Non-stationary Time Series - Economics · and then we can apply the determininstic trend model....

Introduction Comparison of Unit Root and Trend-Stationary Series Asymptotic Theory Trend-Stationary Series

Non-stationary Time Series

Helle Bunzel

ISU

April 8, 2009

Helle Bunzel ISU



Introduction to Non-Stationary Series I

What we have seen so far are series which can be written as

yt = µ+∞

∑j=0

ψj εtj = µ+ ψ (L) εt

where:

ψ0 = 1fεtg is a white noise process with mean 0 and variance σ2.The coe¢ cients are absolutely summable:

∞

∑j=0

ψj < ∞

The roots of ψ (L) = 0 are outside of the unit circle.

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Introduction to Non-Stationary Series II

A couple of important feature of such series are:

E (yt ) = µ

and

lims!∞

yt+s jt = µ

This is not always desirable. Consider, for example US GDP:

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Introduction to Non-Stationary Series III

Clearly there is an upwards trend, which must be accounted for in anymeaningful forecast.

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Modelling Non-Stationarity IThere are are two main approaches to modelling trends like these.The rst approach involves including a deterministic time trend:

yt = µ+ δt + ψ (L) εt

or in the case of the US GDP maybe

yt = µ+ δt + γt2 + ψ (L) εt

The other approach is to model the data as a unit root process:

(1 L) yt = δ+ ψ (L) εt

or

∆yt = δ+ ψ (L) εt

where it is assumed that ψ (1) 6= 0.This is also called a stochastic trend.

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Modelling Non-Stationarity IIWhy did we need ψ (1) 6= 0?

Suppose the original fytg was a stationary series, say

yt = µ+ χ (L) εt

Then, when we di¤erence, we get:

∆yt = (1 L) χ (L) εt ψ (L) εt

where

ψ (L) = (1 L) χ (L)

In this case, we get:

ψ (1) = (1 1) χ (1) = 0

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Modelling Non-Stationarity III

Thus, ruling out ψ (1) = 0, essentially rules out that the original serieswas stationary. I.e. we had to di¤erence in order to obtain a stationaryseries.

Examples of simple unit root processes is the random walk withoutdrift:

yt = yt1 + εt

or the random walk with drift:

yt = δ+ yt1 + εt

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Modelling Non-Stationarity IV

An alternative representation:

yt = α+ δt + ut

where ut is a mean 0 ARMA(p,q) process, such that1 φ1L φ2L

2 ... φpLput =

1+ θ1L+ θ2L2 + ...+ θqLq

εt

where the MA operator1+ θ1L+ θ2L2 + ...+ θqLq

is invertible.

Now, write the AR part as1 φ1L φ2L

2 ... φpLp= (1 λ1L) (1 λ2L) ... (1 λpL)

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Modelling Non-Stationarity V

If jλi j < 1 8i , then we can write

ut =

1+ θ1L+ θ2L2 + ...+ θqLq

(1 λ1L) (1 λ2L) ... (1 λpL)

εt ψ (L) εt

and

yt = α+ δt + ut

is a trend-stationary process.

Now suppose instead that λ1 = 1 and jλi j < 1 for i 2 f2, 3, ..., pg .(The series has a unit root)

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Modelling Non-Stationarity VIThen we can write

(1 L) ut =1+ θ1L+ θ2L2 + ...+ θqLq

(1 λ2L) ... (1 λpL)

εt ψ (L) εt

so

∆ut = ψ (L) εt

where ψ (L) satises the usual properties of stationary sequences.Now consider the original series:

yt = α+ δt + ut

Di¤erence the series:∆yt = yt yt1 = (α+ δt + ut ) (α+ δ (t 1) + ut1)

= δ+ ut ut1 = δ+ ∆ut = δ+ ψ (L) εtIn this situation, yt is a unit root process!

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Some denitions I

When yt contains a unit root it is also said to be integrated of order1, or y I (1) .If yt contains two unit roots, such that

(1 L) (1 L) ut =

1+ θ1L+ θ2L2 + ...+ θqLq

(1 λ3L) ... (1 λpL)

εt

ψ (L) εt ,(1 L)2 ut = ψ (L) εt

then the data must be di¤erenced twice to become stationary:

(1 L) yt = ∆yt = δ+ ∆ut = δ+ (1 L) ut(1 L)2 yt = (1 L)∆yt = (1 L) [δ+ (1 L) ut ]

= (1 L)2 ut = ψ (L) εt

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Some denitions II

In this case we say that yt is integrated of order 2 or y I (2) .In general these process are called autoregressive integrated movingaverage processes or ARIMA (p, d , q) . Here p is the number of ARroots (other than the unit roots), d is the number of unit roots and qis the number of MA roots.

