Today:
- Review
Next Time:
- Final: Monday, March 19th at 8:00 a.m.
Wednesday, March 14, 2012
Tuesday, March 13, 2012
Topics for Final
We covered the following topics in the course:
1. Assumptions required for estimates to be BLUE
2. Hypothesis testing:
a. t- tests (both one-sided and two-sided)
b. Joint hypotheses (F-Tests, Chi-square tests, etc.)i. Exclusion restrictions
ii. Linear combinations of parameters
3. Heteroskedasticity
a. What is heteroskedasticity?
b. How heteroskedasticity occur?
c. The consequences of estimating a heteroskedastic model with OLS
d. Testsi. LaGrange Multiplier Tests (Models 1, 2, and 3)
Model 1:
Model 2:
Model 3:ii. Goldfeld-Quandt test
iii. White's teste. Corrections/Estimation procedures
i. Feasible GLS
Model 1:
Model 2:
Model 3:iii. White’s correction
8. Autocorrelation
a. What is it and why might it occur?
b. Consequences of ignoring serial correlation and estimating with OLSi. Model with serially correlated errors (model 1 in class)
ii. Model including a lagged dependent variable (model 2 in class)
iii. Model with both a lagged dependent variable and serial correlated errors (model 3 in class)c. Tests for serial correlation
i. The Durbin-Watson test.
ii. Durbin's h-test.
iii. The Breusch-Godfrey test for higher order serial correlation.d. Corrections
i. Non-linear estimation
9. Testing for ARCH errors
10. Stochastic Regressors and Measurement Errors
a. Assessing the bias and consistency of an estimator
b. Errors in variablesi. Consequences of estimating with OLS when there are errors in measuring the right-hand side variables (i.e. errors in measuring the independent variables, the X's).
ii. Consequences of estimating with OLS when there are errors in measuring the dependent variable (i.e. in the measurement of Y).
iii. Application of errors in variables: Friedman's Permanent Income Hypothesis.c. Instrumental variable estimation
i. What is an instrument.
ii. How is IV performed?
iii. Show how IV estimation can solve the problem of correlation of the right-hand side variables with the error term.
11. Simultaneous equation models
a. Structural equations (behavioral, identities, equilibrium conditions, technical equations) and reduced form equations. Endogenous, exogenous, and predetermined variables.
b. Consequences of ignoring simultaneity, i.e. demonstrate simultaneity bias.
c. Underidentified models, exactly identified models, and overidentified models
d. Estimation by 2SLS
12. Multicollinearity
a. What is multicollinearity and how does it affect OLS estimates and standard errors?
b. Detection of multicollinearity
c. What to do for perfect and imperfect multicollinearity.
13. Specification tests
a. LM test for adding a variable to a model
b. LM test for adding a variable in a system of equations
c. LM test for serial correlation in a system of equations
d. AIC and SIC criterion
14. Qualitative and limited dependent variables
a. Linear probability model
i. description of model, problems, and estimation
b. Probit model
i. description of model and estimation
c. Logit model
i. description of model, attractive properties, and estimation
d. Limited dependent variables
i. description of the model when the dependent variable is limited, problems with OLS, and estimation
15. Maximum likelihood
a. Brief description of what maximum likelihood estimation does.
b. properties of maximum likelihood estimators.
Solution to Homework 8
1. [Practice problem] (a) After the data have been read in:
Do OLS on
COt = β0 + β1YDt + β2COt-1 + ε1t
by running
with the result
Next, do OLS on
It = β3 + β4Yt + β5rt-1 + ε2t
by running
with the result
(b) Use 2SLS to estimate
COt = β0 + β1YDt + β2COt-1 + ε1t
by running
with the result
Next, use 2SLS to estimate
It = β3 + β4Yt + β5rt-1 + ε2t
by running
with the result
Monday, March 12, 2012
Thursday, March 08, 2012
Wednesday, March 07, 2012
Monday, March 05, 2012
Friday, March 02, 2012
Wednesday, February 29, 2012
Monday, February 27, 2012
Friday, February 24, 2012
Wednesday, February 22, 2012
Monday, February 20, 2012
Wednesday, February 15, 2012
Tuesday, February 14, 2012
Thursday, February 09, 2012
Solution to Homework 4
Economics 421/521
Winter 2012
Solution to Homework #4
1. Perform a Durbin-Watson test at the 5% level of significance for positive first-order autocorrelation using the following regression output (standard errors in parentheses):
Yt = 2.0 + 3.7*X1t - 4.4*X2t, T = 42
(.7) (1.1) (2.8) DW = 1.22
At the 5% level of significance, the critical value of the DW statistic is (approximately, with interpolation) 1.41 at the lower end and 1.61 at the upper end. Since 1.22 is smaller than dL = 1.41, we reject that ρ=0.
