? Chris Brooks 2002 陈磊 2004
7-1
Chapter 7
Modelling long-run relationship
in finance
? Chris Brooks 2002 陈磊 2004
7-2
1 Stationarity and Unit Root Testing
? The stationarity or otherwise of a series can strongly
influence its behaviour and properties - e.g,persistence of
shocks will be infinite for nonstationary series
? Spurious regressions,If two variables are trending over time,
a regression of one on the other could have a high R2 even if
the two are totally unrelated
? If the variables in the regression model are not stationary,
then it can be proved that the standard assumptions for
asymptotic analysis will not be valid,In other words,the
usual,t-ratios” will not follow a t-distribution,so we cannot
validly undertake hypothesis tests about the regression
parameters.
1.1Why do we need to test for Non-Stationarity?
? Chris Brooks 2002 陈磊 2004
7-3Value of R2 for 1000 Sets of Regressions of a Non-
stationary Variable on another Independent Non-
stationary Variable
? Chris Brooks 2002 陈磊 2004
7-4Value of t-ratio on Slope Coefficient for 1000 Sets
of Regressions of a Non-stationary Variable on
another Independent Non-stationary Variable
? Chris Brooks 2002 陈磊 2004
7-5
1.2 Two types of Non-Stationarity
? Various definitions of non-stationarity exist
? In this chapter,we are really referring to the weak form or
covariance stationarity
? There are two models which have been frequently used to
characterise non-stationarity,the random walk model with
drift:
yt = ? + yt-1 + ut (1)
and the deterministic trend process:
yt = ? + ?t + ut (2)
where ut is iid in both cases.
? Chris Brooks 2002 陈磊 2004
7-6
Stochastic Non-Stationarity
? Note that the model (1) could be generalised to the case
where yt is an explosive process:
yt = ? + ? yt-1 + ut
where ? > 1.
? Typically,the explosive case is ignored and we use ? = 1 to
characterise the non-stationarity because
– ? > 1 does not describe many data series in economics
and finance.
– ? > 1 has an intuitively unappealing property,shocks to
the system are not only persistent through time,they are
propagated so that a given shock will have an
increasingly large influence.
? Chris Brooks 2002 陈磊 2004
7-7
Stochastic Non-stationarity,
The Impact of Shocks
? To see this,consider the general case of an AR(1) with no drift:
yt = ? yt-1 + ut (3)
Let ? take any value for now.
? We can write,yt-1 = ? yt-2 + ut-1
yt-2 = ? yt-3 + ut-2
? Substituting into (3) yields,yt = ? (? yt-2 + ut-1) + ut
= ?2yt-2 + ? ut-1 + ut
? Substituting again for yt-2,yt = ?2(? yt-3 + ut-2) + ? ut-1 + ut
= ?3 yt-3 + ?2ut-2 + ? ut-1 + ut
? Successive substitutions of this type lead to:*
yt = ?T y0 + ?ut-1 + ?2ut-2 + ?3ut-3 +,..+ ?T-1u1 + ut
? Chris Brooks 2002 陈磊 2004
7-8The Impact of Shocks for
Stationary and Non-stationary Series
? We have 3 cases:
1,|? | <1 ??T?0 as T??
So the shocks to the system gradually die away.
2,? =1 ??T =1? T
So shocks persist in the system and never die away,We
obtain:
as T??
So just an infinite sum of past shocks plus some starting
value of y0.
3,? >1,Now given shocks become more influential as time goes
on,since if ? >1,? 3 > ? 2 >? etc.
??
?
??
0
0
i
tt uyy
? Chris Brooks 2002 陈磊 2004
7-9
Detrending a Non-stationary Series
? Going back to our 2 characterisations of non-stationarity,the
r.w,with drift:
? yt = ? + yt-1 + ut (1)
and the trend-stationary process
yt = ? + ? t + ut (2)
? The two will require different treatments to induce stationarity.
The second case is known as deterministic non-stationarity and
what is required is detrending去势,
? The first case is known as stochastic non-stationarity,If we let
?yt = yt - yt-1 = yt - L yt =(1-L) yt
then ?yt =(1-L) yt = ? + ut
we have induced stationarity by,differencingonce”.
单位根过程
? Chris Brooks 2002 陈磊 2004
7-10Detrending a Series:
Using the Right Method
? Although trend-stationary and difference-stationary series
are both,trending” over time,the correct approach needs
to be used in each case.
? If we first difference the trend-stationary series,it would
“remove” the non-stationarity,but at the expense on
introducing an MA(1) structure into the errors,p372
? Conversely if we try to detrend a series which has
stochastic trend,then we will not remove the non-
stationarity.
? We will now concentrate on the stochastic non-stationarity
model since deterministic non-stationarity does not
adequately describe most series in economics or finance.
? Chris Brooks 2002 陈磊 2004
7-11
Sample Plots for various Stochastic
Processes,A White Noise Process
-4
-3
-2
-1
0
1
2
3
4
1 40 79 118 157 196 235 274 313 352 391 430 469
? Chris Brooks 2002 陈磊 2004
7-12
Random Walk and
a Random Walk with Drift
-2 0
-1 0
0
10
20
30
40
50
60
70
1 19 37 55 73 91 109 127 145 163 181 199 217 235 253 271 289 307 325 343 361 379 397 415 433 451 469 487
R a n d o m W a l k
R a n d o m W a l k w i t h D ri f t
? Chris Brooks 2002 陈磊 2004
7-13
A Deterministic Trend Process
-5
0
5
10
15
20
25
30
1 40 79 118 157 196 235 274 313 352 391 430 469
? Chris Brooks 2002 陈磊 2004
7-14Autoregressive Processes with
differing values of ? (0,0.8,1)
-2 0
-1 5
-1 0
-5
0
5
10
15
1 53 105 157 209 261 313 365 417 469 521 573 625 677 729 781 833 885 937 989
Ph i = 1
Ph i = 0, 8
Ph i = 0
? Chris Brooks 2002 陈磊 2004
7-15
1.3 Definition of Non-Stationarity
? Consider again the simplest stochastic trend model:
yt = yt-1 + ut or ?yt= ut
We can generalise this concept to consider the case where the
series contains more than one,unit root”,That is,we would
need to apply the first difference operator,?,more than once
to induce stationarity.
Definition
If a non-stationary series,yt must be differenced d times
before it becomes stationary,then it is said to be integrated of
order d,We write yt ?I(d).
So if yt ? I(d) then ?dyt? I(0).
An I(0) series is a stationary series
An I(1) series contains one unit root,e.g,yt = yt-1 + ut
? Chris Brooks 2002 陈磊 2004
7-16
Characteristics of I(0),I(1) and I(2) Series
? An I(2) series contains two unit roots and so would require
differencing twice to induce stationarity,p376
? I(1) and I(2) series can wander a long way from their mean
value and cross this mean value rarely.
? I(0) series should cross the mean frequently,
? The majority of economic and financial series contain a single
unit root,although some are stationary and consumer prices
have been argued to have 2 unit roots.
? Chris Brooks 2002 陈磊 2004
7-17
1.4 Test for a unit root
? 无法使用 acf或 pacf确定序列是否是单位根过程。
? The early and pioneering work on testing for a unit root in
time series was done by Dickey and Fuller (Dickey and
Fuller 1979,Fuller 1976),The basic objective of the test is to
test the null hypothesis that ? =1 in:
yt = ? yt-1 + ut
against the one-sided alternative ? <1,So we have
H0,series contains a unit root
vs,H1,series is stationary,
? We usually use the regression:
?yt = ? yt-1 + ut
a test of ?=1 is equivalent to a test of ?=0 (since ? -1=?).
? Chris Brooks 2002 陈磊 2004
7-18
Different forms for the DF Test
? Dickey Fuller tests are also known as ? tests,?,??,??.
? The null (H0) and alternative (H1) models in each case are
i) H0,yt = yt-1+ut H1,yt = ? yt-1+ut,? <1
This is a test for a random walk against a stationary
autoregressive process of order one (AR(1))
ii) H0,yt = yt-1+ut H1,yt = ? yt-1+?+ut,? <1
This is a test for a random walk against a stationary AR(1) with
drift.
iii) H0,yt = yt-1+ut H1,yt = ? yt-1+?+? t+ut,? <1
This is a test for a random walk against a stationary AR(1) with
drift and a time trend.
? Chris Brooks 2002 陈磊 2004
7-19
Computing the DF Test Statistic
? We can write ?yt=ut
where ? yt = yt- yt-1,and the alternatives may be expressed as
?yt = ? yt-1+?+?t +ut
with ?=?=0 in case i),and ?=0 in case ii) and ?=? -1,In each case,
the tests are based on the t-ratio on the yt-1 term in the estimated
regression of ?yt on yt-1,plus a constant in case ii) and a constant
and trend in case iii),The test statistics are defined as
test statistic =
? The test statistic does not follow the usual t-distribution under the
null,since the null is one of non-stationarity,but rather follows a
non-standard distribution,Critical values are derived from Monte
Carlo experiments in,for example,Fuller (1976),Relevant
examples of the distribution are shown in table 7.1 below
?
?
?
?
?
SE( )
? Chris Brooks 2002 陈磊 2004
7-20
Critical Values for the DF Test
The null hypothesis of a unit root is rejected in favour of the
stationary alternative in each case if the test statistic is more
negative than the critical value.
临界值还与样本容量有关, 参见 675页 表 A2.7。
S ig n if ic a n c e l e v e l 10% 5% 1%
C.V, f o r c o n sta n t
b u t n o t re n d
-2,5 7 -2,8 6 -3,4 3
C.V, f o r c o n sta n t
a n d t re n d
-3,1 2 -3,4 1 -3,9 6
T a b le 4,1, Cri ti c a l V a l u e s f o r DF a n d A D F T e sts (F u ll e r,
1 9 7 6,p 3 7 3 ).
? Chris Brooks 2002 陈磊 2004
7-21
The Augmented Dickey Fuller Test
(ADF)
? The tests above are only valid if ut is white noise,In particular,
ut will be autocorrelated if there was autocorrelation in the
dependent variable of the regression (?yt) which we have not
modelled,The solution is to,augment” the test using p lags of
the dependent variable,The alternative model in case (i) is
now written:
? The same critical values from the DF tables are used as before.
A problem now arises in determining the optimal number of
lags of the dependent variable.
There are 2 ways
- use the frequency of the data to decide
- use information criteria
?
?
?? ?????
p
i
tititt uyyy
1
1 ??
? Chris Brooks 2002 陈磊 2004
7-22
1.5 Testing for Higher Orders of
Integration
? Consider the simple regression,?yt = ? yt-1 + ut
We test H0,? = 0 vs,H1,? < 0.
? If H0 is rejected we simply conclude that yt does not contain a
unit root.
? But what do we conclude if H0 is not rejected? The series
contains a unit root,but is that it? No! What if yt ?I(2)? We
would still not have rejected,So we now need to test
H0,yt ?I(2) vs,H1,yt?I(1)
We now regress ?2yt on ?yt-1 (plus lags of ?2yt if necessary).
