HEC Lausanne - Advanced Econometrics
Christophe HURLIN
Correction Final Exam. January 2014. C. Hurlin
Exercise 1: MLE, parametric tests and the trilogy (30 points)
Part I: Maximum Likelihood Estimation (MLE)
Question 1 (2 points): since the random variables X1 ; ::; XN are i:i:d: (0.5 point), the loglikelihood of the sample x1 ; ::; xN is de…ned as to be:
`N ( ; x) =
N
X
ln fX (xi ; )
(1)
i=1
with (0.5 point)
ln fX (xi ; ) =
1
ln ( )
1
ln (c) +
ln (xi )
(2)
So, we have (1 point):
`N ( ; x) =
N
N ln ( )
ln (c) +
N
X
1
ln (xi )
(3)
i=1
Question 2 (2 points): the ML estimator of
is de…ned as to be (0.5 point):
b = arg max `N ( ; x)
(4)
2R+
The log-likelihood equation (FOC) is the following:
with (0.5 point)
@ ln `N ( ; x)
gN b; x =
@
@ ln `N ( ; x)
@
By solving this system, we get:
=
b
N
N
+ 2 ln (c)
b
b
b = ln (c)
1
=0
(5)
b
N
1 X
ln (xi ) = 0
b2 i=1
N
1 X
ln (xi )
N i=1
(6)
(7)
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The SOC is based on the Hessian number:
HN
2
b; x = @ ln `N ( ; x)
@ 2
b
N
= 2
b
2N
b3
ln (c)
!
N
1 X
ln (xi )
N i=1
(8)
Given the FOC, the Hessian number is equal to (0.5 point):
HN b; x
=
=
=
N
b2
2N
b3
ln (c)
N
2N b
2
b
b3
N
<0
b2
!
N
1 X
ln (xi )
N i=1
(9)
(10)
This number is negative, so we have a maximum. The maximum likelihood estimators of
the parameter is de…ned as to be (0.5 point):
b = ln (c)
N
1 X
ln (Xi )
N i=1
(11)
Question 3 (2 points): The sequence of i:i:d: (0.5 point) random variables ln (X1 ) ; ::; ln (XN )
satisfy E (ln (Xi )) = ln (c)
: Given the WLLN, we have (0.5 point):
N
1 X
p
ln (Xi ) ! E (ln (Xi )) = ln (c)
N i=1
By using the CMP theorem for a function g (z) = ln (c) z (0.5 point), we have:
!
N
1 X
p
g
ln (Xi ) ! g (ln (c)
)
N i=1
(12)
(13)
or equivalently
ln (c)
N
1 X
p
ln (Xi ) ! ln (c)
N i=1
ln (c) +
(14)
As a consequence (0.5 point):
p
b!
The estimator b is (weakly) consistent.
Question 4 (2 points): Since the problem is regular (0.5 point), we have:
p
N b0
d
0
! N 0; I
1
( 0)
(15)
where 0 denotes the true value of the parameter and I( 0 ) the (average) Fisher information matrix for one observation. The Fisher information matrix associated to the sample
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is equal to:
I N ( 0)
= E ( HN ( 0 ; X))
!!
N
2N
1 X
N
ln (c)
ln (Xi )
= E
2 +
3
N i=1
0
0
!
N
N
2N
1 X
=
ln (c)
E (ln (Xi ))
2 +
3
N i=1
0
0
N
=
2
0
N
=
2
0
+
+
2N
3
0
2N
(ln (c)
ln (c) +
0)
(17)
(18)
(19)
0
(20)
3
0
N
=
(16)
(21)
2
0
Since the sample is i:i:d:, we have
I ( 0) =
1
N
IN ( ) =
1
(22)
2
0
As a consequence (1.5 point):
p
or equivalently
d
N b
! N 0;
0
b asy N
0;
2
0
(23)
2
0
(24)
N
Question 5 (2 points). The Fisher information matrix for one observation is equal to:
I ( 0) =
1
(25)
2
0
A …rst natural consistent estimator is given by (0.5 point):
1
b
I b = 2
b
(26)
A second possible estimator is the BHHH estimator based on the cross-product of the
gradients:
N
2
1 X
b
gi xi ; b
(27)
I b =
N i=1
with
gi xi ; b =
1 ln (c) ln (xi )
+
b
b2
(28)
So, a second consistent estimator is given by (0.5 point):
N
1 X
b
I b =
N i=1
1 ln (c) ln (xi )
+
b
b2
2
(29)
Mid-term 2013. Advanced Econometrics. HEC Lausanne. C. Hurlin
The asymptotic variance of b is equal to (0.5 point):
1
Vasy b = I
N
1
( 0) =
4
2
0
(30)
N
So, two alternative estimator of the asymptotic variance of b are given by (0.5 point for
each estimator):
b2
b
b
(31)
Vasy
=
N
! 1
N
2
X
1 ln (c) ln (xi )
b
b
Vasy
=
(32)
+
b
b2
i=1
Part II: Parametric tests
Question 6 (2 points). Given the Neyman Pearson lemma the rejection region of the UMP
test of size is given by (0.5 point):
LN ( 0 ; x)
<K
LN ( 1 ; x)
where K is a constant determined by the size
`N ( 0 ; x)
(33)
or equivalently
`N ( 1 ; x) < ln (K)
(34)
It gives (0.5 point):
N ln ( 0 )
N
ln (c)+
0
1
N
X
1
0
ln (xi )+N ln ( 1 )+
1
1
0
1
0
0 1
1
<
0;
N
X
1
1
1
N
X
ln (xi ) < ln (K)
i=1
ln (xi ) < K2
(35)
i=1
ln ( 1 )) + N ln (c)
1
Since
ln (c)
1
i=1
where K2 = ln (K) + N (ln ( 0 )
N
N
X
1
1
0
1
is a constant term.
