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Lecture 6
Heteroskedasticity
As we have studied in the previous lecture, the desirable properties of OLS are conditional on the validity of assumption. Among these assumptions, one assumption is the homoskedasticity. In this chapter we will study
- What is homoskedasticity
- What happens if homoskedasticity does not exist
- Tests for homoskedasticity
- Remedies for the heteroskedasticity
What is homoskedasticity?
Recall from the previous lecture that an assumption of OLS is
The variance of error term should be fixed i.e. ๐ฃ๐๐(๐๐) = ๐ 2 which is independent of the index i. This property is also called homoskedasticity (antonym: heteroskedasticity)
In the previous lecture we have shown two figures to clears the concept of homoskedasticity. However it is not always possible to judge the existence of homoskedasticity in such a straight forward way especially when we have multiple regressors. We need formal testing to investigate the incidence of homoskedasticity in a model.
What happens if errors are not homoskedastic?
We have learned that OLS has some desirable properties and these properties including
a. Unbiasedness b. Efficiency c. Consistency
In absence of homoskedasticity OLS remain unbiased and consistent but it is no longer efficient.
The loss of efficiency means that there exist some other estimators which give more accurate result than OLS and those estimators should get preference.
Testing For Homoskedasticity
There are many tests for the heteroskedasticity. Two of these tests are:
a. Goldfeld Quandt Test b. Eiker White Test (or White Test)
Dr. Asad zaman alongwith his co-authors have proposed more than one tests. One test was jointly proposed by Dr. Asad Zaman and Mumtaz Ahmed which they call as MZ test.
We will not go into detail of these tests because of the computational difficulty involved. We will learn only the two tests mentioned above.
Goldfeld Quandt Test
The procedure of GQ test is as follows:
Suppose you have data ๐ฆ 1 , ๐ฆ 2 , โฆ. ๐ฆ๐ on the dependent variable and ๐ฅ 1 , ๐ฅ 2 , โฆ. ๐ฅ๐ on the independent variable. We have to test null of homoskedasticity against the alternative of heteroskedasticity i.e.
๐ป0: ๐๐^2 = ๐ 2 ๐๐๐ ๐๐๐ ๐
๐ป0: ๐๐^2 โ ๐ 2 ๐๐๐ ๐๐๐ ๐
- Divide the data into two parts with first part having ๐ฆ 1 , ๐ฆ 2 , โฆ. ๐ฆ๐1 and ๐ฅ 1 , ๐ฅ 2 , โฆ. ๐ฅ๐1 and second part having ๐ฆ๐1+1, ๐ฆ๐1+2, โฆ. ๐ฆ๐ and ๐ฅ๐1+1, ๐ฅ๐1+2, โฆ. ๐ฅ๐, therefore first part contains n observations and second part contains n-n1 observations.
- Estimate regression for part 1 and calculate standard error of regression ๐๏ฟฝ 1
- Estimate regression for part 2 and calculate standard error of regression ๐๏ฟฝ 2
- The GQ statistics is given by ๐บ๐ = ๏ฟฝ ๐๏ฟฝ๐๏ฟฝ^2 1
2
- This test statistics has F-distribution with (n-n1-k, n-k) degree of freedom under the null hypothesis of homoskedasticity. The critical values can be calculated by writing following in MS Excel =FINV(0.05,df2,df1), Where df1 is degree of freedom for numerator (which is n-n1-k in current problem) and df2 is degree of freedom for denominator (which is n1-k in current problem).
- This command will return the critical value of the test statistics. If the GQ statistics is greater than the critical value the null hypothesis will be rejected.
Example:
Consider the following data set, which has data on consumption and income for India
INCOME; GDP per capita constant US$ 2000
CONS: Household Final Consumption per capita constant US$ 2000
GDP CONS
Multiple R 0. R Square 0. Adjusted R Square 0. Standard Error 4. Observations 11 Coefficients Standard Error t Stat P-value Intercept 9.395835 22.10532 0.425049 0. X Variable 1 0.745186 0.101469 7.344001 4.36E- 05
In the above output ๐๏ฟฝ 1 and ๐๏ฟฝ 2 are highlighted which are 4.005 and 1.235 respectively. The GQ statistics
is thus: ๐บ๐ = ๏ฟฝ ๐๏ฟฝ๐๏ฟฝ^21 ๏ฟฝ
2 = ๏ฟฝ 41 ..^005235 ๏ฟฝ
2 = 10.
Here n=23, n1=12 thus
Df1=12-2=10 and df2=11-2=
The critical value is found by writing =FINV(.05,9,10) which is 3.
Decision: since the calculated test statistics is greater than 5% critical value H0 is rejected.
Note about partitioning the data
GQ test requires making two parts of data. These parts may be equal or unequal. If there is some natural breakpoint in the economy, than we can take this point as breakpoint, regardless of whether it is in the middle of data or not. But if we cannot identify any natural break point, than we can make two halves of the data as we have done in the above example (part I with 12 and Part 2 with 11 observations, almost equal parts).
1n 1972, due to the oil price shock, the dynamics of most of economies suddenly changed so it can be taken as break point.
Eiker White Test
The procedure of this test is as follows:
Let Y be the dependent variable and X1, X2, โฆ.Xk are the independent variables;
- Run the regression for Y on all X variables and get the residuals.
- Run the following auxiliary regression; ๐ 2 = ๐ผ + ๐ฝ 1 ๐ฅ 1 + ๐ฝ 2 ๐ฅ 2 + โฏ. +๐พ 1 ๐ฅ 12 +๐พ 2 ๐ฅ 22 + โฏ + ๐พ 1 ๐ฅ 1 ๐ฅ 2 + โฏ + ๐ข. Here the dependent variable is the squared residuals from previous regression and independent variables are all the regressors, their squares and cross product.
