Chapter 4 - Trends

Question 1

Consider the AR(1) model
y_t −μ = φ1(y_(t−1) − μ )+ ε_t, t = 1,2,…,T

When |φ1| < 1 what can you say?

Answer 1

1. y_t has an unconditional mean equal to μ for all t.
2. The unconditional variance is constant and equal to σ^2/(1 − φ1^1)
3. And the autocorrelations of y_t are constant as well.

=> These 3 properties are implied by STATIONARITY.

Question 2

What does stationarity mean?

Answer 2

Shocks have transitory effects.

Question 3

B y recursive substitution, rewrite
y_t −μ = φ1(y_(t−1) − μ )+ ε_t, t = 1,2,…,T

Answer 3

bonne chance :)

Question 4

What does it mean for a time series y to be MEAN-REVERTING?

Answer 4

It means that y_(t+1) is expected to be closer to μ than y_t:

E[yt+1 − μ|Yt] = φ1(yt − μ)

and in fact, for any h>1,
E[yt+h − μ|Y_t] = φ1^h(yt − μ)
which converges to 0 as h increases.

The speed or strength of the mean-reversion depends on the magnitude of φ1.

Question 5

A “simple” AR(1) model can not capture a trend. Two possibilities to incorporate trends in AR models:

Answer 5

1. Insert a deterministic trend.
2. Allow for a unit root in the AR-polynomial φ_p(L) to incorporate a stochastic trend.

In the AR(1) case, the 2nd possibility corresponds with φ1 = 1.

Question 6

What is a deterministic trend in AR models?

Answer 6

with |φ1|<1:

y_t − μ_t = φ1(y_(t−1) − μ_(t-1)) + ε_t,
with μ_t = μ + δt .

=> y_t − μ - δt = φ1(y_(t−1) − μ - δt) + ε_t,

Question 7

To analyse the properties of y_t (deterministic trend in an AR(1) ), it is useful to define A, and write the model B as C.

The model C has already been seen before! We know that D. And the unconditional mean of y_t is equal to E. The unconditional variance and autocorrelation of y_t are the same as those of F.

Answer 7

a. zt ≡ yt − μ − δt
b. y_t − μ - δt = φ1(y_(t−1) − μ - δt) + ε_t
c. z_t = φ1z_(t−1) + εt
d. E[zt] = 0, V[zt] = σ^2/(1−φ1^2) and ρ_k = φ1^k.
e. E[y_t] = μ + δt.
f. z_t

Question 8

What does a it mean when y_t goes back to the slope ?

Answer 8

Trend-stationary. => A model with such a trend is called a deterministic trend model.

Question 9

Derive the 3 cases of a deterministic trend in AR models.

Answer 9

see 17.

Question 10

Write y_t with a linear trend in terms of past shocks and make a conclusion about transitory.

Answer 10

see 18-19

Question 11

From the AR(1) model with a linear trend term, derive the case where there is a stochastic trend. [slide 23].

Answer 11

Start with y_t − μ - δt = φ1(y_(t−1) − μ - δt) + ε_t.

stochastic trend implies φ1=1:
=> yt = δ + yt−1 + εt
=> yt = y0 + δt + SUM( εt, 1:t)

We observe the term SUM( εt, 1:t) ==> Stochastic trend. And the deterministic trend δt appears again.

Question 12

Give some properties of the random walk (with drift)
yt = y0 + δt + SUM( εt, 1:t)

Answer 12

1. The unconditional mean E[y_t] = y0 + δt.
2. The unconditional variance of y_t increases linearly with time: see slide 12.
3. The auto-covariances of y_t increase linearly with time:
   γ1,t = E[(yt − E[yt])(y_(t−1) − E[y_(t−1)])] = (t − 1)σ^2

Combining 12 and 13 result in ρ1,t = (t − 1)/t.
In general, we have ρ_(k,t) = (t−k)/t for any k > 0.

Question 13

A series y_t that can be described by a random walk (with drift):

Answer 13

yt = δ + yt−1 + εt = y0 + δt + SUM( εt, 1:t)

• non-stationarity: shocks have permanent effects and therefore the unconditional mean and variance and autocorrelations are no longer constant over time.

Question 14

How to distinguish between DT and ST models by looking at a slope?

Answer 14

Important differences between deterministic trend [DT] and stochastic trend [ST] models are:
* The impact of shocks is transitory for the DT model and permanent for the ST model.
* The out-of-sample forecast error variances for the ST model are much larger than those for the DT model (see Section 4.4). Hence, ST data may be more difficult to forecast.
* The statistical properties of ST data differ from those of stationary time series. This makes the analysis of models with so-called ‘unit roots’ more complicated. New asymptotic theory had to be derived.

Question 15

how is the Dickey-Fuller test constructed?

Answer 15

Idea: try to distinguish between DT and ST models by testing φ1 =1 versus φ1 <1.

For ease of notation, consider the case where μ = δ = 0. The model in (15) can then be rewritten as
∆1yt = ρy_(t−1) + εt, with ρ=(φ1−1). Thus, φ1 =1⇔ρ=0.

Idea (Dickey-Fuller): use t-statistic of ^ρ as test statistic.
H0 : presence of stochastic trend ρ = 0.

see slide 32.

Question 16

Which deterministic terms to include?

Answer 16

• almost always include an intercept.
• often include a deterministic trend

Question 17

write the AR(1) model
y_t − μ - δt = φ1(y_(t−1) − μ - δt) + ε_t
with φ1(L).

Answer 17

φ1(L)(yt − μ − δt) = εt, with φ1(L) = 1 − φ1L.

Question 18

Which link can you make between φ1(L)(yt − μ − δt) = εt and the ST model?

Answer 18

Note that this reduces to the ST model when z = 1 is a solution to the characteristic equation φ1(z) = 1 − φ1z = 0. Hence, we also say that yt contains a unit root.

Question 19

What holds in general for AR(p) models φ_p(L)(yt − μ − δt) = εt with φ_p(L) = 1−φ1L−···−φpL^p ?

Answer 19

When the characteristic equation φ_p(z) = 0 as a “unit root” solution z = 1, yt has a stochastic trend.