Panel Data Econometrics
Seminar 4: October 23,2007
Solution proposal by:
Aleksander Sahnoun (7th semester)
Problem 4.1: Autoregressive models
Problem 4.1 A:
Dynamic models – models with variables dated in different periods
Panel data handles dynamic models better than cross-section data and time series data. Pure
time series data can’t handle individual-specific heterogeneity (can handle some time specific
heterogeneity if it has to some extend be parameterized). With panel data it is possible to
handle individual-specific heterogeneity and time-specific heterogeneity within the same
framework.
In autoregressive models, with lagging LHS variables as explanatory variables, panel data
offers the possibility to handle time-lags between time indexed variables and individualspecific heterogeneity.
Problem 4.1 B:
(1) yi ,t  c  yi ,t 1  xi ,t    i  ui ,t
ui ,t
IID(o,  2 )
i  1,....., N ; t  1,......., T ,
Cross-section data:
Assume that yi ,t , yi ,t 1, and xi ,t are observed variables.
(2) yi  c  yi ,1  xi    i  ui
(2)' yi    yi ,1  xi   ui
*
i
ui
IID (o,  2 ) ,
i  1,....., N ;
  i  c
*
i
We have N equation but N + 1 +1 coefficients, so it is not possible to estimate all of them. We
are not able to estimate the individual-specific intercept terms from cross section data.
Remark: Given that the model specification is correct, and since the disturbance term is IID,
OLS on (2) '' yi    yi ,1  xi   ui
Would be MVLUE.
Problem 4.1 C:
I assume access to panel data
(3) yi ,t  i*  yi ,t 1  xi ,t   ui ,t
 i*   i  c
 1
ui ,t
IID(o,  2 )
i  1,....., N ;
t  1,....., T ,
uit , xit , yi 0 are independent for all (i,t)
a) Within-individual estimation exploits information from observations within each individual
over time.
If yi ,t 1 were strictly exogenous we could have estimate OLS on (3). The OLS estimator would
coincide with the within-individual estimator and would have given MVLUE1.
t 1
yi ,t   t yi ,0    s ( i*  xi ,t  s   ui ,t  s )
s 0
If the process, for each individual i, have started an infinitely long distance back in time and
  1 , it follows that

yi ,t    s (i*  xi ,t s   ui ,t s ) 
s 0

i*
   s ( xi ,t s   ui ,t s )
1   s 0
We can transform (3) into an equation in departures from the individual-specific means,
(4) (yi ,t  yi , )  ( i*   i* )  ( yi ,t 1  yi ,1 )  ( xi ,t  xi , )   (ui ,t  ui , )
=( yi ,t 1  yi ,1 )  ( xi ,t  xi , )   (ui ,t  ui , )
i  1,....., N ;
t  1,....., T ,
and apply OLS. yi ,t 1 is uncorrelated with ui ,t , but correlated with ui , so we get estimators that
are T consistent. (inconsistent if T is finite) and are asymptotically negatively biased (larger
than ½) when the number of individuals goes to infinity.. We can interpret ˆw , ˆw as a
weighted mean of N individual-specific estimators, all which are consistent when T goes to
infinity.
N
ˆW 
T
 (y
i 1 t 1
N
i ,t
T
 ( y
i 1 t 1
N
ˆW 
 yi , )( yi ,t 1  yi ,1 )
T
 (y
i 1 t 1
N
i ,t
i ,t 1
 yi , )( xi ,t  xi , )
T
 ( x
i 1 t 1
 yi ,1 ) 2
i ,t
 xi , ) 2
b)
1
What do strict exogeneity require?
 WY1,1Y 1WY ,Y 1 
 W X,1X W X ,Y 
1
 45  0, 4592
98
1
 53  0,5047
105
We can transform (3) into an equation where we subtract the individual specific mean with
the global mean. We then exploit the information between individuals.
i  1,....., N ;
(5) ( yi ,  y )  ( i   i )  ( yi ,1  y )  ( xi ,  x )   (ui ,  u )
t  1,....., T ,
If we assume that the individual specific intercepts are uncorrelated with the RHS variables,
we can write the estimators as:
N
ˆB 
T  (yi ,  y )( yi ,1  y )
i 1
N
T  ( yi ,1  y ) 2
 BY11,Y 1 BY ,Y 1 
1
 303  1,3710
221
i 1
N
ˆB 
T  (yi ,  yi )( xi ,  x )
i 1
N
T  ( xi ,  x ) 2
 B X1, X B X ,Y 
1
 240  0,9449
254
i 1
We can see that the estimate of  is violating our assumption   1 .
If the individual specific coefficients are correlated with the RHS variables, the estimators
are2:
1 N
 ( i   )( yi,1  y )
N i 1
ˆB   
1 N T
 ( yi,1  y )2
N i 1 t 1
ˆB   
1
N
N
 (
i 1
1
N
i
N
  )( xi ,  x )
T
 ( x
i 1 t 1
i ,
 x )2
c) Writing the equation in first difference form
(5) (yi ,t  yi ,t 1 )  ( yi ,t 1  yi ,t 2 )  ( xi ,t  xi ,i 1 )   (ui ,t  ui ,t 1 ) i  1,....., N ;
yi ,t 
yi ,t 1  xi ,t   ui ,t
In (5) the lagged RHS variable yi ,t 1 is correlated with ui ,t 1
(5) (yi ,t  yi ,t 1 )  ( yi ,t 1  yi ,t 2 )  ( xi ,t  xi ,i 1 )   (ui ,t  ui ,t 1 )
2
Always biased and inconsistent?
t  2,....., T ,
N
ˆ ,OLS 
T
  (y
i 1 t 11
N
 yi ,t 1 )( yi ,t 1  yi ,t  2 )
i ,t
T
  (y
i 1 t 11
N