If you di¤erence an ARIMA (p, d , q) process d times, you produce astationary ARMA (p, q) process.

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Modelling Economic Series I

Looking at GDP it seems that the best model might be exponentialgrowth, such that

yt = eδt

and

∂yt∂t= δeδt = δyt

This is called "proportional growth" and is common in economics.

In this case we simply take logs:

log (yt ) = δt

and then we can apply the determininstic trend model.Helle Bunzel ISU



Modelling Economic Series II

Similarly it is common to take logs before applying the unit rootmodel. This amounts to constant percentage of growth. For example:

Ination follows a stationary process.Population growth follows a stationary process.

To see this, consider:

∆ log (yt ) = logytyt1

= log

1+

yt yt1yt1

' yt yt1

yt1

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Comparison of Forecasts I

First consider the trend-stationary series:

yt = µ+ δt + ψ (L) εt

Here the forcast is

yt+s jt = µ+ δ (t + s) + ψs εt + ψs+1εt1 + ....

Note that

Eyt+s jt µ+ δ (t + s)

2= E

ψs εt + ψs+1εt1 + ....

2 !s!∞

0

because the ψj are absolutely summable.

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Comparison of Forecasts IINow consider forecasting a unit root series

(1 L) yt = δ+ ψ (L) εt

Since ∆yt is a stationary process, we can apply the usual formulas to∆yt instead of yt .That implies that

∆yt+s jt = δ+ ψs εt + ψs+1εt1 + ....

Now suppose we want to forecast the level of yt+s . To do this, notethat

yt+s = yt+s yt+s1 + yt+s1 yt+s2 + ...+ yt+1 yt + yt= ∆yt+s + ∆yt+s1 + ∆yt+1 + yt

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Comparison of Forecasts III

Then the forecast is

yt+s jt = ∆yt+s jt + ∆yt+s1jt + ∆yt+1jt + yt=

δ+ ψs εt + ψs+1εt1 + ....

+

δ+ ψs1εt + ψs εt1 + ....

+

δ+ ψs2εt + ψs1εt1 + ....

+...

+ fδ+ ψ1εt + ψ2εt1 + ....g+yt

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Comparison of Forecasts IV

This implies that

yt+s jt = δs + yt +ψs + ψs1 + ...+ ψ1

εt

+ψs+1 + ψs1 + ...+ ψ2

εt1

+ψs+2 + ψs1 + ...+ ψ3

εt2 + ......

Note that the deterministic part is similar for the two types ofprocesses.

A few examples of unit root processes and their forcasts:

The random walk with drift:

yt = δ+ yt1 + εt

Here ψj = 0 for all j .

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Comparison of Forecasts VThen

yt+s jt = δs + yt

An ARIMA (0, 1, 1) such that

yt = δ+ yt1 + εt + θεt1

and

ψ1 = θ

This implies that

yt+s jt = δs + yt + θεt

For both types of processes the forcast converges to a line:

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Comparison of Forecasts VIFor the deterministic trend model:


!s!∞

µ+ δ (t + s) = (µ+ δt) + δs

For the unit root model:yt+s jt = δs + yt +

ψs + ψs1 + ...+ ψ1

εt

+ψs+1 + ψs1 + ...+ ψ2

εt1

+ψs+2 + ψs1 + ...+ ψ3

εt2 + ......

!s!∞

δs + yt

The slopes of these lines are identical, but the intercepts are not.

For the deterministic model the intercept is xed.For the unit root model the intercept varies with each new observation.

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Comparison of Forecast Errors I

First consider the trend-stationary process:


and

yt+s = µ+ δ (t + s) + εt+s + ψ1εt+s1 + ....

so

yt+s yt+s jt = fεt+s + ψ1εt+s1 + ....g

ψs εt + ψs+1εt1 + ...

= εt+s + ψ1εt+s1 + ....+ ψs1εt+1

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Comparison of Forecast Errors II

The MSE is:

Ehyt+s yt+s jt

2i=1+ ψ21 + ψ22 + ...+ ψ2s1

σ2

and

lims!∞

Ehyt+s yt+s jt

2i= σ2

1+

∞

∑i=1

ψ2i

!

This is the same as the stationary MA processes.

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Comparison of Forecast Errors III

For the unit root process we get:

yt+s jt = δs + yt

+

ψs εt + ψs+1εt1 + ....

+

ψs1εt + ψs εt1 + ....

+

ψs2εt + ψs1εt1 + ....

+...