? Now we test H0,?yt?I(1) which is equivalent to H0,yt?I(2).
? So in this case,if we do not reject,then yt is at least I(2).
? continue to test for a further unit root until we rejected H0.
? Dickey and Pantula(1987)认为, 检验顺序应从高阶至低阶 。
? Chris Brooks 2002 陈磊 2004
7-23
1.6 The Phillips-Perron Test
? Phillips and Perron have developed a more comprehensive
theory of unit root nonstationarity,The tests are similar to
ADF tests,but they incorporate an automatic correction to
the DF procedure to allow for autocorrelated residuals.
? The tests usually give the same conclusions as the ADF tests,
and the calculation of the test statistics is complex.
? Chris Brooks 2002 陈磊 2004
7-241.7 Criticism of Dickey-Fuller and
Phillips-Perron-type tests
? Main criticism is that the power of the tests is low if the process is
stationary but with a root close to the non-stationary boundary.
e.g,the tests are poor at deciding if ? =1 or ? =0.95,especially
with small sample sizes.
? If the true data generating process (dgp) is
yt = 0.95yt-1 + ut
then the null hypothesis of a unit root should be rejected.
? One way to get around this is to use a stationarity test as well as
the unit root tests we have looked at.
? Chris Brooks 2002 陈磊 2004
7-25
Stationarity tests
? Stationarity tests have
H0,yt is stationary
versus H1,yt is non-stationary
So that by default under the null the data will appear stationary,
? One such stationarity test is the KPSS test (Kwaitowski,Phillips,
Schmidt and Shin,1992).
? Thus we can compare the results of these tests with the ADF/PP
procedure to see if we obtain the same conclusion.
? Chris Brooks 2002 陈磊 2004
7-26
Stationarity tests (cont’d)
? A Comparison
ADF / PP KPSS
H0,yt ? I(1) H0,yt ? I(0)
H1,yt ? I(0) H1,yt ? I(1)
? There are four possible outcomes
1.Reject H0 and Do not reject H0
2.Do not reject H0 and Reject H0
3.Reject H0 and Reject H0
4.Do not reject H0 and Do not reject H0
? Chris Brooks 2002 陈磊 2004
7-27
有关单位根和协整的一些参考书
? 1,董文泉 等,,经济周期波动的分析与预测方法, 第十二章，
吉林大学出版社,1998年。
? 2,李子奈、叶阿忠,,高等计量经济学, 第二章第五节，清
华大学出版社,2000年。
? 3,张晓峒,,计量经济分析,,经济科学出版社,2000年。
? 4,赵国庆 主编,,计量经济学, 第八章，中国人民大学出版
社,2001年。
? 5,王维国 主编,,计量经济学, 第十章，东北财经大学出版
社,2002年。
? 6,Philip H,Franses,,商业和经济预测中的时间序列模型,
第四章，中国人民大学出版社,2002年。
? Chris Brooks 2002 陈磊 2004
7-28
2 Testing for unit roots in
EViews
? p383-385
? Chris Brooks 2002 陈磊 2004
7-29
3 Cointegration,An Introduction
? In most cases,if we combine two variables which are I(1),
then the combination will also be I(1).
? More generally,if we combine variables with differing
orders of integration,the combination will have an order of
integration equal to the largest,i.e.,
if Xi,t ? I(di) for i = 1,2,3,...,k
so we have k variables each integrated of order di.
Let (1)
Then zt ? I(max di)
z Xt i i t
i
k?
??
?,
1
? Chris Brooks 2002 陈磊 2004
7-30
Linear Combinations of Non-
stationary Variables
? Rearranging (1),we can write
where
? This is just a regression equation,
? But the disturbances would have some very undesirable
properties,zt′ is not stationary and is autocorrelated if all of
the Xi are I(1).
? We want to ensure that the disturbances are I(0),Under
what circumstances will this be the case?
? ?? ?i i t tz z i k? ? ? ?1 1 2,',,.,,,
X X zt i i t t
i
k
1
2
,,'? ?
?
? ?
? Chris Brooks 2002 陈磊 2004
7-31
Definition of Cointegration
(Engle & Granger,1987)
? Let zt be a k?1 vector of variables,then the components of zt are
cointegrated of order (d,b) if
i) All components of zt are I(d)
ii) There is at least one vector of coefficients? such that
??zt ? I(d-b) 0<b≤d
? If variables are cointegrated,it means that a linear combination
of them will be stationary.
? There may be up to r linearly independent cointegrating
relationships (where r ? k-1),also known as cointegrating
vectors,r is also known as the cointegrating rank of zt.
? A cointegrating relationship may also be seen as a long term
relationship,Many time series are non-stationary but,move
together” over time.
? Chris Brooks 2002 陈磊 2004
7-32
Cointegration and Equilibrium
? Examples of possible Cointegrating Relationships in finance:
– spot and futures prices
– ratio of relative prices and an exchange rate
– equity prices and dividends
? Market forces arising from no arbitrage conditions should
ensure an equilibrium relationship.
? No cointegration implies that series could wander apart
without bound in the long run.
? Chris Brooks 2002 陈磊 2004
7-33
4 Equilibrium Correction or Error
Correction Models
? When the concept of non-stationarity was first considered,a
usual response was to independently take the first differences
of a series of I(1) variables.
? The problem with this approach is that pure first difference
models have no long run solution.
e.g,Consider yt and xt both I(1).
The model we may want to estimate is
? yt = ??xt + ut
But this collapses to nothing in the long run.
The definition of the long run that we use is where
yt = yt-1 = y; xt = xt-1 = x.
Hence all the difference terms will be zero,i.e,? yt = 0; ?xt = 0.
? Chris Brooks 2002 陈磊 2004
7-34
Specifying an ECM
? One way to get around this problem is to use both first
difference and lagged level terms,e.g.
? yt = ?1?xt + ?2(yt-1-?xt-1) + ut (2)
? yt-1-? xt-1 is known as the error correction term.
? Providing that yt and xt are cointegrated with cointegrating
coefficient ?,then (yt-1-?xt-1) will be I(0) even though the
constituents are I(1),
? We can thus validly use OLS on (2).*
? The Granger representation theorem shows that any
cointegrating relationship can be expressed as an equilibrium
correction model.
? Chris Brooks 2002 陈磊 2004
7-35
5 Testing for Cointegration in Regression
? The model for the equilibrium correction term can be
generalised to include more than two variables:
yt = ?1 + ?2x2t + ?3x3t + … + ?kxkt + ut (3)
ut should be I(0) if the variables yt,x2t,..,xkt are cointegrated.
? So what we want to test is the residuals of equation (3) to see
if they are non-stationary or stationary,We can use the DF /
ADF test on ut.
So we have the regression with vt ? iid.
? However,since this is a test on the residuals of an actual
model,,then the critical values are changed.因为参数估计
的偏差会改变针对残差的检验统计量的分布 。
? ? ?u u vt t t? ??? 1
?ut
? Chris Brooks 2002 陈磊 2004
7-36Testing for Cointegration in
Regression,Conclusions
? Engle and Granger (1987) have tabulated a new set of critical
values （ 见 A2.8） and hence the test is known as the Engle
Granger (E-G) test.
? We can also use the Durbin Watson test statistic（ CRDW）
or the Phillips Perron approach to test for non-stationarity
of,
? What are the null and alternative hypotheses for a test on the
residuals of a potentially cointegrating regression?
H0, unit root in cointegrating regression’s residuals
H1, residuals from cointegrating regression are stationary
如果拒绝 H0,应建立 ECM;
否则, 只能对差分后变量建立回归模型 。
?ut
? Chris Brooks 2002 陈磊 2004
7-37
6 Methods of modelling Cointegrated Systems:
? There are (at least) 3 methods we could use,Engle Granger,
Engle and Yoo,and Johansen.
? The Engle - Granger 2 Step Method
Step 1:
- Make sure that all the individual variables are I(1).
- Then estimate the cointegrating regression using OLS.
- Save the residuals of the cointegrating regression,.
- Test these residuals to ensure that they are I(0).
Step 2:
- Use the step 1 residuals as one variable in the error correction
model e.g.
? yt = ?1?xt + ?2( ) + ut where = yt-1- xt-11??tu
1??tu
?ut
??
? Chris Brooks 2002 陈磊 2004
7-38The Engle-Granger Approach,
Some Drawbacks
1,Unit root and cointegration tests have low power in finite
samples
2,We are forced to treat the variables asymmetrically and to
specify one as the dependent and the other as independent
variables.即使可能不存在理论根据。
3,Cannot perform any hypothesis tests about the actual
cointegrating relationship estimated at stage 1.
- Problem 1 is a small sample problem that should disappear
asymptotically.
- Problem 2 is addressed by the Johansen approach.
- Problem 3 is addressed by the Engle and Yoo approach or the
Johansen approach.
? Chris Brooks 2002 陈磊 2004
7-39
? One of the problems with the EG 2-step method is that we
cannot make any inferences about the actual cointegrating
regression.
? The Engle & Yoo (EY,1987) 3-step procedure takes its first two
steps from EG.
? EY add a third step giving updated estimates of the
cointegrating vector and its standard errors.
? 由于存在针对协整关系及其假设检验的更好估计方法, E-Y方
法在实证研究中很少使用 。
The Engle & Yoo 3-Step Method
? Chris Brooks 2002 陈磊 2004
7-40
? In the general case,where we have more than two variables
which may be cointegrated,there could be more than one
cointegrating relationship.
? In fact there can be up to r linearly independent cointegrating
vectors (where r ? g-1),where g is the number of variables in
total.
? So,in the case where we just had y and x,then r can only be one
or zero.
? And if there are others,how do we know how many there are or
whether we have found the,best”?
? The answer to this is to use a systems approach to cointegration
which will allow determination of all r cointegrating
relationships - Johansen’s method,
多个协整关系
? Chris Brooks 2002 陈磊 2004
7-41
多个协整关系
? 两个非平稳变量之间最多只能有一个线性组合是平稳的,即
最多只存在一个协整关系。
? 一个包括 K个变量的非平稳系统，可以存在最多 K-1个线性
独立的协整关系。
? 前面叙述的回归方法只能发现一个协整关系。
? 如果存在多个协整关系，如何知道是否存在其他的协整关
系？是否已经找到最“好”或最强的协整关系？
? 这需要检验协整关系的系统方法。
? Chris Brooks 2002 陈磊 2004
7-42
7 An Example, Lead-Lag Relationships
between Spot and Futures Prices
Background
? We expect changes in the spot price of a financial asset and its
corresponding futures price to be perfectly contemporaneously
correlated and not to be cross-autocorrelated.
i.e,expect Corr(?ln(Ft),?ln(St))? 1
Corr(?ln(Ft),?ln(St-k))? 0 ? k
Corr(?ln(Ft-j),?ln(St))? 0 ? j
? We can test this idea by modelling the lead-lag relationship
between the two.
? We will consider two papers Tse(1995) and Brooks et al (2001).
? Chris Brooks 2002 陈磊 2004
7-43
Futures & Spot Data
? Tse (1995),1055 daily observations on NSA 东京证券交易所
stock index and stock index futures values from December
1988 - April 1993.
