Christopher Dougherty
EC220 - Introduction to econometrics: past
examinations and marking schemes
2011 marking scheme
Original citation:
Dougherty, C. (2012) EC220 - Introduction to econometrics: past examinations and marking
schemes. [Teaching Resource]
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EC220 INTRODUCTION TO ECONOMETRICS
Marking Scheme for the 2011 Examination
Notes to examiners:
This marking scheme will in due course be posted on the EC220 website as a resource for future
students. For this reason some of the explanations are more detailed than is necessary for a marking
scheme.
Please mark each item in each question to the nearest half mark. Perhaps even one quarter, for small
items.
Examiners are encouraged to award additional marks if the candidate makes points that are not
included in the marking scheme, provided that they are wholly relevant and provided that they are not
just a statement of unnecessary detail. A few possible additional marks have been included in the
marking scheme.
If, contrary to instructions, a candidate has answered more than four questions, all those after the first
four should be disregarded. Candidates have been told that this will be the case.
EC220 examination 2011, marking scheme
2
1 The first question of the examination is usually of this type and it always has a bimodal
distribution of marks. It is pounced on by the stronger candidates who do it first and do it
well. It is also attempted, usually disastrously, as a last resort by weak candidates who are
having trouble finding a fourth question.
The question has deliberately been designed so that an error in one part should not lead to
further errors in the remaining parts. Since candidates know what they need to prove in
parts (a) – (g), examiners should be alert for fudges.
(a) [3 marks] b2* provides an estimate of the effect on Y, in terms of standard deviations of Y, of
a one-standard deviation change in X. [It is, of course, the beta coefficient, but we
have not discussed beta coefficients in the course.] [Give 2 marks if the candidate
says that b2* provides an estimate of the effect on standardized Y of a one unit
change in standardized X. Give 1 if the candidate mechanically states that b2*
provides an estimate of the change in Y* caused by a one unit change in X*.]
(b) [3 marks] We note that, by construction, Y *  X *  0 . So b1*  Y *  b2* X *  0 .
(c) [3 marks] Assuming an intercept,
b2*

 X  X Y  Y    X Y
 X  X 
X
*
i
*
*
* *
i i
*
i
*
i
*2
i
* 2
which is the expression appropriate for the specification without an intercept.
(d) [3 marks] Substituting for Y* and X*,
b2* 

X
X i*Yi*
*2
i

 Xi  X
sX
 

 Yi  Y

 s
 Y
 Xi  X 


 s

X






2

sX
sY
 X  X Y  Y   s
s
 X  X 
i
i
X
2
i
b2
Y
X X 1
s

(e) [3 marks] Yˆi*  b2* X i*  X b2  i
b2  X i  X  .
sY  s X  sY
Yˆi  b1  b2 X i  Y  b2 X   b2 X i  Y  b2 X i  X 


1 ˆ
Yi  Y .
Hence Yˆi* 
sY
(f) [2 marks]


2
(g) [3 marks]


1
Yi  Y   1 Yˆi  Y  1 Yi  Yˆi  1 ei
ei*  Yi*  Yˆi* 
sY
sY
sY
sY
 
s.e. b2* 
1
n

 X
*
i
2
ei*
 X*

2
 1  1
 
ei2
s
n
s
Y
  
 X s.e. b2 
2
sY
 Xi  X 


 s

X




(h) [2 marks] Given the results in (d) and (g), the t statistic for b2* is the same as that for b2.
(i) [3 marks] R2 will be the same.
EC220 examination 2011, marking scheme
*
R2 

