Lecture 16 Unit Root Tests

RS ? EC2 - Lecture 16

Lecture 16 Unit Root Tests

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Autoregressive Unit Root

? A shock is usually used to describe an unexpected change in a variable or in the value of the error terms at a particular time period. ? When we have a stationary system, effect of a shock will die out gradually. But, when we have a non-stationary system, effect of a shock is permanent. ? We have two types of non-stationarity. In an AR(1) model we have: - Unit root: | 1 | = 1: homogeneous non-stationarity - Explosive root: | 1 | > 1: explosive non-stationarity ? In the last case, a shock to the system become more influential as time goes on. It can never be seen in real life. We will not consider them.

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RS ? EC2 - Lecture 16

Autoregressive Unit Root

? Consider the AR(p) process:

(L) yt t

where (L) 1 1L1 L22 .... p Lp

As we discussed before, if one of the rj's equals 1, (1)=0, or

1 2 .... p 1

? We say yt has a unit root. In this case, yt is non-stationary.

Example: AR(1): yt 1 yt 1 t Unit root: 1 = 1.

H0 (yt non-stationarity): 1 = 1

(or, 1 ? 1 = 0)

H1 (yt stationarity): 1 < 1

(or, 1 ? 1 < 0)

? A t-test seems natural to test H0. But, the ergodic theorem and MDS CLT do not apply: the t-statistic does not have the usual distributions.

Autoregressive Unit Root

? Now, let's reparameterize the AR(1) process. Subtract yt-1 from yt:

yt yt yt1 (1 1) yt1 t 0 yt1 t

? Unit root test:

H0: 0 = 1 ? 1 = 0 H1: 0 < 0.

? Natural test for H0: t-test . We call this test the Dickey-Fuller (DF) test. But, what is its distribution?

? Back to the general, AR(p) process: (L) yt t We rewrite the process using the Dickey-Fuller reparameterization::

yt 0 yt1 1yt1 2yt2 .... p1yt( p1) t ? Both AR(p) formulations are equivalent.

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RS ? EC2 - Lecture 16

Autoregressive Unit Root ? Testing

? AR(p) lag (L): (L) 1 1L1 L22 .... p Lp

? DF reparameterization: (1 L ) 0 1 ( L L2 ) 2 ( L2 L3 ) .... p 1 ( L p 1 L p )

? Both parameterizations should be equal. Then, (1)=-0. unit root hypothesis can be stated as H0: 0=0.

Note: The model is stationary if 0< 0 natural H1: 0 < 0.

? Under H0: 0=0, the model is AR(p-1) stationary in yt. Then, if yt has a (single) unit root, then yt is a stationary AR process.

? We have Augmented

a linear regression framework. Dickey-Fuller (ADF) test.

A

t-test

for

H0

is

the

Autoregressive Unit Root ? Testing: DF

? The Dickey-Fuller (DF) test is a special case of the ADF: No lags are included in the regression. It is easier to derive. We gain intuition from its derivation.

? From our previous example, we have:

y t 1y t 1 t 0 y t 1 t

? If 0 = 0, system has a unit root:: H0 :0 0 H1 :0 0 (| 0 | 0)

? We can test H0 with a t-test:

t 1

^ 1 SE ^

? There is another associated test with H0, the -test:. (T-1)(^ -1).

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RS ? EC2 - Lecture 16

Review: Stochastic Calculus

? Kolmogorov Continuity Theorem ? If for all T > 0, there exist a, b, > 0 such that: ? E(|X(t1, ) ? X(t2, )|a) |t1 ? t2|(1 + b) ? Then X(t, ) can be considered as a continuous stochastic process. ? Brownian motion is a continuous stochastic process. ? Brownian motion (Wiener process): X(t, ) is almost surely continuous, has independent normal distributed (N(0,t-s)) increments and X(t=0, ) =0 ("a continuous random walk").

Review: Stochastic Calculus ? Wiener process

? Let the variable z(t) be almost surely continuous, with z(t=0)=0. ? Define (,v) as a normal distribution with mean and variance v. ? The change in a small interval of time t is z

? Definition: The variable z(t) follows a Wiener process if

? z(0) = 0

? z = t ,

where N(0,1)

? It has continuous paths.

? The values of z for any 2 different (non-overlapping) periods of time are independent.

Notation: W(t), W(t, ), B(t).

Example:

WT (r)

1 T

(1 2

3 ... [Tr] );

r [0,1]

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RS ? EC2 - Lecture 16

Review: Stochastic Process: Wiener process

? What is the distribution of the change in z over the next 2 time units? The change over the next 2 units equals the sum of: - The change over the next 1 unit (distributed as N(0,1)) plus - The change over the following time unit --also distributed as N(0,1). - The changes are independent. - The sum of 2 normal distributions is also normally distributed. Thus, the change over 2 time units is distributed as N(0,2).