(
yi ,t )( yi ,t 1 )
i 1 t 11
N
T
(
i 1 t 11
,OLS

T
  (y
i 1 t 11
N
i ,t
yi ,t 1 ) 2
 GY11,Y 1 GY ,Y 1  9, 60 
1
 0,5206
18, 44
 yi ,t 1 )( xi ,t  xi ,t 1 )
T
  (x
i 1 t 11
N

 yi ,t  2 ) 2
T
N
ˆ
i ,t 1
i ,t
 xi ,t 1 ) 2
T
(
i 1 t 11
N
T
yi ,t )( xi ,t )
(
i 1 t 11
xi ,t ) 2
 G X1, X G X ,Y 
1
 5,33  0,5042
10,57
The OLS estimators are inconsistent regardless of whether N,T or both go to infinity even
though the equation is put on difference form. We can alternatively transform the equation, or
estimate it by instrument variables and GMM.
The individual intercept term,
In (a)
ˆ i*  yi ,  yi ,1ˆW  xi , ˆW , ˆi*  ˆi  c


j 1
j 1
ˆ i  ˆ i*   ˆ i* , c   ˆ i*
We have NxT equations and N+1+1 unknown, so we don’t need any additional information.
In (b)
We have N+1+1 unknown, but we have just N linear independent equations, so we need some
additional knowledge/restrictions to estimate i .3
In (c)
The intercept vanish when we take the equation on difference form.
3
Additional conditions?
Problem 4.1 D :
(6) yi ,t  i*  yi ,t 1 1  yi ,t 2 2  xi ,t 1  xi ,t 12  ui,t
i  1,....., N ;
t  1,....., T ,
(6 ')
yi ,t  yi ,t 1 1  yi ,t 2 2  xi ,t 1  xi ,t 1 2  ui ,t
 i*   i  c
ui ,t
IID(o,  2 )
i  1,....., N ;
t  2,....., T ,
uit , xit , yi 0 are independent for all (i,t)
Since we have a Panel Data set, T  2 .
We have seen in problem C that we have to estimate the coefficients with IV to get consistent
and unbiased estimators. Since xi ,t or xi ,t are correlated with yi ,t 1 yi ,t  2 , or yi ,t 1 , yi ,t  2
and uncorrelated with ui ,t , ui ,t , they are valid instruments.
The equations:
yi ,4  yi ,3 1  yi ,2 2  xi ,4 1  xi ,3  2  ui ,4
yi ,5  yi ,4 1  yi ,3 2  xi ,5 1  xi ,4  2  ui ,4
yi ,T  yi ,T 1 1  yi ,T 2 2  xi ,T 1  xi ,T 1 2  ui ,T
Valid instruments:
zi ,3  ( yi ,0 , xi ,4 , xi ,3 ) for ( yi ,2 , yi ,3 , xi ,4 , xi ,3 )
 since cov( y
i ,0
, ui ,2  ui ,1 )  0 while cov( yi ,1 , ui ,2  ui ,1 )  0 
zi ,4  ( yi ,0 , yi ,1 , xi ,5 , xi ,4 ) for ( yi ,3 , yi ,4 , xi ,5 , xi ,4 )
zi ,T 1  ( yi ,0 , yi ,1 ,....., yi ,T 3 xi ,T , xi ,T 1 ) for ( yi ,T 2 , yi ,T 1 , xi ,T , xi ,T 1 ) in the (T-3)'th equation
We are able to estimate the coefficients when T  4 . Then we have enough variables that can
be used as IV’s (uncorrelated with ui ,t )4
4
When are they consistent and unbiased?
Problem 4.1 E:
(7) yi ,t  i*  yi ,t 1  xi ,t   zi ,t  ui ,t
 i*   i  c
ui ,t
IID(o,  2 )
i  1,....., N ;
t  1,....., T ,
Assume:
ui ,t is stochastically independent of xi ,t , yit 1 , zi ,t for all (i,t)
The minimization problem ( within-individual estimation):
min
k ,  , , ,
  (y
i
t
i ,t
 k  yi ,t 1ˆ  xi ,t ˆ  ziˆ  ˆi ) 2
foc :
1.

