Generalized Least Squares Estimation and Heteroskedasticity Testing in Econometrics, Study notes of Introduction to Econometrics

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CHAPTER 8

Exercise Solutions

EXERCISE 8.

When σ (^2) i^ = σ^2

( )

( )

( )

( )

( )

( ) (^ )

2 2 2 2 2 2 2 1 1 1 (^2 2 2 ) 2 2 2 1 1 1 1

N N N i i i i i i i N N N N i i i i i i i i

x x x x x x

x x x x x x x^ x

= = =

= = = =

⎡ (^) − σ ⎤ ⎡ (^) − σ ⎤ σ − ⎣ ⎦ ⎣ ⎦ (^) σ = = = ⎡ (^) − ⎤ ⎡ (^) − ⎤ ⎡ (^) − ⎤ (^) − ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦

∑ ∑ ∑

∑ ∑ ∑ ∑

Exercise 8.2 (continued)

(c) The least squares estimators b 1 and b 2 are functions of the following averages

1 x = (^) N ∑ xi

y = (^) N ∑ yi

N ∑^ x yi^ i

N ∑ xi

For the generalized least squares estimator for βˆ 1 and βˆ 2 , these unweighted averages are replaced by the weighted averages 2 2

i i i

− x −

⎛ (^) σ ⎞ ⎜ ⎟ ⎝ σ ⎠

∑ ∑

2 2

i i i

− y −

⎛ (^) σ ⎞ ⎜ ⎟ ⎝ σ ⎠

∑ ∑

2 2

i i i i

− y x −

⎛ (^) σ ⎞ ⎜ ⎟ ⎝ σ ⎠

∑ ∑

2 2 2

i i i

− x −

⎛ (^) σ ⎞ ⎜ ⎟ ⎝ σ ⎠

∑ ∑

In these weighted averages each observation is weighted by the inverse of the error variance. Reliable observations with small error variances are weighted more heavily than those with higher error variances that make them more unreliable.

EXERCISE 8.

For the model yi = β + β 1 2 xi + ei where var( ei )= σ^2 xi^2 , the transformed model that gives a constant error variance is

y * i^ = β 1 x * i + β 2 + e * i

where y * i = yi xi , xi *^ = 1 xi , and e * i = ei xi. This model can be estimated by least squares with the usual simple regression formulas, but with β 1 and β 2 reversed. Thus, the generalized least squares estimators for β 1 and β 2 are

( )

1 2 2 1 *

• (^2) *

ˆ (^) and ˆ ˆ ( )

i i i i i i

N x y x y y x N x x

β = β = − β −

∑ ∑ ∑ ∑ ∑

Using observations on the transformed variables, we find

∑ yi^ *^ =^7 ,^ ∑ xi *^^ =37 12,^ ∑ x yi *^ * i^^ =47 8,^ ∑(^ xi *^ )^2 =349 144

With N = 5 , the generalized least squares estimates are

(^1 )

ˆ 5(47 8)^ (37 12)(7)

β = = − and ˆ 2 *^ ˆ 1 * (7 5) 2.984 (37 12) 0. 5

β = y − β x = − = −

EXERCISE 8.

(a) The table below displays the 95% confidence intervals obtained using the critical t -value t (0.975,497) (^) = 1.965and both the least squares standard errors and the White’s standard errors. After recognizing heteroskedasticity and using White’s standard errors, the confidence intervals for CRIME , AGE and TAX are narrower while the confidence interval for ROOMS is wider. However, in terms of the magnitudes of the intervals, there is very little difference, and the inferences that would be drawn from each case are similar. In particular, none of the intervals contain zero and so all of the variables have coefficients that would be judged to be significant no matter what procedure is used. 95% confidence intervals Least squares standard errors White’s standard errors Lower Upper Lower Upper CRIME − 0.255 − 0.112 − 0.252 − 0. ROOMS 5.600 7.143 5.065 7. AGE − 0.076 − 0.020 − 0.070 − 0. TAX − 0.020 − 0.005 − 0.019 − 0.

(b) Most of the standard errors did not change dramatically when White’s procedure was used. Those which changed the most were for the variables ROOMS , TAX , and PTRATIO. Thus, heteroskedasticity does not appear to present major problems, but it could lead to slightly misleading information on the reliability of the estimates for ROOMS , TAX and PTRATIO.

