讲之异方差问题.pptVIP

Transformation: divided by the suspected variable (Xi) 1 Xi ?0 + ?1 Yi Xi ^ = 1 Xi ?0 + ?1 Yi Xi ^ = Calculate the C.V. = 0.7467 After transformation divided by the suspected X, the W-statistic indicates there is no heterose\cedasticity ?2(0.05, 2)= 5.9914 ?2(0.10, 2)= 4.60517 W ?2df = not reject Ho After transformation divided by Xi, residuals spread out more stable Procedure 3 4 Refers to Studenmund (2006), Eq.(10.23), pp.373 If | t | tc = reject H0 = heteroscedasticity (ii) White’s general heteroscedasticity test (no cross-term) (Breusch-Pagan test, or LM test) (4) Compare the W and ?2df (where the df is #(q) of regressors in (2)) if W ?2df == reject the Ho H0 : homoscedasticity Var ( ?i ) = ?2 H1 : heteroscedasticity Var ( ?i ) = ?i2 (3) Compute LM=W= n?R2 Or F= R2u / q (1 - R2u) / n-k if F* Fcdf == reject the Ho 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 ^ Yi = ?0 + ?1X1i + ?2X2i + ?3X3i + ?i W= BPG test for a linear model PCON=?0+?1REG+?2Tax+? The W-statistic indicates that the heteroscedasticity is existed. FC(0.05, 5, 44) = 2.45 ?2(0.05, 5) = 11.07 ?2(0.10, 5) = 9.24 Decision rule: W ?2df == reject the Ho FC(0.05, 5, 44) = 2.45 ?2(0.05, 5) = 11.07 ?2(0.10, 5) = 9.24 Decision rule: W ?2df == reject the Ho W= 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 not necessarily remedy the Heterocsedasticity. (4) Compare the W and ?2df (where the df is # of regressors in (2)) if W ?2df == reject the Ho H0 : homoscedasticity Var ( ?i ) = ?2 H1 : heteroscedasticity Var ( ?i ) = ?i2 (3) Compute W (or LM) = n?R2 Test procedures: (1) Run OLS on regression: Yi = ?0 + ?1 X1i + ? 2 X2i + ?i ,