Exercises Chapter 1
Estimation Theory
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University of Orléans
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Exercise 1
Rayleigh distribution
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Problem
We consider two continuous independent random variables X and Y
normally distributed with N 0, σ2 . The transformed variable R de…ned
by:
p
R = X2 + Y2
has a Rayleigh distribution with a parameter σ2 :
R
Rayleigh σ2
with a pdf fR r ; σ2 de…ned by:
r2
r
exp
σ2
2σ2
r
π
E (R ) = σ
V (R ) =
2
fR r ; σ 2 =
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8r 2 [0, +∞[
4
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π
2
σ2
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Problem (cont’d)
We consider an i.i.d. sample fR1 , R2 , .., RN g and an estimator (MLE) of
σ2 de…ned by
1 N 2
b2 =
σ
Ri
2N i∑
=1
b2 is an unbiased estimator of σ2 .
Question 1: Show that σ
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Solution:
We have:
b2 =
σ
1
2N
N
∑ Ri2
i =1
b2 .
We want to calculate E σ
Since the sample fR1 , R2 , .., RN g is i.i.d., we have:
!
1 N 2
1 N
2
b =E
E σ
R
=
∑ E Ri2
i
2N i∑
2N
=1
i =1
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Solution (cont’d):
b
E σ
2
=E
1
2N
N
∑
i =1
Ri2
!
=
1
2N
N
∑E
Ri2
π
σ2
i =1
We know that:
E (Ri ) = σ
r
π
2
V (Ri ) =
4
2
So, we have:
E Ri2
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= V (Ri ) + E (Ri )2
π
4 π
σ2 + σ2
=
2
2
2
= 2σ
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Solution (cont’d):
b2
E σ
=
=
=
1
2N
1
2N
N
∑E
Ri2
i =1
N
∑ 2σ2
i =1
N2σ2
2N
So, we have:
b2 is unbiased.
The estimator σ
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b2 = σ2
E σ
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Remark: Sometimes, the Rayleigh distribution is parametrized by σ. But,
b2 is unbiased than to show that σ
b is unbiased...
it is easier to show that σ
v
u
u 1 N
b=t
Ri2
σ
2N i∑
=1
Then to study the bias, we have to calculate:
0v
1
u
u 1 N
b) = E @t
E (σ
Ri2 A
2N i∑
=1
???
since for a nonlinear function g (.) , E (g (x )) 6= g (E (x )) ... The only
solution is to compute the integral
R∞
b ) = 0 x fσb x; σ2 dx
E (σ
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Problem (cont’d)
b2 is a (weakly) consistent estimator of σ2 . We
Question 2: Show that σ
admit that the raw moments of R are de…ned by:
E Rik
k
= σk 2 2 Γ 1 +
k
2
k2N
and where Γ (.) denotes the gamma function with:
Γ (x ) =
Z∞
tx
1
exp ( t ) dt.
1) !
for x 2 N
0
and
Γ (x ) = (x
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Solution:
b2 . Since the sample fR1 , R2 , .., RN g is i.i.d., we
First, calculate V σ
have:
!
N
1 N
1
2
b2 = V
R
=
V Ri2
V σ
i
2 ∑
2N i∑
4N
=1
i =1
What is the value of V Ri2 ?
V Ri2 = E Ri4
E Ri2
2
We shown that
E Ri2 = 2σ2
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Solution (cont’d):
V Ri2 = E Ri4
E Ri2
2
= E Ri4
4σ4
What is the value of E Ri4 ? For any k 2 N :
E Rik
k
= σk 2 2 Γ 1 +
k
2
For k = 4, we have:
E Ri4 = σ4 22 Γ (3)
with
Γ (3) = (3
1)! = 2! = 2
So, we have:
E Ri4 = σ4 23 = 8σ4
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Solution (cont’d):
The variance of Ri2 is equal to:
V Ri2 = E Ri4
4σ4 = 8σ4
4σ4 = 4σ4
As a consequence
b2
V σ
=
=
=
=
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1
4N 2
1
4N 2
N
∑V
Ri2
i =1
N
∑ 4σ4
i =1
N4σ4
4N 2
σ4
N
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Solution (cont’d):
To sum up:
b2 = σ2 (unbiased estimator)
E σ
b2 =
lim V σ
N !∞
σ4
=0
N !∞ N
lim
b2 is a (weakly) consistent estimator of σ2 :
So, the estimator σ
p
b2 ! σ2
σ
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Problem (cont’d)
Question 3: Sometimes, the Rayleigh distribution is parametrized by σ
(and not by σ2 ) with
R Rayleigh (σ)
Propose a (weakly) consistent estimator for σ.
