Universität Potsdam

Time Series Analysis Exercises

Hans Gerhard Strohe

Potsdam 2005

1

I

Typical exercises and solutions

1

For theoretically modelling the economic development of national economy scenarios the following 2 models for GDP increment are analysed:

a. Yt = Yt-1 + at b. Yt = 1.097 Yt-1 - 0,97 Yt-2 + at ,

where Y

-

GDP increment

a

-

White noise with zero mean and constant variance σ2=100

t

-

Time (quarters starting with Q1, 1993)

Check by an algebraic criterion which one is stationary.

2

Answer:

a)

The process can be written φ(L)Yt = at with the lag polynom φ(L) = 1-L and L being the lag or backshift operator LYt.= Yt-1 From this follows the characteristic equation φ(z) = 1-z = 0 The root, i.e. the solution of it, is z=1. It is not higher than 1, i.e. it is not placed outside the unit circle. The process is not stationary. It is a “unit root” process or more specific a random walk.

b)

The process can be written φ(L)Yt = at with the lag polynom φ(L) = 1 – 1,079L + 0,97 L2. From this follows the characteristic equation φ(z) = 1- 1,097z + 0,97 z2 = 0 The roots (complex numbers) of it are

z1 = 0,57 + 0,84 i z2 = 0,57 - 0,84 i

with

| z1/ 2 |= 0,57 2 + 0,84 2 = 1,015 > 1.

That means that both of the roots are outside the unit circle what is a sufficient condition for the stationarity of the process.

3

2 The variable l (=labour) in the file employees.dat denotes the quantity of labour force, i. e. the number of employees, in a big German company from January 1995 till December 2004.

i.

Display the graph of the time series lt.

ii.

What are the characteristics of a stationary time series? Is the time series lt likely to be stationary? Check it first by the naked eye.

iii.

Test the stationarity of lt by a suitable procedure. Determine the degree of integration.

iv.

Estimate the correlation and the partial correlation function of the process. Give a description and an interpretation.

v.

What type of basic model could fit the time series lt. Why?

vi.

Estimate the parameters concerning the model assumed in question v .

vii.

Estimate alternative models and compare them by suitable indicators.

viii.

Forecast the time series for 2005.

4

ix.

A particular analysis method of time series lt results in the following graph.

1.6

Bartlett 1.4

1.2

Tukey 1.0

0.8 Parzen 0.6 0

1

2

3

4

Circular frequency

Fig. 1: A special diagnostic function

What is the name of this particular analysis method? What can you derive from the curve. What should be the typical shape of the function estimated for a process type assumed in question v?

5

Answers: i.

The graph of the time series lt.

6000 5600 5200

4800 4400 4000 95

96

97

98

99

00

01

02

03

04

L Fig. 2: Graph of the employees time series ii.

The main characteristics of a stationary time series are that the mean µ t and the variance of the stochastic process generating this special time series are independent from time t,

µ t = µ = const σ t2 = σ 2 = const, and that the autocovariances γ t , t depend only on the time difference τ : 1 2

γ t , t = γ t - t = γ τ with τ = t 1 – t 2 1 2 1 2

(Lag)

A check of the graph by the naked eye gives no reason to assume the time series not to be stationary. At the first glance there does not appear any trend or relevant development of the variance.

6

iii.

Test of stationarity by DF Test.

The following Dickey Fuller regression of ∆l on lt-1 produces a t-value – 3,63. Application of an augmented Dickey-Fuller regression is not necessary because the augmented model would have higher values of the Schwarz criterion (SIC, version on the base of the error variance). Table 1: Dickey-Fuller Test Equation Dependent Variable: ∆lt Method: Least Squares Sample (adjusted): 1995M02 2004M12; observations: 119 Variable

Coefficient Std. Error

t-Statistic

Prob.

