ECON0058: Time Series Econometrics
Exam for 2018-2019
Answer all questions. Each question carries equal weight (25 points).
Question 1
Consider the model
Yt = φ1Yt−1 + φ2Yt−2 + εt + εt−1, t = 1, ..., T, (1)
with εt ∼ i.i.d.N(0, σ2).
1. (5 points) Find E[Yt], E [Yt|Ωt−1] and V ar [Yt|Ωt−1].
A: E[Yt]=0, E [Yt|Ωt−1] = φ1Yt−1 + φ2Yt−2 + εt−1 and V ar [Yt|Ωt−1] = σ2.
2. (6 points) Discuss when the model is stationary and when it is invertible.
A: It is stationary when roots of 1 − φ1x − φ2x2 are outside unit circle. It is
not invertible because the MA polynomial 1− x has a unit root.
3. (7 points) SupposeXt = −Yt+vt, with vt ∼ i.i.d.N(0, σ2v), and Correlation(εt, vt) =
ρ. First write down the conditions on the parameters so that Yt and Xt are
cointegrated and then write down the cointegrating vector.
A: Both must have a unit root, so φ1 and φ2 must be such that the roots of
1−φ1x−φ2x2 are outside the unit circle. Then Xt also has a unit root because
it’s the sum of a unit root and a stationary process so it has a unit root. The
two variables are then cointegrated because Xt + Yt = vt which is stationary.
The value of ρ does not matter.
4. (7 points) Suppose φ2 = 0. Find the response of Yt+2 to a one-unit shock in εt.
A: Yt+2 = φ1Yt+1 + εt+2 + εt+1 = φ1(φ1Yt + εt+1 + εt) + εt+2 + εt+1
= φ21(φ1Yt−1 + εt + εt−1) + φ1εt + εt+2 + (1 + φ1)εt+1 so response is φ
2
1 + φ1
CONTINUED
1
Question 2
Consider the model
Yt = c+ ρYt−1 + σtzt︸︷︷︸
εt
, t = 1, ..., T, (2)
σ2t =
{
w1 + φ1ε
2
t−1 if Wt−1 > 0
w2 + φ2ε
2
t−1 if Wt−1 ≤ 0 , (3)
where zt ∼ i.i.d.N(0, 1) and Wt is an observable variable.
1. (5 points) Derive the mean, variance and autocorrelation function of the stan-
dardized residuals, Yt−E[Yt|Ωt−1]
V ar[Yt|Ωt−1] .
A: This questions contained a typo, as the square root is missing from the
denominator of the standardized residual. Everyone was thus awarded 5 points
for this question, whether they answered or not.
2. (6 points) Provide sufficient conditions for stationarity of Yt.
A: |ρ| < 1, w1 = w2, φ1 = φ2 = φ and |φ| < 1
3. (7 points) Give a sufficient condition on the parameters c and ρ so that you can
estimate the model by OLS, and explain how you would do so.
A: if c = ρ = 0 you have t = Yt so you could estimate the model by OLS using
dummy variables: Y 2t = (w1+φ1Y
2
t−1)1
(
Wt−1 > 0) + (w2 + φ2Y 2t−1)1 (Wt−1 ≤ 0) + ut
4. (7 points) Derive the unconditional variance of Yt under the assumptions: 1)
stationarity; 2) Wt has a symmetric distribution (i.e., the probability that Wt >
0 is the same as the probability that Wt < 0); 3) Wt−1 is uncorrelated with 2t−1.
(Hint: use the fact that, for a generic random variable X and a constant c, the
probability that X > c equals E[1(X > c)], where 1 is the indicator function.)
A: It is σ2/(1 − ρ2), where σ2 = E[ε2t ] = E[(w1 + φ1ε2t−1)1(Wt−1 > 0) + (w2 +
φ2ε
2
t−1)1(Wt−1 ≤ 0)] = E[w1+φ1ε2t−1]E[1(Wt−1 > 0)]+E[w2+φ2ε2t−1]E[1(Wt−1 ≤
0)] = (w1 + φ1σ
2)1/2 + (w2 + φ2σ
2)]1/2. Collecting terms, this gives
σ2 =
.5(w1 + w2)
1− .5φ1 − .5φ2 (4)
CONTINUED
2
Question 3
Consider the model
Yt = φYt−1 + εt + θt−1, t = 1, ..., T, (5)
with εt ∼ i.i.d.N(0, σ2).
1. (5 points) Find the optimal 2-step ahead forecast for a quadratic loss.
A: E [Yt+2|Ωt] = φE [Yt+1|Ωt] = φ(φYt + θt) = φ2Yt + φθt
2. (6 points) Find the variance of the optimal 2-step ahead forecast error.
A: The error is t+2 + (φ+ θ)t+1 so its variance is σ
2(1 + (φ+ θ)2)
3. (5 points) Suppose θ = 0. What is the optimal h−step ahead forecast for
h→∞?
A: It equals limh→∞φhYt = 0 (the unconditional mean) when |φ| < 1 and Yt
when φ = 1
4. (8 points) Briefly explain why it is not accurate to perform inference on φ by
regressing Yt on Yt−1 and using a HAC estimator of the asymptotic variance.
Then briefly explain what is the proper way to conduct inference on φ.
A: Because the residuals are correlated with the regressor (i.e., Yt−1 is correlated
with t−1) so the estimator of φ is inconsistent. The correct way is to estimate
the model by ML and do standard inference in models estimated by ML.
CONTINUED
3
Question 4
Suppose Xt = ∆Zt, where Zt = .5Zt−1 + .5Zt−2 + t and Yt = φYt−1 + vt, where εt
and vt are i.i.d.N(0,σ
2) and t, vt are uncorrelated with each other.
1. (6 points) Suppose you estimate the regression Yt = βXt+ut. Explain how you
would perform inference about β (depending on the value of φ).
A: If |φ| < 1 this is a standard regression involving two stationary variables, so
perform standard inference using a HAC estimator. This should indicate that
β = 0. If φ = 1 the regression is unbalanced and the unit root in Yt ends up in
the residuals, so inference becomes non-standard
2. (6 points) Show that Xt is invertible.
A: The model for Xt is (1 + .5L)Xt = t and the root of 1 + .5L is greater than
one in absolute value, so we have invertibility
3. (6 points) Write down the VAR model for (Xt,Yt) and show under which con-
ditions it is stationary.
A: Since Xt = −.5Xt−1 + t it’s a VAR(1) with diagonal coefficient matrix with
elements -.5 and φ, so it is stationary if |φ| < 1
4. (7 points) Use the VAR model your wrote down in question 4.3 to derive the
impulse response functions of Xt+h with respect to shocks in Yt and for Yt+h
with respect to shocks in Xt, for h = 0, 1, 2, ....
A: This only makes sense if the VAR is stationary. The IRFs are all zero,
because we have assumed that t, vt are uncorrelated with each other.
END OF PAPER
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