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Solution:
Since σ > 0, a natural estimator for σ is de…ned by:
v
u
p
u 1 N
t
2
b
b2
σ=
Ri = σ
2N i∑
=1
b2 is a (weakly) consistent estimator of σ2 :
We shown that the estimator σ
p
b2 ! σ2
σ
By applying the Continuous
Mapping Theorem (CMP) for the continuous
p
function g (x ) = x, we get immediately:
p
or
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b2 ! g σ2
g σ
p
b!σ
σ
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Problem (cont’d)
Question 4: For any value of σ2 , what is the …nite sample (or exact
b2 de…ned by
sampling) distribution of the estimator σ
b2 =
σ
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1
2N
N
∑ Ri2
i =1
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Solution:
The estimator is de…ned by
b2 =
σ
1
2N
N
∑ Ri2
i =1
We know that R1 , R2 , .., RN are i.i.d. random variables with a Rayleigh
distribution.
Ri
Rayleigh σ2
Reminder: if X and Y are independent and normally distributed N 0, σ2
p
random variables, the transformed variable R = X 2 + Y 2 has a Rayleigh
distribution.
The exact distribution of R 2 = X 2 + Y 2 is unknown (it is not a χ2 ...),
b2 is unknown.
and as a consequence the …nite sample distribution of σ
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Problem (cont’d)
Question 5: Write a Matlab code in order to approximate the true
b2 for a sample size N = 10, a
(unknown) …nite sample distribution of σ
true value of σ2 = 16 and by using S = 1, 000 simulations.
b2 .
(1) Plot an histogram of the 1,000 realisations of the estimator σ
(2) plot the Kernel estimator of the density fσb2 (x ) by using the Matlab
built-in function ksdensity.
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De…nition (Kernel density estimator)
Let consider a sample X1 , .., XN , where X has a distribution characterized
by the pdf fX (x ) , for x 2 R. A consistent (kernel) estimator of fX (x ) for
any x 2 R is given by:
1
b
fX ( x ) =
λN
N
∑K
i =1
x
xi
λ
where K (.) denotes a kernel function and λ is bandwidth parameter.
p
b
fX ( x ) ! fX ( x )
8x 2 R
For more details (and a discussion on the optimal choice of λ), see:
Lecture notes "Econométrie Non Paramétrique", Hurlin (2008), Master
Econométrie and Statistique Appliquée, Université d’Orléans.
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De…nition (Kernel function)
A kernel function K (u ) satis…es the following properties:
(i ) K (u ) 0
R
(ii ) K (u ) du = 1
(iii ) K (u ) reaches its maximum for u = 0 and decreases with ju j.
(iv ) K (u ) is symmetric, i.e. K (u ) = K ( u ) .
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Some examples of Kernel functions
u2
2
1
Normal : K (u ) = p
exp
2π
Triangular : K (u ) = 1
Quartic or BiWeight : K (u ) =
Epanechnikov : K (u ) =
Triweight : K (u ) =
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u 2 [ 1, 1]
15
1
16
3
1
4
35
1
32
ju j
u2
u2
u2
u2R
3
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2
u 2 [ 1, 1]
u 2 [ 1, 1]
u 2 [ 1, 1]
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250
200
150
100
50
0
5
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10
15
20
25
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30
35
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0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0
0
5
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10
15
20
25
30
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35
40
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Problem (cont’d)
Question 6: For the special case where σ2 = 1, what is the …nite sample
b2 de…ned by
(or exact sampling) distribution of the estimator σ
b2 =
σ
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1
2N
N
∑ Ri2
i =1
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Solution:
The estimator is de…ned by
b2 =
σ
1
2N
N
∑ Ri2
i =1
We know that R1 , R2 , .., RN are i.i.d. random variables with
Ri
Rayleigh (1)
p
Reminder 1 : if Ri
Rayleigh (1), then R = X 2 + Y 2 where X and Y
are independent and standard normally distributed N (0, 1) random
variables.
