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Lecture #10

Lecture #10. Studenmund (2006): Chapter 10. Heteroscedasticity. Heteroskedasticity. Objectives What is heteroscedasticity? What are the consequences? How is heteroscedasticity be identified? How is heteroscedasticity be corrected?. Homoscedasticity Case. Y i. f(Y i ). expenditure. .

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Lecture #10

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  1. Lecture #10 Studenmund (2006): Chapter 10 Heteroscedasticity Heteroskedasticity • Objectives • What is heteroscedasticity? • What are the consequences? • How is heteroscedasticity be identified? • How is heteroscedasticity be corrected?

  2. Homoscedasticity Case Yi f(Yi) expenditure . . . Var(i) = E(i2)= 2 x11=80 x12=90 x13=100 x1i income The probability density function for Yi at three different levels of family income, X1i , are identical.

  3. yi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 xi Homoscedastic pattern of errors The scattered points spread out quite equally

  4. Yi f(Yi) expenditure . . . Var(i) = E(i2)= i2 x1i x11 x12 x13 income Heteroscedasticity Case The variance of Yi increases as family income, X1i, increases.

  5. . yi . . Small i associated with small value of Xi . large i associated with large value of Xi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 xi Heteroscedastic pattern of errors The scattered points spread out quite unequally

  6. Two-variable regression: Yi = 0 +  1 X1i + i  xy  1 = =  wi Yi = wi (0 + 1Xi + i) ^  x2 ^ ^ =>  1 =  1 + wi i E(1) = 1unbiased ^ Var (1) = E (1 - 1)2 = E (wi i)2 ^ = E (w1212 + w2222 + …. + 2w1w2 1 2 + …) = w1212 + w22 22 + … …..+ 0 + ... = i2  wi2 if 12  22  32  … i.e.,heteroscedasticity i2 = xi2 x2 if 12 = 22 = 32 = … i.e.,homoscedasticity = i2 2 ^ (xi2)2 Var (1)= xi2 Var(i) = E(i2) = i22 Definition of Heteroscedasticity: Refer to lecture notes Supplement #03A

  7. ^ 2. Var ( i)s are not minimum. =>not the best=>not efficiency=>notBLUE i2 2 Two-variable case 3. Var ( 1) = instead of Var( 1)= ^ ^ x2 x2 5. t and F statistics are unreliable. Y =  0+  1 X +  ^ i2 SEE =  RSS =  i2 ^ ^ 4. 2= is biased, E(2)  2 ^ ^ n-k-1 Cannot be min. Consequences of heteroscedasticity 1. OLS estimators are stilllinear and unbiased

  8. 1.Graphical method : ^ ^ plot the estimated residual ( i ) or squared (i2 ) against the predicted dependent Variable (Yi) or any independent variable(Xi). ^ Observe the graph whether there is a systematic pattern as:  2 ^ Yes, heteroscedasticity exists Y ^ Detection of heteroscedasticity

  9. 2  2  2  2  2 2 ^ ^ ^ ^ ^ ^ yes yes no heteroscedasticity Y Y Y ^ ^ ^ ^ ^ ^ Y Y Y yes yes yes Detection of heteroscedasticity: Graphical method

  10. Yes, heteroscedasticity Yes, heteroscedasticity no heteroscedasticity Yes, heteroscedasticity

  11. Statistical test: (i)Park test H0 : No heteroscedasticity exists i.e., Var( i ) = 2 (homoscedasticity) H1 : Yes, heteroscedasticity exists i.e., Var( i ) = i2 Park test procedures: 2. Take square and take log : ln ( i2) ^ 1. Run OLS on regression: Yi = 1 + 2 Xi + i , obtain i ^ 3. Run OLS on regression: ln ( i2) = 1* + 2* ln Xi + vi ^ Suspected variable that causes heteroscedasticity 4. Use t-test to test H0 : 2* = 0 (Homoscedasticity) If t* > tc ==> reject H0 ==> heteroscedasticity exists If t* < tc ==> not reject H0 ==> homoscedasticity

  12. Example: Studenmund (2006), Equation 10.21 (Table 10.1), pp.370-1 May misleading Park Test Practical In EVIEWS Procedure 1: PCON: petroleum consumption in the ith state REG: motor vehicle registration TAX: the gasoline tax rate