? Brooks et al (2001),13,035 10-minutely observations on the
FTSE 100 stock index and stock index futures prices for all
trading days in the period June 1996 – 1997.
? Chris Brooks 2002 陈磊 2004
7-44
Methodology
? The fair futures price is given by
where Ft* is the fair futures price,St is the spot price,r is a
continuously compounded risk-free rate of interest,d is the
continuously compounded yield in terms of dividends derived
from the stock index until the futures contract matures,and (T-
t) is the time to maturity of the futures contract,Taking
logarithms of both sides of equation above gives
? First,test ft and st for nonstationarity.
t* t (r - d ) ( T - t)F = S e
t)-d ) ( T-(r s f tt ??*
? Chris Brooks 2002 陈磊 2004
7-45
Dickey-Fuller Tests on Log-Prices and
Returns for High Frequency FTSE Data
F u t u r e s Sp o t
D i c k e y - F u ll e r St a t i s t i c s
f o r L o g - P r i c e D a t a
- 0, 1 3 2 9 - 0, 7 3 3 5
D i c k e y F u l l e r St a t i s t i c s
f o r R e t u r n s D a t a
- 8 4, 9 9 6 8 - 1 1 4, 1 8 0 3
? Chris Brooks 2002 陈磊 2004
7-46
Cointegration Test Regression and Test
on Residuals
? Conclusion,log Ft and log St are not stationary,but ?log Ft
and ?log St are stationary.
? But a model containing only first differences has no long run
relationship.
? Solution is to see if there exists a cointegrating relationship
between ft and st which would mean that we can validly
include levels terms in this framework.
? Potential cointegrating regression:
where zt is a disturbance term.
? Estimate the regression,collect the residuals,,and test
whether they are stationary.
?zt
ttt zfs ??? 10 ??
? Chris Brooks 2002 陈磊 2004
7-47
Estimated Equation and Test for Cointegration
for High Frequency FTSE Data C o i n t e g r a t i n g R e g r e s s i o n
C o e f f i c i e n t
??
0
??
1
E s t i m a t e d V a l u e
0, 1 3 4 5
0, 9 8 3 4
D F T e s t o n r e s i d u a l s
t
z?
T e s t S t a t i s t i c
- 1 4, 7 3 0 3
? Chris Brooks 2002 陈磊 2004
7-48
Conclusions from Unit Root and
Cointegration Tests
? Conclusion,are stationary and therefore we have a
cointegrating relationship between log Ft and log St.
? Final stage in Engle-Granger 2-step method is to use the first
stage residuals,as the equilibrium correction term in the
general equation.
? The overall model is
?zt
?zt
ttttt vFSzS ???????? ??? 111110 lnln?ln ????
? Chris Brooks 2002 陈磊 2004
7-49
Estimated Error Correction Model for
High Frequency FTSE Data
Look at the signs and significances of the coefficients:
? is positive and highly significant
? is positive and highly significant
? is negative and highly significant
C o e f f i c i e n t E s t i m a t e d V a l u e t - r a t i o
?
?
0
9, 6 7 1 3 E - 0 6 1, 6 0 8 3
?
? - 8, 3 3 8 8 E - 0 1 - 5, 1 2 9 8
?
?
1
0, 1 7 9 9 1 9, 2 8 8 6
??
1
0, 1 3 1 2 2 0, 4 9 4 6
1??
1??
??
? Chris Brooks 2002 陈磊 2004
7-50
Forecasting High Frequency FTSE Returns
? Is it possible to use the error correction model to produce
superior forecasts to other models?
Comparison of Out of Sample Forecasting Accuracy E C M E C M - C O C A R I M A VAR
R M SE 0, 0 0 0 4 3 8 2 0, 0 0 0 4 3 5 0 0, 0 0 0 4 5 3 1 0, 0 0 0 4 5 1 0
M A E 0, 4 2 5 9 0, 4 2 5 5 0, 4 3 8 2 0, 4 3 7 8
% C o r r e c t
D i r e c t i o n
6 7, 6 9 % 6 8, 7 5 % 6 4, 3 6 % 6 6, 8 0 %
? Chris Brooks 2002 陈磊 2004
7-51
Can Profitable Trading Rules be Derived from the
ECM-COC Forecasts?
? The trading strategy involves analysing the forecast for the spot return,and
incorporating the decision dictated by the trading rules described below,It is assumed
that the original investment is ￡1000,and if the holding in the stock index is zero,the
investment earns the risk free rate.
– Liquid Trading Strategy - making a round trip trade (i.e,a purchase and sale of
the FTSE100 stocks) every ten minutes that the return is predicted to be positive
by the model.
– Buy-&-Hold while Forecast Positive Strategy - allows the trader to continue
holding the index if the return at the next predicted investment period is positive.
– Filter Strategy,Better Predicted Return Than Average - involves purchasing the
index only if the predicted returns are greater than the average positive return.
– Filter Strategy,Better Predicted Return Than First Decile - only the returns
predicted to be in the top 10% of all returns are traded on
– Filter Strategy,High Arbitrary Cut Off - An arbitrary filter of 0.0075% is
imposed,
? Chris Brooks 2002 陈磊 2004
7-52
Tr a d i ng S t r a t e gy T e r m i n a l
W e a l t h
( ￡ )
R e t ur n ( % )
{A nn ua l i s e d }
T e r m i n a l
W e a l t h ( ￡ )
w i t h sl i ppa ge
R e t ur n ( % )
{A nn ua l i s e d }
w i t h sl i ppa ge
N u m b e r
o f t r a de s
P a ss i ve
I nve st m e nt
1040,92 4,09
{49,08}
1040,92 4,09
{49,08}
1
Liq ui d T r a d i ng 1156,21 15,62
{187,44}
1056,38 5,64
{67,68}
583
B uy - & - H ol d w h i l e
F o r e c a st P o si t i ve
1156,21 15,62
{187,44}
1055,77 5,58
{66,96}
383
F i l t e r I 1144,51 14,45
{173,40}
1123,57 12,36
{148,32}
135
F i l t e r I I 1100,01 10,00
{120,00}
1046,17 4,62
{55,44}
65
F i l t e r I I I 1019,82 1,98
{23,76}
1003,23 0,32
{3,84}
8
Spot Trading Strategy Results for Error Correction
Model Incorporating the Cost of Carry
? Chris Brooks 2002 陈磊 2004
7-53
Conclusions
? The futures market,leads” the spot market because:
? the stock index is not a single entity,so
? some components of the index are infrequently traded
? it is more expensive to transact in the spot market
? stock market indices are only recalculated every minute
? Spot & futures markets do indeed have a long run relationship.
? Since it appears impossible to profit from lead/lag relationships,their
existence is entirely consistent with the absence of arbitrage
opportunities and in accordance with modern definitions of the efficient
markets hypothesis.
? Chris Brooks 2002 陈磊 2004
7-54
? To use Johansen’s method,we need to turn the VAR of the
form
yt = ?1 yt-1 + ?2 yt-2 +...+ ?k yt-k + ut
g× 1 g× g g× 1 g× g g× 1 g× g g× 1 g× 1
into a VECM,which can be written as
?yt = ? yt-k + ?1 ?yt-1 + ?2 ?yt-2 +,.,+ ?k-1 ?yt-(k-1) + ut
where ? = and
? is a long run coefficient matrix since all the ?yt-i = 0.
8 Testing for and Estimating Cointegrating
Systems Using the Johansen Technique
?? ?kj gi I1 )( ? ?? ???
i
j
gji I
1
)( ?
? Chris Brooks 2002 陈磊 2004
7-55
? Let ? denote a g?g square matrix and let c denote a g?1 non-
zero vector,and let ? denote a set of scalars.
? ? is called a characteristic root of ? if we can write
? c = ? c
g?g g?1 g?1
? We can also write
? c = ? Ig c
and hence
( ? -? Ig ) c = 0
where Ig is an identity matrix.
Review of Matrix Algebra
necessary for the Johansen Test
? Chris Brooks 2002 陈磊 2004
7-56
? Since c ? 0 by definition,then for this system to have non-zero
solution,we require the matrix ( ? -? Ig ) to be singular (i.e,to
have zero determinant).
?? -? Ig ? = 0
? For example,let ? be the 2 ? 2 matrix
? Then the characteristic equation is
?? -? Ig ?
This gives the solutions ? = 6 and ? = 3.
Review of Matrix Algebra (cont’d)
? ? ??? ???5 12 4
?
?
?
?
?
?
? ?
?
?
?
?
?
? ?
?
?
?
? ? ? ? ? ? ?
5 1
2 4
1 0
0 1
0
5 1
2 4
5 4 2 9 18
2
?
?
?
? ? ? ?( )( )
? Chris Brooks 2002 陈磊 2004
7-57
? The characteristic roots are also known as Eigenvalues.
? The rank of a matrix is equal to the number of linearly
independent rows or columns in the matrix.
? We write Rank (?) = r
? The rank of a matrix is equal to the order of the largest square
matrix we can obtain from? which has a non-zero determinant.
? For example,the determinant of ? above ? 0,therefore it has
rank 2.
? Some properties of the eigenvalues of any square matrix,
1,the sum of the eigenvalues is the trace
2,the product of the eigenvalues is the determinant
3,the number of non-zero eigenvalues is the rank
Review of Matrix Algebra (cont’d)
? Chris Brooks 2002 陈磊 2004
7-58
? Returning to Johansen’s test,the VECM representation of the
VAR was
?yt = ? yt-k + ?1 ?yt-1 + ?2 ?yt-2 +,.,+ ?k-1 ?yt-(k-1) + ut
? The test for cointegration between the y’s is calculated by
looking at the rank of the ? matrix via its eigenvalues,(To
prove this requires some technical intermediate steps).
? The rank of a matrix is equal to the number of its
characteristic roots (eigenvalues) that are different from zero.
The Johansen Test and Eigenvalues
? Chris Brooks 2002 陈磊 2004
7-59
? The eigenvalues denoted ?i are put in order:
?1??2 ?,.,??g
? If the variables are not cointegrated,the rank of ? will not be
significantly different from zero,so ?i = 0 ? i.
Then if ?i = 0,ln(1-?i) = 0
If the ?’s are roots,they must be less than 1 in absolute value
and positive.
? Say rank (?) = 1,then ln(1-?1) will be negative and ln(1-?i) = 0
? If the eigenvalue i is non-zero,then ln(1-?i) < 0 ? i > 1.
The Johansen Test and Eigenvalues
? Chris Brooks 2002 陈磊 2004
7-60
? The test statistics for cointegration are formulated as
and
where is the estimated value for the ith ordered eigenvalue
from the ? matrix.
?trace tests the null that the number of cointegrating vectors is
less than or equal to r against an unspecified alternative.
?trace = 0 when all the ?i = 0,so it is a joint test.
?max tests the null that the number of cointegrating vectors is
r against an alternative of r+1.
The Johansen Test Statistics
? ?max (,) ln( ? )r r T r? ? ? ? ?1 1 1
?