ln (xi ) < K2
(36)
i=1
we have (1 point):
N
X
ln (xi ) > K3
(37)
i=1
with K3 = K2
0 1= ( 1
0) :
This inequality can be re-expressed as:
ln (c)
with K4 = ln (c)
form (1 point):
N
1 X
ln (xi ) < K4
N i=1
K3 =N: The rejection region of the UMP test of size
n
o
W = x : b (x) < A
(38)
has the general
(39)
where A is a constant determined by the size :
Remark: the expressions of the constant terms K2 ; K3 and K4 are useless (no point for
these expressions).
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Question 7 (2 points). Given the de…nition of the error of type I (0.5 point):
= Pr ( Wj H0 )
(40)
So, here we have (0.5 point):
asy
= Pr b < A b
N
Then, we have:
= Pr
b
A
0
p <
0= N
b
0
p
0= N
0
p
0= N
0;
asy
2
0
(41)
N
!
N (0; 1)
A
=
p
0=
0
N
(42)
where (:) denotes the cdf of the standard normal distribution. From this expression, we
can deduce the critical value of the UMP test of size (1 point):
A=
0
0
+p
N
1
( )
(43)
Question 8 (2 points). Consider the test
H0
H1
with
1
<
0:
:
:
=
=
0
1
The rejection region of the UMP test of size
x : b (x) <
W=
W does not depend on
test (0.5 point):
1
0
0
+p
N
1
is (1 point):
( )
(44)
(0.5 point). It is also the rejection region of the UMP one-sided
H0
H1
:
:
=
<
0
0
Question 9 (2 points). Consider the one-sided tests:
Test A: H
Test B: H
0
0
:
:
=
=
0
0
against H1 :
against H1 :
<
>
0
0
The non rejection regions of the UMP tests of size =2 are (0.5 point):
WA =
WB =
x : b (x) >
x : b (x) <
0
0
0
+p
N
1
0
+p
N
1
So, non rejection region of the two-sided test of size
W=
x:
0
0
+p
N
1
2
< b (x) <
(45)
2
1
(46)
2
is (1 point):
0
0
+p
N
1
1
2
(47)
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Since
1
( =2) =
(1
6
=2) ; this region can be rewritten as:
x : b (x)
W=
0
<p
N
0
The region rejection of the two-sided test of size
x : b (x)
W=
1
(48)
2
is (0.5 point):
0
>p
N
0
1
1
1
(49)
2
Question 10 (2 points). By de…nition of the power function, we have (0.5 point):
P ( ) = Pr ( Wj H1 )
8 6=
Under the alternative:
2
b asy N
;
H1
with
6=
P( )
0:
(50)
0
(51)
N
The power function can be expressed as:
=
1
Pr W H1
=
1
Pr
=
1
Pr b <
=
1
0
+p
N
0
0
+
0
<b<
1
2
0
+p
N
1
p0
N
p
1
1
1
2
= N
0
2
!
0
+p
N
+
1
+ Pr b <
0
+
1
0
2
0
+p
N
1
2
!
1
p0
N
2
p
= N
(52)
The power function is (0.5 point):
0
P( ) = 1
p
= N
+
0
1
1
2
+
0
p
= N
+
0
When N tends to in…nity, two cases have to be considered. If >
point):
lim P ( ) = 1
( 1) + ( 1) = 1
1
2
0,
then we have (0.5
N !1
If
<
0,
(53)
(54)
then we have (0.5 point):
lim P ( ) = 1
N !1
(+1) +
(+1) = 1
1+1=1
(55)
Whatever the value of ; the power function tends to one:
lim P ( ) = 1
N !1
(56)
The test is consistent.