 Y
Yˆi*  Y *
 
2
3
2
Yˆi*
 Y
* 2
*2


 1

 sY
2

 ˆ
 Yi  Y

2

2
 R2
 1 
2
 Yi  Y 
Y 
Alternatively, the candidate may state that F = t2, and so will be the same, and R2 is
the same transformation of F.
i
*
Y
i
  s
EC220 examination 2011, marking scheme
4
2. (a) [4 marks] The marginal effect of a year of training is to raise earnings by 20 percent.
(Literally, by a proportion 0.20.) (2 marks) (Note: This is approximative, but
enough to earn the four marks. The exact figure would be a proportion e0.20 – 1.
Give 0.5 extra mark for this.)
The underlying mathematical model is
Y  e 1   2 X ...
From this, one obtains
Y
  2 e 1   2 X ...   2 Y .
X
Hence
Y
2  Y
X
giving the interpretation. (2 marks). Alternatively, candidates may start with the
regression specification
log Y  1   2 X  ...
take the partial differential with respect to X,
1 Y
 2
Y X
and reach the same result.
(b) [4 marks] Those earning the Diploma tend to earn a proportion 0.30, that is 30 percent, more
than those with no formal training 2 marks). The regression specification implies
the model
Y  e 1   2 X D...  e 1   2 X D... e D e 1   2 X ...
where D is a dummy variable. For the reference category, D = 0 and e D  1 .
When D = 1, e D  e   1   , giving the interpretation (2 marks). (Again, the 30
percent is an approximation. The exact figure is derived from e0.30 – 1. Give 1
extra 0.5 mark for this.)
(c) [3 marks] The coefficient mow provides an estimate of the extra earnings of those with a
Diploma, controlling for the fact that they have had two years of training. In other
words, it is the extra earnings associated with passing the test at the end of the
second year.
(d) [3 marks] Define a new dummy variable NQ (no qualification) equal to 1 for those who do
not possess either the RM Certificate or the RM Diploma, and equal to 0 if RMC =
1 or RMD = 1. Run the regression with NQ and RMD, dropping RMC. Earning a
Certificate is now the reference category and the coefficient of RMD will provide
an estimate of the extra proportional earnings from obtaining a Diploma. One
would perform a t test on the coefficient. Alternatively, one could make earning a
Diploma the reference category and test the coefficient of RMC. Or, one could
merge the Certificate holders and the Diploma holders into one category and
perform an F test on the change in RSS.
(e) [4 marks] In general terms the regression specification is
log Y   1   2 EXP   3TRAINING   RMC RMC   RMD RMD  u
EC220 examination 2011, marking scheme
5
and the hypothesized restriction is
 RMD  2 RMC  0
Define
   RMD  2 RMC
Then
 RMD    2 RMC
Substituting into the regression specification, one has
log Y   1   2 EXP   3TRAINING   RMC RMC  2 RMD   RMD  u
and one performs a t test on the coefficient of RMD.
(Alternatively and equivalently, one may reparameterize the specification as
log Y  1   2 EXP   3TRAINING   RMD 0.5RMC  RMD   0.5RMC  u )
and perform a t test on the coefficient of RMC.
(Alternatively, and equivalently, one may fit the restricted model
log Y   1   2 EXP   3TRAINING   RMC RMC  2 RMD   u
and perform an F test comparing RSS for that specification with RSS for the
unrestricted model.)
(f) [4 marks] In this specification, the dummy variables are capturing the effect of passing the
tests for the award of the qualifications, as opposed to just receiving the training (1
extra mark for this or any other sensible attempt at an interpretation).
F test comparing RSS for specifications (1) and (3). The null hypothesis is that the
coefficients of RMC and RMD in specification (3) are both zero (1 mark). The F
statistic is
F 2,95 
105  95 / 2  5
95 / 95
(2 marks)
The critical value of F(2,100) is 4.82 at the 1 percent level, so one would reject the