lt-1 C

-0.200468 0.055277 1096.460 301.8879

-3.626566 3.632011

0.0004 0.0004

R-squared Adjusted R-squared S.E. of regression Sum squared resid

0.101051 0.093368 88.39903 914283.5

Mean dependent var S.D. dependent var Akaike info criterion(σ) Schwarz criterion(σ)

2.038202 92.83931 11.81826 11.86497

The critical value of the t-statistic for a model with intercept c is -2,89 on the 5 % level: Table 2: Null Hypothesis: l has a unit root Exogenous: Constant Lag Length: 0 (Automatic based on SIC, MAXLAG=12) t-Statistic Test critical values:

1% level 5% level

-3.486064 -2.885863

The t-value measured exceeds the critical value downwards. That means that the null hypothesis of nonstationarity or a unit root can be rejected on the 5 % level. The time series is stationary at least with a probability of 95 %. As lt is stationary its degree of integration is 0 ( I(0)).

7

iv.

Sample correlation and partial correlation functions of the process.

The sample autocorrelation function (AC) for a short series can be calculated using the formulae for the sample autocovariance (9.7) T −τ

cτ =

∑ ( xt − x )( xt +τ − x ) t =1

T

− τ

and the sample autocorrelation (9.8)

rτ = with

x

cτ s x2

and sx being the average and the standard deviation of the time series, respectively.

The partial sample autocorrelation function (PAC) can be obtained by linear OLS regression of xt on xt-1, xt-2,…, xt-τ corresponding to formula (9.9):

xt = φ 0 + φ1τ xt −1 + ... + φττ xt −τ The partial correlation coefficient ρpart(τ) is the coefficient φττ of the highest order lagged variable xt-τ. By calculating the first 5 regressions this way we obtain the following highest order coefficients for each of them, respectively: ф11= 0,799532; ф22= -0,037588;

ф33= 0,075722; ф44= 0,008910; ф55= 0,115970

We can compare these „manually“ calculated coefficient with the partial autocorrelation functions displayed in the following table and figure together with the sample autocorrelation function delivered as a whole in one step by eViews. The differences between the regression results and the eViews output probably are caused by different estimation methods.

8

Table 3:

AC

PAC

τ

AC

PAC

τ

AC

PAC

1

0.794

0.794

9

0.039

-0.008

17

-0.157

-0.066

2

0.616

-0.042

10

0.014

0.049

18

-0.154

0.038

3

0.495

0.052

11

-0.004

-0.027

19

-0.125

0.042

4

0.390

-0.028

12

-0.050

-0.092

20

-0.096

0.038

5

0.348

0.118

13

-0.061

0.069

21

-0.098

-0.083

6

0.298

-0.038

14

-0.100

-0.082

22

-0.142

-0.127

7

0.205

-0.115

15

-0.139

-0.042

23

-0.150

0.046

8

0.109

-0.083

16

-0.137

0.013

24

-0.183

-0.134

τ

Correlogram 1,2 1 0,8 0,6 AC

0,4

PAC

0,2 0 -0,2 0

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

-0,4

Fig. 3: Correlogram The bold curve shows the sample autocorrelation function. The thin one displays the partial autocorrelation, both depending on the lag τ. While the smooth autocorrelation continuously decreases from 1 towards zero and below to a limit of about 0,1 the partial autocorrelation starts as well at the value 1 and has the same value as the autocorrelation for τ=1 but then drops down to zero and remains there oscillating between -0,1 and 0,1.

9

v.

Possible type of basic model fitting the time series l.

The basic model could be a first order autoregressive process , because AC is exponentially decreasing and PAC is dropping down after τ=1. The shape of the AC is typical for the autocorrelation of an AR process with positive coefficients. The partial autocorrelation drops down to and remains close to zero after the lag τ=1 what indicates an AR(1) process. vi.

Estimated parameters concerning the model assumed in answer v.

Table 4: Dependent Variable: lt Method: Least Squares Sample 1995M02 2004M12, observations: 119 Variable

Coefficient Std. Error

t-Statistic

Prob.