+ fψ1εt + ψ2εt1 + ....g

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Comparison of Forecast Errors IV

and

yt+s = ∆yt+s + ∆yt+s1 + ∆yt+1 + yt= δs + yt

+ fεt+s + ψ1εt+s1 + ....g+ fεt+s1 + ψ1εt+s2 + ....g+...

+ fεt+1 + ψ1εt + ....g

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Comparison of Forecast Errors VWe get

yt+s yt+s jt= fεt+s + ψ1εt+s1 + ....g

ψs εt + ψs+1εt1 + ....

+ fεt+s1 + ψ1εt+s2 + ....g

ψs1εt + ψs εt1 + ....

+...

+ fεt+1 + ψ1εt + ....g fψ1εt + ψ2εt1 + ....g=

εt+s + ψ1εt+s1 + ...+ ψs+1εt+1

+ fεt+s1 + ψ1εt+s2 + ....+ ψs εt+1g+...

+εt+1

= εt+s + (1+ ψ1) εt+s1 + (1+ ψ1 + ψ2) εt+s2 + ....

+1+ ψ1 + ψ2 + ...+ ψs1

εt+1

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Comparison of Forecast Errors VI

This implies the MSE:

Ehyt+s yt+s jt

2i= σ2

h1+ (1+ ψ1)

2 + ...+1+ ψ1 + ψ2 + ...+ ψs1

2iThis one does not converge to a xed number. It can be shown that

Ehyt+s yt+s jt

2i 1+ ψ1 + ψ2 + ...+ ψs12

σ2s !s!∞

C

There is no upper bound on MSE!

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Comparison of Dynamic Multipliers I

For the trend-stationary process:

yt+s = µ+ δ (t + s) + εt+s + ψ1εt+s1 + ....

and therefore

∂yt+s∂εt

= ψs

and

lims!∞

∂yt+s∂εt

= 0

Helle Bunzel ISU



Comparison of Dynamic Multipliers IIFor the unit root process:

yt+s = δs + yt+ fεt+s + ψ1εt+s1 + ....g+ fεt+s1 + ψ1εt+s2 + ....g+...

so

∂yt+s∂εt

= 1+ ψs + ψs1 + ...+ ψ1 = 1+s

∑i=1

ψi

and

lims!∞

∂yt+s∂εt

= 1+∞

∑i=1

ψi = ψ (1)

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Comparison of Dynamic Multipliers III

A numerical example of this multiplier:

An ARIMA(4,1,0) was estimated for yt = 100 log (GDPt ) with theestimators:

∆yt = 0.555+ 0.312∆yt1+ 0.122∆yt2 0.116∆yt3 0.081∆yt4+ εt

Rewriting this as and MA process:1 0.312L 0.122L2 + 0.116L3 + 0.081L4

∆yt = 0.555+ εt

Dene φ (L) = 1 0.312L 0.122L2 + 0.116L3 + 0.081L4Check the stationarity of the estimated process: Turns out φ (L) hascomplex roots with norms 1.5 and 2.3.

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Comparison of Dynamic Multipliers IV

The long-run e¤ect of a shock in εt is ψ (1) . Note that

ψ (L) =1

φ (L)

and therefore

lims!∞

∂yt+s∂εt

= ψ (1) =1

φ (1)=

1(1 0.312 0.122+ 0.116+ 0.081) = 1.31

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Transformations to Achieve Stationarity I

Note that if we have a trend-stationary process, then

yt δt = µ+ ψ (L) εt

is a stationary process.

Now, if the data is generated by a random walk with a drift, then

yt = δ+ yt1 + εt

yt δt = y0 + εt + εt1 + ...+ ε1

The mean is stationary, but the variance is σ2t.

Similarly if a trend-stationary process is di¤erenced, the resultingprocess is stationary, but a unit root is introduced in the MA processmaking it non-invertible.

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The Meaning of Unit Root tests I

Because of the inherent di¤erences between trend-stationary and unitroot processes, it is of interest whether an economic process is one orthe other.

Unfortunately, it is not possible to distinguish between the two withnite samples.

The two processes:

yt = yt1 + εt

and

yt = φyt1 + εt , jφj < 1

cannot be distinguished in nite samples, especially for φ very close to1.

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The Meaning of Unit Root tests IIA primary cause is that the processes only di¤er in their implicationsover innite horizons.

Consider the forecasts:

Unit root:

ys+t jt = yt

and

Eyt+s ys+t jt

2= sσ2

Trend-stationary:

ys+t jt = φsyt

Eyt+s ys+t jt

2=1+ φ2 + φ4 + ....+ φ2(s1)

σ2

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The Meaning of Unit Root tests III

Clearly for φ su¢ ciently close to 1, the predictions will be virtuallyidentical.