  i  t ( yi ,t  kˆW  yi ,t 1ˆW  xi ,t ˆW  ziˆW  ˆiW )  0
kW
2.

  i  t ( yi ,t  kˆW  yi ,t 1ˆW  xi ,t ˆW  ziˆW  ˆiW ) yi ,t 1  0
ˆW
3.

  i  t ( yi ,t  kˆW  yi ,t 1ˆW  xi ,t ˆW  ziˆW  ˆiW ) xi ,t  0
ˆ
W
4.

  i  t ( yi ,t  kˆW  yi ,t 1ˆW  xi ,t ˆW  ziˆW  ˆiW ) zi  0
ˆ
W
5.

  t ( yi ,t  kˆW  yi ,t 1ˆW  xi ,t ˆW  ziˆW  ˆiW )  0
ˆiW
i  1,......, N
If the N last equations are satisfied, then the first and the fourth also hold. We thus only have
N+2 linear independent equations, from which we cannot determine N +4 unknowns.
From the N last equations is follows that:
kˆW  ziˆW  ˆ iW  yi ,t 1ˆW  xi ,t ˆWi
i  1,......., N
Inserted in FOC (2,3)

  i  t ( yi ,t  yi ,t 1ˆW  xi ,t ˆW  ( ziˆW  ˆiW  kˆW )) yi ,t 1  0
ˆW
=  i  t ( yi ,t  yi ,t 1ˆW  xi ,t ˆW  yi ,t 1ˆW  xi ,t ˆWi) yi ,t 1  0

  i  t ( yi ,t  yi ,t 1ˆW  xi ,t ˆW  ( ziˆW  ˆiW  kˆW )) xi ,t  0
ˆ

W
=  i  t ( yi ,t  yi ,t 1ˆW  xi ,t ˆW  yi ,t 1ˆW  xi ,t ˆWi) xi ,t  0
Which we can solve for ˆW and ˆW .
The composite term:
kˆ  z ˆ  ˆ  y  y
W
i W
iW
i ,
ˆ  xi , ˆW
i ,1 W
We can only estimate the N composite parameters kˆW  ziˆW  ˆiW and ˆW , ˆW . We are
unable to separate the fixed individual-specific effect ˆ iW from the effect from the individual
specific variable zi since we have a multicollinearity problem. The estimator of the twodimensional variable xi ,t coincides with the one we could have obtained without zi .
Problem 4.2: Model with stochastic slope coefficients
Problem 4.2 A:
The model:
(1) yit   i  i xit  uit
uit
Individual-specific intercept:
 i IID(o,  2 )
i     i
Individual-dependent coefficient
 i IID(o,  2 )
i     i
IID(o,  2 )
i  1,....., N ; t  1,......., T ,
i  1,....., N ;
i  1,....., N ;
(2) yit     i     i  xit  uit
    xit   i xit   i  uit
vit   i xit   i  uit
- Composite disturbance
(2)' yit     xit  vit
i  1,....., N ; t  1,......., T ,
Assume that xit ,  i , uit ,  i are stochastically independent for all i and t.
Properties of the composite disturbance:
E (vit | xit )  E ( i xit   i  uit | xit )  E (i xit | xit )  E ( i | xit )  E (uit | xit )  0
2
E(vit2 | xit )  E i xit   i  uit  | xit 


