¡Descarga Solución de Conjunto de Ejercicios 4: Distribución Asintótica del Estimador Ridge y más Ejercicios en PDF de Econometría solo en Docsity!
Solution to Problem Set 4
Exercise 1 Consider the linear regression model. Recall the ridge estimator is
deÖned as ^^
Ridge = (X^0 X + Ik)
1 X^0 y for some positive constant :
(a) Derive the asymptotic distribution of
p n
^
Ridge
(b) Suppose that dim ( ) = 1 for simplicity. Construct a 99% conÖdence
interval based on the limit distribution in (a).
(c) How would you test a hypothesis that exp (2 ) = 0 against a two-sided
alternative?
Proof. (a)
p n
^
Ridge
n
X^ n
i=
xix 0 i +
n
Ik
! (^1) p n
n
X^ n
i=
xix 0 i +^
p n
X^ n
i=
xi"i
p n
The Örst term 1 n
Pn i=1 xix
0 i +^
1 n Ik !p Q since limn! 1 n Ik = 0 for constant
; and by CMT and Assumption 4,
1 n
Pn i=1 xix
0 i +^
1 n Ik
!p Q ^1 : For the remaining terms, we have that 1 n
Pn i=1 xix
0 i !p^ Q^ ;^ and our usual term p^1 n
Pn i=1 xi"i^ !d^ N^
0 ; ^2 Q
: Putting everything together and applying
Slutsky theorem, we have
p n
^
Ridge
! (^) dQ 1 Q
p n
p n + Q 1 N
2 Q
= N
2 Q 1
So unsurprisingly, the ridge estimator has the same asymptotic distribution as
OLS! In this case we say, the two are asymptotically equivalent.
(b) The 99% asymptotic conÖdence interval for is given by [^^
Ridge
2 : 58 SE
^
Ridge
]; where SE
^
Ridge
is exactly the same as SE
^
OLS
. Note
that the small sample variance of ^^
Ridge was di§ erent and depended on :
(c) To test a hypothesis that exp (2 ) = exp
N U LL
; Örst use the delta
method to obtain that
p n
exp
2 ^^
Ridge
exp (2 )
!d N
2 Q 1 exp (4 )
Therefore,
exp
2 ^^ Ridge exp(2 ) p 4 ^2 Q ^1 exp(4 )=n
N (0; 1) for large n; and we can consistently
estimate the unknown quantities in the denominator, so that
exp
2 ^^ Ridge exp(2 ) dSE
exp
2 ^^ Ridge
N (0; 1) ; with
dSE
exp
2 ^^
Ridge
r
4 b
(^2) b Q ^1 exp
4 b^
ridge =n !p
p 4 ^2 Q ^1 exp (4 ) =n
by CMT and since b^
ridge is consistent.
Finally, we calculate z =
exp
2 ^^ Ridge exp( 2 N U LL) dSE
exp
2 ^^
Ridge^ and reject the null when-
ever jzj > zcrit = 2 with some pre-speciÖed signiÖcance level 1 :
Exercise 2 Use the asymptotic distribution from Problem set 3 of ^
2 to con-
struct a 95% conÖdence interval for =
p ^2 :
Proof. From Problem Set 3, we have that
p n
^
2 2
!d N
0 ; E"
4 1 ^
4
Using the delta method, we know that
p n
p ^
2
p ^2
!d N
0 : 25 ^2
E"
4 1 ^
4
since @
p ^2 @^2 =^ p^0 :^5 ^2
: So a 95% conÖdence interval for
p ^2 is given by
p
^ 2 1 : 96
p n
q (E"^41 ^4 )
Exercise 3 Use the asymptotic distribution you derived in Problem Set 3 for
the sample kth^ moment, to construct a 99% conÖdence interval for the true
unknown population kth^ moment.
Proof. We showed that the asymptotic distribution of the k th sample moment
is given by p n (^k k) !d N
0 ; 2 k 2 k
A 95% conÖdence interval for k is therefore given by ^k 2 : 58
q 2 k ^2 k n
Exercise 4 Consider the model yi = x 0 i +^ "i;^ with the standard assumptions
on " made in the lecture holding. Let zi = (x 0 i ) +^ ui;^ and^ E^ [uijx
0 i] = 0:^ The
aim is to estimate : Consider the following 2 step estimation procedure. Step
1: regress yi on xi and obtain ^^
OLS : Step 2, estimate ; by regressing zi on
x^0 i ^^
OLS : Show that the estimator ^ described in the two step procedure is a consistent
estimator for the true unknown.
Proof. ^^
OLS = (X^0 X)
1 X^0 y = + (X^0 X)
1 X^0 "; and
^ =
^
OLS
X
0 X ^^
OLS
^
OLS
X
0 z;
where z is an n 1 vector containing z 1 ; :::; zn: Plugging in the expression for
^
OLS :
^ =
^
OLS
X
0 X ^^
OLS
^
OLS
X
0 z
= (y 0 Pxy)
1 y 0 Pxz