What are the questions we can answer?

If we have enough data, we can nd out if a shock today has asignicant e¤ect 3 years from now.

While we wont know for sure which process generated the data, thismay be all we really need to know.

If we are willing to restrict ourselves to a specic AR process, e.g.AR(1):

yt = φyt1 + εt

then we can test H0 : φ = 1.

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The Meaning of Unit Root tests IV

It is probably the case that if φ0 = 0.998 we will not reject H0. Thatmay be OK, as the unit root process is a good approximation to thedata and will provide reasonable forecasts etc in nite time.

Note that this provides yet another argument for parsimonious models.

Witin a model, deciding whether to impose the unit root restriction isimportant for the usual reasons. E¢ ciency vs consistency.

Also the asymptotic theory used to test hypotheses di¤ers dependingon whether or not there is a unit root.

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Asymptotic Theory for Dependent processes.

Almost all the asymptotic theory (LLN & CLT) you have seen hasfocussed on iid sequences.

At most you have seen heteroscedastic series.

Clearly this will not provide asymptotic theory for the kind ofprocesses that we have been working with.

This will be a brief introduction, not a thorough investigation.

The material can be found in Hamilton Chapter 7.2.

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LLN for stationary processes I

Assume that we have a covariance stationary process fYtg∞t=1 with

absolutely summable autocovariances.

Let

µ E (Yt )

γj = E [(Yt µ) (Ytj µ)]

Consider the sample mean:

YT =1T

T

∑t=1Yt

and note that

E (YT ) =1T

T

∑t=1

µ = µ

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LLN for stationary processes II

If we can show that V (YT )! 0 as T ! ∞, then YT is consistentfor the mean. Note that weve been using this result.

Find V (YT ) :

V (YT )

= E

24 1T

T

∑t=1Yt µ

!235 = 1T 2E

24 T

∑t=1(Yt µ)

!235=

1T 2E

" T

∑t=1(Yt µ)

! T

∑s=1(Ys µ)

!#

=1T 2E

"T

∑t=1

T

∑s=1(Yt µ) (Ys µ)

#

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LLN for stationary processes III

=1T 2

T

∑t=1

T

∑s=1

E [(Yt µ) (Ys µ)] =1T 2

T

∑t=1

T

∑s=1

γts

=1T 2fTγ0 + 2 (T 1) γ1 + 2 (T 2) γ2 + ...

+2 (T j) γj + ...+ 2 (T (T 1)) γT1

=1T 2

T1∑i=0

(T i) γi 1T 2

T1∑i=0

Tγi 1T

T1∑i=0

jγi j ! 0

We have now demonstrated ergodicity for the mean of a covariancestationary series with absolutely summable autocovariances.

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Some Denitions and Theorems I

Let fYtg∞t=1 be a sequence of random scalars with E (Yt ) = 0 for all

t.

Let Ωt be all the available information at time t.

Denition

If E (Yt jΩt1) = 0 for t = 2, 3, .... then fYtg is a martingale di¤erencesequence with respect to fΩtg .

Note: This is a stronger condition than fYtg being uncorrelated, butweaker than independence.

Example 1: If εt iid N0, σ2

and Yt = εt εt1, then

E (YtYt1) = Eεt ε

2t1εt2

= 0

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Some Denitions and Theorems IINow consider

E (Yt jYt1,Yt2, ...,Y1)

If ε0 is a known constant, then this amounts to

E (Yt jεt1, εt2, ..., ε1) = E (εt εt1jεt1, εt2, ..., ε1)= E (εt ) εt1 = 0

If ε0 N0, σ2

and unknown, then Yt1,Yt2, ...,Y1 does not

provide information on εt1 and

E (Yt jYt1,Yt2, ...,Y1) = E (εt εt1jYt1,Yt2, ...,Y1)= E (εt εt1) = E (εt )E (εt1) = 0

So this sequence is uncorrelated, a martingale di¤erence sequence,but not serially independent.

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Some Denitions and Theorems III

Example 2: If εt iid N0, σ2

and Yt = ε2t εt1, then the series is

uncorrelated because:

E (YtYt1) = Eε2t ε

3t1εt2

= 0

Now consider

E (Yt jεt1, εt2, ..., ε1) = Eε2t εt1jεt1, εt2, ..., ε1

= E

ε2t

εt1 = σ2εt1

This, in general, is not 0, and hence Yt is not a martingale di¤erenceprocess.