null hypothesis at that level (1 mark). [If the F statistic has been computed
incorrectly, give this mark anyway if the conclusion is correct, conditional on F.]
(g) [3 marks] This question could be answered reasonably in various ways, so mark flexibly and
generously. Candidates may focus on any of the three specifications in the table.
If (1) is chosen, define a dummy variable for one sector, say PRIVATE for private
training institutes, and add it with an interactive dummy variable
PRIVATE*TRAINING to the specification. One would then perform t tests on the
coefficients of these two variables and an F test on their joint explanatory power.
Some candidates may propose a Chow test. This would in principle not be
appropriate because a difference in the wage equations could be due to differences
in the experience coefficient. One would have to argue that any such difference
would be attributable to differences in private/public training.
EC220 examination 2011, marking scheme
6
3. Write the original model
Y  1   2 X   3 Z  u
(1)
Then, with
X  0.5V  W  , Z  0.5V  W  ,
the other specifications are
Y   1  0.5 2   3 V  0.5 2   3 W  u
(2)
Y   1   2V  u
(3)
with the implicit restriction  3   2 , and, using X = V – Z,
Y   1   2V   3   2 Z  u
(4)
(a) [20 marks] [Give the marks if the numbers are correct, even if the explanation is vestigial.]
(2) and (4) are reparameterizations of (1), so the measures of fit are unchanged:
E = L = 0.60, F = M = 200. Give 1 mark for each.
Given the relationships among the parameters,
A = 0.70, C = –0.10, J = 0.60, H = 0.20. Give 1 mark for each.
The standard errors B and D cannot be reconstructed because the standard errors of
b2 and b3 cannot be used (on their own) to construct standard errors of linear
combinations (a loose explanation is acceptable because we have hardly touched
on covariances between estimators). Give 1 mark for each.
K = 0.04 since J = coefficient of X in specification (1) (2 marks).
The F statistic for the restriction  3   2 implicit in specification (3) is
F 1,40  
220  200 / 1  4.0
200 / 40
In terms of R2 it would be
F 1,40  
0.60  G  / 1
0.40 / 40
Hence G = 0.56 (4 marks).
A two-sided t test on the coefficient of Z in specification (4) provides an equivalent
test of the restriciton. The t statistic must therefore be
4.0  2.0 and so I = 0.10
(4 marks).
[Note: One may also compute G using the t statistic for the coefficient of V in
specification (3):
G
t2
1  G  / 41
Give the 4 marks also if the candidate proposes this method, even if he or she
cannot do the arithmetic.
EC220 examination 2011, marking scheme
7
Yet another was of computing G is as follows. Since R2 in specification (1) is 0.60,
TSS must be 500, using
R2 1
RSS
TSS
TSS is the same in specification (3). Hence one obtains G = 0.56.]
(b) (i) [2 marks] The estimates of  2   3  and  2   3  obtained from specification (2)
should be relatively precise compared with those of 2 and 3 in specification
(1). As a consequence, their standard errors should be lower. Give 0.5 extra
mark for saying that the coefficient of V in specification (3) will be subject to
omitted variable bias if  3   2 , and a further extra 0.5 mark for arguing that
any such bias is likely to be small. Give 0.5 extra mark for saying that the
standard error of V will be invalid.
(ii) [3 marks] If  2   3 , W is a redundant variable in specification (2), and hence the
estimate of the coefficient of V will be inefficient. Its standard error should be
larger than that in specification (3). Give an extra mark for saying that it
would not be much larger because the correlation was low.
EC220 examination 2011, marking scheme
8
4. [Note: In this answer, CA has been replaced by A.]
(a) (i) [5 marks]
b2 
 D  D Y  Y 
 D  D 
 D  D    
i
i
2
i