(c) As mentioned in parts (a) and (b), the inferences drawn from use of the two sets of standard errors are likely to be similar. However, keeping in mind that the differences are not great, we can say that, after recognizing heteroskedasticity and using White’s standard errors, the standard errors for CRIME , AGE , DIST , TAX and PTRATIO decrease while the others increase. Therefore, using incorrect standard errors (least squares) understates the reliability of the estimates for CRIME , AGE , DIST , TAX and PTRATIO and overstates the reliability of the estimates for the other variables.

Remark: Because the estimates and standard errors are reported to 4 decimal places in Exercise 5.5 (Table 5.7), but only 3 in this exercise (Table 8.2), there will be some rounding error differences in the interval estimates in the above table. These differences, when they occur, are no greater than 0.001.

EXERCISE 8.

(a) ROOMS significantly effects the variance of house prices through a relationship that is

quadratic in nature. The coefficients for ROOMS and ROOMS^2 are both significantly different from zero at a 1% level of significance. Because the coefficient of ROOMS^2 is positive, the quadratic function has a minimum which occurs at the number of rooms for which 2 2 3

e ROOMS ROOMS

= α + α = ∂ Using the estimated equation, this number of rooms is

2 min 3

ROOMS = −α = = α ×

Thus, for houses of 6 rooms or less the variance of house prices decreases as the number of rooms increases and for houses of 7 rooms or more the variance of house prices increases as the number of rooms increases. The variance of house prices is also a quadratic function of CRIME , but this time the quadratic function has a maximum. The crime rate for which it is a maximum is

4 max 5

CRIME

−α = = = α ×

Thus, the variance of house prices increases with the crime rate up to crime rates of around 30 and then declines. There are very few observations for which CRIME ≥ 30 , and so we can say that, generally, the variance increases as the crime rate increases, but at a decreasing rate. The variance of house prices is negatively related to DIST , suggesting that the further the house is from the employment centre, the smaller the variation in house prices.

(b) We can test for heteroskedasticity using the White test. The null and alternative hypotheses are H (^) 0 : α 2 = α 3 = "= α 6 = 0

H (^) 1 : not all α s in H 0 are zero

The test statistic is χ^2 = N × R^2. We reject H (^) 0 if χ 2 > χ (^2) (0.95,5)where χ (^2) (0.95,5) = 11.07. The test value is

χ^2 = N × R^2 = 506 × 0.08467 =42.

Since 42.84 > 11.07, we reject^ H 0 and conclude that heteroskedasticity exists.

Exercise 8.7 (continued)

(d) Variance estimates are given by the predictions σˆ^ i (^2)^ = exp( αˆ z (^) i ) = exp(0.4853 × zi ). These

values and those for the transformed variables

• (^) , * ˆ ˆ

i i i i i i

y x y x

⎝ σ^ ⎠ ⎝ σ ⎠ are given in the following table.

observation σˆ^2 i y * i x i * 1 4.960560 0.493887 − 0. 2 1.156725 − 0.464895 − 2. 3 29.879147 3.457624 0. 4 9.785981 − 0.287700 − 0. 5 2.514531 4.036003 2. 6 27.115325 0.345673 − 0. 7 3.053260 2.575316 1. 8 22.330994 − 0.042323^ − 0.

(e) From Exercise 8.2, the generalized least squares estimate for β 2 is

2 2 2 2 2 (^2 2 ) 2 2

2

i i i i i i i i i i i i i i

y x y x

x x

∗ ∗^ −^ − − − −

∗ − − −

⎛ (^) σ ⎞⎛ (^) σ ⎞ − ⎜ ⎟⎜ ⎟ σ (^) ⎝ σ (^) ⎠⎝ σ ⎠ β = ⎛ (^) σ ⎞ − ⎜ ⎟ σ (^) ⎝ σ ⎠

− × −

∑ ∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑

The generalized least squares estimate for β 1 is 2 2 ˆ 1 2 2 ˆ (^2) 2.193812 ( 0.383851) 1.1242 2. i i i i i i

− (^) y − x − −

σ ⎛^ σ ⎞ β = − (^) ⎜ ⎟β = − − × = σ (^) ⎝ σ ⎠

∑ ∑ ∑ ∑

EXERCISE 8.