Now for a more general denition:

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Some Denitions and Theorems IV

Denition

If we can nd sequences of non-negative deterministic constants fctg∞t=1

and fξmg∞m=1 such that limm!∞ξm = 0 and

E jE (Yt jΩtm)j ctξm , 8t,m

then fYtg is said to follow an L1mixingale with respect to fΩtg .

Note that E (Yt jΩtm) is the m period ahead forecast, and that itmust converge to 0 as m! ∞.Hamilton provides a couple of examples.

With a few additional conditions, LLN for rst and second momentscan be proven for L1mixingales.

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Some Denitions and Theorems V

Furthermore, we have the following two CLTs:

Theorem

Let fYtg∞t=1 be a scalar martingale di¤erence sequence with

YT = 1T ∑T

t=1 Yt . Suppose that (a) EY 2t= σ2t > 0 with

1T ∑T

t=1 σ2t ! σ2 > 0, (b) E jYt jr < ∞ for some r > 2 and all t, and (c)1T ∑T

t=1 Y2t

P! σ2. ThenpTYT ) N

0, σ2

.

Note that Hamilton also provides denitions and theorems for vectors.

Clearly even when we have a scalar martingale di¤erence sequencethere is some work involved in showin that the conditions of the CLThold. The following proposition is useful for that purpose:

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Some Denitions and Theorems VI

Proposition Let Xt be a strictly stationary process withEX 4t= µ4 < ∞. Let Yt = ∑∞

j=0 hjXtj , where∑∞j=0 jhj j < ∞. Then Yt is a strictly stationary process with

E jYsYtYuYv j < ∞ for all s, t, u, v .

Next an example using many of these concepts.

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CLT Example I

Let fYtg∞t=1 be an MA (∞) process, such that Yt = ∑∞

j=0 ψj εtj ,

where ∑∞j=0

ψj < ∞.

Let εt be iid with E (εt ) = 0, Eε2t= σ2 and E

ε4t< ∞.

Now consider

Xt = εtYtk , k > 0

We will assume (c), namely that

1T

T

∑t=1X 2t

P! EX 2t

If you are interested Hamilton shows this, but it is non-trivial.

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CLT Example II

First verify that Xt is a martingale di¤erence sequence.

E (Xt jΩt1) = E (Xt jXt1,Xt2, ...)

Note that εt and Ytk are independent and that Xt1,Xt2, ...contains no information about εt , so

E (Xt jXt1,Xt2, ...) = E (εtYtk jXt1,Xt2, ...)= E (εt )E (Ytk jXt1,Xt2, ...)= 0

Hence Xt is a matingale di¤erence process.

We now need to verify conditions (a) and (b) of the CLT.

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CLT Example III

To verify condition (a) note that

EX 2t= E

ε2tY

2tk= E

ε2tEY 2tk

= σ2E

Y 2tk

We can nd

EY 2tk

= E

24 ∞

∑j=0

ψj εtj

!235 = σ2∞

∑j=0

ψ2j

Note that EY 2t= E

Y 2sfor all t, s.

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CLT Example IV

Since we have assume that ∑∞j=0

ψj < ∞, we also have

σ2 ∑∞j=0 ψ2j < ∞ and therefore

EX 2t= σ2E

Y 2tk

= σ4

∞

∑j=0

ψ2j < ∞

We now need to show that

limT!∞

1T

T

∑t=1EX 2t= lim

T!∞

1T

T

∑t=1

σ2EY 2tk

= σ2

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CLT Example VThis follows easily from the fact that E

Y 2t= E

Y 2sfor all t, s :

limT!∞

1T

T

∑t=1EX 2t= lim

T!∞

1T

T

∑t=1

σ2EY 2tk

= lim

T!∞σ2E

Y 2t= σ4

∞

∑j=0

ψ2j = EX 2t

We can now apply the theorem, so we know that

pT1T

T

∑t=1Xt ) N

0,E

X 2t

or

1pT

T

∑t=1

εtYtk ) N0, σ2E

Y 2t

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CLT Example VI

Now we turn to the properties of series with deterministic andstochastic trends.

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The Simple Trend Model I

Consider the simple time trend model:

yt = α+ δt + εt ,

where εt is a white noise process.

Note: If εt is normal, all the usual nite sample results hold.

Asymptotics are inuenced by the non-stationarity, but nite sampletheory is not.

We will not assume normality and instead develop the asymptotictheory for this case.

Consider estimation with OLS and nd β.