CONST lt-1

1096.460 0.799532

301.8879 0.055277

3.632011 14.46398

0.0004 0.0000

R-squared Adjusted R-squared S.E. of regression Sum squared resid

0.641332 0.638266 88.39903 914283.5

Mean dependent var S.D. dependent var Akaike info criterion Schwarz criterion

5461.386 146.9782 11.81826 11.86497

Thus we get the empirical model: lt = 1096.5+ 0.7995 lt-1 + at Because the t-statistic of both exceeds the 5-% critical value 1,96, both coefficients are significant on this level, at least. The prob. values behind indicate that the even are on the 1 % level.

10

Another way of estimation is to take advantage of the special ARMA estimation procedure of eViews. Here first the mean of the variable is estimated and then the AR(1) coefficient of an AR model of the deviations from the mean: Table 5: Dependent Variable: l Method: Least Squares Sample: 1995M02 2004M12; observations: 119 Convergence achieved after 3 iterations Variable

Coefficient Std. Error

t-Statistic

Prob.

CONST AR(1)

5469.515 0.799532

134.9823 14.46398

0.0000 0.0000

R-squared Adjusted R-squared S.E. of regression Sum squared resid

0.641332 0.638266 88.39906 914284.2

40.52025 0.055277

Mean dependent var S.D. dependent var Akaike info criterion(σ) Schwarz criterion(σ)

5461.386 146.9782 11.81826 11.86497

The equivalent model obtained this way can be written: (lt - 5469,5) = 0,7995 ( lt-1 - 5469.5) + at

11

vii.

As an alternative, an ARMA(2,1) model would fit the data. It can be found by trial and error via several different models and the aim of obtaining significant coefficients and compared by Schwarz criterion.

As the model contains an MA term ordinary least squares is not practicable for estimating the coefficients. Therefore again the iterative nonlinear least squares procedure by eViews is used: Table 6: Dependent Variable: l Method: Least Squares Sample 1995M03 2004M12; observations: 118 Convergence achieved after 7 iterations Backcast: 1994M12 Variable

Coefficient Std. Error

t-Statistic

Prob.

CONST AR(2) MA(1)

5469.200 0.600156 0.856271

144.9444 6.409735 14.23960

0.0000 0.0000 0.0000

R-squared Adjusted R-squared S.E. of regression Sum squared resid

0.646928 0.640787 88.23855 895394.8

37.73310 0.093632 0.060133

Mean dependent var S.D. dependent var Akaike info criterion(σ) Schwarz criterion(σ)

5462.352 147.2253 11.82306 11.89350

Again here is to be considered that the “constant” is the mean and the coefficients belong to a model of derivations from it. Thus the ARMA(2,1) model is to be written: (lt - 5469,2) = 0,6002 ( lt-2 - 5469.5) + 0,8563 at + at

In order to have the model in the explicit form, the mean is subtracted: lt = 5469,2 +

0,6002 ( lt-2 - 5469.2) + 0,8563 at + at

= 5469,2 - 5469.2 × 0,6002 .+

0,6002 lt-2 + 0,8563 at + at

and at last lt =

2186,6 .+ 0,6002 lt-2 + 0,8563 at + at

This model does not show an improvement compared by Akaike and Schwarz criteria with the AR(1) estimated earlier. It has slightly higher values. EViews provides these criteria on the base of the error variance which is to be minimised (in contrast to those on likelihood base such as in Microfit that are to be maximised). 12

viii.

Forecast of the number of employees for 2005:

The somewhat uneasy shape of both models: (lt - 5469,5) = 0,7995 ( lt-1 - 5469.5) + at and (lt - 5469,2) = 0,6002 ( lt-2 - 5469.5) + 0,8563 at + at give us the advantage of easily forecasting the process in the long run because the stochastic limes for time tending to infinity is given by what we called the means 5469,5 and 5469,2, respectively: plim (lt - 5469,5)

= plim (0,7995 ( lt-1 - 5469.5) + at) = plim (0,7995 ( lt-1 - 5469.5) + plim at = 0

That means future.