2. Recall the model from homework 1:
Given data on M2, real GDP, and the T-bill rate, estimate the following regression...:
Mt = β0 + β1RGDPt + β2Tbillt + et
Don't be surprised if the fit is very good - we'll explain why that may be misleading later in the course.
Here's the output:
Does model suffer from serial correlation? Use a Durbin-Watson test to answer the question.
Yes, definitely. The value of dL (5%) is, approximately, 1.63. The test statistic of .02891 is far below this value, so we reject that ρ=0.
Is the fit as good as the R2 and t-statistics indicate?
The t-statistics are biased upward so the fit is not nearly as good as the t-statistics might lead you to believe (which would be evident if we corrected for it). This is due to biased estimates of the residuals. So the the R2 and t-statistics give a misleading picture of how well the model fits the data.
3. Regress the change in the log of real consumption (C) on the change in the log of real disposable income (DI) and test for serial correlation using a Durbin-Watson test. The data are here (the data are quarterly, and span the time period 1947:Q1 - 2007:Q3).
The first step is to log both consumption and disposable income, then difference. The transformed variables are named dlogc and dlogy below (e.g. dlogy = log(di)-log(di(-1)). Then, regress dlogc on a constant and dlogy:
Here is the regression output:
The critical value of the Durbin-Watson statistic is approximately dL = 1.65 and dU = 1.69 (there are tables that go beyond T=100, but the value doesn't change much after 100 observations. I used the value for 100 observations from the table in the text). For a test of negative serial correlation the values are dL = 4.00-1.65 = 2.35 and dU = 4.00-1.69 = 2.31. The null of no positive serial correlation would not be rejected since the test statistic, 2.28, is above 1.65. the rejection point. Similarly, a null of no negative serial correlation cannot be rejected since 2.28 is less than 2.31, but it's a close call.
4. Explain why the Durbin-Watson statistic is always between 0 and 4. Also explain why the Durbin-Watson statistic is between 0 and 2 when there is positive serial correlation, between 2 and 4 when there is negative serial correlation, and equal to 2 when there is no correlation at all.
Start with the demonstration that d, the Durbin-Watson statistic, is approximately (2-2ρ) in large samples:
[Click on figure for larger version]
Now, since ρ must lie between -1 and 1, the Durbin-Watson statistic must lie between 0 and 4 [since 2-2(1)=0 and 2-2(-1)=4].
To see the last part, note that when ρ=0, d=2. Then, as ρ increases from 0 to 1, d moves from 2 to 0, and as ρ moves from 0 to -1, d moves from 2 to 4.
5. Continuing with the model we used in problem 2, test for the presence of fourth order serial correlation.
To do this problem, first regress m2 on a constant, rgdp, and tbillrate:
[Click on figure for larger version]Save the residuals (I saved the resid series as uhat). Regress the estimated residual on four lags of the estimated residual and the other variables in the model (no constant), i.e. regress uhatt on uhatt-1, uhatt-2, uhatt-3, uhatt-4, tbillrate, and rgdp:
[Click on figure for larger version]Finally, form the test statistic (T-P)R2, where T is the number of observations and P is the number of lags. This is distributed χ2(4). The critical value for this test at 5% level of significance is 9.49.
The test statistic is (191)(.972914) = 185.83, so reject that there is no serial correlation in the model.
6. Continuing with the model we used in problem 3, use the AR(1) procedure in EViews to correct the model for the presence of first-order serial correlation.
First, note that the test above did not indicate the presence of serial correlation, so technically there is no need for a correction. However, it's still worthwhile to do as a practice exercise.
To do this problem, just add the ar(1) command to the estimation shown in problem 3:
With the result:
Wednesday, February 08, 2012
Topics for Midterm
[Note: Previous midterms are here.]