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The Simple Trend Model II

Let xt =1t

and β =

αδ

. Then

yt = x 0tβ+ εt

and

β =

T

∑t=1xtx 0t

!1 T

∑t=1xtyt

=

T

∑t=1xtx 0t

!1 T

∑t=1xtx 0tβ+ εt

= β+

T

∑t=1xtx 0t

!1 T

∑t=1xt εt

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The Simple Trend Model III

We get:

β β =

T

∑t=1

1 tt t2

!1 T

∑t=1

εttεt

=

T ∑T

t=1 t∑Tt=1 t ∑T

t=1 t2

1 ∑Tt=1 εt

∑Tt=1 tεt

First consider ∑T

t=1 xtx0t :

T ∑Tt=1 t

∑Tt=1 t ∑T

t=1 t2

=

"T T (T+1)

2T (T+1)

2T (T+1)(2T+1)

6

#

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The Simple Trend Model IV

Note that

T1T ! 1

T2T (T + 1)

2! 1

2

T3T (T + 1) (2T + 1)

6! 1

3In fact, it turns out that

T(ν+1)T

∑t=1tυ ! 1

ν+ 1

Helle Bunzel ISU



The Simple Trend Model V

Therefore, if we want to pre-multiply by some order of T such thatthe whole matrix converges, we need to multiply by T3. In that case:

T3

T ∑Tt=1 t

∑Tt=1 t ∑T

t=1 t2

=

"T3T T3 T (T+1)2

T3 T (T+1)2 T3 T (T+1)(2T+1)6

#!0 00 1

3

This matrix is not invertible. A trick that is used instead is to pre-and post multiply by Υ1T , where

ΥT =

"T

12 00 T

32

#

Helle Bunzel ISU



The Simple Trend Model VI

Then we get:"T

12 0

0 T32

# T ∑T

t=1 t∑Tt=1 t ∑T

t=1 t2

"T

12 0

0 T32

#

=

"T

12 0

0 T32

# "T

12T T

32 ∑T

t=1 tT

12 ∑T

t=1 t T32 ∑T

t=1 t2

#

=

"T

12T

12T T

12T

32 ∑T

t=1 tT

32T

12 ∑T

t=1 t T32T

32 ∑T

t=1 t2

#

=

T1T T2 ∑T

t=1 tT2 ∑T

t=1 t T3 ∑Tt=1 t

2

!1 1

212

13

Helle Bunzel ISU



The Simple Trend Model VII

Now return to the estimator:

β β =

α α

δ δ

=

T ∑T

t=1 t∑Tt=1 t ∑T

t=1 t2

1 ∑Tt=1 εt

∑Tt=1 tεt

It turns out that we can pre-multiply

β β

by ΥT in exactly the

same way we pre-multiplied

β βbypT in the stationary model.

Helle Bunzel ISU



The Simple Trend Model VIII

If we do this, we get:

ΥT

β β

=

"T

12 00 T

32

# α α

δ δ

=

" pT (α α)

T32δ δ

#

= ΥT

T ∑T

t=1 t∑Tt=1 t ∑T

t=1 t2

1 ∑Tt=1 εt

∑Tt=1 tεt

=

T ∑T

t=1 t∑Tt=1 t ∑T

t=1 t2

Υ1T

1 ∑Tt=1 εt

∑Tt=1 tεt

=

ΥTΥ1T

T ∑T

t=1 t∑Tt=1 t ∑T

t=1 t2

Υ1T

1 ∑Tt=1 εt

∑Tt=1 tεt

Helle Bunzel ISU



The Simple Trend Model IX

=

ΥTΥ1T

T ∑T

t=1 t∑Tt=1 t ∑T

t=1 t2

Υ1T

1 ∑Tt=1 εt

∑Tt=1 tεt

=

Υ1T

T ∑T

t=1 t∑Tt=1 t ∑T

t=1 t2

Υ1T

1Υ1T

∑Tt=1 εt

∑Tt=1 tεt

=

Υ1T

T ∑T

t=1 t∑Tt=1 t ∑T

t=1 t2

Υ1T

1 "T

12 ∑T

t=1 εtT

32 ∑T

t=1 tεt

#We know that

Υ1T

T ∑T

t=1 t∑Tt=1 t ∑T

t=1 t2

Υ1T

1!1 1

212

13

1=

4 66 12

Helle Bunzel ISU



The Simple Trend Model XIf we assume that E (εt ) = 0, V (εt ) = σ2 and E

ε2t< ∞, we can

apply a standard CLT:

T12

T

∑t=1

εt ) N0, σ2

The last bit we need is the limit of

T32

T

∑t=1tεt .

For this we will use the CLT for martingale di¤erence sequences. Firstnote that

T32

T

∑t=1tεt =

1pT

T

∑t=1

tT

εt

Helle Bunzel ISU



The Simple Trend Model XINow let xt = t

T εt and note that E (xt jΩt1) = 0.