??
???
g
ri
it r a c e Tr
1
)?1l n ()( ??
i??
? Chris Brooks 2002 陈磊 2004
7-61
? Johansen & Juselius (1990) provide critical values for the 2
statistics,The distribution of the test statistics is non-standard.
The critical values depend on:
1,the value of g-r,the number of non-stationary components
2,whether a constant and / or trend are included in the
regressions.
? If the test statistic is greater than the critical value from
Johansen’s tables,reject the null hypothesis that there are r
cointegrating vectors in favour of the alternative that there are
more than r.
Johansen Critical Values
? Chris Brooks 2002 陈磊 2004
7-62
? The testing sequence under the null is r = 0,1,...,g-1
so that the hypotheses for ?trace are
H0,r = 0 vs H1,0 < r ? g
H0,r = 1 vs H1,1 < r ? g
H0,r = 2 vs H1,2 < r ? g
...,..,..
H0,r = g-1 vs H1,r = g
? We keep increasing the value of r until we no longer reject
the null.
The Johansen Testing Sequence
? Chris Brooks 2002 陈磊 2004
7-63
? But how does this correspond to a test of the rank of the ?
matrix?
? r is the rank of ?.
? ? cannot be of full rank (g) since this would correspond to
the original yt being stationary.
? If ? has zero rank,then by analogy to the univariate case,?yt
depends only on ?yt-j and not on yt-1,so that there is no long
run relationship between the elements of yt-1,Hence there is
no cointegration.
? For 1 < rank (?) < g,there are multiple cointegrating vectors.
Interpretation of Johansen Test Results
? Chris Brooks 2002 陈磊 2004
7-64
Decomposition of the ? Matrix
? For any 1 < r < g,? is defined as the product of two
matrices:
? = ? ??
g?g g?r r?g
? ? contains the cointegrating vectors while ? gives the
“loadings” of each cointegrating vector in each equation.
? For example,if g=4 and r=1,? and ? will be 4?1,and ?yt-k
will be given by:
? ?
kt
y
y
y
y
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
4
3
2
1
14131211
14
13
12
11
????
?
?
?
?
? ? ktyyyy ?
??
?
?
?
?
?
??
?
?
?
?
?
?? 414313212111
14
13
12
11
????
?
?
?
?
? Chris Brooks 2002 陈磊 2004
7-65
Hypothesis Testing Using Johansen
? EG did not allow us to do hypothesis tests on the cointegrating
relationship itself,but the Johansen approach does.
? If there exist r cointegrating vectors,only these linear
combinations will be stationary.
? You can test a hypothesis about one or more coefficients in the
cointegrating relationship by viewing the hypothesis as a
restriction on the ? matrix.
? All linear combinations of the cointegrating vectors are also
cointegrating vectors.
? If the number of cointegrating vectors is large,and the
hypothesis under consideration is simple,it may be possible to
recombine the cointegrating vectors to satisfy the restrictions
exactly.
? Chris Brooks 2002 陈磊 2004
7-66
Hypothesis Testing Using Johansen
? As the restrictions become more complex or more numerous,it
will eventually become impossible to satisfy them by
renormalisation.
? After this point,if the restriction is not severe,then the
cointegrating vectors will not change much upon imposing the
restriction.
? A test statistic to test this hypothesis is given by
??2(m)
are the characteristic roots of the restricted model
are the characteristic roots of the unrestricted model
r is the number of non-zero characteristic roots in the unrestricted
model,and m is the number of restrictions.
?i*
?i
?
?
???? r
i
iiT
1
* ) ]1l n ()1[ l n ( ??
? Chris Brooks 2002 陈磊 2004
7-68
Cointegration Tests using Johansen,
Example 1,Purchasing Power Parity
? PPP states that the equilibrium exchange rate between 2
countries is equal to the ratio of relative prices
? A necessary and sufficient condition for PPP is that the log
of the exchange rate between countries A and B,and the logs
of the price levels in countries A and B be cointegrated with
cointegrating vector [ 1 –1 1] (见 p409).
? Chen (1995) uses monthly data for April 1973-December
1990 to test the PPP hypothesis using the Johansen
approach.
? Chris Brooks 2002 陈磊 2004
7-69
Cointegration Tests of PPP with
European Data
T e st s f o r
c oi n t e g r a t i on be t w e e n
r = 0 r ? 1 r ? 2 ?
1
?
2
FRF – DEM 34.63* 17.10 6.26 1.33 - 2.50
FRF – I T L 52.69* 15.81 5.43 2.65 - 2.52
FRF – NLG 68.10* 16.37 6.42 0.58 - 0.80
F RF – BEF 52.54* 26.09* 3.63 0.78 - 1.15
DE M – I T L 42.59* 20.76* 4.79 5.80 - 2.25
DE M – NLG 50.25* 17.79 3.28 0.12 - 0.25
DE M – BEF 69.13* 27.13* 4.52 0.87 - 0.52
I T L – NLG 37.51* 14.22 5.05 0.55 - 0.71
I T L – BEF 69.24* 32.16* 7.15 0.73 - 1.28
NL G – BEF 64,52* 21.97* 3.88 1.69 - 2.17
Cr i t i c a l v a l u e s 31.52 17.95 8.18 - -
N o te s,FR F - Fr e n ch f r anc; D E M – G er m an M ar k ; N L G – D u tc h g u i ld er ; I T L – I ta lia n lir a; B E F –
B el g ia n f r anc,So u r ce, C h e n ( 1 9 9 5 ), Rep r inted w it h t h e p er m i ssio n o f T a y lo r and Fr anci s L td,
( w w w, ta n d f, co, u k ),
? Chris Brooks 2002 陈磊 2004
7-70Example 2,Are International
Bond Markets Cointegrated?
? Mills & Mills (1991)
? If financial markets are cointegrated,this implies that they
have a,common stochastic trend”.
Data:
? Daily closing observations on redemption yields偿债收益
on government bonds for 4 bond markets,US,UK,West
Germany,Japan.
? For cointegration,a necessary but not sufficient condition
is that the yields are nonstationary,All 4 yields series are
I(1),
? Chris Brooks 2002 陈磊 2004
7-71Testing for Cointegration
Between the Yields
? The Johansen procedure is used,There can be at most 3
linearly independent cointegrating vectors.
? Mills & Mills use the trace test statistic:
?
??
??? g
ri it r a c e
Tr
1
)?1l n ()( ??
Johan sen Test s f or Co i n t e gr at ion b e t w e e n In t e r n a t ion al Bon d Yie ld s
Te st st a ti st ic Criti c a l Va lue s r ( numbe r of c ointe gr a ti n g
ve c tors un de r th e null h y pothe si s) 10% 5%
0 22.06 35.6 38.6
1 10.58 21.2 23.8
2 2.52 10.3 12.0
3 0.12 2.9 4.2
So u r ce, M il ls a n d Mill s ( 1 9 9 1 ), R ep r in ted w it h t h e p er m is s i o n o f B lack w ell P u b li s h er s,
?Conclusion,No cointegrating vectors.
? Chris Brooks 2002 陈磊 2004
7-72Testing for Cointegration
Between the Yields
The paper then goes on to estimate a VAR for the first
differences of the yields,which is of the form
where
They set k = 8,
X
X US
X UK
X WG
X J A P
t
t
t
t
t
i
i i i i
i i i i
i i i i
i i i i
t
t
t
t
t
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
( )
( )
( )
( )
,,?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
11 12 13 14
21 22 23 24
31 32 33 34
41 42 43 44
1
2
3
4
?
?
?
?
?
?
?
? ?????
k
i
titit XX
1
?
? Chris Brooks 2002 陈磊 2004
7-73Variance Decompositions for VAR
of International Bond Yields
Var ia n c e De c o m p osi t io n s f or VA R of In t e r n ation al B on d Yie ld s
Ex pl a in e d b y mov e me nt s in Ex pl a in in g
mov e me nt s i n
Da y s
a he a d US UK Ge r ma n y J a pa n
US 1 95.6 2.4 1.7 0.3
5 94.2 2.8 2.3 0.7
10 92.9 3.1 2.9 1.1
20 92.8 3.2 2.9 1.1
UK 1 0.0 98.3 0.0 1.7
5 1.7 96.2 0.2 1.9
10 2.2 94.6 0.9 2.3
20 2.2 94.6 0.9 2.3
Ge r ma n y 1 0.0 3.4 9 4.6 2.0
5 6.6 6.6 84.8 3.0
10 8.3 6.5 82.9 3.6
20 8.4 6.5 82.7 3.7
J a pa n 1 0.0 0.0 1.4 100,0
5 1.3 1.4 1.1 96.2
10 1.5 2.1 1.8 94.6
20 1.6 2.2 1.9 94.2
So u r ce, M il ls a n d Mill s ( 1 9 9 1 ), R ep r in ted w it h t h e p er m is s i o n o f B lack w ell P u b li s h er s,
? Chris Brooks 2002 陈磊 2004
7-74Impulse Responses for VAR of
International Bond Yields
Im p u ls e Re sp on ses f or VA R of In t e r n ation al B on d Yie ld s
R e spo nse of US t o in nov a ti ons in
Da y s a f te r sho c k US UK Ge r ma n y J a pa n
0 0.98 0.00 0.00 0.00
1 0.06 0.01 - 0,10 0.05
2 - 0.02 0.02 - 0.14 0.07
3 0.09 - 0.04 0.09 0.08
4 - 0.02 - 0.03 0.02 0.09
10 - 0.03 - 0.01 - 0.02 - 0.01
20 0.00 0.00 - 0.10 - 0.01
R e spo nse of UK to i nno v a ti ons in
Da y s a f te r sho c k US UK Ge r ma n y J a pa n
0 0.19 0.97 0.00 0.00
1 0.16 0.07 0.01 - 0.06
2 - 0.01 - 0.01 - 0.05 0.09
3 0.06 0.04 0.06 0.05
4 0.05 - 0.01 0.02 0.07
10 0.01 0.01 - 0.04 - 0.01
20 0.00 0.00 - 0.01 0.00
R e spo nse of Ge r ma n y to in nov a ti ons in
Da y s a f te r sho c k US UK Ge r ma n y J a pa n
0 0.07 0.06 0.95 0.00
1 0.13 0.05 0.11 0.02
2 0.04 0.03 0.00 0.00
3 0.02 0.00 0.00 0.01
4 0.01 0.00 0.00 0.09
10 0.01 0.01 - 0.01 0.02
20 0.00 0.00 0.00 0.00
R e spo nse of J a pa n to inn ovatio ns i n
Da y s a f te r sho c k US UK Ge r ma n y J a pa n
0 0.03 0.05 0.12 0.97
1 0.06 0.02 0.07 0.04
2 0.02 0.02 0.00 0.21
3 0.01 0.02 0.06 0.07
4 0.02 0.03 0.07 0.06
10 0.01 0.01 0.01 0.04
20 0.00 0.00 0.00 0.01
So u r ce, M ill s a n d M ill s ( 1 9 9 1 ), Rep r in ted w it h t h e p er m is s i o n o f B lac k w ell P u b li s h er s,
7-1
Chapter 7
Modelling long-run relationship
in finance
? Chris Brooks 2002 陈磊 2004
7-2
1 Stationarity and Unit Root Testing
? The stationarity or otherwise of a series can strongly
influence its behaviour and properties - e.g,persistence of
shocks will be infinite for nonstationary series
? Spurious regressions,If two variables are trending over time,
a regression of one on the other could have a high R2 even if
the two are totally unrelated
? If the variables in the regression model are not stationary,
then it can be proved that the standard assumptions for
asymptotic analysis will not be valid,In other words,the
usual,t-ratios” will not follow a t-distribution,so we cannot
validly undertake hypothesis tests about the regression
parameters.