1. Assumptions required for estimates to be BLUE
2. Hypothesis testing:
a. t- tests (both one-sided and two-sided)
b. Joint hypotheses (F-Tests, Chi-square tests, etc.)i. Exclusion restrictions
ii. Linear combinations of parameters
3. Heteroskedasticity
a. What is heteroskedasticity?
b. How heteroskedasticity occur?
c. The consequences of estimating a heteroskedastic model with OLS
d. Testsi. LaGrange Multiplier Tests (Models 1, 2, and 3)
Model 1:
Model 2:
Model 3:ii. Goldfeld-Quandt test
iii. White's teste. Corrections/Estimation procedures
i. Feasible GLS
Model 1:
Model 2:
Model 3:iii. White’s correction
8. Autocorrelation
a. What is it and why might it occur?
b. Consequences of ignoring serial correlation and estimating with OLSi. Model with serially correlated errors (model 1 in class)
ii. Model including a lagged dependent variable (model 2 in class)
iii. Model with both a lagged dependent variable and serial correlated errors (model 3 in class)c. Tests for serial correlation
i. The Durbin-Watson test.
ii. Durbin's h-test.
iii. The Breusch-Godfrey test for higher order serial correlation.d. Corrections
i. Non-linear estimation
9. Testing for ARCH errors
10. Stochastic Regressors and Measurement Errors
a. Assessing the bias and consistency of an estimator
b. Errors in variablesi. Consequences of estimating with OLS when there are errors in measuring the right-hand side variables (i.e. errors in measuring the independent variables, the X's).
ii. Consequences of estimating with OLS when there are errors in measuring the dependent variable (i.e. in the measurement of Y).
Monday, February 06, 2012
Sunday, February 05, 2012
Solution to Homework 3
Economics 421/521
Winter 2012
Solution to Homework 3
1. Using this data set, repeat the example from class for the first of the three cases we discussed, i.e. first regress the log of salary on a constant and the two variables proxying for experience, years and years2:
log(salary) = β0 + β1*years + β2*years2 + ut
Then, form the estimated residual squared (resid2) and perform the LM test for heteroskedasticity (note: resid is the estimated value of ut).
To do this problem, first read the data in from the Excel spread sheet:
Then, transform the data to get the log of salary and years squared:
Regress the log of salary on a constant, years, and years squared:
To do the test, we need the square of the residuals from this regression:
Regress the squared residuals on a constant, years, and years squared:
The results are:
The test statistic is
NR2 = 222*.0747 = 16.59
This is distributes Chi-Square with 2 degrees of freedom, so the critical value (5%) is 5.99. Therefore, the null of no heteroskedasticity is rejected.
2. To do this problem, first regress the log of salary on a constant, years, and years squared:
Next, square the residuals to get an estimate of the variance for each observation, uhatsq:
Regress this on a constant, years, and years squared (this is the model of the variance):
Next, we need the predicted value of the variance. To get the predicted value, we can simply subtract the estimated residuals from the left-hand side variables (this uses that actual Y = predicted Y + predicted error):
Or, you can get exactly the same values by using the forecast button on the regression output:
The next step is to ensure that all the estimates of the variance are positive. If any are negative, they should be replaced by their absolute values:
We need the square root of this value to use in transforming the original data:
Use this value to transform the data:
Use the transformed values to obtain BLUE estimates
The result is:
3. The test statistic is N*R2. In this case, N=30 and R2=.9878 so that the test statistic is 29.624. The 5% critical value for this test is 11.07 (the test is Chi-Square with 5 degrees of freedom, 5 because if the error is homoskedastic, then the coefficients on G, Y, their squared values, and their cross-product must all be zero). Because the test statistic exceeds the critical value, the null of homoskedasticity is rejected.
4. Here is the uncorrected regression from the last homework:
To do White's test, click the view button, then as follows:
Here's the result:
Looking at the line with Obs*R-squared, we see that the probability is les than .05, the significance level, hence the null of no heteroskedasticity is rejected.
To correct the model, run a regression as usual:
Click on options to bring up this screen and click the boxes as shown:
Hit OK to return to this screen:
Hit OK to get this output:
Notice that the coefficient estimates are identical (compare to the original values given above), White's correction fixes the standard errors after the regression is estimated, but it doesn't change the estimates.
