Summer 2013 examination

EC402 E onometri s

Suitable for all andidates

Instru tions to andidates

Time allowed: 3 hours This paper ontains THREE se tions. Answer all questions in Se tion A, ONE question from Se tion B, and ONE question from Se tion C. Question 1 arries 60% of the overall mark; Se tions B and C ea h arry 20% of the overall mark. You are supplied with:

 Di key-Fuller Statisti al Tables  Murdo h and Barnes Statisti al Tables (2nd-4th editions)  Table A5 Durbin-Watson d-statisti

EC402  Summer Exam

E onometri s

SECTION A This se tion arries 60% of the total marks. Answer all questions. [Questions 1.5. from o-examiners.℄ 6. [8 points℄ Suppose W = { Wt : t = 0, 1, . . . } follows

Wt = ǫt + θǫt−1 ,

ǫ w.n. (0, ν2 ), |θ| > 1.

(a) Find W 's ovariogram GW . (b) Suppose Y = { Yt : t = 0, 1, . . . } follows

Yt = ηt + φηt−1 ,

η w.n. (0, σ2 ), φ = θ−1 =⇒ |φ| < 1.

Under what (if any) restri tions on σ2 will the ovariogram of Y equal that of W ? ( ) Suppose that in W the w.n. ǫ is iid N(0, ν2 ). Show that if 2 def

λ =

s=∞ X

GW (s)

s=−∞

then −1 −1/2

λ

T

T X

L

Wt −→ N(0, 1)

t=1

ANS

for T → ∞.

(a) Sin e EWt = 0 we an dire tly al ulate

GW (0) = ν2 + θ2 ν2 = (1 + θ2 )ν2  θν2 for s = ±1; GW (s) = 0 elsewhere. (b) Sin e EYt = 0 we an dire tly al ulate

GY (0) = (1 + φ2 )σ2 = (1 + θ2 )σ2 /θ2  φσ2 = σ2 /θ for s = ±1; GY (s) = 0 elsewhere. Provided σ2 = θ2 ν2 then GY = GW .

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( ) To prove the statement, rst write T X

Wt = ǫ1 + θǫ0 + ǫ2 + θǫ1 + · · · + ǫT + θǫT −1

t=1

= θǫ0 +

TX −1



(1 + θ)ǫt + ǫT .

t=1

This is a sum of independent normal random variables and so is itself normally distributed. It has mean zero and varian e equal to   (T − 1)(1 + θ)2 + 1 + θ2 ν2 so that

Var



−1 −1/2

λ

T

T X t=1

Wt



=



T − 1 (1 + θ2 )ν2 + T λ2 T

→ 1 as T → ∞.

Therefore −1 −1/2

λ

T

T X t=1

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L

Wt −→ N(0, 1)

for T → ∞.

Q.E.D.

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E onometri s

7. [8 points℄ Suppose the true model is

Xt = Xt−1 + ut , with

ut = ρut−1 + ǫt ,

ǫ w.n. (0, ν2 ), |ρ| < 1.

A resear her onsiders the regression model (7.1)

Xt = βXt−1 + ut

under the null hypothesis β0 = 1. She proposes to estimate equab T from observations tion (7.1) using OLS, yielding the estimator β X0 , X1 , . . . , XT . A referee of the resear h noti es that in the tted model the disturban es are serially orrelated. Sin e the regressor is a lagged dependent variable, the referee says the resulting estimate is in onsistent. P (a) Dening σ2 = (1 − ρ2 )−1 ν2 and λ2 = ∞ s=−∞ E[u0 u−s ], prove that the OLS estimator on equation (7.1) satises:   1 2 2 2 2 λ B(1) − σ L b T · (βT − 1) −→ . R1 λ2 0 B(r)2 dr with B standard Brownian motion.

(b) Respond to the referee. ANS

(a) Dene the approximant: [rT ]

WT (r) = T −1/2 X[rT ] = T −1/2

X

ǫt

for r ∈ [0, 1].

t=1

By FCLT we know that

λ−1 WT −→ B. L

Next onsider the OLS estimator.

