ECON 3350/7350 Deterministic and Stochastic Trends
Eric Eisenstat
The University of Queensland
Tutorial 5
Eric Eisenstat
(School of Economics) ECON3350/7350 Week 5 1 / 13
Trend Stationary
yt =a0+a2t+a1yt−1+εt;|a1|<1
yt is trend stationary because if we take the trend out the new process is
stationary.
Implies a2 is the average or long run growth rate.
Eric Eisenstat (School of Economics) ECON3350/7350 Week 5 2 / 13
Difference Stationary
An alternative form of trend comes from a2 = 0 and a1 = 1 yt =a0 +a2t+a1yt−1 +εt
yt = a0 + yt−1 + εt or ∆yt = a0 + εt
a0 is a trend and can be interpreted as the average or long run growth
rate..
Eric Eisenstat (School of Economics) ECON3350/7350 Week 5
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Stochastic Trends
Consider the stationary AR(1)
yt = a1yt−1 + εt with |a1| < 1
=aty +ε +a ε +···+at−1ε 10t1t−1 11
The effect of shocks or innovations dies out since at1 → 0 as t increases. Contrast this with the Random Walk Model
yt = yt−1 + εt or ∆yt = εt t
= y0 + εt−i i=0
All past shocks have a permanent effect.
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ACF when Stochastic Trends in the data
Eric Eisenstat (School of Economics) ECON3350/7350 Week 5 5 / 13
Random Walk with Drift
The original model
Thus
yt =a0 +yt−1 +εt
=a0 +μ+yt−2 +εt−1 +εt
.
=a0t+y0 +εt +εt−1 +···ε1 E(yt|y0) = y0 + a0t
yt = a0 + yt−1 + εt or ∆yt = a0 + εt
The expected change is E(∆yt) = a0; If a0 > 0, ytwill tend to drift upwards. Hence this is called the Random Walk with Drift.
Substituting back
And, the variance also has a trend in it.
Eric Eisenstat (School of Economics) ECON3350/7350 Week 5 6 / 13
Difference Stationary or Integrated Process
yt ∼ I(1) integrated of order 1
An AR(p) for yt ∼ I(1) implies an AR(p − 1) for ∆yt ∼ I(0).
Difference of yt, ∆yt ∼ I(0). Thus yt is difference stationary. Generally, if yt has d unit roots,
yt ∼ I(d), ∆yt ∼I(d−1),
.
∆dyt ∼ I(0)
Eric Eisenstat (School of Economics) ECON3350/7350 Week 5
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The Dickey-Fuller Test Equations
p−1
∆yt =a0 +a2t+γyt−1 +βi∆yt−i +εt
i=1
p−1
∆yt =a0 +γyt−1 +βi∆yt−i +εt
i=1
p−1
∆yt = γyt−1 + βi∆yt−i + εt
i=1
Use testing procedure in the next slide and Tables A and B pp
488-489 Enders
(1)
(2)
(3)
Eric Eisenstat (School of Economics) ECON3350/7350 Week 5
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Testing Procedure
Eric Eisenstat (School of Economics) ECON3350/7350 Week 5 9 / 13
KPSS Test
Tests of unit roots, e.g., H0 : yt ∼ I(1) against H1 : yt ∼ I(0) ; often suffer from low power. That is, the probability they lead us to reject the null is low. This may lead us to conclude there are unit roots where there are not.
One way to circumvent this problem is to test the null there is no unit root against the alternative that there is a unit root.
That is, test
H0 : yt ∼I(0)againstH1 : yt ∼I(1)
Eric Eisenstat (School of Economics) ECON3350/7350 Week 5 10 / 13
KPSS test (cont.)
The Kwiatkowski–Phillips–Schmidt–Shin (KPSS) tests test is one such test. The steps are:
1 Estimate yt = a0 + εt and save the residuals, et
2 Compute St = ts=1 es for t = 1,2,…,T.
3 Compute KPSS = T−2 T St2 t=1 2
σ
4 Compare this to the critical values. – See Table
Eric Eisenstat (School of Economics) ECON3350/7350 Week 5 11 / 13
KPSS trend stationary
If we assume the process is trend stationary under the null, the test becomes:
H0 : yt ∼ trend stationary against H1 : yt ∼ I(1)
The steps are:
1 Estimate yt = a0 + a2t + εt and save the residuals, et
2 Compute St = ts=1 es for t = 1,2,…,T.
3 Compute KPSS = T−2 T St2 t=1 2
σ
4 Compare this to the critical values. – See Table
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Identification of ARIMA(p, d, q)
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