(a) The regression results with standard errors in parenthesis are

n

( ) ( ) ( )

(se) 3586.64 2.1687 35.

PRICE = + SQFT − AGE

These results tell us that an increase in the house size by one square foot leads to an increase in house price of $63.39. Also, relative to new houses of the same size, each year of age of a house reduces its price by $217.84.

(b) For SQFT = 1400 and AGE = 20

n PRICE (^) = 5193.15 + 68.3907 × 1400 − 217.8433 × 20 =96,

The estimated price for a 1400 square foot house, which is 20 years old, is $96,583. For SQFT = 1800 and AGE = 20 n PRICE (^) = 5193.15 + 68.3907 × 1800 − 217.8433 × 20 =123,

The estimated price for a 1800 square foot house, which is 20 years old, is $123,940.

(c) For the White test we estimate the equation

2 2 2 e ˆ i = α + α 1 2 SQFT + α 3 AGE + α 4 SQFT + α 5 AGE + α 6 SQFT × AGE + vi

and test the null hypothesis H (^) 0 : α 2 = α 3 = " = α 6 = 0. The value of the test statistic is

χ^2 = N × R^2 = 940 × 0.0375 =35.

Since χ (^2) (0.95,5) = 11.07, the calculated value is larger than the critical value. That is, 2 2 χ > χ(0.95,5). Thus, we reject the null hypothesis and conclude that heteroskedasticity exists.

(d) Estimating the regression (^) log( e ˆ i^2 )= α + α 1 2 SQFT + vi gives the results

α =ˆ^1 16.3786, αˆ 2 =0.

With these results we can estimate σ i 2 as

ˆ (^2) exp(16.3786 0.001414 ) σ i (^) = + SQFT

EXERCISE 8.

(a) (i) Under the assumptions of Exercise 8.8 part (a), the mean and variance of house prices for houses of size SQFT = 1400 and AGE = 20 are

E PRICE ( ) = β + 1 1400 β 2 + 20 β 3 var( PRICE )= σ^2

Replacing the parameters with their estimates gives

E PRICE ( ) = 96583 var( PRICE ) =22539.63^2

Assuming the errors are normally distributed,

( )

( )

P PRICE P Z

P Z

where Z is the standard normal random variable Z ∼ N (0,1). The probability is depicted as an area under the standard normal density in the following diagram.

The probability that your 1400 square feet house sells for more than $115,000 is 0.207.

(ii) For houses of size SQFT = 1800 and AGE = 20 , the mean and variance of house prices from Exercise 8.8(a) are

E PRICE ( ) = 123940 var( PRICE ) =22539.63^2

The required probability is

( )

( )

P PRICE P Z

P Z

< = ⎛^ <^ − ⎞

The probability that your 1800 square feet house sells for less than $110,000 is 0.268.

Exercise 8.9 (continued)

(b) (i) Using the generalized least squares estimates as the values for β 1 ,β 2 and β 3 , the

mean of house prices for houses of size SQFT = 1400 and AGE = 20 is, from Exercise 8.8(f), E PRICE ( ) = 96196. Using estimates of α 1 and α 2 from Exercise 8.8(d), the variance of these house types is

1 2

8

2

var( ) exp( 1.2704 1400) exp(16.378549 1.2704 0.00141417691 1400)

3.347172 10

(18295.3)

PRICE = α + + α × = + + ×

= ×

=

Thus,

( )

( )

P PRICE P Z

P Z

The probability that your 1400 square feet house sells for more than $115,000 is 0.152.

(ii) For your larger house where SQFT = 1800 , we find that E PRICE ( ) = 122326 and

1 2

8

2

var( ) exp( 1.2704 1800)

exp(16.378549 1.2704 0.00141417691 1800)

5.893127 10

(24275.8)

PRICE = α + + α ×

= + + ×

= ×

=

Thus,

( )

( )

P PRICE P Z

P Z

The probability that your 1800 square feet house sells for less than $110,000 is 0.306.