Part III: The trilogy
Question 11 (3 points). The LR test statistic is de…ned as to be (0.5 point):
LR =
2 `N bH 0 ; x
`N bH 1 ; x
(57)
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where bH 0 and bH 1 respectively denote the ML estimator of the parameter
null and the alternative. We know that under the alternative
under the
N
1 X
ln (xi )
N i=1
bH (x) = ln (c)
1
(58)
As a consequence, we have (0.5 point):
N
X
ln (xi ) = N ln (c)
i=1
bH (x) = 100 (4
1
2) = 200
(59)
The log-likelihood of the sample under the alternative is equal to (0.5 point):
! N
X
bH
1
N
1
ln (c) +
ln (xi )
`N bH 1 ; x
=
N ln bH 1
bH
bH
1
=
100 ln (2)
=
369:31
Under the null, the parameter
is equal to (0.5 point)
`N bH 0 ; x
i=1
1
100
2
4+
1
2
200
2
(60)
is known ( = 1:5) and the log-likelihood of the sample
=
N
N ln ( 0 )
ln (c) +
1
0
0
=
100 ln (1:5)
=
373:87
0
100
1:5
4+
1
N
X
ln (xi )
i=1
1:5
1:5
200
(61)
The LR test statistic (realisation) is equal to (0.5 point):
LR (x)
=
=
=
2 `N bH 0 ; x
`N bH 1 ; x
2 ( 373:87 + 369:31)
9:13
(62)
Under some regularity conditions and under the null, we have:
d
LR !
2
(1)
(63)
since there is only one restriction imposed. The critical region for a 5% signi…cance level
is:
W = x : LR (x) > 20:95 (1) = 3:8415
(64)
Conclusion: for a 5% signi…cance level, we reject the null H0 :
= 1:5 (0.5 point).
Question 12 (3 points). In this case, the Wald test statistic (realisation) is equal to (1.5
point):
2
2
bH
bH
2
1:5
1:5
1
1
(2 1:5)
=
=
= 6:25
(65)
Wald (x) =
2
22 =100
b asy bH
b
V
=N
1
H1
Under some regularity conditions and under the null, we have (0.5 point):
d
Wald !
2
(1)
(66)
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since there is only one restriction imposed. The critical region for a 5% signi…cance level
is (0.5 point):
W = x : Wald (x) > 20:95 (1) = 3:8415
(67)
Conclusion: for a 5% signi…cance level, we reject the null H0 :
= 1:5 (0.5 point).
Question 13 (4 points). The LM test statistic is de…ned as to be (0.5 point):
s2N ( c ; xi )
b
I N ( c)
LM =
(68)
where sN ( c ; x) is the score of the unconstrained model evaluated at
score is de…ned by (0.5 point):
sN ( ; X) =
N
@ ln `N ( ; X)
=
@
The realisation of the score evaluated at
sN ( c ; x)
=
N
=
N
2
N
1 X
ln (c)
2
In our case, the
ln (Xi )
(69)
i=1
c
+
is equal to (1 point)
N
c
=
+
c:
2
c
N
1 X
ln (c)
100
100
+
1:5
1:52
22:2222
2
c i=1
4
ln (xi )
1
1:52
200
(70)
The estimate of the Fisher information number associated to the sample b
I N ( c ) is equal
to (0.5 point):
N
100
b
I N ( c) = 2 =
= 44:4444
(71)
1:52
c
So, the realisation of the LM test statistic is (0.5 point):
LM (x) =
22:22222
s2N ( c ; x)
=
= 11:1111
b
44:4444
I N ( c)
(72)
Under some regularity conditions and under the null, we have (0.5 point):
d
LM !
2
(1)
(73)
since there is only one restriction imposed. The critical region for a 5% signi…cance level
is:
W = x : LM (x) > 20:95 (1) = 3:8415
(74)
Conclusion: for a 5% signi…cance level, we reject the null H0 :
= 1:5 (0.5 point).