1
i

  3 Ai  u i    1   2 D   3 A  u
2 Di
 D  D 
 D  D A  A    D  D u  u 
 D  D 
 D  D 

2
i
 2  3
i
i
i
i
2
2
i
i
(Give 2 marks for this decomposition.) Hence
b2   2   3
 d A
i
  d u
A 
i
i
i
u
where
di 
Di  D
 D
j
D

2
Hence
E b2    2   3
 Ed A
i
i
  Ed u
A 
i
i
 u 
Now, since the assignment to the course was random, D is distributed
independently of both A and u, and hence
 



E d i Ai  A  E d i E Ai  A  0
and
Ed i u i  u   E d i E u i  u   0
Hence b2 is an unbiased estimator of 2.
(Give 3 marks for the proof of unbiasedness. Be alert for false explanations.
Candidates were told that the estimator is unbiased.)
(ii) [3 marks] The researcher is nearly correct. Given the random selection of the sample, A
will be distributed independently of D and so it can be treated as part of the
disturbance term and the standard error will remain valid. The requirement
that A have a normal distribution is too strong, since the expression for the
standard error does not depend on it. However, if the standard error is to be
used for t tests, then it is important that the enlarged disturbance term should
have a normal distribution, and this will be the case if an only if A has a
normal distribution (assuming that u has one). If both A and u have normal
distributions, a linear combination will also have one.
(iii) [3 marks] The commentator is correct for the reasons explained in (ii).
EC220 examination 2011, marking scheme
9
(b) (i) [3 marks] Add an interactive term (slope dummy):
Y   1   2 D   3 A   4 DA  u
where DA is the product of D and A.
(ii) [5 marks] Abstracting from the effect of the disturbance term, 1 is literacy test score of
an individual not assigned to the extra course possessing zero ability (which
may not be possible, depending on the way that ability is measured, but
candidates are not expected to say this) (1 mark); 2 is the effect of the extra
course on the literacy test score, for those with zero ability, sign positive (1
mark); 3 is the effect on the test score of a unit increase in ability, for those
who were not assigned to the extra course, sign positive (1 mark). Writing the
model as
Y   1   2   4 AD   3 A  u
4 may be interpreted as the increase in the effect of the extra course for every
unit increase in the measure of ability (1 mark), or rather decrease, since the
sign will be negative if the researcher is correct (1 mark). The model may
also be rewritten
Y   1   2 D   3   4 D A  u
so that 4 may also be interpreted as the difference in the effect of a unit
increase in ability for those who took the course compared with those who did
not (give 1 extra mark),
(iii) [4 marks] Extending the decomposition of b2,
b2   2   3
 d A
i
i

 A  4
 d DA
i
i
  d u
 DA 
i
i
 u .
DA will be positively correlated with D because DA = 0 when D = 0 and DA is
some positive number when D = 1. Hence b2 will be a biased estimator of 2
(3 marks). Since DA will be positively correlated with D and the coefficient
of DA is likely to be negative, the bias will be downwards (1 mark).
(c) [2 marks] The decision to take the course is now endogenous and likely to be correlated with
the disturbance term. In particular, it would be rational for lower ability children to
take the course. This should increase the estimated slope coefficient (2 marks.
This is not the only possible answer. Give the marks for any sensible alternative
and give up to 2 extra marks for a particularly good answer. One alternative
answer might be that there could be a pushy parent effect, with parents of wellperforming children making their children take the course, even though they do not
need it. This could lead to a bias in the other direction.)
EC220 examination 2011, marking scheme
10
5. (a) (i) [1 mark] Yi   2 Z i  vi   2  X i  wi   vi   2 X i  u i where u i  vi   2 wi .
(ii) [1 mark] wi is a component of both Xi and Yi, and therefore of both the numerator and
the denominator of the expression for b2OLS . As a consequence it is not
possible to derive a closed-form expression for the expectation.
(iii) [5 marks]
n
b2OLS


n
X u
X i  2 X i  u i 
i 1
n
X
i i
i 1
n
 2 
X
2
i
i 1
i 1
 X i  X u i  u   nXu
n

i 1
 X i  X 
n
2
i
2
 nX
2
i 1

1
n
 X
n
i
 X u i  u   Xu
i 1
1
n
 X
n
 X  X2
2
i
i 1
Now
1
plim 
n
 X
n
i 1
i

 X u i  u   Xu   cov X .u    X  u  cov X .u 

 covZ  w, v   2 w    2 w2
assuming that Z, v and w are distributed independently, and
1
plim 
n
 X
n
i 1
i

2
 X   X 2   var X    X2   Z2   u2   Z2

Hence
plim b2OLS   2 
 w2
 Z2   w2   Z2
(iv) [2 marks] (Mark this flexibly. Give the 2 marks for any sensible answer.) For fixed  Z2 ,
 v2 . and  w2 , the observations in the sample form a cloud around the point
Z , Y .
The larger is the value of Z, the further this cloud is away from the
origin, and the clearer will be the relationship between Y and Z. As a
consequence, the bias will attenuate. (The variance will also decrease, but the
candidate is not expected to state this. Give an extra mark if it is stated.)
(v) [4 marks]
Y 2 X  u
u

