高级计量经济学试题final_2017a

2 E xi u0 i "i 0 E ui ui "2 i
The CLT applies if wi "i is iid, mean zero, and has …nite second moments. The vector wi "i is iid because it is assumed to be a random sample. It is mean zero since E (xi "i ) = 0 by part (a) and E (ui "i ) = 0 by the control function projection. The variable wi "i has …nite second moment if the observations have …nite fourth moments.
2
Together, we obtain p n b !d Q
1 1
In addition, this requires that Q 3. GMM criterion (a) Evaluated at the true value
0,
b
N (0; ) = N 0; Q
1
Q
1
:
exists.
mn (
Econometrics 710 Final Exam, Spring 2017 Sample Answers 1. IV Regression (a) b = (Z 0 X )
1
(b) Write
P 0 Z 0Y = ( n i=1 zi xi ) b E b j Z; X
1
P ( n i=1 zi yi )
0) =
1X zi e i n
i=1
n
which is the average of iid random vectors with mean zero (by assumption). Standardized, the CLT implies n 1 X p zi ei !d Z N (0; ) n
0 0
E z i ei E ( zi e i )
u0 i
0
+ E (ui "i ) + E (ui "i )
E zi u0 i
The three components are each zero. First, E (zi ei ) = 0 by the IV assumption. Second, E (zi u0 i ) = 0 by the reduced form projection equation. Third, E (ui "i ) = 0 by the control function projection equation. 0 0 (b) Derive the asymptotic distribution of ( b ; b) . First, write wi = (x0 i ; ui ) so that we can write the estimators as ! 1 n ! n X X b 0 = wi wi wi yi : b
i=1
e ei
= yi
n 1X z i zi 0 e ei 2 n i=1
xi 0 b
That is, e bi is calculated using the bootstrap estimate b In either case, we obtain B replications of the statistic Jnb by simulation. The p-value for the test is then calculated as pn =
C = f : Jn ( )
(f) For a bootstrap test, sample (yi ; xi ; zi ) iid from the data. If the estimator e was used, 3
f e( ) 1 = where c is the 1 quantile of the distribution of 2 ` . When W ( ) = P 2 n 1 0 x0 and the con…dence region i ) then the weight matrix depends on i=1 zi zi (yi n is n o 0 0 C= : (Y 0 Z X Z )0 e ( ) 1 Z 0 Y Z 0 X nc which is not an ellipse.
j Z; X A ! !
1
n X i=1
0 x i zi n X i=1
!
1
0 x i zi 1
!
j Z; X A
1
1
E
0 zi e i e i zi
j zi ; xi
0 x i zi
0 zi z i
2 i
!
n X i=1
1
n X i=1
0 x i zi
!
= E e2 i j zi ; xi
Notice that this is the variance conditional on both z and x 2. Control function regression. (a) The reduced form equation for xi is xi = E (xi "i ) = E =
1
!
b
0 1
Leabharlann Baidu
j Z; X zi ei
n X i=1
z i x0 i
!
0 @
n X i=1
!
n X i=1
0 e i zi
!
=
= where
n X i=1
n X i=1
z i x0 i zi x 0 i
!
!
1
n X n X i=1 j =1
E
0 zi ei ej zj
1
n X i=1 2 i
n X i=1
1
then the limit distribution is Z 0
Z 0X
1
X 0 Z (Z 0 Z )
1
Z 0 Y . This does not take advantage of H0 .
> cjH0 ) ! P
2 `
>c =
is the set
for which the test does not reject. It is cg
0 0z i
+ ui so
0
zi + ui "i
E (zi "i ) + E (ui "i ) : u0 i . Substituting into the …rst
The de…nition for "i is from ei = u0 i + "i so "i = ei expression on the right side, we …nd E (xi "i ) = =
i=1 i=1 0 Centered and standardized, since yi = x0 i + ui + "i
p By the WLLN
n
b
b
=
1X 0 wi wi n
i=1
n
!
1
1 X p w i "i n
i=1
n
!
:
1X 0 0 wi wi !p E wi wi = n
i=1
n
0 E (xi x0 i ) E (xi ui ) 0 E (ui x0 i ) E (ui ui )
i=1
where
0 2 = E zi zi ei
It follows that Jn ( (b) If W =
1 0)
=
p nmn (
0)
0
p W nmn (
1Z
0) 2 `
!d Z 0 W Z
(d) An asymptotic test rejects H0 in favor of H1 at level if Jn ( 0 ) > c where c is the 1 quantile of the distribution of 2 since ` . This test has asymptotic level of P (Jn ( (e) A 1 con…dence region for
n X i=1
=
zi x0 i
!
1
Then
n X i=1
zi ei
! ! 1
= E@ =
0
n X i=1
zi x 0 i
1
!
1
n X i=1 n X i=1
zi x0 i zi x0 i
! !
n X i=1
zi e i
1
= = 0
n X i=1
n X i=1
j Z; X A ! !
zi E (ei j Z; X )
zi E (ei j zi ; xi )
Thus by the law of iterated expectations E b =
and b is unbiased for . (c) Since E b j Z; X = var b jX; Z = E 0
= E@ =
b
n X i=1
zi x0 i
=Q
say. The WLLN applies since wi are iid, if xi and ui have …nite second moments. By the CLT n 1 X p wi "i !d N (0; ) n
i=1
where
0 2 = E wi wi "i =
2 E xi x0 i "i 0 E ui xi "2 i
0)
where ` is the dimension of zi . P n 1 0 e2 where e (c) An estimator which takes advantage of H0 is e = n ei = yi x0 i i 0 i=1 zi zi e 1 f e e f and W = : The estimator is unbiased for , and W is consistent W = Pn for 1 under H . This is di¤erent than the standard estimator b = 1 0e 2 and z z b 0 i i i i=1 n b and b is a consistent estimator of , for example b = c = b 1 where e W bi = yi x0 i X 0 Z (Z 0 Z )
4
B 1 X 1 (Jnb > Jn ( B b=1 0)
b
e bi
= yi =
X 0Z
xi 0 b
Z 0Z
1
Z0 X
1
X0 Z
Z0 Z
1
Z0 Y
The bootstrap test rejects H0 at level
if pn < , otherwise it does not reject H0 :
then we set Jn = nmn 0 e 1 mn n 1X mn = zi yi xi 0 b n
i=1
e
=
where b is either 2SLS or GMM on the original sample. This is appropriate because b is the analog of 0 in the bootstrap distribution. If the estimator b was used then we would alternatively set Jn = nmn 0 b 1 mn n X b = 1 zi zi 0 e bi 2 n