ECON 7710, 2010

10.1

Heteroskedasticity

• What is heteroskedasticity?

• What are the consequences?

• How is heteroskedasticity identified?

• How is heteroskedasticity corrected?

Objectives

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Main empirical model for Unit 10:

foodexpi = 0 + 1incomei + i.

foodexp: Family food expenditureincome : Family income

Least squares estimates, US data (UE_Tab0301)

Is this the best estimated equation?

40. N 31710

Income12807740oodexpf

2

i03101422

i

,.

..ˆ ***

..

Rse

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1. The Nature of Heteroskedasticity

In a regression about firms, for the same mistake,

millionbillion

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Heteroskedasticity is a problem that occurs when the error term does not have a constant variance.

CLRM: Each error term comes from the same probability distribution.

Assumption CLRM.5 is violated!

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Yi = 0 + 1X1i + 2X2i + i

Regression Model

E(i|X1i,X2i) = 0

var(i|X1i,X2i) = 2

zero mean:

homoskedasticity:

cov(i, j|X1i,X2i,X1j,X2j) = i = j

no autocorrelation:

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Identical distributions for observations i and j

Distribution for i

Distribution for j

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X2

..

X1

..

X3 X4 X

Yf(Y)

0

HomoskedasticityYi = 0 + 1Xi + i var(i|Xi) = 2 for all i

Conditional Distribution

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HeteroskedasticityYi = 0 + 1Xi + i var(i|Xi) = i

2 for all i

Conditional Distribution

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Pure heteroskedasticity

Different variances of the error term.

Correctly specified PRF.

Impure heteroskedasticity

Different variances of the error term.

Specification error.

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2. Detecting Heteroscedasticity

2.1 Graphical Method

Plotting foodexp against income (for one regressor)

Example 1:Food expenditure, US Data (UE_Tab0301)

40

80

120

160

200

240

280

200 400 600 800 1,000 1,200

income

foo

de

xp

Scatter Diagram of Regressing foodexp on income

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Example 1: Food expenditure, US Data, UE_Tab0301

-80

-40

0

40

80

120

200 400 600 800 1,000 1,200

income

resi

dual

0

1,000

2,000

3,000

4,000

5,000

6,000

7,000

200 400 600 800 1,000 1,200

income

squa

red

resi

dual

Plotting e against income.

Plotting e2 against income.

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Example 2: textbook data, (Woody3)*** *** *** **

se

2

Y 102,192 9,075 N 0.35 P 1.29 I

R 0.6182, N = 33.

-30,000

-20,000

-10,000

0

10,000

20,000

30,000

40,000

0 50,000 100,000 150,000 200,000 250,000

Population

resi

du

al

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3.2 Park Test

Model

Yi = 0 + 1X1i + … + KXKi + t i = 1,…,N (*)

Suppose it is suspected that var(i) depends on Zi in the form of

var(i) = i2 = 2Zi

1evi

lni2 = ln2 + 1lnZki + vi

Ho: 1 = 0 (Homoskedastic errors);HA: 1 0 (Heteroskedastic errors).

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Step 1: Estimate the equation (*) with OLS and obtain the residuals.

0 1 1ˆ ˆ ˆˆ

i i i i i K Kie Y Y Y X X

Step 2: Regress the natural log of squared residuals on the natural log of a possible proportionality factor

ln(ei2) = 0 + 1lnZi + vi

where vi is an error term satisfying all classical assumptions.

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Step 3

If the coefficient of lnZ is significantly different from zero, then it would suggest that there is heteroscedastic pattern in the residuals with respect to Z. Otherwise, homoscedastic errors cannot be rejected.

Example 3: Park Test: US data (UE_Tab0301) ^ ln(e2) = -7.46 + 2.07** ln(income) t (2.28) p-value (0.0284)

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Advantages of the Park test:

a. The test is simple.

b. It provides information about the variance structure.

Limitations of the Park test:a. The distribution of the dependent variable is

problematic.b. It assumes a specific functional form.c. It does not work when the variance depends on two or

more variables.

d. The correct variable with which to order the observations must be identified first.

e. It cannot handle partitioned data.

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3.3 White’s Test

Model

Yi = 0 + 1X1i + 2X2i + i i = 1,…,N (*)

Suppose it is suspected there may be heteroskedasticity but we are not sure of its functional form.

Ho: The conditional variance of i is constant.

HA: The conditional variance of i is not constant.

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Step 1: Estimate the equation (*) with OLS and obtain the residuals.

0 1 1 2 2ˆ ˆ ˆˆ

i i i i i ie Y Y Y X X

Step 2: Regress the squared residuals on all explanatory variables, all cross product terms and the square of each explanatory variable.

ei2 = 0 + 1X1i + 2X2i

+ 3X1i2 + 4X2i

2

+ 5X1iX2i + vi

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Step 3: Test the overall significance of the equation in Step 2. (df = number of regressors)

Reject the hypothesis of homoskedasticity if NR2er

r > cv.