- Get the R-square from this auxiliary regression and
- Test statistic EW=n*R-square
- Under the null hypothesis of homoskedasticity, this EW statistics has chi-square distribution.
Example
Consider the following data set on consumption, GDP and CPI for Australia
CONS GDP CPI 9.8 16.3 10. 10.0 17.1 10. 10.2 17.2 10. 10.3 17.3 10. 10.2 17.3 11. 10.2 16.9 11. 10.3 17.0 11. 10.4 17.1 11. 10.7 17.6 11. 10.9 18.0 11. 11.0 18.3 11. 11.2 18.7 11. 11.4 19.3 11. 11.5 19.2 11. 12.0 20.3 11. 12.1 20.6 11. 12.3 20.8 11.
Run the following regression ๐ถ๐ก = ๐ผ + ๐ฝ 1 ๐ฆ๐ก + ๐ฝ 2 ๐๐ก + ๐๐ก and test the residual for heteroskedasticity using EW test
Solution
The output of regression is as follows:
SUMMARY OUTPUT
Regression Statistics Multiple R 0. R Square 0. Adjusted R Square 0.
๐๐ก^2 GDP CPI GDP^2 CPI^2 GDP*CPI
Taking first column as dependent variable, the regression output is
ANOVA
- 1961 192 185.
- 1962 195 181.
- 1963 206 188.
- 1964 216 193.
- 1965 232 210.
- 1966 239 218.
- 1967 245 232.
- 1968 255 240.
- 1969 262 236.
- 1970 283 283.
- 1971 275 261.
- 1972 269 224.
- 1973 279 221.
- 1974 280 247.
- 1975 282 234.
- 1976 288 238.
- 1977 290 248.
- 1978 304 260.
- 1979 306 280.
- 1980 327 293.
- 1981 344 278.
- 1982 356 280.
- 1960 to 1971 containing 12 observations and second part from 1972 to 1982 containing The data starts form 1960 and ends at 1982, and it forms 23 observations. Let me make first part from
- SUMMARY OUTPUT FOR PART observations. The regression output is as follows:
- Multiple R 0.
- R Square 0.
- Adjusted R Square 0.
- Standard Error 1.
- Observations
- Intercept 84.04524 6.539493 12.85195 1.53E- Coefficients Standard Error t Stat P-value
- X Variable 1 0.391994 0.034091 11.49859 4.36E-
- 0.0206 16.3 10.5 265.7 110.3 171.
- 0.0294 17.1 10.7 292.4 114.5 183.
- 0.0054 17.2 10.8 295.8 116.6 185.
- 0.0057 17.3 10.9 299.3 118.8 188.
- 0.0507 17.3 11.0 299.3 121.0 190.
- 0.0003 16.9 11.0 285.6 121.0 185.
- 0.0009 17.0 11.0 289.0 121.0 187.
- 0.0061 17.1 11.0 292.4 121.0 188.
- 0.0142 17.6 11.0 309.8 121.0 193.
- 0.0125 18.0 11.0 324.0 121.0 198.
- 0.0032 18.3 11.0 334.9 121.0 201.
- 0.0024 18.7 11.0 349.7 121.0 205.
- 0.0038 19.3 11.0 372.5 121.0 212.
- 0.0081 19.2 11.0 368.6 121.0 211.
- 0.0009 20.3 11.1 412.1 123.2 225.
- 0.0012 20.6 11.0 424.4 121.0 226.
- 0.0001 20.8 11.1 432.6 123.2 230.
- Multiple R 0. Regression Statistics
- R Square 0.
- Adjusted R Square -0.
- Standard Error 0.
- Observations
- Regression 5 0.000634 0.000127 0. df SS MS F
- Residual 11 0.00217 0.
- Total 16 0.
- Intercept 7.158684 13.18587 0.542906 0. Coefficients Standard Error t Stat P-value
- GDP 0.189207 0.496347 0.381199 0.
- CPI -1.60404 2.774231 -0.57819 0.
- GDP^2 -0.00073 0.003247 -0.22604 0.
- CPI^2 0.085013 0.150458 0.565027 0.
- GDP*CPI -0.01494 0.049852 -0.29965 0.
Therefore EW stat is 18*0.226=4.
This EW statistics has Chi-square distribution with df=6 (number of variables in the auxiliary regression)
The critical Value can be found by writing =CHIINV(0.05,6) in MS excel which is 12.
EW statistics is smaller than the 5% critical value therefore the null hypothesis of homoskedasticity cannot be rejected.
REMEDIES FOR HETEROSKEDASTICITY
We have learned that in presence of heteroskedasticity the OLS is not efficient. Therefore to get efficient estimate of the model, following remedies can be suggested.
- Take the data in log transform, this usually removes the problem of heteroskedasticity in macroeconomic data
- Take the variables in growth rate, this works like log transform
- Use Generalized Least Square which is efficient for heteroskedastic data
- Use heteroskedasticity consistent standard errors.
Here we will show how the first remedy works
Consider the data on Indian Cons-GDP where we found the problem of heteroskedasticity. We will take log transform of the data the results are as follows:
GDP CONS LogGDP LogCONS 1960 186 194.7 5.2257 5. 1961 192 185 5.2575 5. 1962 195 181.7 5.2730 5. 1963 206 188.2 5.3279 5. 1964 216 193.8 5.3753 5. 1965 232 210.6 5.4467 5. 1966 239 218.2 5.4765 5. 1967 245 232.3 5.5013 5. 1968 255 240.5 5.5413 5. 1969 262 236.7 5.5683 5. 1970 283 283.1 5.6454 5.
2 = 1.
Now the GQ statistics is much smaller than 5% critical value which was 3.02, so the null of heteroskedasticity is not rejected.
