1 / 24

Returning to Consumption

Returning to Consumption. More on Consumption. We return to the consumption problem to illustrate the issue of heteroscedasticity It turns out that OLS may NOT give us the best estimate of the MPC

alda
Download Presentation

Returning to Consumption

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Returning to Consumption

  2. More on Consumption • We return to the consumption problem to illustrate the issue of heteroscedasticity • It turns out that OLS may NOT give us the best estimate of the MPC • The reason is that one of the assumptions of the GM theorem is probably violated in the consumption model • The data is probably not heteroscedastic • Var(ui|xi) ≠ 2

  3. Homoscedastic

  4. Heteroscedastic

  5. Characteristics of Heteroscedasticty • Systematic pattern exists in variance of residuals: • Var(ui|xi) = i2 = 2.f(xi) • i2 = f(1+2Z2+…+ kZk) • variance has different values for different observations or groups of observations • Intuition: if random bit come from roll of dice then homo is with same dice and hetero is with different dice • Evident in cross-section data or time series

  6. Consequences • OLS is unbiased • OLS is consistent • OLS is no longer efficient • Variance formula used previously is incorrect • significance test, confidence intervals etc. cannot be used • Aside: a corrected formula can be used • Stata: regress y x, robust  • We don’t bother with this because can do better with alternative estimator

  7. Testing for Heteroskedasticity • Plot of residuals • Sort the residuals by explanatory variable and plot against that variable , look for pattern, do this for each explanatory variable • Not a formal test but can give an idea of what's going on • Can use it to reject idea of Het

  8. An example of Het

  9. Consumption Example

  10. GoldfeldQuandt test • Used for i2 = 2.f(xi) i.e. related to one variable only • State Hypothesis • Test H0: i2 = 2. H1: i2 ≠ 2. • Note: the null is homoscedasticity • Sort residuals by ascending order of xi • Omit middle 20% observations: (n-c) observations remain • Estimate the original model separately for two samples • first (n-c)/2 obs (keep RSS1 ) • last (n-c)/2 obs (keep RSS2) • Compute: g = SSR2/SSR1 • If g > Fc(df,df) => reject null hypothesis of homoscedasticity at a significance level • Test can be carried out for each xi

  11. Intuition of GQ • If het does exist then we can split sample into a low variance and high variance bit • Run the regression separately for the two samples • Calculate the ratio of variances of the residual (remember s2=RSS/df) • If this ratio is 1 then they are equal and the data is homoscedastic • So reject null of homoscedasticity if bigger than 1 • How much bigger? Bigger than F critical value

  12. Consumption Example • Test it for nmwage • State Hypothesis • Test H0: i2 = 2H1: i2 ≠ 2. • Note: the null is homoscedasticity • Sort residuals by ascending order of nmwagei Stata command: sort nmwage • Omit middle 20% observations: (n-c) observations remain • Two sample: 1..550 781..1330 • Estimate the original model separately • Compute: g = SSR2/SSR1 • g= 92733079.9/ 91033129.2= 1.018674 • If g > Fc(df,df) => reject null hypothesis of homoscedasticity at a significance level • 5% sig level F(550,550)=1.15 • So cannot reject the null at 5% significance level • Test can be carried out for each xi

  13. sort nmwage . regress cons nmwage if _n<=550 Source | SS df MS Number of obs = 550 -------------+------------------------------ F( 1, 548) = 49.12 Model | 8312140.6 1 8312140.6 Prob > F = 0.0000 Residual | 92733079.9 548 169220.949 R-squared = 0.0823 -------------+------------------------------ Adj R-squared = 0.0806 Total | 101045220 549 184053.225 Root MSE = 411.36 ------------------------------------------------------------------------------ cons | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- nmwage | .7304001 .1042153 7.01 0.000 .5256897 .9351104 _cons | 67.60298 49.48125 1.37 0.172 -29.59315 164.7991 ------------------------------------------------------------------------------ . regress cons nmwage if _n>780 Source | SS df MS Number of obs = 550 -------------+------------------------------ F( 1, 548) = 118.87 Model | 19747100.1 1 19747100.1 Prob > F = 0.0000 Residual | 91033129.2 548 166118.849 R-squared = 0.1783 -------------+------------------------------ Adj R-squared = 0.1768 Total | 110780229 549 201785.481 Root MSE = 407.58 ------------------------------------------------------------------------------ cons | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- nmwage | .7270654 .0666855 10.90 0.000 .596075 .8580558 _cons | 94.26419 75.29303 1.25 0.211 -53.63408 242.1625 ------------------------------------------------------------------------------