? Properties of Wiener processes:

? Mean of z is 0

? Variance of z is t

? Standard deviation of z is t

? Let N=T/t, then

z (T

)

z(0)

n

i 1

i

t

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Review: Stochastic Calculus ? Wiener process

Example:

WT (r)

1 T

(1 2

3 ... [Tr] )

1 T

S[Tr ] ;

r [0,1]

? If T is large, WT(.) is a good approximation to W(r); r [0,1], defined: W(r) = limT WT(r) E[W(r) ] =0 Var[W(r) ] =t

? Check Billingsley (1986) for the details behind the proof that WT(r) converges as a function to a continuous function W(r).

? In a nutshell, we need - t satisfying some assumptions (stationarity, E[|t|q < for q>2, etc.) - a FCLT (Functional CLT). - a Continuous Mapping Theorem. (Similar to Slutzky's theorem). 10

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RS ? EC2 - Lecture 16

Review: Stochastic Calculus ? Wiener process

? Functional CLT (Donsker's FCLT) If t satisfies some assumptions, then

WT(r) DW(r), where W(r) is a standard Brownian motion for r [0, 1].

Note: That is, sample statistics, like WT(r), do not converge to constants, but to functions of Brownian motions.

? A CLT is a limit for one term of a sequence of partial sums {Sk}, Donsker's FCLT is a limit for the entire sequence {Sk} instead of one term.

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Review: Stochastic Calculus ? Wiener process

? Example: yt = yt-1 + t (Case 1). Get distribution of (X'X/T2)-1 for yt.

T

T t1

T

T2 (yt1)2 T2 [ ti y0]2 T2 [St1 y0]2

t1

t1 i1

t1

T

T2 [(St1)2 2y0St1 y02]

t1

2

T t1

St1 T

2T 1

2y0T1/2

T t1

St1 T

T 1

T 1y02

2

T t1

t /T (t1)T

1 T

S[Tr] 2dr 2y0T 1/2

T t1

t /T (t1)T

1 T

S[Tr] drT 1y02

1

1

2 XT (r)2dr 2y0T1/2 XT (r) drT1y02

0

0

1

d 2 W(r)2dr, T .

0

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RS ? EC2 - Lecture 16

Review: Stochastic Calculus ? Ito's Theorem

? The integral w.r.t a Brownian motion, given by Ito's theorem (integral):

f(t,) dB = f(tk,) Bk

where tk* [tk,tk + 1) as tk + 1 ? tk 0.

As we increase the partitions of [0,T], the sum p to the integral.

? But, this is a probability statement: We can find a sample path where the sum can be arbitrarily far from the integral for arbitrarily large partitions (small intervals of integration).

? You may recall that for a Rienman integral, the choice of tk* (at the start or at the end of the partition) is not important. But, for Ito's integral, it is important (at the start of the partition).

? Ito's Theorem result: B(t,) dB(t,) = B2(t,)/2 ? t/2.

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Autoregressive Unit Root ? Testing: Intuition

? We continue with yt = yt-1 + t (Case 1). Using OLS, we estimate :

T

T

T

yt yt 1

( yt 1 yt 1 ) yt 1

yt 1yt 1

^

t 1 T

t 1

T

1

t 1 T

y2 t 1

t 1

y2 t 1

t 1

y2 t 1

t 1

? This implies:

T

T

yt 1yt 1

( yt 1 / T )(t / T )

T (^ 1) T

t 1 T

y2 t 1

t 1

t 1

1 T

T

( yt 1 /

t 1

T )2

? From the way we defined WT(.), we can see that yt/sqrt(T) converges to a Brownian motion. Under H0, yt is a sum of white noise errors.

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RS ? EC2 - Lecture 16

Autoregressive Unit Root ? Testing: Intuition

? Intuition for distribution under H0: - Think of yt as a sum of white noise errors. - Think of t as dW(t).

Then, using Billingley (1986), we guess that T( ^-1) converges to

1

W (t)dW (t)

T (^ 1) d 0 1

W (t)2dt

0

? We think of t as dW(t). Then, k=0 to t k, which corresponds to 0to(t/T) dW(s)=W( s/T) (for W(0)=0). Using Ito's integral, we have

T

(^

1)

d

1 2

W (1)2 1

1

W (t)2dt

0

Autoregressive Unit Root ? Testing: Intuition

?

T (^ 1)

d

1 2

W

1

(1) 2

1

W (t)2dt

0

Note: W(1) is a N(0,1). Then, W(1)2 is just a 2(1) RV.

? Contrary to the stable model the denominator of the expression for the OLS estimator ?i.e., (1/T)txt2-- does not converge to a constant a.s., but to a RV strongly correlated with the numerator.

? Then, the asymptotic distribution is not normal. It turns out that the limiting distribution of the OLS estimator is highly skewed, with a long tail to the left.

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