Statistic = NR2white ~ 2

df

Critical value (cv) = 2df,

Example 4: White test: US data (UE_Tab0301) ^ e2 = 1924 – 7.4 income + 0.0088income2*

R2 = 0.3646, N = 40, NR2 = 14.58 cv = 2(2, 0.01) = 9.21.

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Advantages of the White test:

a. It does not assume a specific functional form.

b. It is applicable when the variance depends on two or more variables.

Limitations of the White test:

a. It is an large-sample test.

b. It provides no information about the variance structure.

c. It loses many degrees of freedom when there are many regressors.

d. It cannot handle partitioned data.

e. It also captures specification errors.

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3. Consequences of Heteroskedasticity

If heteroskedasticity appears but OLS is used for estimation, how are the OLS estimates affected?

Unaffected: OLS estimators are still linear and unbiased because, on average, overestimates ar

e as likely as underestimates.

K,1,,0k ˆE kk

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3.1 OLS estimators are inefficient.

Some fluctuations of the error term are attributed to the variation in independent variables.

There are other linear and unbiased estimators that have smaller variances than the OLS estimator.

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3.2 Unreliable Hypothesis Testing

khetero

kols ˆvarˆvar

kˆse biased

unreliable testing conclusion

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4. Remedies

4.1 Heteroskedasticity-Corrected Standard Errors

Yi = 0 + 1X1i + 2X2i + i

heteroskedasticity: var(i) = i2

OLS estimators are unbiased.

The standard errors of OLS are biased.

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A heteroskedasticity-consistent (HC) standard error of an estimated coefficient is a standard error of an estimated coefficient adjusted for heteroskedasticity.

a. HC standard errors are consistent for any type of heteroskedasticity.

b. Hypothesis tests are valid with HC standard errors in large samples.

c. Typically, HC se > OLS se

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incorrect variance formula:

2i

2

1XX

ˆvar

Example 5:

Yi = 0 + 1Xi + i, var(i|Xi) = i.

correct variance formula:

22

i

2i

2i

1

XX

XXˆvar

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HC estimator of the variance of the slope coefficient in a simple regression model

22

i

2i

2i

1

XX

XXeˆvar .est

Example 6: HC Standard Errors, US data (UE_Tab0301)

***i i

22.14 0.031ols se24.32 0.039hc se

2

foodexp = 40.77 0.13 income

0.3171, N = 40.R

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Yi = 0 + 1X1i + 2X2i + i

i2 = c Zi

2

The variance is assumed to be proportional to the value of Zi

2

var(i) = i2E(i) = 0 cov(t, s) = 0 t = s

4.2 Weighted Least Squares

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Step 1: Decide which variable is proportional to the heteroskedasticity.

Step 2: Divide all terms in the original model by that variable (divide by Zi ).

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Step 3: Run least squares on the transformed model which has new variables. Note that the transformed model have an intercept only if Z is one of the explanatory variables.

For example, if Zi = X2i, then

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Example 7: WLS: US data (UE_Tab0301)

What are values of the estimated coefficients of the original model?

Has the problem of heteroskedasticity solved?

***

0.02342 14.0380

2

foodexp 10.1577 21.2858

0.0570, N = 40.

se

income income

R

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0 1

OLS estimate 40.77 0.128***

OLS se 22.14 0.031

HC se 24.32 0.039

WLS estimate 21.28 0.158***

WLS se 14.03 0.023

Comparing different estimates: US data (UE_Tab0301)

The WLS estimates have improved upon those of OLS.

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Other possibilities

• var(i) = cZi

• var(i) = cZi

• var(i) = c(a1X1i + a2X2i)

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In large samples HC standard errors are consistent measures for any type of heteroscedasticity. CI & t-test are valid.

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4.3 Re-specifying the Regression Model

4.3.1 Use another functional form E.g., Double-log: Less variation

The heteroskedasticity may be impure.

Example 8: US data (UE_Tab0301)

***

0.90 0.14se

2

ˆln foodexp 0.30 0.69 ln

0.4014, N = 40.

income

R

The hypothesis of constant variance can be rejected.

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** ***

i50.86 0.078se

2

foodexp 94.21 0.44 totexp

0.3698, N = 55.R

Empirical model: foodexpi = 0 + 1totexpi + i.

Example 9: India data (Food_India55)

The hypothesis of homoskedasticity can be rejected by the Park and White tests.

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** ***i

50.86 0.078ols se43.26 0.074hc se

foodexp 94.21 0.44 totexp.