  14. White’s Test • More general test that allows for more than one varibles to influence the variance of the residuals • Estimate model yi = 1 + 2 x2i + 3 x3i + ui • Run auxiliary regression: sq’d residuals on squares and cross products of X variables: • ei2 =1 +2 x2i +3 x3i + 4 x2i2+ 5 x3i2+ 6 x2i x3i + vi • Null hypothesis is homoscedastic errors i.e. • 2 = 3 = 4 = 5 = 6 = 0 i.e. ei2 = constant • calculate nR2 ~ df2 test • nR2 > df2critical value reject null hypothesis • Comment: why not an F-test

  15. Consumption Example • Test it for nmwage • State Hypothesis • Test H0: i2 = 2H1: i2 ≠ 2. • Note: the null is homoscedasticity • Estimate the Model and generate residuals squared • Regress residual squared on all of the variables that may cause heteroscdasticity • Form the test statistic: NR2=0.266 • Find critical value: chi-sq, df=2 alpha=0.05=5.99 • We cannot reject the null at the 5% significance level

  16. predict u, residual gen u2=u^2 gen nmwage2=nmwage^2 regress u2 nmwage nmwage2 Source | SS df MS Number of obs = 1330 -------------+------------------------------ F( 2, 1327) = 0.11 Model | 1.1824e+10 2 5.9121e+09 Prob > F = 0.8921 Residual | 6.8688e+13 1327 5.1762e+10 R-squared = 0.0002 -------------+------------------------------ Adj R-squared = -0.0013 Total | 6.8700e+13 1329 5.1693e+10 Root MSE = 2.3e+05 ------------------------------------------------------------------------------ u2 | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- nmwage | -13.03642 58.98887 -0.22 0.825 -128.758 102.6852 nmwage2 | .0109503 .0325921 0.34 0.737 -.0529874 .0748881 _cons | 163843.1 24661.71 6.64 0.000 115462.9 212223.3 ------------------------------------------------------------------------------

  17. Efficient Estimation • If we find heteroscsadstity we know that OLS will be inefficient • Remember why this might be a problem (see over) • Can we do better? • Yes. There is an efficient estimator called Generalised Least Squares (GLS) • Two steps • Remove the heteroscedasticity from the data • Do OLS on the transformed data

  18. Prob of error is lower for efficient estimator at any sample size Same sample size, different estimator

  19. The GLS Procedure • Assume that i2 is known: • Basic model: Yi = 1 + 2 Xi + ui , E(ui2) = i2 (not constant) • Create new data with each observations weighted by the heteroscedastic standard deviation

  20. The GLS Procedure • Then run the regression on the transformed data • The slope estimates are the BLUE of the coefficients of the original model • Note the intercept tem is slightly different (it has now become coefficient on a variable)

  21. How it Works • GLS eliminates heteroscedasticity • To see this note that • var(ui*) = E(ui*)2 = E(ui/i)2 = 1/i2.E(ui2) = (1/i2).i2 = 1 • varof transformed error term is homoskedastic: it is constant • NB This model does not have a constant now: it has two explanatory variables: 1/i and Xi/I • Cannot apply GLS if the exact type of hetero is unknown. So do FGLS (Feasible GLS) and replace i with an estimate of I • From White’s test

  22. The Consumption Example • Transform the data to eliminate the heteroscedasticty • Use the estimate of from White’s test • Stata command • Predict white • generate c=cons/sqrt(white) • generate y=nmwage/sqrt(white) • The GLS of the MPC is given by the regression on the transformed data

  23. predict white (option xb assumed; fitted values) . gen sigma=white^0.5 . gen c=cons/sigma . gen y=nmwage/sigma . gen invsigma=1/sigma . reg c y invsigma, noconstant Source | SS df MS Number of obs = 1330 -------------+------------------------------ F( 2, 1328) = 1959.33 Model | 3924.62819 2 1962.3141 Prob > F = 0.0000 Residual | 1330.01976 1328 1.00152091 R-squared = 0.7469 -------------+------------------------------ Adj R-squared = 0.7465 Total | 5254.64796 1330 3.95086313 Root MSE = 1.0008 ------------------------------------------------------------------------------ c | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- y | .7564394 .031111 24.31 0.000 .6954072 .8174715 invsigma | 62.35942 26.05967 2.39 0.017 11.23681 113.482 ------------------------------------------------------------------------------

  24. Conclusion • The example didn’t appear to have heteroscedasticity. • When het does exist the difference between GLS and OLS can be substantial • Both are unbiased and consistent • GLS is preferable because it is efficient so there is a lower probability of substantial error

More Related