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i. First, T X t=1

2

Xt−1 = T =T

T X t=1 T X t=1

=⇒ T

−2

T X t=1

2

= T2

Z1

L

2

WT

"Z t

t−1 T

T

Z1 0

2 2

1 r= t− T

r=0

Xt−1 −→ λ



#

WT (r) dr × T

WT (r)2 dr B(r)2 dr.

ii. Then taking the square on both sides of Xt = Xt−1 + ǫt , we have  1  2 Xt−1 ǫt = Xt − X2t−1 − ǫ2t 2 " # T X X 1 2 2 XT − =⇒ Xt−1 ǫt = ǫt 2

=⇒ T −1

X

Xt−1 ǫt = L

1 2

−→

" 1 2

t=1

WT (1)2 − T −1

T X t=1

ǫ2t

#

 2  λ B(1)2 − σ2 .

b T behaves as iii. From these we see that the OLS estimator β   −1  X X −2 −1 2 b T · (βT − β0 ) = T T Xt−1 Xt−1 ǫt   1 2 B(1)2 − σ2 λ 2 L . −→ R1 λ2 0 B(r)2 dr

Q.E.D.

(b) The referee is in orre t. Under the statement just proven, T times the regression deviation is a well-behaved random variable. Therefore, the regression deviation taken just by itself onverges in probability to 0, i.e., OLS is onsistent. But not only is OLS

onsistent, it is a tually onsistent at the faster rate T , not the standard regression analysis rate of T 1/2 . The reason for the referee's error is that he is applying standard regression theory and intuition to a situation that is non-standard. In this unit roots regression, the regressor Xt−1 is, roughly speaking, growing

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E onometri s

so qui kly, and thus a

umulating so mu h information ontent, that it dominates the onfounding inuen e of the residual being

orrelated with the regressor.

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SECTION B This se tion arries 20% of the total marks. Answer ONE question. 8. Is this the next Question? ANS

If so, then this must be the next Answer.

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9. Is this the next Question? ANS

If so, then this must be the next Answer.

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SECTION C This se tion arries 20% of the total marks. Answer ONE question. 10. [20 points℄ Consider the regression model

yt = Xt′ β0 + ut

(10.1)

where (i) E[Xt−s ut ] = general;

0

for s =

0

but will otherwise dier from zero, in

(ii) u follows

ut = ρut−1 + ǫt ,

|ρ| < 1, ǫ w.n. (0, ν2 ).

Taking equation (10.1) and onditions (i)(ii) as given, answer the following. (a) Write down additional onditions under whi h OLS on equab T onsistent for the true value tion (10.1) gives the estimator β β0 and asymptoti ally normal.

(b) Provide a onsistent  estimator for the varian e- ovarian e matrix b T − β0 . of T 1/2 β ( ) A resear her spe ulates he might improve e ien y of the estimator by using a variant of GLS. He proposes al ulating

et = yt − ρyt−1 y et = Xt − ρXt−1 X e t = ut − ρut−1 u

and then performing OLS on the transformed model

e′ β + u et = X et . y t

Des ribe the possible advantages and disadvantages of this pro edure for the model above. (Be as expli it as you an: among other things, explain why the resear her might think this transformation a good idea and explain whether this pro edure a hieves what the resear her intends.) ANS

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(a) The following onditions will make OLS onsistent and asymptoti ally normal: i. E[Xt ut ] = 0, i.e., E[Xt · (yt − Xt′ β)] = 0 at β = β0 ; P ii. plimT →∞ T −1 Tt=1 Xt Xt′ = QX full rank, nite; P L iii. T −1/2 Tt=1 Xt ut −→ N(0, Ω), Ω full rank, nite.

Noti e that the rst of these simply repeats what is already given in equation (10.1) and assumption (i). Also, sin e nothing is said about the serial orrelation in u, these onditions allow the kind of dependen e des ribed in assumption (ii).   1 −1 1 / 2 b βT − β0 is Q− (b) The varian e- ovarian e matrix of T X ΩQX . A onsistent estimator for this uses



T

−1

T X

Xt Xt′

t=1

−1

Pr

1 −→ Q− X

and the HAC form

b T =T Ω

−1

X T t=1

k  X j=1

Zt Zt′ +

j 1 − k+1

 X T

′ [Zt Zt−j

t=j+1

+

Zt−j Zt′ ]



.

to estimate Ω onsistently (where the lag length k goes to innity with T but more slowly). Combining these gives a onsistent 1 −1 estimator for the varian e- ovarian e matrix Q− X ΩQX .

e serially un orrelated and thus the ( ) The transformed model has u varian e- ovarian e matrix of the residuals will now be diagonal. The hope is that this makes OLS on the transformed model e ient for estimating β0 . However, noti e that: et u e t ] = E[(Xt − ρXt−1 )(ut − ρut−1 )] E[X

= E[Xt ut + ρ2 Xt−1 ut−1 − ρXt−1 ut − ρXtut−1 ] = −ρE[Xtut−1 ] 6= 0.

Thus, OLS on the transformed system will be in onsistent.

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11. Is this the next Question? ANS

If so, then this must be the next Answer.

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