** ***

37.9435 0.0632

se

foodexp 176.5439 0.4650 .

totexp totexp

***

0.78 0.12se

2

ˆln foodexp 1.15 0.74 ln totexp

0.4125, N = 55.R

Double-log

HC

WLS

Which model is the best?

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4.3.2 Other reformulations

E.g., take average of variables related to the size of observed units, adding more variables

Example 10: Data set “Concert”The concert tour of a singer in the US

revenue = 0 + 1adv + 2stad + 3cd

+ 4radio + 5weekend + .

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revenue 73 3.15adv 34.66stad 8.30cd

300radio 356weekend

se

revenue 1 adv cd81 2.10 50.20 stad 7.53

stad stad stad stad

radio weekend176 293

stad stad

revenue adv stad cd22 2.21 109 7.93

pop pop pop pop

2.53radio 4.28weekend

(1)

(2)

(3)

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Remarks:

•The variable Z is difficult to identify. The functional relationship between the error and Z is not known. Use WLS at last.

•With correct WLS, we expect the standard errors of the regression coefficients will be smaller than the OLS counterparts.

•A log transformation usually reduces the degree of heteroskedasticity.

•The hypothesis of homoskedasticity should not be rejected in the new model.

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5. A Complete Example

Sources: Section 8.2.2 (pp. 255 – 256) Section 10.5 (pp. 369 – 376)

pconi = 0 + 1regi + 2taxi + 3uhmi + i.

Empirical regression model

pconi1: petroleum consumption in the ith state

regi : motor vehicle registrations in the ith state (‘000)

taxi : the gasoline tax rate in the ith state(cents per gallon)

uhm : urban highway miles wihtin the ith state

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pcon = 389.57*** – 0.061reg – 36.47***tax + 60.76***uhm se, vif (0.04, 24.3) (13.15, 1.1) (10.26, 24.9)

Adj. R2 = 0.9192, N = 50.

^

Equation 1

Equation 2

pcon = 551.69*** + 0.19***reg – 53.59***tax se (0.012) (16.86)

Adj. R2 = 0.8607, N = 50.

^

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-800

-400

0

400

800

1,200

0 5,000 10,000 15,000 20,000

REG

resi

dual

Graphical investigation

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Park test

White test

ln(e2) = 1.65 + 0.95***ln(REG) R2 = 0.1657, N = 50 se (0.3083)

^

e2 = 11,098,291 + 140REG – 0.0005REG2 – 12.84REGTAX – 237,873TAX + 12347TAX2.

R2 = 0.6645, N = 50, NR2 = 33.22.

^

Checking for other specifications:

Double log, quadratic

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50N ,1989.0R

tax0103.0pop

reg1082.01684.0

pop

oncp

2

***

00349.007159.0

**

se

pcon = 551.69*** + 0.19***reg – 53.59***taxhc se (0.022) (23.90)

R2 = 0.8664, N = 50.

^

50N ,3600.0R

reg

tax3890.17

reg

1539.2181678.0

reg

oncp

2

6822.4

***

1033.48

******

01367.0

se

(4)

(5)

(6)

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Selected Exercises

Ch. 10: Q. 1, 3, 4, 5, 8, 10, 12, 14

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Yi = 0 + 1X1i + 2X2i + i

Regression Model

E(i|X1i,X2i) = 0

var(i|X1i,X2i) = 2

zero mean:

homoskedasticity:

cov(i, j|X1i,X2i,X1j,X2j) = i = j

no autocorrelation:

heteroskedasticity: var(i|X1i,X2i) = i2

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.

X1 X2

.

X3

.

X

Yf(Y)

HeteroskedasticityYi = 0 + 1Xi + i var(i|Xi) = i

2 for all i

Conditional Distribution

0

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Step 3: Test the overall significance of the equation in Step 2. (df = number of regressors)

Reject the hypothesis of homoskedasticity if NR2er

r > cv.

Statistic = NR2err ~ 2

df

Critical value (cv) = 2df,

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Step 1: Decide which variable is proportional to the heteroskedasticity.

Step 2: Divide all terms in the original model by that variable (divide by Zi ).

*i

*2i2

*1i1

*0i0

*i

i

i

i

i22

i

1i1

i0

i

i

XβXβXβY

ZZ

X

Z

Xβ

Z

1β

Z

Y

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Step 3: Run least squares on the transformed model which has new variables. Note that the transformed model have an intercept only if Z is one of the explanatory variables.

*i2

*1i1

*0i0

*i

i

i2

i

1i1

i0

i

i

βXβXβY

ZZ

Xβ

Z

1β

Z

Y

For example, if Zi = X2i, then

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In large samples HC standard errors are consistent measures for any type of heteroscedasticity. CI & t-test are valid.

WLS

HC se’s

OLS estimator Improve No

Standard errors Improve Improve

Specific form Yes No

Large sample No Yes