Nonstationary Time Series Midterm Exam Kaiji...

Nonstationary Time Series3rd Quarter, 2017

Midterm ExamOctober 24, 2017

Kaiji MotegiKobe University

Problem-1: Consider random walk with drift and noise (RW-DN):

yt = c+ yt−1 + ϵt +∆ηt, (1)

where {ϵt} is white noise with E[ϵ2t ] = σ2ϵ > 0; ∆ηt = ηt − ηt−1; {ηt} is white noise with

E[η2t ] = σ2η > 0; E[ϵtηs] = 0 for any t and s; y0 and η0 are non-stochastic initial values.

(a) Show that yt = ct+ y0 +∑t

j=1 ϵj + ηt.

(b) Show that E[yt] = ct+ y0.

(c) Show that V ar[yt] ≡ E[(yt − E[yt])2] = σ2

ϵ t+ σ2η.

(d) Show that Cov[yt, yt−h] ≡ E[(yt − E[yt])(yt−h − E[yt−h])] = σ2ϵ (t− h) for h ≥ 1.

(e) Is {yt} covariance stationary? Explain why or why not.

(f) Suppose that we simulate a RW-DN process with c = 1, y0 = 0, η0 = 0, ϵti.i.d.∼ N(0, σ2

ϵ ),

and ηti.i.d.∼ N(0, σ2

η). We consider four cases for the variance terms:

Case A. (σ2ϵ , σ

2η) = (1, 1).

Case B. (σ2ϵ , σ

2η) = (1, 30).

Case C. (σ2ϵ , σ

2η) = (15, 1).

Case D. (σ2ϵ , σ

2η) = (15, 30).

Figure 1 plots simulated sample paths with sample size n = 100. Panels 1-4 of Figure

1 match Cases A-D but possibly with a different order. Answer with a brief reason

which panel matches which case.

1




Figure 1: Random Walk with Drift and Noise

20 40 60 80 100-50

0

50

100

150

200

1. Which case?

20 40 60 80 100-50

0

50

100

150

200

2. Which case?

20 40 60 80 100-50

0

50

100

150

200

3. Which case?

20 40 60 80 100-50

0

50

100

150

200

4. Which case?

Case A: (σ2ϵ , σ

2η) = (1, 1). Case B: (σ2

ϵ , σ2η) = (1, 30). Case C: (σ2

ϵ , σ2η) = (15, 1). Case D:

(σ2ϵ , σ

2η) = (15, 30).

2




Problem-2: In Assignment #1, we ran Monte Carlo simulations on detrending. In this

problem we extend that in order to get further implications. Suppose that a true data

generating process is a trend stationary process:

yt = α0 + δ0 × t+ ϵt, ϵti.i.d.∼ N(0, σ2

0). (2)

Fix sample size n ∈ {50, 100, 200} and true values α0 = 2, δ0 = 2, and σ20 = 80. We fit either

Model 1 or Model 2:

Model 1: yt = α+ δ × t+ ut.

Model 2: yt = α+ δ × t+ ϕyt−1 + ut.

Model 1, which was considered in Assignment #1, is exactly specified relative to the true

DGP (2). Model 2, which is newly considered in this problem, is correctly specified but

has a redundant regressor yt−1. We investigate how this redundancy affects the speed of

convergence of OLS estimators.

For each sample size n, simulation procedures with Model 1 are as follows. Model 2 is

covered in the same way.

Step 1. Generate {yt}nt=1 according to DGP (2).

Step 2. Run OLS for Model 1 to get α̂n and δ̂n.

Step 3. Repeat Steps 1-2 J = 10000 times to get a set of OLS estimates {α̂(1)n , . . . , α̂

(J)n }

and {δ̂(1)n , . . . , δ̂(J)n }.

Step 4. Compute the bias, variance, and MSE for each parameter.

Table 1 summarizes simulation results. Comparing the results for Model 1 and those for

Model 2, comment on the impact of adding the redundant regressor yt−1 on MSE.

3




Table 1: Bias, Variance, and MSE of Each Estimator

Model 1: yt = α+ δ × t+ ut

Bn Vn MSEn

n = 50 n = 100 n = 200 n = 50 n = 100 n = 200 n = 50 n = 100 n = 200

α̂n 0.049 0.019 0.005 6.556 3.249 1.640 6.559 3.250 1.640

δ̂n 6.3× 10−4 9.5× 10−4 2.1× 10−5 0.008 9.5× 10−4 1.2× 10−4 0.008 9.5× 10−4 1.2× 10−4

ϕ̂n - - - - - - - - -

Model 2: yt = α+ δ × t+ ϕyt−1 + ut

Bn Vn MSEn

n = 50 n = 100 n = 200 n = 50 n = 100 n = 200 n = 50 n = 100 n = 200

α̂n 0.018 0.015 0.006 7.703 3.489 1.675 7.703 3.490 1.675

δ̂n 0.080 0.041 0.021 0.085 0.041 0.020 0.091 0.043 0.020

ϕ̂n -0.040 -0.020 -0.010 0.019 0.010 0.005 0.021 0.010 0.005

Problem-3: Briefly explain what spurious regression is, why it occurs, and how to avoid it.

Use all of the following keywords in your explanation. (Instruction: Five or six sentences in

total should be enough. Try to be concise.)

Keywords: nonstationary, OLS estimator, t-statistic, R2, residual.

4

Nonstationary Time Series Midterm Exam Kaiji...

Documents

Transcript of Nonstationary Time Series Midterm Exam Kaiji...