2
2
2
 E ( i xit | xit )  E ( i | xit )  E (uit2 | xit )  2 E ( i xit ( i  uit ) | xit )  E 2(uit  i | xit )
 xit 2 2
 2
 2
0
 xit 2 2   2   2
E (vit vis | xit xis )  E  i xit   i  uit  i xis   i  uis  | xis 
 E ( i 2 xit xis | xit xis )  E ( i2 | xit xis )  E (uit uis | xit xis )
 xit xis 2
0
 xit xis    
2
2
0
ts
0
Variance – covariance matrix
 xit 2 2   2   2

2
2
 xit xis    
cov(vit v js | xit x js )  
0
0

for j  i, t  s 

for j  i , t  s 
 i, j  1,....., N ; t , s  1,......., T ,
for j  i, t  s 
for j  i, t  s 
The composite disturbance is heteroskedastic both across individuals and periods (different
variance across individuals and periods). E (vit2 | xit )  xit 2 2   2   2
Further, the disturbance is autocorrelated over periods, bur not across individuals (correlation
between different periods, but not between individuals).
E (vit vis | xit xis )  xit xis 2   2
ts
The composite disturbance term is uncorrelated with the equation’s right hand side variable.
Using the law of iterated expectations:
cov(vit , xit )  E(vit xit )  Ex  Ev (vit xit | xit )  Ex  xit Ev (vit | xit )  0
But the equation (2) RHV is not stochastic independent of the composite disturbance, since
the conditional and marginal distribution differs.
Using the law of iterated expectations:
E (vit2 | xit )  xit 2 2   2   2
Ex [ Ev (vit2 | xit )]  E (vit2 )  E ( xit 2 2   2   2 )  E ( xit 2 ) 2   2   2
E (vit2 )  E ( xit 2 ) 2   2   2
Together with:
E (vit2 xit )  Ex [ Ev (vit2 xit | xit )]  Ex [ xit Ev (vit2 | xit )]  E  xit ( 2  2   2   2 )   E ( xit 3 ) 2  E ( xit ) 2  E ( xit ) 2
E (vit2 xit )  E ( xit 3 ) 2  E ( xit ) 2  E ( xit ) 2
This implies that:
cov(vit2 , xit )  E (vit2 xit )  E (vit2 xit )  E (vit2 ) E ( xit )
 E ( xit 3 ) 2  E ( xit ) 2  E ( xit ) 2  E ( xit )  E ( xit 2 ) 2   2   2 
  E ( xit 3 )  E ( xit ) E ( xit )   2   E ( xit )  E ( xit )   2   E ( xit )  E ( xit )   2
 cov( xit2 , xit ) 2
So  2  0 , xit is uncorrelated with vit , but correlated with vit2
In matrix notation (2’) can be written as:
(3) y i  eT   xi  v i
v i  eT i  xi i  ui

ii  e e    x i x     I T
(4) E(v i , v 'j | x i , x j )  
ij  0TT

We can also write (3) as:
(5) y  Xβ  v
Where β  ( ,  )
'
T T
2
'
i
2
2
i  1,....., N ;
j=i 

 i,j=1,....,N,
j=i 

The variance – covariance matrix is then:
 11

(6) E(vv | X)  
0
 TT
0TT 

 =
NN 
'
Since the variance – covariance matrix has nonzero elements outside the diagonal GLS will
be MVLUE. (The GLS estimator is more efficient than the OLS estimator).
CASE 1)
2
If the variances of the stochastic elements are know (  2 ,  2 ,  112 ,......., NN
) , the GLS
estimator is:
1
β*   X'1X  X'1y 
The estimator is then weighted by the variance and covariances.
Let β*i denote the individual GLS estimator
1
β*i   X'i ii-1 Xi   X'i ii-1y i 
i  1,....., N
CASE 2)
2
If  2 ,  2 ,  112 ,......., NN
are unknown, we can estimate them from their empirical counterpart
̂ and then estimate by GLS using ̂ . This is the method of FGLS/EGLS/Two Step.
First estimate ˆi and ˆi by OLS when they are considered as unknown constants. Then find
the residual vector uˆ  y  e ˆ  x ˆ .
i
i
T
i
i
i
1
1 N 2
ˆ j and estimate  2 by it’s empirical counterpart ˆ2 

