# answers2008final - p(1 − ˆ p − 1 ª − 1 = ˆ p(1 −...

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–1– Econ 220B Answers to 220B Winter 2008 Final Exam 1a.) ( X 0 X ) 1 X 0 y b.) ( Rb ) 0 h R s 2 ( X 0 X ) 1 R 0 i 1 ( Rb ) R = £ 1 2 ¤ F (1 ,T 2) c.) Rb p R s 2 ( X 0 X ) 1 R 0 t ( T k ) d.) Replace s 2 ( X 0 X ) 1 with ( X 0 X ) 1 T X t =1 e 2 t x t x 0 t ( X 0 X ) 1 e.) Let g ( b )=log b 1 2log b 2 g 0 = £ 1 b 1 2 b 2 ¤ g ( b ) 2 / £ g 0 s 2 ( X 0 X ) 1 g ¤ χ 1 (1) 2a.) Factors such as education might be correlated with both ε i and s i b.) E ¡ x i £ 1 s i ¤¢ has rank 2. To test, regress s i on x i and test whether coe cients on x i area l lthesame c.) E ( x i ε i )= 0 , could test with GMM test of overidentifying restrictions d.) Step 1: regress s i on x i ,save f tted values ˆ s i Step 2: regress y i on constant and ˆ s i 3a.) L ∂p = N p T N 1 p =0 or ˆ p = N/T b.) T p p ) L N (0 ,p (1 p )) c.) h t = y t p 1 y t 1 p = y t p p (1 p ) d.) E ( h t )= E ( y t ) p p (1 p ) = p p p (1 p

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–2– Econ 220B e.) S = E ( h 2 t )= E · y 2 t 2 py t + p 2 p 2 (1 p ) 2 ¸ = · p 2 p 2 + p 2 p 2 (1 p ) 2 ¸ = 1 p (1 p ) f.) ∂h t ∂p = y t p 2 (1 y t ) (1 p ) 2 ∂g ∂p = T 1 T X t =1 · y t p 2 + (1 y t ) (1 p ) 2 ¸ ∂g ∂p ¯ ¯ ¯ ¯ p p = · 1 ˆ p + 1 1 ˆ p ¸ = 1 ˆ p (1 ˆ p ) g.) ˆ V = © p (1 ˆ p )] 1 p (1 ˆ p )][ˆ
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Unformatted text preview: p (1 − ˆ p )] − 1 ª − 1 = ˆ p (1 − ˆ p ) which is same answer as for (b) h.) Yes because the moment condition E ( h t ) = 0 would still hold i.) For b, plim is same but distribution is now √ T (ˆ p − p ) L → N (0 , Q ) for Q = ∞ X j = −∞ E ( y t − p )( y t − j − p ) For (e) we have S = ∞ X j = −∞ E · y t − p p (1 − p ) ¸· y t − j − p p (1 − p ) ¸ = · 1 p 2 (1 − p ) 2 ¸ Q and V = ( [ p (1 − p )] − 1 · Q p 2 (1 − p ) 2 ¸ − 1 [ p (1 − p )] − 1 ) − 1 = Q again the same answer....
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answers2008final - p(1 − ˆ p − 1 ª − 1 = ˆ p(1 −...

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