(c) In part (a) where the heteroskedastic nature of the error term was not recognized, the same standard deviation of prices was used to compute the probabilities for both house types. In part (b) recognition of the heteroskedasticity has led to a standard deviation of prices that is smaller than that in part (a) for the case of the smaller house, and larger than that in part (a) for the case of the larger house. These differences have in turn led to a smaller probability for part (i) where the distribution is less spread out and a larger probability for part (ii) where the distribution has more spread.

EXERCISE 8.

The results are summarized in the following table and discussed below.

part (a) part (b) part (c) βˆ 1 81.000 76.270 81. se( βˆ 1 ) 32.822 12.004 33. βˆ 2 10.328 10.612 10. se( βˆ 2 ) 1.706 1.024 1. χ 2 = N × R^2 6.641 2.665 6.

The transformed models used to obtain the generalized estimates are as follows.

(a) (^) 0.25 1 0.25 2 0. i^1 i i i i i

y x e x x x

⎛⎜ ⎞⎟ (^) = β ⎛⎜ ⎞⎟ (^) + β ⎛⎜ ⎞⎟+ ∗ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

where (^) i 0.25^ i i

e e x

(b) i^ 1 1 2 i i i i i

y x e x x x

⎛⎜ ⎞⎟ (^) = β ⎛⎜ ⎞⎟ (^) + β ⎛⎜ ⎞⎟+ ∗ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

where (^) i i i

e e x

(c) (^1 )

ln( ) ln( ) ln( )

i i i i i i

y x e x x x

⎛⎜ ⎞⎟ (^) = β ⎛⎜ ⎞⎟ (^) + β ⎛⎜ ⎞⎟+ ∗ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠

where ln( )

i i i

e e x

In each case the residuals from the transformed model were squared and regressed on income and income squared to obtain the R^2 values used to compute the χ^2 values. These equations were of the form 2 2 e ˆ (^) 1 2 x (^) 3 x v ∗ (^) = α + α + α +

For the White test we are testing the hypothesis H (^) 0 : α 2 = α 3 = 0 against the alternative hypothesis H 1 (^) : α 2 ≠ 0 and/or α 3 ≠ 0.The critical chi-squared value for the White test at a 5% level of significance is (^) χ^2 (0.95,2) = 5.991. After comparing the critical value with our test statistic values, we reject the null hypothesis for parts (a) and (c) because, in these cases, 2 2 χ > χ(0.95,2). The assumptions var( )^2 ei = σ xi and var( ) 2 ln( ) ei = σ xi do not eliminate heteroskedasticity in the food expenditure model. On the other hand, we do not reject the null hypothesis in part (b) because χ^2 < χ^2 (0.95,2). Heteroskedasticity has been eliminated with the assumption that var( ei )= σ^2 xi^2.

In the two cases where heteroskedasticity has not been eliminated (parts (a) and (c)), the coefficient estimates and their standard errors are almost identical. The two transformations have similar effects. The results are substantially different for part (b), however, particularly the standard errors. Thus, the results can be sensitive to the assumption made about the heteroskedasticity, and, importantly, whether that assumption is adequate to eliminate heteroskedasticity.

EXERCISE 8.

(a) This suspicion might be reasonable because richer countries, countries with a higher GDP per capita, have more money to distribute, and thus they have greater flexibility in terms of how much they can spend on education. In comparison, a country with a smaller GDP will have fewer budget options, and therefore the amount they spend on education is likely to vary less.

(b) The regression results, with the standard errors in parentheses are

n

( ) ( )

(se) 0.0485 0.

i i i i

EE GDP

P P

⎜⎜ ⎟⎟ = −^ + ⎜ ⎟

⎝ ⎠ ⎝^ ⎠

The fitted regression line and data points appear in the following figure. There is evidence of heteroskedasticity. The plotted values are more dispersed about the fitted regression line for larger values of GDP per capita. This suggests that heteroskedasticity exists and that the variance of the error terms is increasing with GDP per capita.

-0.