+ 0.

plim lt = 5469,5 and we can take this constant as suitable forecast for the farther

In the same way we obtain from the second model the forecast 5469,2 that does not differ very much. The following graph displays the forecast for the years 2004 and 2005. In 2004 we obtain an irregular curve, because the one-month ahead forecast for every month can be calculated on the base of the varying values of the previous month. But in 2005, there is a smooth exponential curve because the forecasts can be calculated only on the base of the previous forecast instead of the real data.

13

5800 5700 5600 5500 5400 5300 5200 5100 5000 2004M01

2004M07

2005M01

2005M07

Forecast of the number of employees

Fig. 4: Forecast

ix.

Spectral analysis is the name of this method.

Basically, the main peaks of spectral density occur at the circular frequencies ω1= 0,71

ω2 = 1,86.

These correspond to frequencies f1 = 0,114

f2 = 0,295

and to periods of average length p1 = 8,8

p2 = 3,4 month, respectively.

But these periodicities are not of any practical importance. They superimpose to the point of being unrecognizable. They are not visible in the original time series and the correlogram. Taking standard errors of the spectral estimates into consideration, peaks and troughs of the spectral density curve does not significantly differ. The typical shape of the spectral density function for an following

AR(1) process is similar to the

14

Estimates of spectral density of an AR(1) process 8

Bartlett 6

4

Tukey

2

Parzen 0 0

1

2

3

4

Circular frequency

Fig. 5: Estimates of spectral density of an AR(1) process In case of an additional seasonal component the spectral density would show a peak over the frequency f = 1/12 or the circular frequency ω = 2π/12 = 0,52.

15

3

The file dax_j95_a04.txt contains the daily closing data of the main German stock price index DAX from January 1995 by August 2004, i.e. 2498 values. i.

Display the graph of the time series dax.

ii.

Determine the order of integration of dax.

iii.

Generate the series of the growth rate or rate of return r in the direct way and in the logarithmic way. Display both graphs of r.

iv.

Compare the r data. Characterize the general patterns of both graphs of r

v.

Check whether or not r is stationary.

vi.

Estimate the correlation function and the partial correlation function till a lag of 20. Characterize generally the extent of correlation in this series.

vii.

Estimate an AR(8) model to r that contains only coefficients significant at least on the five percent level

viii. Estimate an ARIMA(8,d,8) model to r that contains only coefficients significant at least on the one percent level. ix.

Test the residuals of this model for autoregressive conditional heteroscedasticity by a rough elementary procedure

x.

Make visible conditional heteroscedasticity by estimating a 5-day moving variance of the residuals

xi.

Estimate an ARCH(1) model on the base of the ARIMA(8,d,8) model estimated in viii. You can change the ARMA part in order to keep only coefficients significant on the one percent level.

xii.

Estimate a GARCH(1,1) model on the base of the simple model r=const.

16

Answers:

i.

The graph of dax.

9000 8000 7000 6000 5000 4000 3000 2000 1000 0 500

1000

1500

2000

DAX Fig. 6: The graph of the German share price index This graph indicates a non-stationary process, perhaps a random walk. It is characterized by changing stochastic trends and increasing variance.

17

ii.

The order of integration of dax.

