Alastair Hall
1. We have:
ECON61001: Econometric Methods
Solutions to Problem Set for Tutorial 5
Semester 1, 2020-21
. .
X′X = T
T(T + 1)/2
T(T+1)/2 T(T + 1)(2T + 1)/6
1.(b) In this case, we have
T−2X′X = T−1 (T+1)/2T .
(T + 1)/2T (T + 1)(2T + 1)/6T
Since limT →∞T −1 = 0 and limT →∞(T + 1)/2T = 0.5 the (1, 1), (1, 2) and (2, 1) elements of T−2X′X converge to finite constants. However, the (2,2) element of T−2X′X diverges and so the matrix also diverges.
1.(c) In this case, we have
T−3X′X = T−2 (T+1)/2T2 .
(T + 1)/2T2 (T + 1)(2T + 1)/6T2
Since limT→∞T−2 = 0 and limT→∞(T + 1)/2T2 = 0 the (1,1), (1,2) and (2,1) elements of T−3X′X converge to zero. Further, limT→∞(T + 1)(2T + 1)/6T2 = 1/3 and so T−3X′X converges to a matrix of constants but this limit matrix is singular.
(To develop a limiting distribution theory for the OLS estimator in this case, it is necessary to scale the elements of βˆT −β0 by different functions of T, and this leads to different elements of X′X being scaled by different functions of T; see Lecture Notes Section 3.5.)
2.(a) To assess whether ut is weakly stationary, it is necessary to derive its first two moments. Since we are given that ut = εt, a white noise process, it follows immediately that E[ut] = 0 for all t, Var[ut] = σ2 for all t and Cov[ut,us] = 0 for all t ̸= s. Using Definition 3.7 in the Lecture Notes, it can re recognized that ut is a weakly stationary process.
2.(b) The first two moments of vt can be derived as follows. 1
1.(a) Therefore, it follows that:
T−1X′X = 1
(T+1)/2
(T + 1)(2T + 1)/6
(T + 1)/2
and so as T → ∞ T−1X′X diverges, and so does not converge to a matrix of finite constants.
• E[vt]: Substituting for vt and using E[εt] = 0 we have,
E[vt] = E[(−1)tεt] = (−1)tE[εt] = 0.
• V ar[vt]: Using the previous result it follows that V ar[vt] = E[vt2]. So substituting for vt, it follows that
Var[vt] = E[vt2] = E{(−1)tεt}2 = (−1)2tE[ε2t]=σ2,
where the last equality uses the fact that εt is white noise.
• Cov[vt, vs]: Using E[vt] = 0 for all t, we have Cov[vt, vs] = E[vtvs]. Substituting for vt,
it follows that
Cov[vt,vs] = E[vtvs] =E[(−1)t(−1)sεtεs] = (−1)t+sE[εtεs] = 0,
where the last identity uses the fact that εt is white noise and so Cov[εtεs] = E[εtεs] = 0. Therefore, vt is a weakly stationary process because E[vt] = 0 and V ar[vt] are independent
of t, and Cov[vt, vs] = 0 depends on |t − s| but not t or s individually (t ̸= s).
2.(c) Note that the event t = 10 is not random and so I(t = 10) can be treated as a constant.
Substituting for wt, we have:
E[wt] = E[I(t=10) + εt] = E[I(t=10)] + E[εt] = I(t=10) + E[εt].
Therefore, using E[εt] = 0, it follows that E[wt] = 0 for all t ̸= 10 and E[w10] = 1. So the mean of wt depends on t and wt is not weakly stationary.
2.(d) All we know about vt are the properties of its first two moments for which we can deduce that it is weakly stationary. But in the absence of additional information about the underlying probability distribution, there is no way to assess whether the process is strongly stationary. So the answer is that it may or may not be strongly stationary we lack the information to tell.
3.(a) We are given that εt is white noise and so E[εt] = 0 for all t. Thus, it follows that:
E[yt] = E[εt + φεt−1]
= E[εt] + φE[εt−1]
= 0, b/c E[εt] = 0.
3.(b) Since εt is white noise, we have E[ε2t] = Var[εt] = σ2 and E[εtεs] = Cov[εt,εs] = 0 for all s ̸= t. Using these properties and part(a), it follows that:
E (εt + φεt−1)2 , b/c E[yt] = 0
= E ε2t + φ2ε2t−1 + 2φεtεt−1
= E[ε2t ] + φ2E[ε2t−1] + 2φE[εtεt−1]
= σ2 + φ2σ2, b/c E[εtεt−1] = 0,
= (1 + φ2)σ2.
2
V ar[yt] =
3.(c) Using part(a), we have:
Cov[yt, yt−s] = E [(εt + φεt−1) (εt−s + φεt−s−1)]
= E εt εt−s + φεt−1 εt−s + φεt εt−s−1 + φ2 εt−1 εt−s−1
= E[εtεt−s] + φE[εt−1εt−s] + φE[εtεt−s−1] + φ2E[εt−1εt−s−1]. From the previous equation we obtain (using E[εtεn] = 0 for all t ̸= n): for s = 1,
Cov[yt,yt−s] = φE[ε2t−1] = φσ2, Cov[yt,yt−s] = 0.
and for s > 1,
3.(d) By inspection of the results in parts (a)-(c), it can be seen that yt is a weakly stationary
process because E[yt] and V ar[yt] do not depends on t and Cov[yt, yt−s] only depends on s.
