Alastair Hall
1.(a) We have
Using the WLLN, we have
Semester 1, 2020-21
(1)
( 2 )
ECON61001: Econometric Methods
Solutions to Problem Set for Tutorial 7
1/2 T −1/2 Tt=2 utut−1 T ρˆT = T−1T u2 .
t=1 t
T
T − 1 u 2t →p E [ u 2t ] = σ ε2 .
t=1
Now consider T−1/2 Tt=2 utut−1. Since we are given that ut is an independent sequence with mean zero, it follows that E[utut−1] = Cov[ut, ut−1] = 0. Therefore, using the CLT for time series, we have
T
T−1/2utut−1 →d N(0,ω) (3)
t=2
where ω = γ0 + 2∞i=1γi and γi = E[utut−1ut−iut−i−1]. Using the fact that ut is an independent sequence it follows that: (i) γ0 = E[u2t]E[u2t−1] = σε4; (ii) for i > 0, γi = E[ut]E[ut−1ut−iut−i−1] = 0 as E[ut] = 0. Therefore ω = σε4.
Using Lemma 3.5 in the Lecture Notes (in Section 3.1), it follows from (1)-(3) that: T 1 / 2 ρˆ T →d N ( 0 , 1 ) .
1.(b) We have
T − 1 Tt = 2 u t u t − 1
ρˆT = T−1T u2 . (4)
t=1 t
From Tutorial 5 Question 4 parts (b) and (c) respectively it follows that, it follows from the
WLLN that
Combining (4-(6) and using Slutsky’s theorem, it follows that: ρˆT →p θ.
1.(c) We have
T − 1 Tt = 2 u t u t − 1
ρˆT = T−1T u2 . (7)
t=1 t 1
Tp σ2
T−1u2t→E[u2t]= ε, (5)
t=1 1 − θ2
− 1 T p θ σ ε 2
T utut−1 → E[utut−1] = 1−θ2. (6) t=2
From Tutorial 5 Question 3 parts (b) and (c) respectively it follows that, it follows from the WLLN that
T
T − 1 u 2t
t=1 T
T−1utut−1 t=2
→p E [ u 2t ] = σ ε2 ( 1 + φ 2 ) , →p E[utut−1] = σε2φ.
( 8 ) (9)
(10)
( 1 1 ) (12)
Combining (4-(6) and using Slutsky’s theorem, it follows that: ρˆT→p φ.
1.(d) We have
From the WLLN, it follows that
1 + φ2
T − 1 Tt = 2 u t u t − 1
ρˆT = T−1T u2 . t=1 t
T
T − 1 u 2t
t=1 T
→p E [ u 2t ] ,
→p E[utut−1].
T−1 utut−1 t=2
To evaluate these expectations, we substitute in for ut and use the given properties of εt as follows.
E[u2t ] = E[(εt + φεt−2)2]
= E[ε2t + 2φεtεt−2 + φ2ε2t−2]
= E[ε2t ] + 2φE[εtεt−2] + φ2E[ε2t−2]
= σε2(1+φ2), (13)
using the given information that E[ε2t] = σε2, and that εt is independent of εt−2 so that E[εtεt−2] = E[εt]E[εt−2] = 0 as E[εt] = 0.
E [utut−1 ] = E [(εt + φεt−2 )(εt−1 + φεt−3 )]
= E[εtεt−1 + φεt−1εt−2 + φεtεt−3 + φ2εt−2εt−3]
= E[εtεt−1] + φE[εt−1εt−2] + φE[εtεt−3] + φ2E[εt−2εt−3]
=0 (14)
using the given information that εt is independent of εs for all t ̸= s so that E[εtεs] = E[εt]E[εs] = 0 as E[εt] = 0. From (10)-(14) and using Slutsky’s Theorem, it follows that ρˆT →p 0.
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2. The assumed model specification is:
yt =x′tβ0+ut
ut =ρut−1+εt (15)
If the assumed specification is correct then the strategy makes sense.