For testing whether or not dax is stationary first the original date are tested by the Dickey-Fuller test of the levels. The following table shows the estimation of the coefficient of the DickeyFuller Test regression of ∆daxt on daxt-1 : Table 7: Dickey-Fuller Test Equation Dependent Variable: ∆daxt Method: Least Squares Sample (adjusted): 2 2498; observations: 2497 Variable

Coefficient Std. Error

t-Statistic

Prob.

daxt-1 C

-0.001453 0.000919 6.970114 4.213823

-1.580936 1.654107

0.1140 0.0982

R-squared Adjusted R-squared S.E. of regression Sum squared resid

0.001001 0.000600 71.26043 12669731

Mean dependent var S.D. dependent var Akaike info criterion(σ) Schwarz criterion(σ)

0.701418 71.28183 11.37136 11.37602

The t-value of the slope coefficient is -1.580936. As the following table demonstrates it is not less than the 5%-critical t-value for a model with a constant, i.e. -2.862497. Table 8: Null Hypothesis: DAX has a unit root Exogenous: Constant Lag Length: 0 (Automatic based on SIC, MAXLAG=26)

Augmented Dickey-Fuller test statistic Test critical values: 1% level 5% level

t-Statistic

Prob.*

-1.580936 -3.432775 -2.862497

0.4922

That means that dax is on the 5% level not stationary. The next step is to check the first differences of dax in the same way.

18

The Dickey-Fuller test equation of the first differences is a regression of the second differences ∆2daxt on the lagged first differences ∆daxt-1 . Table 9: Augmented Dickey-Fuller Test Equation Dependent Variable: D(DAX,2) = ∆2daxt Method: Least Squares Sample: 3 2498; observations: 2496 Variable

Coefficient Std. Error

t-Statistic

Prob.

D(DAX(-1))= ∆daxt-1 C

-0.996234 0.020026 0.702902 1.427408

-49.74669 0.492432

0.0000 0.6225

R-squared Adjusted R-squared S.E. of regression Sum squared resid

0.498061 0.497860 71.30959 12682133

Mean dependent var S.D. dependent var Akaike info criterion(σ) Schwarz criterion(σ)

-0.017540 100.6319 11.37274 11.37741

The t-value -49.74669 obviously exceeds all thinkable critical values. Thus the first differences are stationary and dax itself is first order integrated I(1).

iii.

The growth rate or rate of return r can be generated as the ratio

rt =

( dax t − dax t − 1 )

dax t − 1

That is displayed in the following graph:

19

.08 .04

.00 -.04

-.08 -.12 500

1000

1500

2000

R _ R A T IO

Fig. 7: Graph of the growth rate of DAX

Another way of calculating the rate of return mostly used in financial market analysis is the logarithmic way:

rt = ln dax t − ln dax t −1 The next figure shows the graph of this logarithmically generated rate of return r. .08 .04 .00 -.04 -.08 -.12 500

1000

1500

2000

R _ LO G

Fig. 8: Graph of the logarithmic return rate of DAX

20

iv.

Both curves are very similar to each other and can hardly be distinguished by the naked eye. An enlarged display of the differences as shown in the next figure demonstrates that the ratio always slightly exceeds the logarithmic approximation. In the following analysis the latter will be used.

Differences between alternative series of dax return rate r

.005 .004 .003 .002 .001 .000 500

1000

1500

2000

D IF F _ R

Fig. 9: Graph of the differences between both ways of calculating return rates

Both graphs of r show certain common general patterns. These are significant clusters of high variability or volatility separated by quieter periods. This changing behaviour of variance is typical for ARCH or GARCH processes.

v.

Stationarity test for r.

Particularly for later fitting an ARCH or GARCH model, the time series should be stationary. The following Dickey Fuller Regression of ∆rt on rt-1 produces a negative t-value of -50.377 that lies below of all possible critical values: The return rate r proves stationary.

21

Table 10: Dickey-Fuller Test Equation Dependent Variable: ∆rt Method: Least Squares Sample (adjusted): 3 2498; observations: 2496 Variable

Coefficient Std. Error

t-Statistic

Prob.

rt-1 C

-1.008850 0.020026 0.000250 0.000320

-50.37696 0.780708

0.0000 0.4350

R-squared Adjusted R-squared S.E. of regression Sum squared resid

0.504356 0.504157 0.015967 0.635829

vi.

