Problem Set 9 Heteroskedasticty Answers

Transcription

1 Problem Set 9 Heteroskedasticty Answers /* INVESTIGATION OF HETEROSKEDASTICITY */ First graph data. u hetdat2. gra manuf gdp, s([country].) xlab ylab manufacturing output (US$ miilio korea switzerl mexico belgium netherla turkey sweden ireland singapor malaysia finland denmark chile portugal slovenia hungary israel greece hong norway kon slovakia kuwait syria spain canada e e+06 gdp (US $ million) uk italy france Data are tightly packed with exception of a few outliers (Italy, UK and France) Sometimes outliers can lead to heteroskedasticity. (Similar levels of GDP, big variation in level of manufacturing output).. reg manuf gdp Source SS df MS Number of obs = F( 1, 26) = Model e e+11 Prob > F = Residual e R-squared = Adj R-squared = Total e e+09 Root MSE = manuf Coef. Std. Err. t P> t [95% Conf. Interval] gdp _cons As a first check for heteroskedasticity, inspect pattern of residuals. If H.is present should see residual variance increasing (or decreasing) with value of gdp.. predict res, resid /* this command gets residuals from previous regression & saves them with the name res */

2 . gra res gdp, s([country]) yline(0) xlab ylab korea uk Residuals 0 ireland singapor malaysia slovenia chile finland turkey switzerl slovakia kuwait hungary portugal belgium israel syria denmark sweden greece norway hong kon netherla spain canada france mexico italy e e+06 gdp (US $ million) Graph is a little unclear. There is an increase in the residual spread with GDP, but with the exception of France. Need a more formal test to get round the ambiguity. 1) Goldfeld-Quandt Test. sort gdp /* sort data in ascending order */ Omit middle c observations (approx. 20% of total sample. In this example c=4) So N-c/2 = 28-4/2 = 12. reg manuf gdp if _n<=12 /* regression on first 12 observations. These should have smallest residual variance, if residual variance increases with level of GDP. Source SS df MS Number of obs = F( 1, 10) = Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE = manuf Coef. Std. Err. t P> t [95% Conf. Interval] gdp _cons Now repeat regression for top 12 in the data set

3 . reg manuf gdp if _n>=17 Source SS df MS Number of obs = F( 1, 10) = Model e e+10 Prob > F = Residual e e+09 R-squared = Adj R-squared = Total e e+09 Root MSE = manuf Coef. Std. Err. t P> t [95% Conf. Interval] gdp _cons Goldfeld-Quandt Test = RSShigh/RSSlow ~ F(N-c-2k/2, N-c-2k/2) Where N= full sample size C= number observations dropped from middle K = number parameters in estimated equation = 1.358E10/1.57E8 ~ F(28-4-2(2)/2, (2)/2) = 86.5 ~ F(11,11) From tables, critical value at 5% level for F(11,11) = 2.82 So estimated F > Fcritical So reject null that data are homoskedastic. 2) Breusch-Pagan Test Idea is to regress a proxy for the residual variance on the variables thought to be causing the Het. problem, (or if not sure which variables, include all of them in this auxiliary regression). Hence this is a more general test for Het. than Goldfeld-Quandt test. Since residual variance s 2 = RSS/N-k = Σ u /N-k, then square of estimated ^ 2 residual for each observation, u, is a good proxy for the residual variance. ^ 2 Hence a regression of u i = d 0 + d 1 GDP i + e I This will suggest Het. if the coeffcicient on GDP in this auxiliary regression is significantly different from zero (residual variance correlated with rhs variable).. predict reshat, resid /* so 1 st save residuals */. g res2=reshat^2 /* square them */ Then regress square of residuals on all rhs variables from original regression (in this case just GDP). 2 ^

4 . reg res2 gdp Source SS df MS Number of obs = F( 1, 26) = 3.12 Model e e+18 Prob > F = Residual e e+18 R-squared = Adj R-squared = Total e e+18 Root MSE = 1.5e+09 res2 Coef. Std. Err. t P> t [95% Conf. Interval] gdp _cons 1.17e e e e+08 Breusch-Pagan Test is N*R 2 auxiliary = 28*0.107 = 3.00 This has a Chi-squared distribution with degrees of freedom equal to the number of right hand side variables in the auxiliary regression excluding the constant, (in this case 1). From tables, critical value at 5% level is 3.84 So estimated χ 2 < χ 2 critical, so accept null that residuals are not heteroskedastic Note that an equivalent test is to take the F test of goodness of fit of the model as a whole from the auxiliary regression)

5 If the exact form of H. is not known (and this is likely to be the case for most estimations. H. will vary across more than one variable -so these tests and methods above are not valid and it is unlikely that can say with certainty what the true functional form of H. looks like) In this case use White Adjusted standard errors (see lecture notes), which fix up the biased OLS estimates. (Note that these are not the same as if you knew the true form of H. but they are better than the unadjusted OLS ones). Original regression. reg manuf gdp Source SS df MS Number of obs = F( 1, 26) = Model e e+11 Prob > F = Residual e R-squared = Adj R-squared = Total e e+09 Root MSE = manuf Coef. Std. Err. t P> t [95% Conf. Interval] gdp _cons Now with adjusted standard errors (you should know what the adjustment does see lecture notes).. reg manuf gdp, robust Regression with robust standard errors Number of obs = 28 F( 1, 26) = Prob > F = R-squared = Root MSE = Robust manuf Coef. Std. Err. t P> t [95% Conf. Interval] gdp _cons consistent standard error is now larger (.108 compared with.013) and hence the t value is lower and the confidence interval larger than in the unadjusted regression. The reason you might worry is that this adjustment is only valid asymptotically (as the sample size gets very large). With only 28 observations this is far from being valid. So the adjusted standard errors and t stats. may be as wrong as the original OLS ones.

