are used in Ordinary Least Squares (OLS) regression to evaluate model fit: R-squared, the overall F-test, and the Root Mean Square Error (RMSE).  i) Using F-test Equation one-Female From the regression table we observe that the F-computed is 257.6> 1.88260439 (F-critical), we thus reject the null hypothesis that all regression coefficients are equal to zero. This shows that F-test is significant indicating that the observed R-squared is reliable, and is not a spurious result of oddities in the data set. Also it shows that the proposed relationship between the response variable and the set of predictors is statistically reliable, and can be useful when the research objective is either prediction or explanation. Equation two-male From the regression table we observe that the F-computed is 356.47> 1.88266598 (F-critical), we thus reject the null hypothesis that all regression coefficients are equal to zero. This shows that F-test is significant indicating that the observed R-squared is reliable, and is not a spurious result of oddities in the data set. Also it shows that the proposed relationship between the response variable and the set of predictors is statistically reliable, and can be useful when the research objective is either prediction or explanation. Table 1: Regression output table for the female Table 2: Regression output table for the male ii) Using R-squared and Adjusted R-squared Equation one-Female The value of R-Squared is given as 0.4032, this implies that about 40.32% variation in the dependent variable (lhwage) is explained by the independent variables in the model. Equation two-male The value of R-Squared is given as 0.4887, this implies that about 48.87% variation in the dependent variable (lhwage) is explained by the independent variables in the model. iii) Using Root MSE Equation one-Female The value of Root MSE is given as 0.39466, which is very small this indicates better fit of the model. Equation two-male The value of Root MSE is given as 0.3892, which is very small this indicates better fit of the model. c) Interpret the results of the test of overall significance for each equation. Do not forget to state the decision rule for the test, the level of significance you are using, and the critical value of the test statistic. SOLUTION Equation one-Female The p-value for the overall model is 0.0000, a value less than 5%, we thus reject the null hypothesis. The null hypothesis states that the coefficient is equal to zero (no effect). We therefore conclude that the model is appropriate and that there is significant effect on the dependent variable (lhwage) by the independent variables. The independent variables reliably predict the dependent variable Equation two-male Similar to the first equation, the p-value for the overall model is 0.0000, a value less than 5%, we thus reject the null hypothesis. The null hypothesis states that the coefficient is equal to zero (no effect). We therefore conclude that the model is appropriate and that there is significant effect on the dependent variable (lhwage) by the independent variables.  The independent variables reliably predict the dependent variable d) Interpret the results of individual tests of significance for the coefficients of exp, exp2, and educ for both men and women. Do not forget to state the decision rule for the test, the level of significance you are using, and the critical value of the test statistic. SOLUTION Equation one-Female The t value and 2 tailed p-value are used in testing the null hypothesis that the coefficient/parameter is 0. We compare each p-value to the value of alpha (5%).  Coefficients having p-values less than alpha are significant.  The coefficient for exp is significantly different from 0 using alpha of 0.05 because its p-value of 0.000 is smaller than 0.05.  The coefficient for exp2 (-0.000541) is significantly different from 0 because its p-value (0.000) is definitely smaller than 0.05 and even 0.01. The coefficient for educ is significantly different from 0 using alpha of 0.05 because its p-value of 0.000 is smaller than 0.05. Equation two-male Similar to the first equation for the female, the t value and 2 tailed p-value are used in testing the null hypothesis that the coefficient/parameter is 0. We compare each p-value to the value of alpha (5%).  Coefficients having p-values less than alpha are significant.  The coefficient for exp is significantly different from 0 using alpha of 0.05 because its p-value of 0.000 is smaller than 0.05.  The coefficient for exp2 (-0.0007905) is significantly different from 0 because its p-value (0.000) is definitely smaller than 0.05 and even 0.01. The coefficient for educ is significantly different from 0 using alpha of 0.05 because its p-value of 0.000 is smaller than 0.05. e) Carry out a hypothesis test using the test command that will help you answer this question. SOLUTION Looking at the table below, we observe that the p-value is 0.000 Read More