1.1Why do we need to test for Non-Stationarity?
? Chris Brooks 2002 陈磊 2004
7-3Value of R2 for 1000 Sets of Regressions of a Non-
stationary Variable on another Independent Non-
stationary Variable
? Chris Brooks 2002 陈磊 2004
7-4Value of t-ratio on Slope Coefficient for 1000 Sets
of Regressions of a Non-stationary Variable on
another Independent Non-stationary Variable
? Chris Brooks 2002 陈磊 2004
7-5
1.2 Two types of Non-Stationarity
? Various definitions of non-stationarity exist
? In this chapter,we are really referring to the weak form or
covariance stationarity
? There are two models which have been frequently used to
characterise non-stationarity,the random walk model with
drift:
yt = ? + yt-1 + ut (1)
and the deterministic trend process:
yt = ? + ?t + ut (2)
where ut is iid in both cases.
? Chris Brooks 2002 陈磊 2004
7-6
Stochastic Non-Stationarity
? Note that the model (1) could be generalised to the case
where yt is an explosive process:
yt = ? + ? yt-1 + ut
where ? > 1.
? Typically,the explosive case is ignored and we use ? = 1 to
characterise the non-stationarity because
– ? > 1 does not describe many data series in economics
and finance.
– ? > 1 has an intuitively unappealing property,shocks to
the system are not only persistent through time,they are
propagated so that a given shock will have an
increasingly large influence.
? Chris Brooks 2002 陈磊 2004
7-7
Stochastic Non-stationarity,
The Impact of Shocks
? To see this,consider the general case of an AR(1) with no drift:
yt = ? yt-1 + ut (3)
Let ? take any value for now.
? We can write,yt-1 = ? yt-2 + ut-1
yt-2 = ? yt-3 + ut-2
? Substituting into (3) yields,yt = ? (? yt-2 + ut-1) + ut
= ?2yt-2 + ? ut-1 + ut
? Substituting again for yt-2,yt = ?2(? yt-3 + ut-2) + ? ut-1 + ut
= ?3 yt-3 + ?2ut-2 + ? ut-1 + ut
? Successive substitutions of this type lead to:*
yt = ?T y0 + ?ut-1 + ?2ut-2 + ?3ut-3 +,..+ ?T-1u1 + ut
? Chris Brooks 2002 陈磊 2004
7-8The Impact of Shocks for
Stationary and Non-stationary Series
? We have 3 cases:
1,|? | <1 ??T?0 as T??
So the shocks to the system gradually die away.
2,? =1 ??T =1? T
So shocks persist in the system and never die away,We
obtain:
as T??
So just an infinite sum of past shocks plus some starting
value of y0.
3,? >1,Now given shocks become more influential as time goes
on,since if ? >1,? 3 > ? 2 >? etc.
??
?
??
0
0
i
tt uyy
? Chris Brooks 2002 陈磊 2004
7-9
Detrending a Non-stationary Series
? Going back to our 2 characterisations of non-stationarity,the
r.w,with drift:
? yt = ? + yt-1 + ut (1)
and the trend-stationary process
yt = ? + ? t + ut (2)
? The two will require different treatments to induce stationarity.
The second case is known as deterministic non-stationarity and
what is required is detrending去势,
? The first case is known as stochastic non-stationarity,If we let
?yt = yt - yt-1 = yt - L yt =(1-L) yt
then ?yt =(1-L) yt = ? + ut
we have induced stationarity by,differencingonce”.
单位根过程
? Chris Brooks 2002 陈磊 2004
7-10Detrending a Series:
Using the Right Method
? Although trend-stationary and difference-stationary series
are both,trending” over time,the correct approach needs
to be used in each case.
? If we first difference the trend-stationary series,it would
“remove” the non-stationarity,but at the expense on
introducing an MA(1) structure into the errors,p372
? Conversely if we try to detrend a series which has
stochastic trend,then we will not remove the non-
stationarity.
? We will now concentrate on the stochastic non-stationarity
model since deterministic non-stationarity does not
adequately describe most series in economics or finance.
? Chris Brooks 2002 陈磊 2004
7-11
Sample Plots for various Stochastic
Processes,A White Noise Process
-4
-3
-2
-1
0
1
2
3
4
1 40 79 118 157 196 235 274 313 352 391 430 469
? Chris Brooks 2002 陈磊 2004
7-12
Random Walk and
a Random Walk with Drift
-2 0
-1 0
0
10
20
30
40
50
60
70
1 19 37 55 73 91 109 127 145 163 181 199 217 235 253 271 289 307 325 343 361 379 397 415 433 451 469 487
R a n d o m W a l k
R a n d o m W a l k w i t h D ri f t
? Chris Brooks 2002 陈磊 2004
7-13
A Deterministic Trend Process
-5
0
5
10
15
20
25
30
1 40 79 118 157 196 235 274 313 352 391 430 469
? Chris Brooks 2002 陈磊 2004
7-14Autoregressive Processes with
differing values of ? (0,0.8,1)
-2 0
-1 5
-1 0
-5
0
5
10
15
1 53 105 157 209 261 313 365 417 469 521 573 625 677 729 781 833 885 937 989
Ph i = 1
Ph i = 0, 8
Ph i = 0
? Chris Brooks 2002 陈磊 2004
7-15
1.3 Definition of Non-Stationarity
? Consider again the simplest stochastic trend model:
yt = yt-1 + ut or ?yt= ut
We can generalise this concept to consider the case where the
series contains more than one,unit root”,That is,we would
need to apply the first difference operator,?,more than once
to induce stationarity.
Definition
If a non-stationary series,yt must be differenced d times
before it becomes stationary,then it is said to be integrated of
order d,We write yt ?I(d).
So if yt ? I(d) then ?dyt? I(0).
An I(0) series is a stationary series
An I(1) series contains one unit root,e.g,yt = yt-1 + ut
? Chris Brooks 2002 陈磊 2004
7-16
Characteristics of I(0),I(1) and I(2) Series
? An I(2) series contains two unit roots and so would require
differencing twice to induce stationarity,p376
? I(1) and I(2) series can wander a long way from their mean
value and cross this mean value rarely.
? I(0) series should cross the mean frequently,
? The majority of economic and financial series contain a single
unit root,although some are stationary and consumer prices
have been argued to have 2 unit roots.
? Chris Brooks 2002 陈磊 2004
7-17
1.4 Test for a unit root
? 无法使用 acf或 pacf确定序列是否是单位根过程。
? The early and pioneering work on testing for a unit root in
time series was done by Dickey and Fuller (Dickey and
Fuller 1979,Fuller 1976),The basic objective of the test is to
test the null hypothesis that ? =1 in:
yt = ? yt-1 + ut
against the one-sided alternative ? <1,So we have
H0,series contains a unit root
vs,H1,series is stationary,
? We usually use the regression:
?yt = ? yt-1 + ut
a test of ?=1 is equivalent to a test of ?=0 (since ? -1=?).
? Chris Brooks 2002 陈磊 2004
7-18
Different forms for the DF Test
? Dickey Fuller tests are also known as ? tests,?,??,??.
? The null (H0) and alternative (H1) models in each case are
i) H0,yt = yt-1+ut H1,yt = ? yt-1+ut,? <1
This is a test for a random walk against a stationary
autoregressive process of order one (AR(1))
ii) H0,yt = yt-1+ut H1,yt = ? yt-1+?+ut,? <1
This is a test for a random walk against a stationary AR(1) with
drift.
iii) H0,yt = yt-1+ut H1,yt = ? yt-1+?+? t+ut,? <1
This is a test for a random walk against a stationary AR(1) with
drift and a time trend.
? Chris Brooks 2002 陈磊 2004
7-19
Computing the DF Test Statistic
? We can write ?yt=ut
where ? yt = yt- yt-1,and the alternatives may be expressed as
?yt = ? yt-1+?+?t +ut
with ?=?=0 in case i),and ?=0 in case ii) and ?=? -1,In each case,
the tests are based on the t-ratio on the yt-1 term in the estimated
regression of ?yt on yt-1,plus a constant in case ii) and a constant
and trend in case iii),The test statistics are defined as
test statistic =
? The test statistic does not follow the usual t-distribution under the
null,since the null is one of non-stationarity,but rather follows a
non-standard distribution,Critical values are derived from Monte
Carlo experiments in,for example,Fuller (1976),Relevant
examples of the distribution are shown in table 7.1 below
?
?
?
?
?
SE( )
? Chris Brooks 2002 陈磊 2004
7-20
Critical Values for the DF Test
The null hypothesis of a unit root is rejected in favour of the
stationary alternative in each case if the test statistic is more
negative than the critical value.
临界值还与样本容量有关, 参见 675页 表 A2.7。
S ig n if ic a n c e l e v e l 10% 5% 1%
C.V, f o r c o n sta n t
b u t n o t re n d
-2,5 7 -2,8 6 -3,4 3
C.V, f o r c o n sta n t
a n d t re n d
-3,1 2 -3,4 1 -3,9 6
T a b le 4,1, Cri ti c a l V a l u e s f o r DF a n d A D F T e sts (F u ll e r,
1 9 7 6,p 3 7 3 ).
? Chris Brooks 2002 陈磊 2004
7-21
The Augmented Dickey Fuller Test
(ADF)
? The tests above are only valid if ut is white noise,In particular,
ut will be autocorrelated if there was autocorrelation in the
dependent variable of the regression (?yt) which we have not
modelled,The solution is to,augment” the test using p lags of
the dependent variable,The alternative model in case (i) is
now written:
? The same critical values from the DF tables are used as before.
A problem now arises in determining the optimal number of
lags of the dependent variable.
There are 2 ways
- use the frequency of the data to decide
- use information criteria
?
?
?? ?????
p
i
tititt uyyy
1
1 ??
? Chris Brooks 2002 陈磊 2004
7-22
1.5 Testing for Higher Orders of
Integration
? Consider the simple regression,?yt = ? yt-1 + ut
We test H0,? = 0 vs,H1,? < 0.
? If H0 is rejected we simply conclude that yt does not contain a
unit root.
? But what do we conclude if H0 is not rejected? The series
contains a unit root,but is that it? No! What if yt ?I(2)? We
would still not have rejected,So we now need to test
H0,yt ?I(2) vs,H1,yt?I(1)
We now regress ?2yt on ?yt-1 (plus lags of ?2yt if necessary).