  13. Graphical detection

  14. Procedure 2: Obtain the residuals, take square and take log

  15. Scatter plot Horizontal variable

  16. Procedure 3 & 4 If | t | > tc => reject H0 => heteroscedasticity Refers to Studenmund (2006), Eq.(10.23), pp.373

  17. (ii)Breusch-Pagan test, or LMtest H0 : homoscedasticity Var ( i ) = 2 H1 : heteroscedasticity Var ( i ) = i2 Test procedures: (1) Run OLS on regression: Yi = 0 + 1X1i + 2X2i +...+ qXqi + i , obtain the residuals, i ^ (2)Run the auxiliary regression: i2 = 0 + 1 X1i + 2X2i +… +qXqi + vi ^ R2u / q Or F= (1 - R2u) / n-k if F*> Fcdf ==> reject the Ho (3) Compute LM=W= nR2 (4) Compare the Wand 2df (where the df is #(q) of regressors in (2)) if W > 2df ==> reject the Ho

  18. Yi = 0 + 1X1i + 2X2i + 3X3i + i

  19. FC(0.05, 5, 44) = 2.45 2(0.05, 5) = 11.07 2(0.10, 5) = 9.24 W= Decision rule:W > 2df ==> reject the Ho BPG test for a linear model PCON=0+1REG+2Tax+ The W-statistic indicates that the heteroscedasticity is existed.

  20. FC(0.05, 5, 44) = 2.45 2(0.05, 5) = 11.07 2(0.10, 5) = 9.24 W= Decision rule:W > 2df ==> reject the Ho The BPG test for a transformed log-log model: log(PCON)=0+1log(REG)+2log(Tax)+ The W-statistic indicates that the heteroscedasticity is still existed. Therefore, a double-log transformation may notnecessarily remedy the Heterocsedasticity.

  21. H0 : homoscedasticity Var ( i ) = 2 H1 : heteroscedasticity Var ( i ) = i2 Test procedures: (1) Run OLS on regression: Yi = 0 + 1 X1i +  2 X2i + i , obtain the residuals, i ^ (2)Run the auxiliary regression: i2 = 0 + 1 X1i + 2 X2i + 3 X21i + 4 X22i + vi ^ (iiia) White’s general heteroscedasticity test (no cross-term) (The White Test) (3) Compute W (or LM) = nR2 (4) Compare the W and 2df (where the df is # of regressors in (2)) if W > 2df ==> reject the Ho

  22. (iiib) White’s general heteroscedasticity test (with cross-terms) (The White Test) H0 : homoscedasticity Var ( i ) = 2 H1 : heteroscedasticity Var ( i ) = i2 Test procedures: (1) Run OLS on regression: Yi = 0 + 1 X1i +  2 X2i + i , obtain the residuals, i ^ (2)Run the auxiliary regression: i2 = 0 + 1 X1i + 2 X2i + 3 X21i + 4 X22i + 5 X1i X2i + vi ^ Cross-term (3) Compute W (or LM) = nR2 (4) Compare the W and 2df (where the df is # of regressors in (2)) if W > 2df ==> reject the Ho

  23. With cross-term No cross-term

  24. FC(0.05, 5, 44) = 2.45 2(0.05, 5) = 11.07 2(0.10, 5) = 9.24 W= Decision rule: W > 2df ==> reject the Ho White test for a linear model PCON=0+1REG+2Tax+ The W-statistic indicates that the heteroscedasticity is existed.

  25. FC(0.05, 5, 44) = 2.45 2(0.05, 5) = 11.07 2(0.10, 5) = 9.24 W= Decision rule: W > 2df ==> reject the Ho The White test for a transformed log-log model: log(PCON)=0+1log(REG)+2log(Tax)+ The W-statistic indicates that the heteroscedasticity is still existed. Therefore, a double-log transformation may notnecessarily remedy the Heterocsedasticity.

  26. Another example 8.4 (Wooldridge(2003), pp.258) W = 2(0.05, 9) = 16.92 2(0.10, 9) = 14.68 Decision rule: W > 2df => reject the H0 The White test for a linear model PCON=0+1REG+2Tax+ The test statistic indicates heteroscedasticity is existed.

  27. Testing the log-log model 2(0.05, 9) = 16.92 2(0.10, 9) = 14.68 Decision rule:W < 2df => not reject H0 The White test for a log-log model The test statistic indicates heteroscedasticity is not existed Using the log-log transformation in some cases may remedy the heteroscedasticity, (But not necessary). W=

  28. If Var(i2)=2Zi Yi 1 X1i X2ii Then each term divided by Zi =0 + 1 + 2 + ZiZi Zi Zi Zi Remedy :Weighted Least Squares(WLS) Suppose : Yi = 0 + 1 X1i + 2 X2i + i E(i) = 0, E(i j )= 0 i  j Vqr (i2) = i2 = 2 [f(ZX2i)] = 2Zi2 If all Zi = 1 (or any constant), homoscedasticity returns. But Zican be any value, and it is the proportionality factor. In the case of 2 was known:Tocorrect the heteroscedasticity Transform the regression: => Y* = 0 X0* + 1 X1* + 2 X2* + i*