0 2 4 6 8 10 12 14 16 18 GDP per capita

(c) For the White test we estimate the equation

2 2 ˆ (^1 2 ) i i i i i i

GDP GDP

e v P P

= α + α (^) ⎜ ⎟ + α (^) ⎜ ⎟ + ⎝ ⎠ ⎝ ⎠

This regression returns an R 2 value of 0.29298. For the White test we are testing the hypothesis H (^) 0 : α 2 = α 3 = 0 against the alternative hypothesis H 1 (^) : α 2 ≠ 0 and/or α 3 ≠0. The White test statistic is

χ^2 = N × R^2 = 34 × 0.29298 =9.

The critical chi-squared value for the White test at a 5% level of significance is 2 χ(0.95,2) = 5.991. Since 9.961 is greater than 5.991, we reject the null hypothesis and conclude that heteroskedasticity exists.

EXERCISE 8.

(a) For the model C 1 (^) t = β + β 1 2 Q 1 (^) t + β 3 Q 1^2 t + β 4 Q 1^3 t + e 1 t , where var (^) ( e 1 (^) t ) = σ 2 Q 1 t , the generalized

least squares estimates of β 1 , β 2 , β 3 and β 4 are:

estimated coefficient

standard error β 1 93.595 23. β 2 68.592 17. β 3 −10.744 3. β 4 1.0086^ 0.

(b) The calculated F value for testing the hypothesis that β 1 = β 4 = 0 is 108.4. The 5% critical value from the F (2,24) distribution is 3.40. Since the calculated F is greater than the critical F , we reject the null hypothesis that β 1 = β 4 = 0. The F value can be calculated from

( ) ( )

( ) ( )

R U U

SSE SSE

F

SSE

(c) The average cost function is given by

1 2 1 2 3 1 4 1 1 1 1

t^1 t t t t t t

C e Q Q Q Q Q

= β (^) ⎜ ⎟+ β + β + β + ⎝ ⎠

Thus, if β = β 1 4 = 0 , average cost is a linear function of output.

(d) The average cost function is an appropriate transformed model for estimation when heteroskedasticity is of the form var (^) ( e 1 (^) t ) = σ^2 Q 12 t.

EXERCISE 8.

(a) The least squares estimated equations are

( ) ( ) ( )

2 3 2 1 1 1 1 1 1

(se) 23.655 4.597 0.2721 7796.

C Q Q Q

SSE

= + − + σ =

( ) ( ) ( )

2 3 2 2 2 2 2 2 2

(se) 28.933 6.156 0.3802 20343.

C Q Q Q

SSE

= + − + σ =

To see whether the estimated coefficients have the expected signs consider the marginal cost function 2 2 2 3 3 4 MC dC Q Q dQ

= = β + β + β

We expect MC > 0 when Q = 0; thus, we expect β 2 > 0. Also, we expect the quadratic MC function to have a minimum, for which we require β 4 > 0. The slope of the MC function is d MC ( ) dQ = 2 β + 3 6 β 4 Q. For this slope to be negative for small Q (decreasing MC ), and positive for large Q (increasing MC ), we require β 3 < 0. Both our least-squares estimated equations have these expected signs. Furthermore, the standard errors of all the coefficients except the constants are quite small indicating reliable estimates. Comparing the two estimated equations, we see that the estimated coefficients and their standard errors are of similar magnitudes, but the estimated error variances are quite different.

(b) Testing H (^) 0 :σ 1 2 = σ^22 against H (^) 1 :σ 12 ≠ σ 22 is a two-tail test. The critical values for

performing a two-tail test at the 10% significance level are F (0.05,24,24) (^) = 0.0504 and F (0.95,24,24) (^) = 1.984. The value of the F statistic is 2 2 2 1

F

σ = = = σ

Since F > F (0.95,24,24), we reject H 0 and conclude that the data do not support the proposition that σ 12 = σ 22.

(c) Since the test outcome in (b) suggests σ 1 2 ≠ σ^22 , but we are assuming both firms have the

same coefficients, we apply generalized least squares to the combined set of data, with the observations transformed using σˆ 1 and σˆ 2. The estimated equation is

( ) ( ) ( )

ˆ 67.270 89.920 15.408 2 1.3026^3

(se) 16.973 3.415 0.

C = + Q − Q + Q

Remark: Some automatic software commands will produce slightly different results if the transformed error variance is restricted to be unity or if the variables are transformed using variance estimates from a pooled regression instead of those from part (a).