Recall the theorem:

Theorem

Let fYtg∞t=1 be a scalar martingale di¤erence sequence with

YT = 1T ∑T

t=1 Yt . Suppose that (a) EY 2t= σ2t > 0 with

1T ∑T

t=1 σ2t ! σ2 > 0, (b) E jYt jr < ∞ for some r > 2 and all t, and (c)1T ∑T

t=1 Y2t

P! σ2. ThenpTYT ) N

0, σ2

.

We know that Ex2t= E

tT

2ε2t

= tT

2σ2 and that

1T

T

∑t=1

tT

2σ2 = σ2

1T 3

T

∑t=1t2 ! σ2

3

Helle Bunzel ISU



The Simple Trend Model XII

Also since we assumed that εt has nite fourth moments,

Ex4t= E

tT

4ε4t

!=

tT

4Eε4t< ∞

which satises (b).

Finally we need to show that

1T

T

∑t=1x2t

P! σ2

3

Helle Bunzel ISU



The Simple Trend Model XIII

Now, we know that

E

(1T

T

∑t=1x2t

)! σ2

3

so the proof is done if we can verify that

V

(1T

T

∑t=1x2t

)! 0

Helle Bunzel ISU



The Simple Trend Model XIV

Using the usual formula

V

(1T

T

∑t=1x2t

)

= E

24 1T

T

∑t=1x2t

1T

T

∑t=1Ex2t!235

= E

24 1T

T

∑t=1

tT

2ε2t

1T

T

∑t=1

tT

2σ2

!235= E

24 1T

T

∑t=1

tT

2 ε2t σ2

!235Helle Bunzel ISU



The Simple Trend Model XV

V

(1T

T

∑t=1x2t

)

= E

24 1T

T

∑t=1

tT

2 ε2t σ2

!235=

1T 2

T

∑t=1

tT

4Eh

ε2t σ22i

=Eh

ε2t σ22i

T1T 5

T

∑t=1t4 ! 0 1

5= 0

Helle Bunzel ISU



The Simple Trend Model XVIThis veries that

1T

T

∑t=1x2t

P! σ2

3

and hence we can use the CLT, so

pTxT ) N

0,

σ2

3

or

1pT

T

∑t=1

tT

εt = T

32

T

∑t=1tεt ) N

0,

σ2

3

Now a quick re-cap:

Helle Bunzel ISU



The Simple Trend Model XVIIThe model is the simple time trend model:

yt = α+ δt + εt ,

We are considering the OLS estimator of the parameters.

So far we have shown that" pT (α α)

T32δ δ

#

=

Υ1T

T ∑T

t=1 t∑Tt=1 t ∑T

t=1 t2

Υ1T

1 "T

12 ∑T

t=1 εtT

32 ∑T

t=1 tεt

#where

ΥT =

"T

12 00 T

32

#Helle Bunzel ISU



The Simple Trend Model XVIII

and Υ1T

T ∑T

t=1 t∑Tt=1 t ∑T

t=1 t2

Υ1T

1!1 1

212

13

1We also know the limiting distributions of T

12 ∑T

t=1 εt andT

32 ∑T

t=1 tεt . Some additional work is required to show their jointlimiting distribution (which we need).

It is fairly easy to show that this is Gaussian and that the covariance

matrix is1 1

212

13

σ2.

Helle Bunzel ISU



The Simple Trend Model XIX

From this, we get that" pT (α α)

T32δ δ

# )1 1

212

13

1N0,1 1

212

13

σ2

= N

0,1 1

212

13

1σ2

!

= N0,4 66 12

σ2

Helle Bunzel ISU



Testing in the Simple Trend Model I

Now that we have the distribution of the parameter estimates we canconsider testing hypotheses about the parameters.

First note thatpT (α α)) N

0, 4σ2

, so as long as the standard

estimate of the variance works, the standard t test for hypotheseson α should work. This turns out to be the case.

Now consider the standard t test for hypotheses on δ :

tT =

δ δ

hs2Tn(X 0TXT )

1o22

i 12

Helle Bunzel ISU



Testing in the Simple Trend Model II

where xt =1t

,

XT =

2664x 01x 02:x 0T

3775and fgij signies the ij 0th entry of the matrix fg .