? Now we test H0,?yt?I(1) which is equivalent to H0,yt?I(2).
? So in this case,if we do not reject,then yt is at least I(2).
? continue to test for a further unit root until we rejected H0.
? Dickey and Pantula(1987)认为, 检验顺序应从高阶至低阶 。
? Chris Brooks 2002 陈磊 2004
7-23
1.6 The Phillips-Perron Test
? Phillips and Perron have developed a more comprehensive
theory of unit root nonstationarity,The tests are similar to
ADF tests,but they incorporate an automatic correction to
the DF procedure to allow for autocorrelated residuals.
? The tests usually give the same conclusions as the ADF tests,
and the calculation of the test statistics is complex.
? Chris Brooks 2002 陈磊 2004
7-241.7 Criticism of Dickey-Fuller and
Phillips-Perron-type tests
? Main criticism is that the power of the tests is low if the process is
stationary but with a root close to the non-stationary boundary.
e.g,the tests are poor at deciding if ? =1 or ? =0.95,especially
with small sample sizes.
? If the true data generating process (dgp) is
yt = 0.95yt-1 + ut
then the null hypothesis of a unit root should be rejected.
? One way to get around this is to use a stationarity test as well as
the unit root tests we have looked at.
? Chris Brooks 2002 陈磊 2004
7-25
Stationarity tests
? Stationarity tests have
H0,yt is stationary
versus H1,yt is non-stationary
So that by default under the null the data will appear stationary,
? One such stationarity test is the KPSS test (Kwaitowski,Phillips,
Schmidt and Shin,1992).
? Thus we can compare the results of these tests with the ADF/PP
procedure to see if we obtain the same conclusion.
? Chris Brooks 2002 陈磊 2004
7-26
Stationarity tests (cont’d)
? A Comparison
ADF / PP KPSS
H0,yt ? I(1) H0,yt ? I(0)
H1,yt ? I(0) H1,yt ? I(1)
? There are four possible outcomes
1.Reject H0 and Do not reject H0
2.Do not reject H0 and Reject H0
3.Reject H0 and Reject H0
4.Do not reject H0 and Do not reject H0
? Chris Brooks 2002 陈磊 2004
7-27
有关单位根和协整的一些参考书
? 1,董文泉 等,,经济周期波动的分析与预测方法, 第十二章，
吉林大学出版社,1998年。
? 2,李子奈、叶阿忠,,高等计量经济学, 第二章第五节，清
华大学出版社,2000年。
? 3,张晓峒,,计量经济分析,,经济科学出版社,2000年。
? 4,赵国庆 主编,,计量经济学, 第八章，中国人民大学出版
社,2001年。
? 5,王维国 主编,,计量经济学, 第十章，东北财经大学出版
社,2002年。
? 6,Philip H,Franses,,商业和经济预测中的时间序列模型,
第四章，中国人民大学出版社,2002年。
? Chris Brooks 2002 陈磊 2004
7-28
2 Testing for unit roots in
EViews
? p383-385
? Chris Brooks 2002 陈磊 2004
7-29
3 Cointegration,An Introduction
? In most cases,if we combine two variables which are I(1),
then the combination will also be I(1).
? More generally,if we combine variables with differing
orders of integration,the combination will have an order of
integration equal to the largest,i.e.,
if Xi,t ? I(di) for i = 1,2,3,...,k
so we have k variables each integrated of order di.
Let (1)
Then zt ? I(max di)
z Xt i i t
i
k?
??
?,
1
? Chris Brooks 2002 陈磊 2004
7-30
Linear Combinations of Non-
stationary Variables
? Rearranging (1),we can write
where
? This is just a regression equation,
? But the disturbances would have some very undesirable
properties,zt′ is not stationary and is autocorrelated if all of
the Xi are I(1).
? We want to ensure that the disturbances are I(0),Under
what circumstances will this be the case?
? ?? ?i i t tz z i k? ? ? ?1 1 2,',,.,,,
X X zt i i t t
i
k
1
2
,,'? ?
?
? ?
? Chris Brooks 2002 陈磊 2004
7-31
Definition of Cointegration
(Engle & Granger,1987)
? Let zt be a k?1 vector of variables,then the components of zt are
cointegrated of order (d,b) if
i) All components of zt are I(d)
ii) There is at least one vector of coefficients? such that
??zt ? I(d-b) 0<b≤d
? If variables are cointegrated,it means that a linear combination
of them will be stationary.
? There may be up to r linearly independent cointegrating
relationships (where r ? k-1),also known as cointegrating
vectors,r is also known as the cointegrating rank of zt.
? A cointegrating relationship may also be seen as a long term
relationship,Many time series are non-stationary but,move
together” over time.
? Chris Brooks 2002 陈磊 2004
7-32
Cointegration and Equilibrium
? Examples of possible Cointegrating Relationships in finance:
– spot and futures prices
– ratio of relative prices and an exchange rate
– equity prices and dividends
? Market forces arising from no arbitrage conditions should
ensure an equilibrium relationship.
? No cointegration implies that series could wander apart
without bound in the long run.
? Chris Brooks 2002 陈磊 2004
7-33
4 Equilibrium Correction or Error
Correction Models
? When the concept of non-stationarity was first considered,a
usual response was to independently take the first differences
of a series of I(1) variables.
? The problem with this approach is that pure first difference
models have no long run solution.
e.g,Consider yt and xt both I(1).
The model we may want to estimate is
? yt = ??xt + ut
But this collapses to nothing in the long run.
The definition of the long run that we use is where
yt = yt-1 = y; xt = xt-1 = x.
Hence all the difference terms will be zero,i.e,? yt = 0; ?xt = 0.
? Chris Brooks 2002 陈磊 2004
7-34
Specifying an ECM
? One way to get around this problem is to use both first
difference and lagged level terms,e.g.
? yt = ?1?xt + ?2(yt-1-?xt-1) + ut (2)
? yt-1-? xt-1 is known as the error correction term.
? Providing that yt and xt are cointegrated with cointegrating
coefficient ?,then (yt-1-?xt-1) will be I(0) even though the
constituents are I(1),
? We can thus validly use OLS on (2).*
? The Granger representation theorem shows that any
cointegrating relationship can be expressed as an equilibrium
correction model.
? Chris Brooks 2002 陈磊 2004
7-35
5 Testing for Cointegration in Regression
? The model for the equilibrium correction term can be
generalised to include more than two variables:
yt = ?1 + ?2x2t + ?3x3t + … + ?kxkt + ut (3)
ut should be I(0) if the variables yt,x2t,..,xkt are cointegrated.
? So what we want to test is the residuals of equation (3) to see
if they are non-stationary or stationary,We can use the DF /
ADF test on ut.
So we have the regression with vt ? iid.
? However,since this is a test on the residuals of an actual
model,,then the critical values are changed.因为参数估计
的偏差会改变针对残差的检验统计量的分布 。
? ? ?u u vt t t? ??? 1
?ut
? Chris Brooks 2002 陈磊 2004
7-36Testing for Cointegration in
Regression,Conclusions
? Engle and Granger (1987) have tabulated a new set of critical
values （ 见 A2.8） and hence the test is known as the Engle
Granger (E-G) test.
? We can also use the Durbin Watson test statistic（ CRDW）
or the Phillips Perron approach to test for non-stationarity
of,
? What are the null and alternative hypotheses for a test on the
residuals of a potentially cointegrating regression?
H0, unit root in cointegrating regression’s residuals
H1, residuals from cointegrating regression are stationary
如果拒绝 H0,应建立 ECM;
否则, 只能对差分后变量建立回归模型 。
?ut
? Chris Brooks 2002 陈磊 2004
7-37
6 Methods of modelling Cointegrated Systems:
? There are (at least) 3 methods we could use,Engle Granger,
Engle and Yoo,and Johansen.
? The Engle - Granger 2 Step Method
Step 1:
- Make sure that all the individual variables are I(1).
- Then estimate the cointegrating regression using OLS.
- Save the residuals of the cointegrating regression,.
- Test these residuals to ensure that they are I(0).
Step 2:
- Use the step 1 residuals as one variable in the error correction
model e.g.
? yt = ?1?xt + ?2( ) + ut where = yt-1- xt-11??tu
1??tu
?ut
??
? Chris Brooks 2002 陈磊 2004
7-38The Engle-Granger Approach,
Some Drawbacks
1,Unit root and cointegration tests have low power in finite
samples
2,We are forced to treat the variables asymmetrically and to
specify one as the dependent and the other as independent
variables.即使可能不存在理论根据。
3,Cannot perform any hypothesis tests about the actual
cointegrating relationship estimated at stage 1.
- Problem 1 is a small sample problem that should disappear
asymptotically.
- Problem 2 is addressed by the Johansen approach.
- Problem 3 is addressed by the Engle and Yoo approach or the
Johansen approach.
? Chris Brooks 2002 陈磊 2004
7-39
? One of the problems with the EG 2-step method is that we
cannot make any inferences about the actual cointegrating
regression.
? The Engle & Yoo (EY,1987) 3-step procedure takes its first two
steps from EG.
? EY add a third step giving updated estimates of the
cointegrating vector and its standard errors.
? 由于存在针对协整关系及其假设检验的更好估计方法, E-Y方
法在实证研究中很少使用 。
The Engle & Yoo 3-Step Method
? Chris Brooks 2002 陈磊 2004
7-40
? In the general case,where we have more than two variables
which may be cointegrated,there could be more than one
cointegrating relationship.
? In fact there can be up to r linearly independent cointegrating
vectors (where r ? g-1),where g is the number of variables in
total.
? So,in the case where we just had y and x,then r can only be one
or zero.
? And if there are others,how do we know how many there are or
whether we have found the,best”?
? The answer to this is to use a systems approach to cointegration
which will allow determination of all r cointegrating
relationships - Johansen’s method,
多个协整关系
? Chris Brooks 2002 陈磊 2004
7-41
多个协整关系
? 两个非平稳变量之间最多只能有一个线性组合是平稳的,即
最多只存在一个协整关系。
? 一个包括 K个变量的非平稳系统，可以存在最多 K-1个线性
独立的协整关系。
? 前面叙述的回归方法只能发现一个协整关系。
? 如果存在多个协整关系，如何知道是否存在其他的协整关
系？是否已经找到最“好”或最强的协整关系？
? 这需要检验协整关系的系统方法。
? Chris Brooks 2002 陈磊 2004
7-42
7 An Example, Lead-Lag Relationships
between Spot and Futures Prices
Background
? We expect changes in the spot price of a financial asset and its
corresponding futures price to be perfectly contemporaneously
correlated and not to be cross-autocorrelated.
i.e,expect Corr(?ln(Ft),?ln(St))? 1
Corr(?ln(Ft),?ln(St-k))? 0 ? k
Corr(?ln(Ft-j),?ln(St))? 0 ? j
? We can test this idea by modelling the lead-lag relationship
between the two.
? We will consider two papers Tse(1995) and Brooks et al (2001).
? Chris Brooks 2002 陈磊 2004
7-43
Futures & Spot Data
? Tse (1995),1055 daily observations on NSA 东京证券交易所
stock index and stock index futures values from December
1988 - April 1993.