ˆ j
N j 0
N  1 j 0
And symmetric for ˆ2 .
Form ˆi  ˆi 
N
1 N
1 N ˆ2
Form ˆi  ˆi   ˆ j and estimate  2 by it’s empirical counterpart ˆ2 
j
N j 0
N  1 j 0
Problem 4.2 B
(7) yit  i  i xit  uit uit
where i , i are constants.
IID(o,  2 )
 2

0
cov(uit u js | xit xis )  
0
0

i  1,....., N ; t  1,......., T ,
for j  i, t  s 

for j  i, t  s 

for j  i, t  s 
for j  i, t  s 
We can estimate the within estimator of i .
ˆ W  ( X' B X ) 1 ( X' B y )  W 1 W
i
i
T
i
i
T
i
XXi
1
XYi
1
 WXX  0  B XX   WXY  0  B XY 
We use the variance within the individual to estimate ˆiW .
The GLS estimator in A of  is:
1
ˆ GLS   X' (I N  BT )X   B X' (I N  BT )X    X' (I N  BT )y   B X' (I N  B T )y 
1
  WXX   B B XX   WXY   B B XY 
Where :  B 
1
2
 2  T ( 2   2 )
In problem A we estimate the common slope coefficient  . The GLS estimator is a weighted
estimator, weighted with the inverse of the variance covariance matrix. If i , i are constants,
we extract the within variance (and sets  B  0 ).
Problem 4.2 C
Test of heterogeneity in slope coefficient i . Assuming unknown and constant i , i (FE).
Full individual heterogeneity:
H A : 1 ,.......,  N and 1 ,.......,  N are unrestricted
Individual intercept heterogeneity, homogeneous slopes:
H B : 1 ,.......,  N and 1   2     N are unrestricted
Full homogeneity:
H C : 1   2     N and 1   2     N are unrestricted
We make the assumption ui IIN (0T,1 ,  2IT ) ( replacing ui
IIN => Independent, identically normally distributed.
IID(0T,1 ,  2IT ) )
This normality assumption ensure that the F-test is valid.
The sum of squared residuals are:
N
T
SSRA   i 1  t 1 uˆ for model H A
it
SSRB   i 1  t 1 uˆ it for model H B
N
T
SSRC   i 1  t 1 uˆ it for model H C
N
T
We want to test the null hypothesis – homogeneous slope coefficient - against the alternative
hypothesis – heterogeneous slope coefficient.
The test can in general be written as
SSR0  SSR1
Number of restrictions under H 0
F01 
SSR 1
Numbers of degrees of freedom under H1
If we assume individual intercept, the test would be:
SSRB  SSRA
SSRB  SSRA N (T  K  1)
K ( N  1)
FBA 

SSR A
SSR A
K ( N  1)
N (T  K  1)
Which is F-distributed with K(N-1) and N(T-K-1) degrees of freedom under H B .
If we assume a random effect model, we must use the Breusch-Pagan tests5.
5
Lecture note 7
Problem 4.2 D
(8) yit     xit   i   i xit  uit
i
i     i
i
i     i
uit  xit ,  i ,  i
IID(o,  2 )
uit
IID(o,  2 )
i  1,....., N ;
IID(o,  2 )
i  1,....., N ;
i  1,....., N ; t  1,......., T ,
Expanding the model allowing for correlation between i , i and xit , i  1,......, N . This may
reflect that a latent taste variable for household i -  i ,  i - , is correlated with household’s
income - the observable explanatory variable xit .
First we formalize a correlation between  i and xit .
 i   xi  E( xi )   wi
wi
IID(o,  w2 )
If  i is correlated with xit then it can’t be  i
IID(o,  2 ) , so we do the same formalization
for  i :
i   xi  E( xi ) i
i
IID(o,  2 )
(9) yit  (   x  )   xit  xi   xi xit  wi  xiti  uit
 k   xit  xi   xi xit  Qit
where E ( xi )   x
where Qit  wi  xiti  uit , k     x 
We have now controlled for the problem with latent heterogeneity, but we now have a
problem with the composite disturbance term.
So we transform the equation as we did in problem 4.2 a to get a covariance-variance matrix
that has zero outside the diagonal.
0TT 
 11