Mean dependent var S.D. dependent var Akaike info criterion(σ) Schwarz criterion(σ)

-3.68E-06 0.022675 -5.435790 -5.431125

Sample correlation function and partial correlation functions of r

Correlogram 1,2 1 0,8 0,6

AC

0,4

PAC

0,2 0 -0,2

0

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Fig. 10: Graph of sample correlation and partial correlation function

22

Table 11: τ

AC

PAC

τ

AC

PAC

τ

AC

PAC

1 2 3 4 5 6 7 8 9 10 11 12

-0.009 -0.005 -0.032 0.027 -0.041 -0.042 -0.006 0.042 -0.005 -0.007 0.012 0.014

-0.009 -0.005 -0.032 0.026 -0.040 -0.043 -0.006 0.038 -0.005 -0.007 0.011 0.010

13 14 15 16 17 18 19 20 21 22 23 24

-0.017 0.071 0.027 -0.024 -0.038 0.001 -0.048 0.023 0.025 0.005 -0.002 0.033

-0.015 0.076 0.028 -0.026 -0.030 -0.001 -0.048 0.029 0.031 -0.008 -0.004 0.033

25 26 27 28 29 30 31 32 33 34 35 36

0.005 -0.048 -0.046 0.021 0.056 0.025 0.020 0.010 -0.001 -0.014 -0.021 0.019

0.004 -0.048 -0.036 0.014 0.048 0.033 0.030 0.001 -0.001 -0.006 -0.022 0.017

Let us try a general characterisation of the extent of correlation in this series: The sample correlation and partial correlation function do not differ very much. The graphs of both of them coincide. There is not any significant correlation value. This could be the functions of a white noise. Anyway, because of the little peaks at 5, 6 and 8 it could be sensible try to fit an AR(8) model.

vii.

Example: AR(8) model coefficients significant on the five percent level.

Several trials with AR coefficient up to the order 8 finally resulted in three significant coefficients and a missing constant. The criterion for rejecting other variants was nonsignificance of coefficients. Table 12: Dependent Variable: r Method: Least Squares Sample (adjusted): 10 2498; observations: 2489 Variable

Coefficient Std. Error

t-Statistic

Prob.

rt-5 rt-6 rt-8

-0.039842 0.020023 -0.041799 0.020013 0.039960 0.020028

-1.989799 -2.088571 1.995155

0.0467 0.0368 0.0461

Mean dependent var 0.000244 S.E. of regression 0.015951 Sum squared resid 0.632487

S.D. dependent var Akaike info criterion Schwarz criterion

0.015982 -5.437444 -5.430429

23

Thus we obtain by OLS regression the AR(8) model rt = -0.039842 rt-5 - 0.041799 rt-6 + 0.039960 rt-8 + et , where et is the error term of the process estimated. We get exactly the same results, if we consider this AR model as a special case of an ARMA model estimated by iterative nonlinear least squares

Table 13: Dependent Variable: r Method: Least Squares Sample (adjusted): 10 2498; observations: 2489 Convergence achieved after 3 iterations Variable

Coefficient Std. Error

t-Statistic

Prob.

AR(5) AR(6) AR(8)

-0.039842 0.020023 -0.041799 0.020013 0.039960 0.020028

-1.989799 -2.088571 1.995155

0.0467 0.0368 0.0461

Mean dependent var S.E. of regression Sum squared resid

0.000244 0.015951 0.632487

viii.

S.D. dependent var Akaike info criterion(σ) Schwarz criterion(σ)

0.015982 -5.437444 -5.430429

Example: ARIMA(8,d,8) model for r that contains only coefficients significant at least on the one percent level.

Now we try to improve the model by including a moving average term. In this case we must use the iterative nonlinear least squares method. Now the aim is to obtain coefficients, significant on the 1% level. Again after a series of trials with varying ARMA(8,8) models we got finally the following results by omitting non significant coefficients:

24

Table 14: Dependent Variable: r Method: Least Squares Sample (adjusted): 10 2498; observations: 2489 Convergence achieved after 15 iterations Variable

Coefficient Std. Error

t-Statistic

Prob.