• if ρ = 0 then both OLS and FGLS inferences are valid, but those based on OLS are likely more reliable as FGLS is based on a model that is overfit (i.e. we estimate ρ but its true value is zero and so this parameter could have been omitted.)
• if ρ ̸= 0 then FGLS inferences are valid, but those based on OLS are not as we have the wrong variance estimator. Note that from Question 1 part (b) and the hint, the large sample behaviour of the test statistic is determined by |T1/2θ|. Therefore, as T → ∞, |T 1/2ρˆT | → ∞ with probability one and so we reject with probability one as T → ∞. Therefore, inferences are based on the FGLS estimator in this case with probability one asT →∞. (IfatestrejectswithprobabilityoneasT →∞thenitissaidtobe consistent. This is, perhaps, an unfortunate choice of words and is not to be confused with the concept of consistency of an estimator.)
However, using the results from Question 1 part (c) and the hint, it is clear that if the errors are generated by a MA(1) model – instead of an AR(1) model – then the test also rejects with probability one as T → ∞. In this case, the FGLS inferences would be invalid.
Similarly, if the errors are generated by the process in part (d) then this type of dependence may not be detected by the test and so the failure to reject does not necessarily mean that the errors are serially uncorrelated. Using similar arguments to Tutorial 5 Question 3 part (e), we can use the CLT to deduce that
T
T−1/2utut−1 →d N(0,ω1),
t=2
where the exact form of ω1 need not concern us, and from Question 1(d) we have
T
T−1u2t →p E[u2t]=σε2(1+φ2);
t=1 and so using Lemma 2.11, we have
T1/2ρˆT→d N(0,ωρ),
where ωρ = ω1/{σε2(1 + φ2)}2. Therefore, if the errors are generated by the MA(2) process in Question 1 part (d) then the probability H0 is not rejected in the limit as T → ∞ is P(|ξ| ≤ 1.96) where ξ ∼ N(0, ωρ); this probability is non-zero.
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The following terminology is used to underscore the differences between the null and alter- native hypotheses of the test and the process for which the test does not reject. Here the nominal null hypothesis is that ut is generated via (15) with ρ = 0 and the nominal alterna- tive hypothesis is that ut is generated via (15)with ρ ̸= 0. The implicit alternative hypothesis consists of all process for ut for which the rejects with probability one in the limit (in our context here), and the implicit nominal null hypothesis consists of processes for which there is non-zero probability of failing to reject H0 as T → ∞.1
The only protection against the issues highlighted here is to perform a joint test for serial cor- relation at multiple lags, and then, if this test rejects, to engage in a more general investigation of the model specification.
3. As discussed in lectures, the key condition for the consistency of the OLS estimator is E[xtut] = 0. This case, we have xt = [1, yt−1]. Consider E[yt−1ut]. Since
yt−1 = β0,1 + β0,2yt−2 + εt−1 + φεt−2 ut = εt + φεt−1,
and
so E[yt−1ut] = φV ar[εt−1] ̸= 0. Therefore E[xtut] ̸= 0 and so OLS is not consistent.
4. Define:
βˆ T , 2
τˆ2 = Vˆsc,2,2/T
where Vˆsc,2,2 is the (2, 2)th element of Vˆsc,
Vˆsc/T = ΩˆHAC =
Γˆj= where {et}Tt=1 are the OLS residuals,
T(X′X)−1ΩˆHAC(X′X)−1, T−1
Γˆ0+ω(i,T)Γˆi+Γˆ′i, i=1
T
T−1 xtx′t−jetet−j,
t=j+1
ω(i,T) = 1−ai forai ≤1 0 for ai > 1
forai =i/(bT +1)andbT ∝T1/3.
1We note that the precise definition of the implicit null and alternative can differ in the statistics literature although the spirit is always the same. We have adapted the concepts to fit our setting here in which we deal with large sample properties against what are known as “fixed alternatives”.
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A suitable decision rule is to reject H0 : β0,2 = 0 at the (approximate) 100α% significance level in favour of HA : β0,2 ̸= 0 if |τˆ2| > z1−α/2 where z1−α/2 is the 1001 − α/2th percentile of the standard normal distribution.
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