Reminder 2: if X
N (0, 1) , then X 2
χ2 (1)
Reminder 3: if X
χ2 (v1 ) and Y
χ2 (v2 ) , and X and Y are
2
independent, then X + Y
χ (v1 + v2 )
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Solution (cont’d):
So, if X and Y are independent and standard normally distributed
N (0, 1) random variables
p
Ri = X 2 + Y 2 Rayleigh (1)
Ri2 = X 2 + Y 2
χ2 (2)
The sum of independent chi-squared distributed random variable has a
chi-squared distribution.
N
∑ Ri2
χ2 (2N )
i =1
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Solution (cont’d):
b2 =
2N σ
N
∑ Ri2
χ2 (2N )
i =1
b2 has an
In the special case where σ2 = 1, the transformed variable 2N σ
exact sampling (…nite sample) distribution that corresponds to a
chi-squared distribution with 2N degrees of freedom.
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Problem (cont’d)
Question 7: Write a Matlab code in order to approximate the true
b2 for
(unknown) …nite sample distribution of the transformed variable 2N σ
a sample size N = 10 in the special case where σ2 = 1 by using
S = 10, 000 simulations.
(1) Plot the Kernel estimator of the density f2N σb2 (x ) by using the Matlab
built-in function ksdensity.
(2) Compare this estimated density function to the pdf of a chi-squared
distribution with 2N degrees of freedom.
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0.07
Estimated finite sample pdf
Theoretical pdf of a chi-squared
0.06
0.05
0.04
0.03
0.02
0.01
0
0
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10
20
30
40
50
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60
70
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Problem (cont’d)
b2 ?
Question 8: What is the asymptotic distribution of the estimator σ
b2 =
σ
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1
2N
N
∑ Ri2
i =1
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Solution:
We know that R12 , R22 , .., RN2 are i.i.d. variables with
E Ri2 = 2σ2
V Ri2 = 4σ4
Step 1: By applying the Lindberg-Levy univariate Central Limit
Theorem (CLT), we get:
!
p
1 N 2
d
2
N
Ri
2σ
! N 0, 4σ4
N i∑
=1
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Solution (cont’d):
Step 2: By de…nition, we have
1
b =
σ
2N
2
p
N
1
N
N
∑
N
∑
Ri2
=g
i =1
Ri2
i =1
1
N
2σ
2
!
N
∑
i =1
Ri2
!
d
! N 0, 4σ4
with g (x ) = x /2. So, g (.) is a continuous and continuously di¤erentiable
function with g 2σ2 6= 0 and not involving N, then the delta method
implies
!
!
!
2
p
1 N 2
∂g
x
(
)
d
Ri
g 2σ2
! N 0,
N g
4σ4
N i∑
∂x
2
2σ
=1
g 2σ2 =
2σ2
= σ2
2
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∂g (x )
∂x
=
2σ2
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∂x /2
∂x
=
2σ2
1
2
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Solution (cont’d):
p
N
g
1
N
N
∑
i =1
Ri2
!
g 2σ
2
!
d
!N
0,
1
2
2
4σ
4
!
b2 is asymptotically normally distributed
The estimator σ
p
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b2
N σ
d
σ2 ! N 0, σ4
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Problem (cont’d)
b2 ?
Question 9: What is the asymptotic variance of the estimator σ
b2 =
σ
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1
2N
N
∑ Ri2
i =1
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Solution:
We know that:
p
or equivalently
b2
N σ
b2
σ
asy
d
σ2 ! N 0, σ4
N
σ2 ,
b2 is equal to:
The asymptotic variance of σ
b2 =
Vasy σ
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σ4
N
σ4
N
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Problem (cont’d)
Question 10: Write a Matlab code in order to approximate the
asymptotic distribution of the transformed variable
p
Z =
b2
N σ
σ2
σ2
for a sample size N = 10, 000, a true value of σ2 = 16 by using
S = 10, 000 simulations.
(1) Plot the Kernel estimator of the density fZ (x ) by using the Matlab
built-in function ksdensity.
(2) Compare this estimated density function to the pdf of a standard
normal distribution.