'
E(vv | X)  
 =
0
NN 
 TT
The transformed model can be written:
(10) yit  k   xit  xi   xi xit  Qit
OLS on 10 will give the MVLUE estimators.
Problem 4.3 Measurement error models
Problem 4.3 A
(1)
yit  k  it   uit
(2) xit  it  it
(3) it , uit ,it are independent for all i,t
ˆB 
  x  x  y  y 
 x  x
i
i
IID (o,  2 )
i  1,....., N ;
it
IID (o,  2 )
t  1,......., T ,
ˆC 
i
2
i
uit
i
  x  x  y  y 
 x  x
i
t
t
2
i
t
(a)
   xi  x  yi  y  

p lim ˆB  p lim  i
2

T 
T  
x

x



i

i


1
p lim 
T   N

 x
1
p lim 
T   N

  x
i
1

p lim   i  xi  x  yi  y  
N
T  

=

1
2


Slutsky
p lim   i  xi  x  
T   N


 x    xi  x    (i   )   (ui  u )  

2
1
p lim   i  xi  x  
T   N

i

 x  xi  x     xi  x  (i   )    xi  x  (ui  u  

2
1
p lim   i  xi  x  
T   N

1

1

1
p lim   i  xi  x  xi  x    p lim   i   xi  x  (i   )     p lim 
T   N
 T   N
 T   N

2
1
p lim   i  xi  x  
T   N

1

p lim   i i  i      (i   )  
T   N


2
1
p lim   i  xi  x  
T   N

1

1

p lim   i i    (i   )    p lim   i (i   )(i   ) 
T   N
 T   N
 

2
1
p lim   i  xi  x  
T   N

i
i

1
p lim 
T   N
since p lim  1

T   N


   (i   )    0, i    i  


1

lim   i (i2  2i   2 ) 
 i (i  )(i  )   pT 
N


 
i
i
  E(i2 )  2 E (i )  E ( 2 )   0, by assumption E(i2 )  E (i )  E ( 2 )
 x
i
i

 x  (ui  u ) 

b)
1

1

1
p lim   i  i    (i   )   p lim   i  (i   )(i   ) 
 2 
N   T
 N   T

T
p lim ˆB   

2
1
1
1
1
N 






bN*   2
p lim   i i      p lim   i (i   ) 2   p lim   i i    (i   ) 
T
N   T
 N   T
 N   T

2
1
bN*  p lim   i i    
N   T

since:
1

1

p lim   i  (i   )(i   )   p lim   i  (i2  2i   2 ) 
N   T
 N   T

1
 E i  E i    2 E i  E     E  2     2 , (since T is finite) and by assumption E i   0
T
c)
   xt  x  yt  y  

p lim ˆC  p lim  t
2

N 
N  
x

x




t
t


1

p lim   t  xt  x  yt  y  
N   T

=

2
1
Slutsky
p lim   t  xt  x  
N   T

1

1

1

p lim   t  xt  x  xt  x    p lim   t   xt  x  (t   )     p lim   t  xt  x  (ut  u ) 
N   T
 N   T
 N   T


2
1
p lim   t  xt  x  
N   T

1

1

p lim   t  t    (t   )   p lim   t  (t   )(t   ) 
N   T
 N   T
 

2
1
p lim   t  xt  x  
N   T

d)
1

1

1
p lim   t  t    (t   )   p lim   t  (t   )(t   ) 
 2 
T   N
 T   N

N
p lim ˆC   

2
1
1
1
1
T 






cT*   2
p lim   t t      p lim   t (t   )2   p lim   t t    (t   ) 
N
T   N
 T   N
 T   N