5. What are the consequences of estimating an autoregressive model using OLS?
The coefficients remain unbiased, but OLS is inefficient, and OLS results in biased estimates of the standard errors (so the test statistics, e.g. t's and F's, are wrong)
Wednesday, February 01, 2012
Monday, January 30, 2012
Solution to Homeworks 1 and 2
Economics 421/521
Winter 2012
Solution to Homework #1
Part I. Hypothesis Testing
1. Suppose that you estimate a model of house prices to determine the impact of having beach frontage on the value of a house. You do some research, and you decide to use the size of the lot instead of the size of the house for a number of theoretical and data availability reasons. Your results (standard errors in parentheses) are:
PRICEi = 40 + 35.0 LOTi – 2.0 AGEi + 10.0 BEDi – 4.0FIREi + 100 BEACHi
(29) (5.0) (1.1) (10.0) (3.0) (9.0)
n = 30, R2 = .63
where,
PRICEi = the price of the ith house (in thousands of dollars)
LOTi = the size of the lot of the ith house (in thousands of square feet)
AGEi = the age of the ith house in years
BEDi = the number of bedrooms in the ith house
FIREi = a dummy variable for a fireplace (1 = yes for the ith house)
BEACHi = a dummy for having beach frontage (1 = yes for the ith house)
a) You expect the variables LOT, BED, and BEACH to have positive coefficients. Create and test the appropriate hypotheses to evaluate these expectations at the 5 percent level.
For LOT;
Ho: βLOT = 0
Ha: βLOT > 0t – score: (35.0) / (5.0) = 7.0
t-critical: 1.711 because d.f. is 24 and 5% level of significance.
Since 7.0 > 1.711, we can reject the null hypothesis that the true coefficient of LOT is not positive.
For BED;
Ho: βBED = 0
Ha: βBED > 0t – score: (10.0) / (10.0) = 1.0
t-critical: 1.711 because d.f. is 24 and 5% level of significance.
Since 1.0 < 1.711, we cannot reject the null hypothesis that the true coefficient of BED is not positive.
For BEACH;
Ho: βBEACH = 0
Ha: βBEACH > 0t – score: (100) / (0.9) = 11.1
t-critical: 1.711 because d.f. is 24 and 5% level of significance.
Since 10.0 > 1.711, we can reject the null hypothesis that the true coefficient of BEACH is not positive.
b) You expect AGE to have a negative coefficient. Create and test the appropriate hypothesis to evaluate these expectations at the 10 percent level.
For AGE;
Ho: βAGE = 0
Ha: βAGE < 0t – score: ( - 2.0) / (1.1) = - 1.81
t-critical: 1.318 because d.f. is 24 and 10% level of significance.
Since | - 1.81 | > |1.318|, we can reject the null hypothesis that the true coefficient of AGE is not negative.
c) At first you expect FIRE to have a positive coefficient, but one of your friends says that fireplaces are messy and are a pain to keep clean, so you are not sure. Run a two-sided t-test around zero to test these expectations at the 5 percent level.
For FIRE;
Ho: βFIRE = 0
Ha: βFIRE ≠ 0t – score: ( - 4.0 ) / (3.0) = - 1.3
t-critical:: 2.064 because d.f. is 24 and 5% level of significance.
Since | - 1.3 | < 2.064, we cannot reject the null hypothesis that the true coefficient of FIRE is not different from zero.
2. Consider the following regression:
log(Qci) = 921.6 – 1.3log(Pci) + 0.7log(Pai) + 11.4log(Inci)
(121) (0.3) (0.05) (2.8)
n = 30, R2 = 0.82
where,
Qci = the total sales of CAMRY in the ith city, the year of 2003
Pci = the price of a CAMRY in the ith city, the year of 2003 (in thousands)
Pai = the price of a ACCORD in the ith city, the year of 2003 (in thousands)
Inci = the average income in the ith city, the year of 2003 (in thousands)
Numbers in the parentheses are standard errors.
a) What does the constant term (= 921.6) mean?
The constant term means is the value of the dependent variable when all the independent variables are zero.
b) How would you interpret the coefficient on log(Pci). Be explicit and explain, in terms of economic theory, the importance of its magnitude.
Totally differentiate the estimated equation w.r.t. Qci and Pci to obtain the elasticity:
Elasticity. |-1.3| > 1, so elastic. If the price of CAMRY increases by 1%, then the total sales of CAMRY decreases by 1.3%. (Thus, the coefficient gives the elasticity.)
c) Get t-values of each of all coefficients in the regression. Are all of our coefficients statistically significant at the 5% level of significance? How about at the 1% level of significance?