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Exercise 2: Logit model and MLE (15 points)
Question 1 (2 points). the conditional distribution of the dependent variable Yi is a Bernoulli
distribution with a (conditional) success probability equal to pi = Pr ( yi = 1j xi ) : (0.5
point)
yi j xi Bernoulli (pi )
N
Since the variables fyi ; xi gi=1 are i:i:d: (0.5 point), the conditional log-likelihood of the
N
sample fyi ; xi gi=1 corresponds to the sum of the (log) probability mass functions associated to the conditional distributions of yi given xi for i = 1; ::; N (0.5 point):
`N ( ; yj x) =
N
X
i=1
ln f yjx ( yi j xi ; )
with
(75)
yi
f yjx ( yi j xi ; ) = pyi i (1
(76)
pi )
N
As a consequence, the (conditional) log-likelihood of the sample fyi ; xi gi=1 is equal to
(0.5 point):
`N ( ; yj x) =
N
X
yi ln
(x|i ) +
N
X
(1
(x|i ))
yi ) ln (1
(77)
i=1
i=1
or equivalently
`N ( ; yj x) =
N
X
exp (x|i )
1 + exp (x|i )
yi ln
i=1
+
N
X
(1
exp (x|i )
1 + exp (x|i )
yi ) ln 1
i=1
(78)
Question 2 (2 points). By de…nition of the score vector, we have (1 point):
sN ( ; yj x)
@`N ( ; yj x)
@
N
X
(x|i ) @x|i
=
yi
(x|i ) @
i=1
=
N
X
=
i=1
Since
(x) =
sN ( ; yj x)
(x) (1
=
N
X
=
(x|i )
xi
(x|i )
yi
=
yi )
i=1
N
X
(1
yi )
1
i=1
(x|i ) (1
(x|i ))
xi
|
(xi )
yi (1
(x|i )) xi
i=1
N
X
(1
1
(x|i ) @x|i
(x|i ) @
(x|i )
xi
(x|i )
(79)
(x)) ; this expression can be simpli…ed as follows (0.5 point):
i=1
N
X
yi
N
X
N
X
(1
N
X
(1
yi )
i=1
yi )
(x|i ) (1
(x|i ))
xi
|
1
(xi )
(x|i ) xi
i=1
(yi
yi (x|i )
(x|i ) + yi (x|i )) xi
i=1
=
N
X
i=1
(yi
(x|i )) xi
(80)
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So, the score vector of the logit model is simply de…ned by (0.5 point):
sN ( ; yj x) =
N
X
xi (yi
(x|i ))
(81)
i=1
Question 3 (2 points). The Hessian matrix is a K
HN ( ; yj x) =
K matrix given by (0.5 point):
@sN ( ; yj x)
@ 2 `N ( ; yj x)
=
|
@ @
@ |
(82)
So, we have (0.5 point):
HN ( ; yj x) =
N
X
xi (x|i )
i=1
@x|i
@ |
(83)
The Hessian matrix is de…ned as to be (1point) :
HN ( ; yj x) =
N
X
xi (x|i ) x|i
(84)
(x|i ) xi x|i
(85)
i=1
or equivalently (since
(x|i ) is a scalar)
HN ( ; yj x) =
N
X
i=1
Question 4 (4 points). The …rst step consists in writing a function Matlab to de…ne the
opposite of the log-likelihood. The following syntax (2 points) is proposed, cf. Figure 1:
Figure 1: Log-Likelihood function
The second step consists in using the numerical optimisation algorithm of Matlab by using
the following syntax, cf. Figure 2 (2 points).
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Figure 2: Main program
Question 5 (2 points). Since the problem is regular, the asymptotic variance covariance
matrix of the MLE estimator b is (0.5 point):
1
Vasy b = I
N
1
b =I 1 b
N
(86)
A consistent estimator for average Fisher information matrix, based on the Hessian matrix,
is (0.5 point):
N
1 X
1
b
Hi b ; yi j xi =
HN b ; yj x
I b =
(87)
N i=1
N
So, an estimator of the asymptotic variance covariance matrix of the MLE estimator b is
given by (0.5 point):
b asy b = 1 b
I
V
N
1
In this case, we have (0.5 point):
b asy b =
V
b =
N
X
i=1
HN 1 b ; yj x
x|i b
xi x|i
!
(88)
1
Question 6 (3 points). The following syntax is proposed, cf. Figure 3:
(89)
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Figure 3: Asymptotic standard errors
Exercise 3: OLS and heteroscedasticity (5 points)
The aim of this code is to compute the OLS estimate of the parameter of a linear model,
denoted beta in the code (0.5 point) and the corresponding White consistent standard errors
(1 point). The matrix M denotes the estimator de…ned by (0.5 point):
M=
N
1 X 2
b
" xi x|i
N i=1 i
(90)
where b
"i denotes the OLS residual for the ith unit and xi the corresponding vector of explicative
variables. The code uses a …nite sample correction (0.5 point) for M (Davidson and MacKinnon,
1993). The matrix V corresponds to the White consistent estimate of the asymptotic variance
covariance matrix of the OLS estimator (1 point). The vector std corresponds to the robust
asymptotic standard errors (0.5 point).