 2 
X
X
X
Hence
plim
Y
0
 2 
 2
Z
X
(vi) [2 marks] (Mark this flexibly. Give the 2 marks for any sensible answer.) As the size of
the sample becomes large, the impact of the disturbance term in the
determination of Y attenuates and Y approaches  2 Z . At the same time, the
EC220 examination 2011, marking scheme
11
impact of the measurement error in the determination of X attenuates and X
approaches Z . As a consequence, the estimator approaches 2.
Q  2Z  v
(b) (i) [5 marks]
X i  Z i  wi
Yi  Qi   2 wi
Hence
Yi   2 wi   2  X i  wi   v i
Yi   2 X i  v i
There is no violation of any regression model assumption and the estimator
will be both unbiased and consistent.
(ii) [2 marks] Variations in the observations on X and Y due to variations in the
measurement error will be indistinguishable from those due to variations in Z.
Hence there will be no inconsistency. Note: A variety of answers, not
necessarily as above, may be offered. Examiners should mark generously.
(iii) [3 marks] [Note: The variance expression given in the examination question is wrong.
There should have been an additional term nX 2 in the denominator. However,
since X  Z , the following answer remains approximately correct.]
Given that the distributions of Z and w are independent, the MSD of X will
tend to be larger than that of Z and so the variance of the estimator will be
smaller. From part (b) (i) we can see that the estimator will be unbiased as
well as consistent. Hence the only effect will be a benefit—a reduction in the
variance.
EC220 examination 2011, marking scheme
12
6. (a) (i) [3 marks] The IV estimator will be
b2IV 
 Z
 Z
t

 Z Y


Y 
 Z  Ct  C
t
t

2
 Z
 Z
t
t
 Z u t  u 
 Z Yt  Y 
It is not possible to obtain a closed form expression for the expectation of the
error term since ut influences both the numerator explicitly and the
denominator implicitly through Yt. Instead we take plims. We know that
plim
1
n
 Z
t
 Z u t  u   covZ , u   0
plim
1
n
 Z
t
 Z Yt  Y   covZ , Y   0
and
so

plim 



1
  Z t  Z u t  u  





Z
Z
u
u


t
 t
  plim  n


1
 Z t  Z Yt  Y  
  Z t  Z Yt  Y  

n
1
n

1
plim
n
plim

0
 Z
 Z
covZ , Y 
t
 Z u t  u 
t
 Z Yt  Y 
0
covZ , Y   0 because Z is a component of Y. b2IV is therefore a consistent
estimator of 2.
(ii) [3 marks] Note to examiners: Although this is standard bookwork, it has not previously
appeared in an examination. The marks should be given for evidence of
conceptual understanding and one should not expect the mathematical
precision of the answer provided below. In future years, of course, having
seen it once, the students would all be ready with expert mathematical
regurgitation if this item reappeared on an examination in any guise or form.
The approximation derives from the use of a central limit theorem to establish
that
 2
1
d
T b2IV   2 
N  0, u2  2
 
 Y ,Z
Y







where Y, Z is the population correlation between Y and Z. The expression
applies only in the limit. However, dividing by
approximation,
b
IV
2