2
1
cT*  p lim   t t    
T   N

Problem 4.3 B
(1)
yit  k  it    i   t  uit
(2)
xit  it  i   t  it
i  1,....., N ;
(3)
it , uit ,it ,i , t , ,i , t are independent for all i,t
t  1,......., T ,
(4)
yit  k  xit    i   t  uit  i   t   it 
uit
IID (o,  2 )
i
IID(o,  2 )
t
IID(o, 2 )
it
IID (o,  2 )
i
IID(o,  2 )
t
IID(o,  2 )
a)
   xi  x  yi  y  

p lim ˆB  p lim  i
2

T 
T  
x

x


 i i


1
p lim 
T   N

 x
1
p lim 
T   N

  x
i
1

p lim   i  xi  x  yi  y  
T   N

=

2
1
Slutsky
p lim   i  xi  x  
N
T  


 x    xi  x     i     ui   (i   )  (i   )   

1
2


p lim   i  xi  x  
T   N

i

 x   xi  x     (i   )  xi  x    i    xi  x    xi  x  (i   )    xi  x  ui  

1
2


p lim   i  xi  x  
T   N

1

1

p lim   i  xi  x  xi  x    p lim   i    xi  x i  
T   N
 T , N   N


2
1
p lim   i  xi  x  
T   N

1

1

1

p lim   i  xi  x  ui   p lim   i   xi  x  i   p lim   i   xi  x   i 
N
N
N
T , N  
 T , N  
 T , N  

2
1
p lim   i  xi  x  
T   N

1

1

p lim   i i    i    i    (i   )    p lim   i i    i    i    i  
N
T   N
T





2
1
p lim   i  xi  x  
T   N

1

1

p lim   i (i   )(i   )   p lim   i (i   )(i   ) 
N
T   N
T



 

2
1
p lim   i  xi  x  
T   N

i
i

So:
p lim ˆB  
T , N 



   xt  x  yt  y  

p lim ˆC  p lim  t
2

N 
N  
x

x




t
t


1

p lim   t  xt  x  yt  y  
N   T

=

1
2


Slutsky
p lim   t  xt  x  
N   T

1

1

1

p lim   t  xt  x  xt  x    p lim   t   xt  x  (t   )     p lim   t  xt  x  (ut  u ) 
N   T
 N   T
 N   T


2
1
p lim   t  xt  x  
N   T

1

1

p lim   t  xt  x  ( t   )   p lim   t   xt  x  ( t  )   
N   T
 N   T

2
1
p lim   t  xt  x  
N   T

1

1

p lim   t  t    (t   )     p lim   t   ( t  )( t  )  
N   T
 N   T
 

2
1
p lim   t  xt  x  
N   T

so :
p lim ˆ  
N ,T 
C
b)
T is finite:
1

1

1
p lim   i  i    (i   )   p lim   i  (i   )(i   ) 
  2   2 
T
T
N

N





T
p lim ˆB   

2
2
1
2
N 
1
1
1
*
bN   2   2 
p lim   i i      p lim   i i      p lim   i i    
T
N   T
 N   T
 N   T

2
1
bN*  p lim   i i    
N   T

p lim ˆC  
N 
c)
N is Finite:
p lim ˆB  
T 
   xt  x  yt  y  

p lim ˆC  p lim  t
2

T 
T  
x

x


 t t


1
p lim 
T   N

1
p lim 
T   N
 x
t
t
 x
t
t
1

p lim   t  xt  x  yt  y  
T   N

=

1
2


Slutsky
p lim   t  xt  x  
T   N


1
 x  xt  x    p lim 
 T   N

1
 x  (t   )     p lim 
 T   N
2
1
p lim   t  xt  x  
T   N

  x
t
t
 x
t
t

 x  (ut  u ) 


1

 x  ( t   )   p lim   t   xt  x  ( t  )   
N
T




2
1
p lim   t  xt  x  
T   N

1

1

1
p lim   t  t    (t   )     p lim   t   ( t  )( t  )  
 ( 2  2 )
T   N
 T   N
 
N

2
1
2
2
1
1
1
*
cT   2  2 
p lim   t t      p lim   t  t     p lim   t t    
N
T   N
 T   N
 T   N

1
cT*  p lim 
T   N
 
t
t
2
  
