Heteroskedasticity

Paulo Fagandini

Nova SBE

Review

Review

1. Testing for Heteroskedasticity
2. Breusch-Pagan test
3. White test
4. Lazy White test
5. Weighted Least Squares (WLS)
6. Robust Standard Errors
7. Feasible Generalized Least Squares

Exercises

https://moodle.novasbe.pt/mod/folder/view.php?id=4021

Exercise 7.1

Which of the following are consequences of heteroskedasticity? a) The OLS estimators, \(\hat{\beta}_j\), are inconsistent. b) The usual F statistics no longer has an \(F\) distribution. c) The OLS estimators are no longer BLUE.

1. MLR5 is not needed for consistency, only MLR1-MLR4
2. If MLR5 fails, usual SEs are wrong and \(t\) and \(F\) statistics do not have \(t\) and \(F\) distributions, respetively, thus inference as usual is invalid.
3. MLR1-MLR5 need to hold for OLS to be BLUE

Exercise 7.3

Consider a linear model to explain omnthly beer consumption:

\[beer = \beta_0+\beta_1 inc + \beta_2 price + \beta_3 educ + \beta_4 female + u\]

\(E[u|X]=0\), \(V[u|X]=\sigma^2 inc^2\).

Write the transformed equation that has a homoskedastic error term.

In this case we know the expression of the variance of the model. Therefore we can use \(WLS\) where: \[V[u|X]=\sigma^2 inc^2=\sigma^2 h(x)\] The transformed equation is: \[beer/\sqrt{inc^2}=\beta_0/\sqrt{inc^2}+... + u/\sqrt{inc^2}\] The new error is homoskedastic: \(V[u/inc|X]=\sigma^2 inc^2/inc^2=\sigma^2\)

Exercise 7.7

From a sample of 88 observations the following regression was obtained to explain the price of housing in a particular US state:

\[ \hat{p}_i = \underset{\underset{[37.14]}{(29.48)}}{-21.77}+\underset{\underset{[0.0013]}{0.0006}}{0.002}ls_i+\underset{\underset{[0.018]}{0.013}}{0.123}hs_i+\underset{\underset{[8.48]}{(9.01)}}{13.85} nr_i \]

where, for house \(i\), \(p\) represents price (in thousands), \(ls\) is the lot size (in \(m^2\)), \(hs\) is the size of the house (in \(m^2\)) and \(nr\) the number of rooms. The usual and heteroskedasticity-robust standard errors are in parenthesis and squared brackets, respetively.

Exercise 7.7

1. Consider the following regression

\[ \hat{u}_i=-5,522.79+0.202 ls_i + 1.69 hs_i + 1,041.76 nr_i \]

with \(R^2=0.160\) and \(F=5.339\)

How many degrees of freedom has the \(F\) distribution from where you obtain the critical value? Given the output above, analyze whether the errors of the initial model are heteroskedastic. Use both \(F\) and \(LM\) versions of the test to derive your conclusion.

The output above is the Breush-Pagan test for heteroskedasticity: \(H_0:V[u|X]=\sigma^2\), \(H_a:V[u|X]=\sigma^2_i\).

\(F=[(R^2)/K]/[(1-R^2)/(N-K-1)]=[(0.16)/(3)]/[(1-0.16)/(88-3-1)]\approx 5.339>2.71=F^c_{3,88-3-1}\)

\(LM=nR^2=88\cdot 0.16=14.88>7.815=\chi_3^2\) Reject \(H_0\)! There is evidence of heteroskedasticity.

Exercise 7.7

2. Explain why the model is estimated with heteroskedasticity-robust standard errors. Compare these results with those obtained with the usual standard errors.

We found evidence of heteroskedasticity. Therefore, usual standard errors are wrong. But the coefficients are still unbiased if MLR1-MLR4 hold. Thus, we can use estimated heteroskedasticity-robust standard errors to conduct inference.

\(ls\) - usual \(t=3.33^*\) - robust \(t=1.54\) \(hs\) - usual \(t=9.46^*\) - robust \(t=6.83^*\) \(nr\) - usual \(t=1.54\) - robust \(t=1.63\)

\(ls\) turns out to be not statistically significant at the 5% level. \(hs\) is the only statistically significant variable.

Exercise 7.7

3. One of the econometricians in charge suggested that, given his knowledge of the market and the problem detected in (a), it would make sense to estiamte a model in which the elasticities price-lot size and price-house size were constant. The results of such model are:

\[ \widehat{ln(p_i)}=\underset{(0.65)}{-1.30}+\underset{(0.038)}{0.168}\ln (ls_i)+\underset{(0.093)}{0.70}\ln (hs_i) + \underset{(0.028)}{0.037}nr_i \]

with \(R^2=0.643\).

Explain why these auxiliary regressions were estimated. What do you conclude?

Taking the log of the dependent variable usually reduces heteroskedasticity because the spread of the distribution is usually reduced and the influence of outliers is mitigated.

Exercise 7.7

4. The following auxiliary regressions were also estimated, using the residuals of the last model:

\[ \begin{aligned} \hat{u}_i^2 =& 12.21 - 1.27 \ln (ls_i) - 1.76 \ln (hs_i) + 0.29 nr_i + 0.02 \ln (ls_i)^2 +\\ +& 0.04 \ln (hs_i)^2-0.005 nr_i^2 + 0.12 [\ln (ls_i)\times \ln (hs_i)]\\ -&0.03[\ln (ls_i)\times nr_i] - 0.001[\ln (hs_i)\times nr_i] \end{aligned} \]

With \(R^2=0.109\) and

\[ \hat{u}_i^2=5.05-1.71\ln \widehat{\ln (p_i)}+0.145\widehat{\ln (p_i)}^2 \] With \(R^2=0.039\).

Explain why these auxiliary regressions were estimated. What do you conclude?

These auxiliary regressions were estimated to perform a White test and a “Lazy”/Alternative White test. \(H_0: V[u|X]=\sigma^2\), \(H_a:V[u|X]=\sigma_i^2\)

\(LM_1=n\cdot R^2=88\cdot 0.109 = 9.592<16.92=\chi_9^2\) \(LM_2=n\cdot R^2=88\cdot 0.0309 = 3.432<5.99=\chi_2^2\) There is no evidence of heteroskedasticity anymore!