6 2. Sometimes heteroskedasticty exists within sub-samples of your data amongst variables not included in your regression. In this question you test whether the residual variances differ across the male and female subsamples of the data. In this case there are equal numbers of men and women in the data set so ok to use Goldfeld-Quandt test */ 1 st do Breusch-Pagan test again /* read in data set hetwage.dta */. u hetwage. reg logpay exper exper2 Source SS df MS Number of obs = F( 2, 967) = Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE = logpay Coef. Std. Err. t P> t [95% Conf. Interval] exper exper _cons predict res, resid /* save residuals and square them */. g res2=res^2. reg res2 exper exper2 Source SS df MS Number of obs = F( 2, 967) = 1.63 Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE = res2 Coef. Std. Err. t P> t [95% Conf. Interval] exper exper _cons Breusch-Pagan Test is N*R 2 auxiliary = 970* = 3.20 This has a Chi-squared distribution with degrees of freedom equal to the number of right hand side variables in the auxiliary regression excluding the constant, (in this case 1). From tables, critical value at 5% level is 3.84 So estimated χ 2 < χ 2 critical, so accept null that residuals are not heteroskedastic

7 Sometimes heteroskedastcity can be cause by variables that do not appear in the original regression (but perhaps should).. reg logpay exper exper2 /* pooled regression on men and women together */ Source SS df MS Number of obs = F( 2, 967) = Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE = logpay Coef. Std. Err. t P> t [95% Conf. Interval] exper exper _cons Regression suggests that pay rises with work experience but at a decreasing rate, (the coefficient on the quadratic is negative) Now do separate regressions for a) women. reg logpay exper exper2 if female==1 Source SS df MS Number of obs = F( 2, 482) = Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE = logpay Coef. Std. Err. t P> t [95% Conf. Interval] exper exper _cons b) men. reg logpay exper exper2 if female==0 Source SS df MS Number of obs = F( 2, 482) = Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE =.5212 logpay Coef. Std. Err. t P> t [95% Conf. Interval] exper exper _cons

8 In this case Goldfeld-Quandt Test = RSShigh/RSSlow ~ F(N-c-2k/2, N-c-2k/2) Becomes RSSwomen/RSSmen (since women are the high variance sub-sample look at the RSS in the output above) Where N= full sample size = 970 C= number observations dropped from middle (zero in this case ) K = number parameters in estimated equation (3) = 135.0/130.9 ~ F(970-2(2)/2, 970-2(2)/2) = 1.03 ~ F(483, 483) From tables, critical value at 5% level for F(483,483) = 1.00 So estimated F > Fcritical So again reject null that data are homoskedastic, (though this is very borderline and in most cases will not make much difference if pool the data) Fix up combined regression using White-adjusted standard errors.. reg logpay exper exper2, robust Regression with robust standard errors Number of obs = 970 F( 2, 967) = Prob > F = R-squared = Root MSE = Robust logpay Coef. Std. Err. t P> t [95% Conf. Interval] exper exper _cons Note sample size is 970, so do not have to worry too much about small sample effects on White adjusted standard errors.

9 3. You have data on firm size, N, and the level of profits, PROF, measured in million, for 118 firms. You estimate the following regression: ^ (1) PROF = 18, ,000 N - 1.0N 2 R 2 = 0.35 (12,000) (100) (0.25) RSS = 20,000 where the numbers in brackets are estimated standard errors You then run the regression twice more, first with the 50 firms with the lowest profits, then with the 50 firms with the highest profits. The residual sums of squares in the two regressions are 10,000 and 35,000 respectively. i) Interpret the results. At what firm size are profits maximised? Should recognise that firm size variables are statistically significant (t values >2) Constant is not. Constant gives notional amount of profit when firm size is zero. R-squared suggests 35% of variation in profits is explained by right hand side variables. Coefficients suggest profits rise then fall with firm size Profits are maximised when dprof/dn = 0 (1 st order condition for a max) Where (from question) dprof/dn = *1.0N = 0 2N = 1000 N= 500 ii) What do you understand by the term heteroskedasticity? What causes heteroskedasticity? What are the implications for OLS estimation if heteroskedasticity exists? - H. means that is residual variance not constant. Can be caused by incorrect functional form in dependent or dependent variables (try taking logs). Otherwise nature of behavioural agents (firms, individuals) means that variation in sample populations often depends on the values of the right hand side variables under study. Consequences. OLS estimates still unbiased. (see lecture notes for proof). Standard errors are biased (and therefore t and F values and confidence intervals also biased) iii) Perform a test for heteroskedasticity in the model. Comment on your results. Given information in the question the only test you could do is the Goldfeld-Quandt test for heteroskedasticity = RSShigh variance sample/rsslow variance sample ~ F(N-c-2k/2, N-c-2k/2)

10 Where N= full sample size C= number observations dropped from middle K = number parameters in estimated equation = 35000/10000 ~ F( (3)/2, (3)/2) = 3.5 ~ F(47,47) From tables, critical value at 5% level for F(16,16) = 2.32 So estimated F > Fcritical So reject null hypothesis of no heteroskedasticity