? Brooks et al (2001),13,035 10-minutely observations on the
FTSE 100 stock index and stock index futures prices for all
trading days in the period June 1996 – 1997.
? Chris Brooks 2002 陈磊 2004
7-44
Methodology
? The fair futures price is given by
where Ft* is the fair futures price,St is the spot price,r is a
continuously compounded risk-free rate of interest,d is the
continuously compounded yield in terms of dividends derived
from the stock index until the futures contract matures,and (T-
t) is the time to maturity of the futures contract,Taking
logarithms of both sides of equation above gives
? First,test ft and st for nonstationarity.
t* t (r - d ) ( T - t)F = S e
t)-d ) ( T-(r s f tt ??*
? Chris Brooks 2002 陈磊 2004
7-45
Dickey-Fuller Tests on Log-Prices and
Returns for High Frequency FTSE Data
F u t u r e s Sp o t
D i c k e y - F u ll e r St a t i s t i c s
f o r L o g - P r i c e D a t a
- 0, 1 3 2 9 - 0, 7 3 3 5
D i c k e y F u l l e r St a t i s t i c s
f o r R e t u r n s D a t a
- 8 4, 9 9 6 8 - 1 1 4, 1 8 0 3
? Chris Brooks 2002 陈磊 2004
7-46
Cointegration Test Regression and Test
on Residuals
? Conclusion,log Ft and log St are not stationary,but ?log Ft
and ?log St are stationary.
? But a model containing only first differences has no long run
relationship.
? Solution is to see if there exists a cointegrating relationship
between ft and st which would mean that we can validly
include levels terms in this framework.
? Potential cointegrating regression:
where zt is a disturbance term.
? Estimate the regression,collect the residuals,,and test
whether they are stationary.
?zt
ttt zfs ??? 10 ??
? Chris Brooks 2002 陈磊 2004
7-47
Estimated Equation and Test for Cointegration
for High Frequency FTSE Data C o i n t e g r a t i n g R e g r e s s i o n
C o e f f i c i e n t
??
0
??
1
E s t i m a t e d V a l u e
0, 1 3 4 5
0, 9 8 3 4
D F T e s t o n r e s i d u a l s
t
z?
T e s t S t a t i s t i c
- 1 4, 7 3 0 3
? Chris Brooks 2002 陈磊 2004
7-48
Conclusions from Unit Root and
Cointegration Tests
? Conclusion,are stationary and therefore we have a
cointegrating relationship between log Ft and log St.
? Final stage in Engle-Granger 2-step method is to use the first
stage residuals,as the equilibrium correction term in the
general equation.
? The overall model is
?zt
?zt
ttttt vFSzS ???????? ??? 111110 lnln?ln ????
? Chris Brooks 2002 陈磊 2004
7-49
Estimated Error Correction Model for
High Frequency FTSE Data
Look at the signs and significances of the coefficients:
? is positive and highly significant
? is positive and highly significant
? is negative and highly significant
C o e f f i c i e n t E s t i m a t e d V a l u e t - r a t i o
?
?
0
9, 6 7 1 3 E - 0 6 1, 6 0 8 3
?
? - 8, 3 3 8 8 E - 0 1 - 5, 1 2 9 8
?
?
1
0, 1 7 9 9 1 9, 2 8 8 6
??
1
0, 1 3 1 2 2 0, 4 9 4 6
1??
1??
??
? Chris Brooks 2002 陈磊 2004
7-50
Forecasting High Frequency FTSE Returns
? Is it possible to use the error correction model to produce
superior forecasts to other models?
Comparison of Out of Sample Forecasting Accuracy E C M E C M - C O C A R I M A VAR
R M SE 0, 0 0 0 4 3 8 2 0, 0 0 0 4 3 5 0 0, 0 0 0 4 5 3 1 0, 0 0 0 4 5 1 0
M A E 0, 4 2 5 9 0, 4 2 5 5 0, 4 3 8 2 0, 4 3 7 8
% C o r r e c t
D i r e c t i o n
6 7, 6 9 % 6 8, 7 5 % 6 4, 3 6 % 6 6, 8 0 %
? Chris Brooks 2002 陈磊 2004
7-51
Can Profitable Trading Rules be Derived from the
ECM-COC Forecasts?
? The trading strategy involves analysing the forecast for the spot return,and
incorporating the decision dictated by the trading rules described below,It is assumed
that the original investment is ￡1000,and if the holding in the stock index is zero,the
investment earns the risk free rate.
– Liquid Trading Strategy - making a round trip trade (i.e,a purchase and sale of
the FTSE100 stocks) every ten minutes that the return is predicted to be positive
by the model.
– Buy-&-Hold while Forecast Positive Strategy - allows the trader to continue
holding the index if the return at the next predicted investment period is positive.
– Filter Strategy,Better Predicted Return Than Average - involves purchasing the
index only if the predicted returns are greater than the average positive return.
– Filter Strategy,Better Predicted Return Than First Decile - only the returns
predicted to be in the top 10% of all returns are traded on
– Filter Strategy,High Arbitrary Cut Off - An arbitrary filter of 0.0075% is
imposed,
? Chris Brooks 2002 陈磊 2004
7-52
Tr a d i ng S t r a t e gy T e r m i n a l
W e a l t h
( ￡ )
R e t ur n ( % )
{A nn ua l i s e d }
T e r m i n a l
W e a l t h ( ￡ )
w i t h sl i ppa ge
R e t ur n ( % )
{A nn ua l i s e d }
w i t h sl i ppa ge
N u m b e r
o f t r a de s
P a ss i ve
I nve st m e nt
1040,92 4,09
{49,08}
1040,92 4,09
{49,08}
1
Liq ui d T r a d i ng 1156,21 15,62
{187,44}
1056,38 5,64
{67,68}
583
B uy - & - H ol d w h i l e
F o r e c a st P o si t i ve
1156,21 15,62
{187,44}
1055,77 5,58
{66,96}
383
F i l t e r I 1144,51 14,45
{173,40}
1123,57 12,36
{148,32}
135
F i l t e r I I 1100,01 10,00
{120,00}
1046,17 4,62
{55,44}
65
F i l t e r I I I 1019,82 1,98
{23,76}
1003,23 0,32
{3,84}
8
Spot Trading Strategy Results for Error Correction
Model Incorporating the Cost of Carry
? Chris Brooks 2002 陈磊 2004
7-53
Conclusions
? The futures market,leads” the spot market because:
? the stock index is not a single entity,so
? some components of the index are infrequently traded
? it is more expensive to transact in the spot market
? stock market indices are only recalculated every minute
? Spot & futures markets do indeed have a long run relationship.
? Since it appears impossible to profit from lead/lag relationships,their
existence is entirely consistent with the absence of arbitrage
opportunities and in accordance with modern definitions of the efficient
markets hypothesis.
? Chris Brooks 2002 陈磊 2004
7-54
? To use Johansen’s method,we need to turn the VAR of the
form
yt = ?1 yt-1 + ?2 yt-2 +...+ ?k yt-k + ut
g× 1 g× g g× 1 g× g g× 1 g× g g× 1 g× 1
into a VECM,which can be written as
?yt = ? yt-k + ?1 ?yt-1 + ?2 ?yt-2 +,.,+ ?k-1 ?yt-(k-1) + ut
where ? = and
? is a long run coefficient matrix since all the ?yt-i = 0.
8 Testing for and Estimating Cointegrating
Systems Using the Johansen Technique
?? ?kj gi I1 )( ? ?? ???
i
j
gji I
1
)( ?
? Chris Brooks 2002 陈磊 2004
7-55
? Let ? denote a g?g square matrix and let c denote a g?1 non-
zero vector,and let ? denote a set of scalars.
? ? is called a characteristic root of ? if we can write
? c = ? c
g?g g?1 g?1
? We can also write
? c = ? Ig c
and hence
( ? -? Ig ) c = 0
where Ig is an identity matrix.
Review of Matrix Algebra
necessary for the Johansen Test
? Chris Brooks 2002 陈磊 2004
7-56
? Since c ? 0 by definition,then for this system to have non-zero
solution,we require the matrix ( ? -? Ig ) to be singular (i.e,to
have zero determinant).
?? -? Ig ? = 0
? For example,let ? be the 2 ? 2 matrix
? Then the characteristic equation is
?? -? Ig ?
This gives the solutions ? = 6 and ? = 3.
Review of Matrix Algebra (cont’d)
? ? ??? ???5 12 4
?
?
?
?
?
?
? ?
?
?
?
?
?
? ?
?
?
?
? ? ? ? ? ? ?
5 1
2 4
1 0
0 1
0
5 1
2 4
5 4 2 9 18
2
?
?
?
? ? ? ?( )( )
? Chris Brooks 2002 陈磊 2004
7-57
? The characteristic roots are also known as Eigenvalues.
? The rank of a matrix is equal to the number of linearly
independent rows or columns in the matrix.
? We write Rank (?) = r
? The rank of a matrix is equal to the order of the largest square
matrix we can obtain from? which has a non-zero determinant.
? For example,the determinant of ? above ? 0,therefore it has
rank 2.
? Some properties of the eigenvalues of any square matrix,
1,the sum of the eigenvalues is the trace
2,the product of the eigenvalues is the determinant
3,the number of non-zero eigenvalues is the rank
Review of Matrix Algebra (cont’d)
? Chris Brooks 2002 陈磊 2004
7-58
? Returning to Johansen’s test,the VECM representation of the
VAR was
?yt = ? yt-k + ?1 ?yt-1 + ?2 ?yt-2 +,.,+ ?k-1 ?yt-(k-1) + ut
? The test for cointegration between the y’s is calculated by
looking at the rank of the ? matrix via its eigenvalues,(To
prove this requires some technical intermediate steps).
? The rank of a matrix is equal to the number of its
characteristic roots (eigenvalues) that are different from zero.
The Johansen Test and Eigenvalues
? Chris Brooks 2002 陈磊 2004
7-59
? The eigenvalues denoted ?i are put in order:
?1??2 ?,.,??g
? If the variables are not cointegrated,the rank of ? will not be
significantly different from zero,so ?i = 0 ? i.
Then if ?i = 0,ln(1-?i) = 0
If the ?’s are roots,they must be less than 1 in absolute value
and positive.
? Say rank (?) = 1,then ln(1-?1) will be negative and ln(1-?i) = 0
? If the eigenvalue i is non-zero,then ln(1-?i) < 0 ? i > 1.
The Johansen Test and Eigenvalues
? Chris Brooks 2002 陈磊 2004
7-60
? The test statistics for cointegration are formulated as
and
where is the estimated value for the ith ordered eigenvalue
from the ? matrix.
?trace tests the null that the number of cointegrating vectors is
less than or equal to r against an unspecified alternative.
?trace = 0 when all the ?i = 0,so it is a joint test.
?max tests the null that the number of cointegrating vectors is
r against an alternative of r+1.
The Johansen Test Statistics
? ?max (,) ln( ? )r r T r? ? ? ? ?1 1 1
?
??
