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Unformatted text preview: r. You may need to consider
2
plotting jbi j against xji or ln xji or ...
ε Note that the relationship may be over many xji . You may need to
consider plotting jbi j against x1i and x2i and ... or against yi
ε Lecture 7 (heteroscedasticity) EMET2007/6007 24 th April 2013 10 / 34 An example: Explaining wages wage =
[ 3.96 + 0.595educ + 0.268exper b = wage
ε 0.005exper 2 wage the residuals
[ We plot the bi against educi and then we plot the jbi j against educi
ε
ε Lecture 7 (heteroscedasticity) EMET2007/6007 24 th April 2013 11 / 34 15
10
Residuals
5
0
5 0 Lecture 7 (heteroscedasticity) 5 10
years of education EMET2007/6007 15 20 24 th April 2013 12 / 34 15
10
arr
5
0 0 Lecture 7 (heteroscedasticity) 5 10
years of education EMET2007/6007 15 20 24 th April 2013 13 / 34 How can I tell if I have it? Formal evidence from Testing for
heteroscedasticity
It is important to know whether there is heteroscedasticity because then
OLS may not be the most e¢ cient linear estimator anymore
BreuschPagan test for heteroscedasticity
H0 : Var (εi jx1 , x2 , , xk ) = Var (εi jxi ) = σ2 That is, the null hypothesis states that Var (εi jxi ) = Var (εi ) is not a
function of xi
Under MLR.4:
Var (εi jxi ) = E ε2 jxi
[E (εi jxi )]2 = E ε2 jxi
i
i
) E ε2 jxi = E ε2 = σ2
i
i The mean of ε2 must not vary with x1 , x2 ,
i
Lecture 7 (heteroscedasticity) EMET2007/6007 , xk
24 th April 2013 14 / 34 BreuschPagan test for heteroscedasticity (cont.)
The alternative hypothesis: H1
H1 : Var (εi jx1 , x2 , , xk ) = Var (εi jxi ) = f (xi ) That is, the alternative hypothesis states that the variance changes
with xi (or Var (εi jxi ) 6= Var (εi ))
The mean of ε2 may vary with x1 , x2 ,
i Lecture 7 (heteroscedasticity) EMET2007/6007 , xk 24 th April 2013 15 / 34 BreuschPagan test for heteroscedasticity (cont.)
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This note was uploaded on 06/15/2013 for the course EMET 2007 taught by Professor Strachan during the Two '13 term at Australian National University.
 Two '13
 strachan

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