 u2
1
  2 ~ N  0,

 T MSDY  r 2
Y ,Z


T , we have, as an




EC220 examination 2011, marking scheme
13
in a finite sample, and hence

 u2
1
b2IV ~ N   2 ,
 2

T MSDY  rY , Z





How large the sample must be for the approximation to be ‘good’ is usually
established by simulation. (Give 1 extra mark if the answer is precise.)
(iii) [3 marks] Obviously, one consequence of the approximation is that the estimate of the
standard error is only an approximation (1 mark). Probably more important in
practice is the fact that the distribution will only approximately be normal and
the distortions in the tails may be serious, undermining the tests (2 marks).
Give an extra mark for saying that the size of the test is likely to differ from
the nominal size, and another extra mark in the unlikely event that, even
allowing for this, asymmetry in the tails may cause the power of the test to be
suboptimal.
Yt 
(iv) [1 mark]
1
1  I t  Gt  u t 
1 2
(v) [2 marks] The asymptotic variance of an IV estimator in a simple regression is inversely
proportional to the square of the correlation between the instrument and the
variable for which it is acting. If we fit the reduced form equation, the
coefficients of I and G will be chosen so as to maximize the correlation
between the actual and fitted values of Yt. The fitted values, which are linear
combinations of I and G, will therefore have higher correlations with Y than I
or G individually. and consequently the asymptotic variance of the TSLS
estimator will be smaller than those of the IV estimators.
(vi) [3 marks] The assertion is incorrect. The correlation between Y and Z will, in general,
be higher than the correlations between Y and I and Y and G. Therefore the
TSLS estimator will be more efficient than the use of either I or G on its own.
(vii) [2 marks] Nonsense.
(viii)[3 marks] Correct. We are now using TSLS with the extra benefit of the imposition of
the theoretical restriction that the coefficients of It and Gt should be the same.
(b) (i) [3 marks] The estimator in (a) (i) would be inconsistent.
Z t  Yt  C t  1   2 Yt  1  u t
Hence
plim
plim
1
n
and so
1
n
 Z
 Z
t
t
 Z u t  u   1   2  covY , u   varu    u2


 Z Yt  Y  1   2  varY   covu , Y   1   2  varY 
EC220 examination 2011, marking scheme

plim 


14

1
  Z t  Z u t  u  





Z
Z
u
u


t
 t
  plim  n


1
 Z t  Z Yt  Y  
  Z t  Z Yt  Y  

n
1
n

1
plim
n
plim

 Z
 Z
t
 Z u t  u 
t
 Z Yt  Y 
 u2
varY 
and the IV estimator will be downwards biased in large samples.
(ii) [2 marks] OLS is unbiased and efficient if Y is exogenous.
EC220 examination 2011, marking scheme
15
7. (a) (i) [2 marks] The ensemble distribution is the limiting distribution of the cross-section of
possible realizations of the process at time t.
(ii) [2 marks] By construction, there are no transitory initial effects and so the ensemble
distribution will apply equally to Xt and Xt–1. Hence E  X t   E  X t 1  and,
from the process
E  X t    2 E  X t 1   E  t   0
(iii) [3 marks] For the same reason, var X t   var X t 1  , so
var X t    22 var X t 1   var t  
 2
1   22


(iv) [3 marks] The disturbance term t is a determinant of the explanatory variable X s 1 in
observation s
X s   2 X s 1   s
for all s  t . The regression model assumption that the all values of the
disturbance term in the sample be distributed independently of all values of
the explanatory variables in the sample is therefore violated. As a
consequence, the estimator will be biased in finite samples. However, it will
be consistent because the disturbance term is contemporaneously independent
of the regressor.
(b) (i) [1 mark] The slope coefficient is
 X t  X Yt  Y 
T
b2 
t 1
 X
T
 X
 X
T
 2 
2
t
t
 X  t   
t 1
t 1
 X
T
 X
2
t
t 1
t is a component of Xt and hence the error term is the ratio of two related
stochastic terms. Consequently, it is not possible to obtain a closed-form
expression for the expectation.
(ii) [3 marks] 
1 T

X t  X  t    
n

plim b2   2  plim  t 1 T

1
2

X t  X  

n
t 1




Now
1
plim 
n
 X
T
t
t 1

 X  t     cov X t ,  t    2

from the DGP for Xt, and
1
plim 
n
2
 X t  X     X2 
T
t 1

 2
1   22


Hence the limits of the numerator and denominator of the error term both exist
and so
EC220 examination 2011, marking scheme
16
plim
plim b2   2 
1
n
 X
T
t
 X  t   
t 1
1
plim
n
 X
T
 X
 2 
2
t
 2