???
g
ri
it r a c e Tr
1
)?1l n ()( ??
i??
? Chris Brooks 2002 陈磊 2004
7-61
? Johansen & Juselius (1990) provide critical values for the 2
statistics,The distribution of the test statistics is non-standard.
The critical values depend on:
1,the value of g-r,the number of non-stationary components
2,whether a constant and / or trend are included in the
regressions.
? If the test statistic is greater than the critical value from
Johansen’s tables,reject the null hypothesis that there are r
cointegrating vectors in favour of the alternative that there are
more than r.
Johansen Critical Values
? Chris Brooks 2002 陈磊 2004
7-62
? The testing sequence under the null is r = 0,1,...,g-1
so that the hypotheses for ?trace are
H0,r = 0 vs H1,0 < r ? g
H0,r = 1 vs H1,1 < r ? g
H0,r = 2 vs H1,2 < r ? g
...,..,..
H0,r = g-1 vs H1,r = g
? We keep increasing the value of r until we no longer reject
the null.
The Johansen Testing Sequence
? Chris Brooks 2002 陈磊 2004
7-63
? But how does this correspond to a test of the rank of the ?
matrix?
? r is the rank of ?.
? ? cannot be of full rank (g) since this would correspond to
the original yt being stationary.
? If ? has zero rank,then by analogy to the univariate case,?yt
depends only on ?yt-j and not on yt-1,so that there is no long
run relationship between the elements of yt-1,Hence there is
no cointegration.
? For 1 < rank (?) < g,there are multiple cointegrating vectors.
Interpretation of Johansen Test Results
? Chris Brooks 2002 陈磊 2004
7-64
Decomposition of the ? Matrix
? For any 1 < r < g,? is defined as the product of two
matrices:
? = ? ??
g?g g?r r?g
? ? contains the cointegrating vectors while ? gives the
“loadings” of each cointegrating vector in each equation.
? For example,if g=4 and r=1,? and ? will be 4?1,and ?yt-k
will be given by:
? ?
kt
y
y
y
y
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
4
3
2
1
14131211
14
13
12
11
????
?
?
?
?
? ? ktyyyy ?
??
?
?
?
?
?
??
?
?
?
?
?
?? 414313212111
14
13
12
11
????
?
?
?
?
? Chris Brooks 2002 陈磊 2004
7-65
Hypothesis Testing Using Johansen
? EG did not allow us to do hypothesis tests on the cointegrating
relationship itself,but the Johansen approach does.
? If there exist r cointegrating vectors,only these linear
combinations will be stationary.
? You can test a hypothesis about one or more coefficients in the
cointegrating relationship by viewing the hypothesis as a
restriction on the ? matrix.
? All linear combinations of the cointegrating vectors are also
cointegrating vectors.
? If the number of cointegrating vectors is large,and the
hypothesis under consideration is simple,it may be possible to
recombine the cointegrating vectors to satisfy the restrictions
exactly.
? Chris Brooks 2002 陈磊 2004
7-66
Hypothesis Testing Using Johansen
? As the restrictions become more complex or more numerous,it
will eventually become impossible to satisfy them by
renormalisation.
? After this point,if the restriction is not severe,then the
cointegrating vectors will not change much upon imposing the
restriction.
? A test statistic to test this hypothesis is given by
??2(m)
are the characteristic roots of the restricted model
are the characteristic roots of the unrestricted model
r is the number of non-zero characteristic roots in the unrestricted
model,and m is the number of restrictions.
?i*
?i
?
?
???? r
i
iiT
1
* ) ]1l n ()1[ l n ( ??
? Chris Brooks 2002 陈磊 2004
7-68
Cointegration Tests using Johansen,
Example 1,Purchasing Power Parity
? PPP states that the equilibrium exchange rate between 2
countries is equal to the ratio of relative prices
? A necessary and sufficient condition for PPP is that the log
of the exchange rate between countries A and B,and the logs
of the price levels in countries A and B be cointegrated with
cointegrating vector [ 1 –1 1] (见 p409).
? Chen (1995) uses monthly data for April 1973-December
1990 to test the PPP hypothesis using the Johansen
approach.
? Chris Brooks 2002 陈磊 2004
7-69
Cointegration Tests of PPP with
European Data
T e st s f o r
c oi n t e g r a t i on be t w e e n
r = 0 r ? 1 r ? 2 ?
1
?
2
FRF – DEM 34.63* 17.10 6.26 1.33 - 2.50
FRF – I T L 52.69* 15.81 5.43 2.65 - 2.52
FRF – NLG 68.10* 16.37 6.42 0.58 - 0.80
F RF – BEF 52.54* 26.09* 3.63 0.78 - 1.15
DE M – I T L 42.59* 20.76* 4.79 5.80 - 2.25
DE M – NLG 50.25* 17.79 3.28 0.12 - 0.25
DE M – BEF 69.13* 27.13* 4.52 0.87 - 0.52
I T L – NLG 37.51* 14.22 5.05 0.55 - 0.71
I T L – BEF 69.24* 32.16* 7.15 0.73 - 1.28
NL G – BEF 64,52* 21.97* 3.88 1.69 - 2.17
Cr i t i c a l v a l u e s 31.52 17.95 8.18 - -
N o te s,FR F - Fr e n ch f r anc; D E M – G er m an M ar k ; N L G – D u tc h g u i ld er ; I T L – I ta lia n lir a; B E F –
B el g ia n f r anc,So u r ce, C h e n ( 1 9 9 5 ), Rep r inted w it h t h e p er m i ssio n o f T a y lo r and Fr anci s L td,
( w w w, ta n d f, co, u k ),
? Chris Brooks 2002 陈磊 2004
7-70Example 2,Are International
Bond Markets Cointegrated?
? Mills & Mills (1991)
? If financial markets are cointegrated,this implies that they
have a,common stochastic trend”.
Data:
? Daily closing observations on redemption yields偿债收益
on government bonds for 4 bond markets,US,UK,West
Germany,Japan.
? For cointegration,a necessary but not sufficient condition
is that the yields are nonstationary,All 4 yields series are
I(1),
? Chris Brooks 2002 陈磊 2004
7-71Testing for Cointegration
Between the Yields
? The Johansen procedure is used,There can be at most 3
linearly independent cointegrating vectors.
? Mills & Mills use the trace test statistic:
?
??
??? g
ri it r a c e
Tr
1
)?1l n ()( ??
Johan sen Test s f or Co i n t e gr at ion b e t w e e n In t e r n a t ion al Bon d Yie ld s
Te st st a ti st ic Criti c a l Va lue s r ( numbe r of c ointe gr a ti n g
ve c tors un de r th e null h y pothe si s) 10% 5%
0 22.06 35.6 38.6
1 10.58 21.2 23.8
2 2.52 10.3 12.0
3 0.12 2.9 4.2
So u r ce, M il ls a n d Mill s ( 1 9 9 1 ), R ep r in ted w it h t h e p er m is s i o n o f B lack w ell P u b li s h er s,
?Conclusion,No cointegrating vectors.
? Chris Brooks 2002 陈磊 2004
7-72Testing for Cointegration
Between the Yields
The paper then goes on to estimate a VAR for the first
differences of the yields,which is of the form
where
They set k = 8,
X
X US
X UK
X WG
X J A P
t
t
t
t
t
i
i i i i
i i i i
i i i i
i i i i
t
t
t
t
t
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
( )
( )
( )
( )
,,?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
11 12 13 14
21 22 23 24
31 32 33 34
41 42 43 44
1
2
3
4
?
?
?
?
?
?
?
? ?????
k
i
titit XX
1
?
? Chris Brooks 2002 陈磊 2004
7-73Variance Decompositions for VAR
of International Bond Yields
Var ia n c e De c o m p osi t io n s f or VA R of In t e r n ation al B on d Yie ld s
Ex pl a in e d b y mov e me nt s in Ex pl a in in g
mov e me nt s i n
Da y s
a he a d US UK Ge r ma n y J a pa n
US 1 95.6 2.4 1.7 0.3
5 94.2 2.8 2.3 0.7
10 92.9 3.1 2.9 1.1
20 92.8 3.2 2.9 1.1
UK 1 0.0 98.3 0.0 1.7
5 1.7 96.2 0.2 1.9
10 2.2 94.6 0.9 2.3
20 2.2 94.6 0.9 2.3
Ge r ma n y 1 0.0 3.4 9 4.6 2.0
5 6.6 6.6 84.8 3.0
10 8.3 6.5 82.9 3.6
20 8.4 6.5 82.7 3.7
J a pa n 1 0.0 0.0 1.4 100,0
5 1.3 1.4 1.1 96.2
10 1.5 2.1 1.8 94.6
20 1.6 2.2 1.9 94.2
So u r ce, M il ls a n d Mill s ( 1 9 9 1 ), R ep r in ted w it h t h e p er m is s i o n o f B lack w ell P u b li s h er s,
? Chris Brooks 2002 陈磊 2004
7-74Impulse Responses for VAR of
International Bond Yields
Im p u ls e Re sp on ses f or VA R of In t e r n ation al B on d Yie ld s
R e spo nse of US t o in nov a ti ons in
Da y s a f te r sho c k US UK Ge r ma n y J a pa n
0 0.98 0.00 0.00 0.00
1 0.06 0.01 - 0,10 0.05
2 - 0.02 0.02 - 0.14 0.07
3 0.09 - 0.04 0.09 0.08
4 - 0.02 - 0.03 0.02 0.09
10 - 0.03 - 0.01 - 0.02 - 0.01
20 0.00 0.00 - 0.10 - 0.01
R e spo nse of UK to i nno v a ti ons in
Da y s a f te r sho c k US UK Ge r ma n y J a pa n
0 0.19 0.97 0.00 0.00
1 0.16 0.07 0.01 - 0.06
2 - 0.01 - 0.01 - 0.05 0.09
3 0.06 0.04 0.06 0.05
4 0.05 - 0.01 0.02 0.07
10 0.01 0.01 - 0.04 - 0.01
20 0.00 0.00 - 0.01 0.00
R e spo nse of Ge r ma n y to in nov a ti ons in
Da y s a f te r sho c k US UK Ge r ma n y J a pa n
0 0.07 0.06 0.95 0.00
1 0.13 0.05 0.11 0.02
2 0.04 0.03 0.00 0.00
3 0.02 0.00 0.00 0.01
4 0.01 0.00 0.00 0.09
10 0.01 0.01 - 0.01 0.02
20 0.00 0.00 0.00 0.00
R e spo nse of J a pa n to inn ovatio ns i n
Da y s a f te r sho c k US UK Ge r ma n y J a pa n
0 0.03 0.05 0.12 0.97
1 0.06 0.02 0.07 0.04
2 0.02 0.02 0.00 0.21
3 0.01 0.02 0.06 0.07
4 0.02 0.03 0.07 0.06
10 0.01 0.01 0.01 0.04
20 0.00 0.00 0.00 0.01
So u r ce, M ill s a n d M ill s ( 1 9 9 1 ), Rep r in ted w it h t h e p er m is s i o n o f B lac k w ell P u b li s h er s,