  29. Why the WLS transformation can remove the heteroscedasticity? In the transformed equation where i 1 (i) E ( ) = E (i) = 0 Zi Zi i 1 1 (ii) E ( )2 = E (i2) = Zi22 = 2 Zi Zi2 Zi2 i j 1 (iii) E ( ) = E ( i j ) = 0 ZiZj Zj Zi These three results satisfy the assumptions of classical OLS.

  30. If the residuals plot against X2i are as following : 2 ^ ^ i + 0 X2 - X2 Hence, the transformed equation becomes Therefore, we might expect Zi2 = X2i2 =>Zi =X2i i2 = Zi22 Yi 1 X1i X2ii = 0+1+ 2 + X2i X2i X2i X2i X2i Now this becomes the intercept coefficient These plots suggest variance is increasing proportional to X2i2. The scattered plots spreading out as nonlinear pattern. => Yi* = 1 X0* + 2 X1* + 3 + * Where *i satisfies the assumptions of classical OLS

  31. Example: Studenmund (2006), Eq. 10.24, pp.374 C.V.=0.3392

  32. The correction is built in EVIEWS Use the weight (1/REG) to remedy the heteroscedasticity

  33. WLS result OLS result Refers to Studenmund (2006), Eq.(10.28), pp.376

  34. W= FC(0.05, 5, 44) = 2.45 2(0.05, 5) = 11.07 2(0.10, 5) = 9.24 Decision rule: W < 2df ==> not reject Ho Now, after the reformulation The test statistic value indicates that the heteroscedasticity is not existed.

  35. Alternative remedy of heteroscedasticity: reformulate with per capita

  36. FC(0.05, 5, 44) = 2.45 2(0.05, 5) = 11.07 2(0.10, 5) = 9.24 W= Decision rule: W < 2df ==> not reject Ho Now, after the reformulation The test statistic value indicates that the heteroscedasticity is not existed.

  37. W < 2df => not reject Ho

  38. If the residuals plot against X2i are as following : ^ i 2 ^ + 0 X2 - X2 i2 = Zi2 Therefore, we might expect hi2 = X2i =>hi = X2i The transformed equation is Yi 1 X1i X2ii = 0+ 1 + 2 + X2i X2i X2i X2i X2i => Yi* = 1 X0* + 2 X1* + 3 X2* + * This plot suggests a variance is increasing proportional to X2i. The scattered plots spreading out as a linear pattern

  39. Transformation: divided by the squared root term --“@sqrt(X)” Yi 1 X1i X2ii = 0+ 1 + 2 + X2i X2i X2i X2i X2i

  40. 2(0.05, 5) = 11.07 2(0.10, 5) = 9.24 W < 2df => not reject Ho calculate: the C.V. = 0.36928 Compare to the transformation divided by the REG, the CV of That one is smaller.

  41. Example: Gujarati (1995), Table 11.5, pp.388 Simple OLS result : R&D = 192.99 + 0.0319 Sales SEE = 2759 (0.194) (3.830) C.V. = 0.9026

  42. White Test for heteroscedasticity W= 2(0.05, 2)= 5.9914 2(0.10, 2)= 4.60517

  43. Observe the shape pattern of residuals:linear or nonlinear?

  44. Transformation equations: Yi ^ 1 ( ) = -246.67 + 0.036 Xi Xi Xi (-0.64) (5.17) 1. ^ =>(1) R&Di = -246.67 + 0.036 SalesSEE = 7.25 (-0.64) (5.17) C.V. = 0.8195 Yi 1 Xi 2. Compare the C.V. To determine which weight is appropriated ( ) = 0 + 1 Xi Xi Xi ^ (2)R&D = -243.49 + 0.0369 Sales SEE = 0.021 (-1.79) (5.52) C.V. = 0.7467

  45. Transformation: divided by the squared root term --“@sqrt(X)” Yi ^ 1 +1 = 0 Xi Xi Xi

  46. Yi ^ 1 +1 = 0 Xi Xi Xi calculate: the C.V. = 0.8195

  47. After transformation by @sqrt(x), the W-statistic indicates there is no heteroscedasticity 2(0.05, 2)= 5.9914 2(0.10, 2)= 4.60517 W < 2df => not reject Ho

  48. After transformation by @sqrt(Xi), residuals still spread out

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