 2 / 1   22

2

 1   22
t 1
(c) (i) [3 marks] When 2 = 0, Xt = t and
Yt  5  2 X t   t  5  3 X t
Hence there is an exact linear relationship between the values of Y and those
of X, with slope coefficient 3.
(ii) [2 marks] The large sample bias should be 0.75 when 2 = 0.5 and zero when 2 = 1.
The distributions support the analysis.
(iii) [3 marks] If 2 = 1, Xt is a random walk. Since Yt is a linear function of Xt, Yt will also
be nonstationary. The linear relationship is by construction the cointegrating
relationship.
(iv) [3 marks] When 2 = 0.5, the distribution should be T consistent, the variance being
inversely proportional to T, the standard deviation being inversely
proportional to
T , and hence, approximately, the height of the distribution
proportional to T . Thus one would expect h100/h25 = h400/h100 = 2. This
appears to be the case.
When 2 = 1 and the relationship is a cointegrating relationship, the
distribution of the slope coefficient should be superconsistent, the variance
being inversely proportional to T2, the standard deviation being inversely
proportional to T, and hence, approximately, the height of the distribution
proportional to T. Thus one would expect h100/h25 = h400/h100 = 4. Again, this
appears to be the case.

EC220 examination 2011, marking scheme
17
8. (a) (i) [3 marks] Let the fitted model be
Yˆi  b1
Then ei, the residual in observation i, is given by
ei  Yi  Yˆi  Yi  b1
Hence RSS, the residual sum of squares, is given by
RSS 
n
 Y
i
 b1 
2
i 1
Differentiating this, the first-order condition is
n
dRSS
 2 Yi  b1   0
db1
i 1

from which we have b1  Y . The second derivative is 2n, which is positive,
confirming that we have reached a minimum. (Give the 3 marks even if the
second-order condition is not mentioned. Give an extra 0.5 mark for
checking it.)
(ii) [1 mark] Y   1  u . E u   0 , so E Y    1
(iii) [1 mark] plim u = 0 since E u   0 . Hence plim Y = 1. (This is sufficient, given the
instruction on page 2 of the examination paper. Alternatively one could argue
that Y is a consistent estimator because it is unbiased and its variance tends to
zero as n becomes large. The variance of Y is the variance of u , and this is
 u2 / n .)
Yt   1  u t
(b) (i) [2 marks]
Hence
Yt 1   1   u t 1
Subtracting from the previous equation and rearranging.
Yt   1 1     Yt 1   t (1 mark)
The point is that the specification is now free from autocorrelation (1 mark)
(ii) [2 marks] t is a determinant of the explanatory variable Ys 1 in observation s for all
s  t . The regression model assumption that the all values of the disturbance
term in the sample be distributed independently of all values of the
explanatory variables in the sample is therefore violated. As a consequence,
the estimator will be biased in finite samples. However, it will be consistent
because the disturbance term is contemporaneously independent of the
regressor.
(iii) [3 marks] Following the usual analysis, the OLS estimator of  can be decomposed as
EC220 examination 2011, marking scheme
18

1
T
 Y
T
t 1
 Yt 1  t   t 
t 2
1
T
 Y
T
 Yt 1 
2
t 1
t 2
Since Yt is a stationary process,
plim
1
T
 Y
T
t 1
 Yt 1  t   t   covYt 1 ,  t   0
t 2
and
plim
1
T
 Y
T
 Yt 1    Y2t
2
t 1
t 2
Hence the OLS estimator of  is consistent.
(iv) [2 marks] Suppose that when Yt is regressed on Yt–1, the fitted relationship is
Yˆt  A  BYt 1
Then plim B =  and plim A =  1 1    and
A
will be a consistent
1 B
estimator of 1:
plim
 1   
plim A
A

 1
 1
1  B 1  plim B
1 
(v) [2 marks] plim Y   1  plim u  1 since E u   0 and plim u  E u  .
(vi) [2 marks] It is not possible to determine which is preferable, at least without a
simulation. Both estimators are consistent. Y is also unbiased, but for a finite
sample it might have a larger variance than the other estimator.
(c) (i) [1 mark] The equation
Yt   1 1     Yt 1   t
becomes
Yt  Yt 1   t
and so 1 is washed out.
(ii) [3 marks]
Y  1 
1
T
T
 T  1  t 
t
t 1
and E Y    1 (1 mark). The variance of Y is an increasing function of T
and so the estimator is inconsistent (2 marks).
(iii) [1 mark] Y1   1   1 . Hence E Y1    1 and its variance is  2 It is an inconsistent
estimator because the variance is independent of the size of the sample and
does not tend to zero.
(iv) [2 marks] Y1 will have a smaller variance if T > 1.