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0.45
0.4
Estimated finite sample pdf
Theoretical pdf of a standard normal
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-5
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0
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5
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Problem (cont’d)
b of
Question 11: What is the asymptotic distribution of the estimator σ
the parameter σ de…ned by:
v
u
u 1 N
b=t
σ
Ri2
2N i∑
=1
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Solution:
Step 1 : We know that:
p
b > 0, we have:
Since σ
d
b2
N σ
v
u
u 1
b=t
σ
2N
σ2 ! N 0, σ4
N
∑ Ri2 =
i =1
p
b2 = g σ
b2
σ
p
where g (x ) = x is a continuous and continuously di¤erentiable function
with g σ2 6= 0 and that does not depend on N
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Solution (cont’d):
Step 2: We have
p
d
b2
N σ
σ2 ! N 0, σ4
p
The delta method for g (x ) = x implies
p
with
b
N g σ
g σ
2
2
=σ
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g σ2
∂g (x )
∂x
d
!N
σ2
∂g (x )
∂x
0,
p
∂ x
=
∂x
σ2
2
σ4
σ2
!
1
1
= p =
2σ
2 σ2
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Solution (cont’d):
Step 2 (cont’d):
p
b2
N g σ
g σ
So, we have:
2
d
g σ2
=σ
p
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!N
∂g (x )
∂x
b
N (σ
σ2
d
0,
p
∂ x
=
∂x
σ) ! N
∂g (x )
∂x
0,
σ2
2
σ4
σ2
!
1
1
= p =
2
2σ
2 σ
σ2
4
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Exercise 2
CAPM
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Problem (CAPM)
The empirical analogue of the CAPM is given by:
= αi + βi
rit rft
| {z }
excess return of security i for time t
market excess return for time t
where εt is an i.i.d. error term. We assume that
e
rit = rit
E ( εt ) = 0
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+ εt
(rmt rft )
| {z }
rft
e
rmt = rmt
V ( εt ) = σ 2
rft
rmt ) = 0
E ( εt j e
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Problem (CAPM, cont’d)
Consider the model
e
rit = αi + βi e
rmt + εt
Data: Microsoft, SP500 and Tbill (closing prices) from 11/1/1993 to
04/03/2003
0.10
0.08
RMSFT
0.05
0.04
0.00
0.00
-0.05
-0.04
-0.10
-0.06 -0.04 -0.02 0.00 0.02 0.04 0.06 0.08
-0.08
500
RSP500
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1000
RSP500
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1500
2000
RMSFT
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Problem (CAPM, cont’d)
We consider the CAPM model rewritten as follows
e
rit = xt> β + εt
t = 1, ..T
where xt = (1 e
rmt )> is 2 1 vector of random variables, β = (αi βi )> is
2 1 vector of parameters, and where the error term εt satis…es
rmt ) = 0.
E (εt ) = 0, V (εt ) = σ2 and E ( εt j e
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Problem (CAPM, cont’d)
Question 1: show that the OLS estimator
! 1
T
b = ∑ xt xt>
β
t =1
satis…es
p
b
T β
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d
T
∑ xt erit
t =1
β0 ! N 0, σ2 E
1
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!
xt> xt
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Solution:
1
let us rewrite the OLS estimator as:
! 1
!
T
T
>
b = ∑ xt x
β
∑ xt erit = β0 +
t
t =1
2
b
T β
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∑
t =1
b
Normalize the vector β
p
T
β0 =
xt xt>
t =1
!
1
T
∑ xt εt
t =1
!
β0
1
T
T
∑ xt xt>
t =1
!
1
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p
1
T
T
T
∑ xt εt
t =1
!
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Solution (cont’d):
3. Using the WLLN and the CMP:
1
T
T
∑
xt xt>
t =1
!
1
p
! E
1
xt xt>
4. Using the CLT:
p
T
1
T
T
∑ xt εt
t =1
E (xt εt )
!
d
! N (0, V (xt εt ))
with E ( εt j e
rmt ) = 0 and E ( εt j 1) = E (εt ) = 0 =) E (xt εt ) = 0
and
V (xt εt ) = E xt εt εt xt> = E E xt εt εt xt> xt
= E xt V ( εt j xt ) xt> = σ2 E xt xt>
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Solution (cont’d):
So, we have
1
T
p
T
1
T
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T
∑
xt xt>
t =1
T
∑ xt εt
t =1
!
!
1
p
! E
1
xt xt>
d
! N 0, σ2 E xt xt>
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Solution (cont’d):
By using the Slutsky’s theorem (for a convergence in distribution), we
have:
! 1
!