Helle Bunzel ISU



Testing in the Simple Trend Model III

Re-writing the statistic:

tT =

δ δ

hs2Tn(X 0TXT )

1o22

i 12

=

δ δ

s2T0 1

(X 0TXT )

101

12

=T

32δ δ

s2Th0 T

32

i(X 0TXT )

10T

32

12

Helle Bunzel ISU



Testing in the Simple Trend Model IV

=T

32δ δ

s2T0 1

ΥT (X 0TXT )

1 ΥT

01

12

=T

32δ δ

s2T0 1

Υ1T X

0TXTΥ1T

1 01

12

=T

32δ δ

24s2T 0 1 " T

T 3T (T+1)2T 3

T (T+1)2T 3

T (T+1)(2T+1)6T 3

#1 01

35 12

Helle Bunzel ISU



Testing in the Simple Trend Model V

=) 1sσ20 1

1 12

12

13

1 01

N 0, 12σ2

=1s

σ20 1

4 66 12

01

N 0, 12σ2

=1p12σ2

N0, 12σ2

= N (0, 1)

The result is that even though the parameter converges faster thatususal, we can still use our standard t test.

Helle Bunzel ISU



Testing in the Simple Trend Model VI

DenitionWhen a parameter estimate converges to its true value at a rate fasterthan

pT , we say that the estimate is super-consistent.

Now we will consider a single hypothesis involving both parameters.We wish to test:

H0 : r1α+ r2δ = r

Helle Bunzel ISU



Testing in the Simple Trend Model VIIGenerally we would test hypotheses like this one using the square rootof the F test. This implies that

tT =

r1 α+ r2 δ r

s2Tr1 r2

(X 0TXT )

1r1r2

12

=

r1 α+ r2 δ r

s2Tr1 r2

Υ1T ΥT (X 0TXT )

1 ΥTΥ1T

r1r2

12

=

r1 α+ r2 δ r

"s2ThT

12 r1 T

32 r2

i Υ1T X

0TXTΥ1T

1 " T 12 r1

T32 r2

## 12

Helle Bunzel ISU



Testing in the Simple Trend Model VIII

tT =

r1 α+ r2 δ r

"s2ThT

12 r1 T

32 r2

i Υ1T X

0TXTΥ1T

1 " T 12 r1

T32 r2

## 12

=

pTr1 α+ r2 δ r

s2Tr1 T1r2

Υ1T X

0TXTΥ1T

1 r1T1r2

12

Now, under the null hypothesis,pTr1 α+ r2 δ r

=

pTr1 α+ r2 δ (r1α+ r2δ)

= r1

pT (α α) + r2

pTδ δ

Helle Bunzel ISU



Testing in the Simple Trend Model IX

Now,

pTδ δ

P! 0

andpT (α α)) N

0, 4σ2

This implies that

pTr1 α+ r2 δ r

) r1N

0, 4σ2

= N

0, 4r21 σ2

Helle Bunzel ISU



Testing in the Simple Trend Model XTo nd the limit of the demonator, note that

s2TP! σ2r1 T1r2

!r1 0

and

Υ1T X0TXTΥ1T

1 ! 4 66 12

implying that

s2Tr1 T1r2

Υ1T X

0TXTΥ1T

1 r1T1r2

! σ2

r1 0

4 66 12

r10

= σ24r21

Helle Bunzel ISU



Testing in the Simple Trend Model XI

such that

tT =

pTr1 α+ r2 δ r

s2Tr1 T1r2

Υ1T X

0TXTΥ1T

1 r1T1r2

12

=) 1σ24r21

N0, 4r21 σ2

= N (0, 1)

As a result we can, again, use the standard t test for this type ofhypotheses.

Note that this test has the same distribution under the null as itwould have had if we used the true value of δ instead of an estimate.This is due to the super-consistency. The slowest convergingparameter estimate dominates the asymptotic distribution.

Helle Bunzel ISU



Testing in the Simple Trend Model XIINow consider a joint test of separate hypotheses about α and δ :

H0 :

αδ

=

α0δ0

or

H0 : β = β0

For this we use a Wald test. The standard test statistic for thishypothesis is:

W =

β β00 hs2TX 0TXT

1i1 β β0

=

β β0

0ΥThs2TΥT

X 0TXT

1 ΥTi1

ΥT

β β0

Helle Bunzel ISU



Testing in the Simple Trend Model XIIIBecause

s2TΥTX 0TXT

1 ΥT ! σ24 66 12

and

ΥT

β β0) N

0,4 66 12

σ2

we have

W =

β β00

ΥThs2TΥT

X 0TXT

1 ΥTi1

ΥT

β β0) χ2 (2)

We have thus established that for the simple regression model with adeterministic trend, all the usual test statistics are valid.

Helle Bunzel ISU



Testing in the Simple Trend Model XIV

The rest of Hamilton Chapter 16 is spent on more complicated trendmodels with ARMA errors. The proofs are more complicated, but theend results are the same: We can apply the usual asymptotic testsand the have the usual asymptotic distributions.

Helle Bunzel ISU


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