3.(e) The long run variance is defined in Section 3.3.1 of the lecture notes. By convention, we use γi to denote the ith autocovariance of a scalar and so the the long run covariance of yt is given by: γ0 + ∞i=1(γi + γi′) where γi = Cov[yt, yt−i]. Since the transpose of a scalar is itself, the formula can be equivalently written as:
∞
Ωy =γ0+2γi. (1) i=1
Using the results from parts (b)-(c), we have:
Ωy = σ2(1+φ2 +2φ) = σ2(1+φ)2.
4.(a) By similar arguments to the answer to Question 3.(a), we have:
∞ E[yt] = E θiεt−i
i=0 ∞
= θiE[εt−i] i=0
= 0, b/c E[εt] = 0. 4.(b) By similar arguments to the answer to Question 3.(b):
∞
V ar[ θiεt−i]
V ar[yt] =
= E[( θiεt−i)2] since E[ θiεt−i] = 0
i=0 i=0 3
i=0 ∞∞
Notice further that
and so we have:
∞ s−1 ∞
θiεt−i = θiεt−i + θiεt−i. i=0 i=0 i=s
∞∞
θiεt−i = θs θiεt−s−i, i=s i=0
∞ s−1 ∞
θiεt−i = θiεt−i + θs θiεt−s−i. i=0 i=0 i=0
Substituting (3) into (2), it follows that
Cov[yt, yt−s]
= E =E
∞ 2
∞
= E[ θ2iε2t−i]
i=0 ∞
= σ2θ2i i=0
since E[εtεs] = 0 ∀t ̸= s
= σ2/(1−θ2)
where the last equality uses the hint (note that |θ| < 1 implies 0 ≤ θ2 < 1.)
4.(c) Assume s > 0. Using the MA(∞) representation and part(a), we have
∞∞ Cov[yt,yt−s] = Cov[θiεt−i,θiεt−s−i]
i=0 i=0
∞ ∞
= E θiεt−i θiεt−s−i . (2) i=0 i=0
At this point it is convenient to write ∞i=0 θiεt−i as the sum of two components,
s−1 ∞
θiεt−i +θsθiεt−s−i
i=0 i=0
∞ θiεt−s−i
i=0
+ E θs θiεt−s−i (4)
i=0
i=0
it can be seen that the result is a linear combination of terms of the form εtεn where t ̸= n.
Since the expectation of a linear combination of rv’s is the corresponding linear combination 4
s−1
θiεt−i
∞
θiεt−s−i
i=0
Consider each of these two terms in turn. Multiplying out
∞
θiεt−s−i
s−1
θiεt−i i=0
i=0
(3)
of the expectations and E[εtεn] = 0 for all t ̸= n (as εt is white noise), it follows that
s−1 ∞ E θiεt−i θiεt−s−i
i=0 i=0
= 0. (5)
Now consider the other term. Since θs is a constant, it follows that
∞ 2 ∞ 2
Eθs θiεt−s−i = θsE θiεt−s−i , (6) i=0 i=0
and so using the same argument as for part (b), we have
∞ 2
θsE θiεt−s−i = θsσ2/(1−θ2). (7)
i=0
The desired result then follows from (4)-(7).
4.(d) Inspection of parts (a)-(c) reveals that E[yt] and V ar[yt] do not depend on t, and Cov[yt, yt−s] depends on s but neither t or t − s. Therefore the process is weakly stationary. (The “stationarity” condition in ARMA models ensures that the process is weakly stationary.)
4.(e) As yt is a scalar, we evaluate (1). Using parts (b)-(c), it follows that:
σ2 σ2 σ2∞
Ωy = 1−θ2 +21−θ2 θ+θ2 +… = 1−θ2 1+2 θi . (8)
i=1
Now, using the hint, we have: ∞∞21+θ
1+2θi = 2θi −1 =1−θ −1 = 1−θ, i=1 i=0
and substituting this into (8), we obtain:
σ2 1+θ σ2
Ωy = 1−θ2 1−θ = (1−θ)2.
5. As can be seen from Figures 1 and 2, the pdf is still bell-shape but the bell has been elongated along the negative 45-degree line. Most of each contour lies in in the North-West and South- East quadrants of the plot. This reflects that if u1 and u2 are negatively correlated then positive values of u1 are more likely to occur with negative values of u2 and vice versa. The shape of the plots is different from those presented in Figures 4.1-4.6.
5
Figure 1: Probability density function with non-spherical errors due to negative correlation
Figure 2: Probability contours with non-spherical errors due to negative correlation
6
6. A suitable program is contained in the file Tut5timetrend.m where you need to make the
comparison on the screen; this program calls the file ols.m used in the solutions to Tutorial 2.
It is common practice to include a time trend in time series regression models. This exercise
demonstrates a general result: if we regress yt on an intercept, a time trend and a vector of
variables zt then the estimated coefficients on zt are the same as those obtained by regressing
̃
y ̃ on z ̃ where ( · ) denotes the linearly de-trended value of ( · ). tt
7