T
p 1 T
p
1
d
>
b β =
T β
xt xt
T ∑ xt µt ! N (Π, Ω)
0
T t∑
T
=1
t =1
with
Π=E
Ω=E
1
xt xt>
1
xt xt>
σ2 E xt xt>
0=0
E
1
xt xt> = σ2 E
1
xt xt>
Finally, we have:
p
b
T β
Christophe Hurlin (University of Orléans)
d
β0 ! N 0, σ2 E
1
xt xt>
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Problem (CAPM, cont’d)
Question 2: What is the asymptotic variance-covariance matrix of the
b?
OLS estimator β
b=
β
Christophe Hurlin (University of Orléans)
T
∑
t =1
xt xt>
!
1
T
∑ xt erit
t =1
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!
November 2013
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Solution:
We shown that
p
or equivalently
b
T β
b
β
d
β0 ! N 0, σ2 E
asy
N
β0 ,
σ2
E
T
1
1
xt xt>
xt xt>
b is equal to:
The asymptotic variance-covariance matrix of β
2
b = σ E
Vasy β
T|
Christophe Hurlin (University of Orléans)
1
xt xt>
{z
}
2 2
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November 2013
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Remarks:
1
2
The asymptotic variance covariance matrix is a 2
matrix:
0
2
Vasy (b
α)
σ
b =
E 1 xt xt> = @
Vasy β
T
β, b
α
cov b
Since xt = (1 e
rmt )> , we have:
E xt xt> = E
Christophe Hurlin (University of Orléans)
1 e
rmt
2
e
rmt e
rmt
=
2 symmetric
α, b
β
cov b
β
Vasy b
1
A
1
E (e
rmt )
2
E (e
rmt ) E e
rmt
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Problem (CAPM, cont’d)
Question 3: Let us consider a consistent estimator of σ2 de…ned by:
b2 =
σ
1
T
T
∑ bε2t =
2 t =1
1
T
T
2 ∑
t =1
e
rit
b
xt> β
2
b ?
Propose a consistent estimator of the asymptotic variance Vasy β
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Solution:
We know that
2
b = σ E
Vasy β
T
Using the LLN, if xt is i.i.d., we get:
1
T
T
p
xt xt>
1
∑ xt xt> ! E
xt xt>
!
1
t =1
Using the CMP:
1
T
Christophe Hurlin (University of Orléans)
T
∑
t =1
xt xt>
1
p
!E
xt xt>
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November 2013
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Solution (cont’d):
So, we have
2
b = σ E
Vasy β
T
and
1
T
T
∑
xt xt>
t =1
!
1
xt xt>
1
p
!E
1
xt xt>
p
By using the Slutsky’s theorem:
b2
σ
T
1
T
T
∑
xt xt>
t =1
Christophe Hurlin (University of Orléans)
!
b2 ! σ2
σ
1
p
!
σ2
E
T
1
b
xt xt> = Vasy β
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Solution (cont’d):
b is de…ned by
A consistent estimator of the asymptotic variance Vasy β
b asy
V
Or equivalently by
b2
b = σ
β
T
1
T
b =σ
b asy β
b2
V
with
b2 =
σ
1
T
Christophe Hurlin (University of Orléans)
T
∑ bε2t =
2 t =1
1
T
T
∑
xt xt>
t =1
T
∑
xt xt>
t =1
!
T
2 ∑
t =1
!
e
rit
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1
1
b
xt> β
2
November 2013
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Remark:
Since xt = (1 e
rmt )> :
b =σ
b asy β
b2
V
T
∑ xt xt> =
t =1
Christophe Hurlin (University of Orléans)
T
∑
t =1
T
rmt
∑Tt=1 e
xt xt>
!
1
rmt
∑Tt=1 e
2
rmt
∑Tt=1 e
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November 2013
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Problem (CAPM, cont’d)
Question 4: Using the excel …le capm.xls, write a Matlab code to
estimate the beta and the alpha for MSFT. Compare your results with the
following table of estimation results (Eviews).
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November 2013
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Christophe Hurlin (University of Orléans)
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Perfect....
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November 2013
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Problem (CAPM, cont’d)
Question 5: Using the excel …le capm.xls, write a Matlab code
(1) to estimate the variance of the error term εt
b
(2) to estimate the asymptotic standard errors of the estimators β
(3) Compare your results with the table of estimation results (Eviews).
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Advanced Econometrics - HEC Lausanne
November 2013
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Christophe Hurlin (University of Orléans)
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Perfect too....
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End of Exercices - Chapter 1
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Christophe Hurlin (University of Orléans)
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