Pci Pai Inci
t-score -4.3 14. 4.08Two-tailed test at the 5% level of significance, t-critical = 2.056 because d.f. is 26.
Two-tailed test at the 1% level of significance, t-critical = 2.779 because d.f. is 26.Therefore, all of our coefficients statistically significant at the both 1% and 5% level of significances.
d) Interpret R2. What does it mean?
R2 tells how well the sample regression line fits the data.
(TSS = ESS + RSS)
Thus, it is the ratio of the explained variation in the dependent variable divided by the total variation and hence measures the percentage of the total variation that is explained by the variable sin the model.
Part II. Short Answer
1. State the Gauss-Markov Theorem and explain the term BLUE.
Given the classical assumptions (model is linear and correctly specified, X's are exogenous, no perfect multicollinearity, error has zero mean, homoskedasticity, errors are independent, errors and X's are uncorrelated, errors normally distributed), the OLS estimator is the minimum variance estimator from among the set of all linear unbiased estimators. OLS is BLUE. This means it is the Best (minimum variance) Linear Unbiased Estimator.
Part III. Estimation
1. Given data on M2, real GDP, and the T-bill rate, estimate the following regression and test whether the coefficients differ from zero. Do the coefficients have the expected signs?:
Mt = β0 + β1RGDPt + β2Tbillt + et
To do this problem, first create a new workfile by following these steps (some of the figures are pop-ups that bring up larger, clearer versions):
Next, read in the data set (save it to your computer first by right clicking on the link in the homework set):
Finally, run the regression of M2 on a constant, RGDP and the TBillRate:
Running this regression gives:
The critical value at the 1% level is 2.576, thus the coefficients are significant. They also have the expected sign: People tend to demand more money as income (RGDP) increases. They also tend to demand less money as the T-bill rate increases – this is because as the T-bill rate goes up, the opportunity cost to holding money also goes up.
Economics 421/521
Winter 2012
Solution to homework #2
1. Using the EAEF data set, regress LGEARN on S, EXP, and ASVABC. Use F-tests to determine whether the coefficients on S and EXP are (a) jointly significant, and (b) equal. [Parts (a) and (b) are two separate tests.]
First, read in the data:
Then, take the log of earnings:
Next, estimate the unrestricted model This is needed for both parts (a) and (b). To estimate the UR model, regress lnearnings on a constant, exper, and asvabc:
The results are:
(a) The restricted model for the null hypothesis that the coefficients on s and exper are zero is:
F= [(149.31-130.35)/2]/[130.35/(540-4) =38.98
The critical value for this test is F(2,536)= 3.05 (approx. for 536), so reject that both coefficients are zero.
(b) For this part, the null hypothesis is that the coefficients on s and exper are equal. To impose this on the model, start with:
lnsalary = b1 + b2*s + b3*exper + b4*asvabc + u
Then impose that b2=b3:
lnsalary = b1 + b2*s + b2*exper + b4*asvabc + u
Group terms:
lnsalary = b1 + b2*(s + exper) + b4*asvabc + u
This is the model we need to estimate. Thus, the first step is to obtain data on the sum of s and exper:
Run the restricted regression:
Here are the results:
Finally, use these to calculate the F-statistic:
F= [(138.29-130.35)/1]/[130.35/(540-4) = 32.65
The critical value for this test is F(1,536)= 3.90 (approx.), so reject that the coefficients are equal.
2. Problem 7.1 in the text.
In this case the F statistic is simply the ratio of the residual sum of squares (more generally it is the rss divided by n-k, i.e. F = [rss2/(n2-k)]/ [rss1/(n1-k)], where n1 and n2 are the number of observations in each sample, but when n1=n2 the terms cancel):
F=28,101/321 = 87.54
The critical value (5%) for and F(8,8) = 3.44, so the null of homoskedasticity is rejected.
3. Moved to the next homework.
Friday, January 27, 2012
Wednesday, January 25, 2012
LM Tests for Heteroskedasticity
Here is an outline of the LM tests for Heteroskedasticity:
Monday, January 23, 2012
Thursday, January 19, 2012
Monday, January 16, 2012
Wednesday, January 11, 2012
Monday, January 09, 2012
Economics 421
Mark Thoma
Department of Economics
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