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hws07 - Economics 241B Answers to Exercise 7 1(Final...

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Economics 241B Answers to Exercise 7 1. (Final Examination, Economics 241B, Winter 2004) a. For the classic linear regression model, identi°cation of ° 1 follows from mulitiplying both sides of the regression by X t : X t Y t = °X 2 t + X t U t : Taking the expectation of both sides identi°es ° if E ( X t U t ) = 0 (because we have sample analogs for E ( X t Y t ) and E ( X 2 t ) . With measurement error, the resultant equation is X t Y t = °X 2 t + X t ( U t ° °V t ) : Because E ( X t V t ) 6 = 0 , ° = E ( X t Y t ) E ( X 2 t ) ° E ( X t V t ) . Because there is no sample analog for E ( X t V t ) , ° is not identi°ed. b. For Z t to be a valid instrument, it must be the case that n ° 1 n X t =1 Z t Y t P ! °Q ZX ; where Q ZX = lim n !1 n ° 1 P n t =1 Z t X t . To check n ° 1 n X t =1 Z t Y t = n ° 1 n X t =1 Z t ( °X t + U t ° °V t ) : Because Z t is a function of X t and X t is independent of U t : n ° 1 n X t =1 Z t U t P ! 0 : Next n ° 1 n X t =1 Z t V t = n ° 1 n X t =1 V t 1 ( X t ± med ( X )) ° n ° 1 n X t =1 V t 1 ( X t < med ( X )) ; which converges in probability to E ( V j X ± m ) P ( X ± m ) ° E ( V j X < m ) P ( X < m ) ;
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where m is the population median of X . Clearly P ( X ± m ) = P ( X < m ) = 1 2 . Thus Z is a valid instrument if E ( V j X ± m ) = E ( V j X < m ) : Such a result would hold if V and X were independent, but they are clearly dependent. Because V and X are dependent, the two conditional expecta- tions are not generally equal, and Z is not a valid instrument. In fact, if X is also symmetrically distributed around its mean (then mean=median) and E ( V j X ± m ) = ° E ( V j X < m ) , so that n ° 1 P n t =1 Z t V t P ! E ( V j X ± m ) 6 = 0 . c. The assumption that V t is symmetrically distributed yields additional mo- ment information. To exploit this, we set up the moment conditions. Because the model is in deviation-from-means form, the °rst moments are zero. The second moments yield EX 2 t = E ( X ± t ) 2 + ± 2 V EY 2 t = ° 2 E ( X ± t ) 2 + ± 2 U E ( X t Y t ) = °E ( X ± t ) 2 ; which is a system of 3 equations in 4 unknowns. For this system ° is not identi°ed. To estimate the parameters, we need further moment conditions. Because the distributions of U t and V t are symmetric, the third moments are zero, which yields EX 3 t = E ( X ± t ) 3 EY 3 t = ° 3 E ( X ± t ) 3 : We now have a system of 5 equations in 5 unknowns, so ° is identi°ed. We solve via the method of moments. In particular B MM = ° P n t =1 Y 3 t P n t =1 X 3 t ± 1 3 : 2. (Fall 1990 Final Examination) a) To correct for heteroskedasticity we transform the model to obtain an error that is homoskedastic. In the case at hand, we know V ( U t ) = ± 2 Z 2 t with Z t known, so the transformation is simple. From the de°nition of the variance, V ² U t Z t ³ = ± 2 , and so U t Z t is a homoskedastic error. The transformation is Y t Z t = ² + ° X t Z t + U t Z t :
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Why the presence of an intercept? Because P n t =1 U t = 0 does not imply P n t =1 U t Z t = 0 . To remove the intercept, we would need to remove the mean of Y t Z t and X t Z t .
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