AR(3) AR(8) MA(3) MA(6) MA(8)

-0.339258 -0.613633 0.296335 -0.075689 0.654681

-13.38956 -23.73905 10.89544 -5.427549 32.59034

0.0000 0.0000 0.0000 0.0000 0.0000

Mean dependent var S.E. of regression Sum squared resid

0.000244 0.015874 0.625950

The

model

resulting

0.025338 0.025849 0.027198 0.013945 0.020088

S.D. dependent var Akaike info criterion(σ) Schwarz criterion(σ)

from

this

0.015982 -5.446226 -5.434536

estimation

is

the

following:

rt = - 0.339 rt-3 - 0.614 rt-8 + et + 0.296 et-3 - 0.076 et-6 + 0.655 et-8 where et is the error term of the estimated ARMA process. Now the goodness of fit of both models should be compared. Besides considering the significance level of the coefficients in both models a powerful mean of comparison are the Akaike and Schwarz criteria, which are to be minimised. Here both criteria are very close to each other. But the slightly smaller (negative!) values of both criteria in the second case give a certain preference to the ARMA model. Because we know from v. that r is stationary, that means integrated of order 0, this model is an ARIMA(1,0,1) for r at the same time.

25

ix.

Rough and preliminary test of the residuals for autoregressive conditional heteroscedasticity by OLS regression

A rough check for first order autoregressive conditional heteroscedasticity (ARCH(1)) is the estimation of a regression of the squared residuals et2 on et-12 : Table 15: Dependent Variable: et2 Method: Least Squares Sample (adjusted): 11 2498; observations: 2488 Variable

Coefficient Std. Error

t-Statistic

Prob.

C et-12

0.000208 0.175073

17.59238 8.866325

0.0000 0.0000

R-squared Adjusted R-squared S.E. of regression Sum squared resid

0.030652 0.030263 0.000535 0.000710

1.18E-05 0.019746

Mean dependent var S.D. dependent var Akaike info criterion (σ) Schwarz criterion (σ)

0.000252 0.000543 -12.22961 -12.22493

For testing this dependence we cannot simply use the usual t-test because the t-values estimated do not meet an exact student distribution. Anyway, because here the t-values immensely exceed the 5-percent and 1-percent critical values, we can with great practical confidence assume, that there exist a highly significant relationship between et2 and et-12 i.e. high conditional heteroscedasticity.

x.

Visualisation of conditional heteroscedasticity by moving 5-day residual variance

The existence of conditional heteroscedasticity can be visualised by smoothing the series of squared residuals that means by moving averages of the et2 or moving variances.

26

.004

.003

.002

.001

.000 500

1000

1500

2000

5 -d a y m o v in g re s id u a l v a ria n c e Fig. 11: At the figure, you can see typical clusters of higher variance, i.e. volatility, changing with intervals of lower variance. The similarity of neighbouring variances is one more indicator for conditional heteroscedasticity: On the base of knowing the variance at time t (i. e. conditionally) you can forecast the variance at time t+1.

xi.

ARCH(1) model on the base of the AR(8) model

Because of the latter analytical results it would be worthwhile to estimate an ARCH(1) model. Then the conditional variance of the error et is

ht2 = var(et | et −1 ) = E(et2 | et −1 ) = λ0 + λ1et2−1 , where practically ht2 is estimated by et2 In the software used this is a special case of the more general GARCH model. An ARCH(1) there corresponds to a GARCH(0,1) model. We assume here the time series to follow an AR(8) process as estimated earlier without ARCH: 27

Table 16: Dependent Variable: r Method: ML – ARCH Sample (adjusted): 10 2498; observations: 2489 Included after adjustments GARCH = C(4) + C(5)*RESID(-1)^2

AR(5) AR(6) AR(8)

Coefficient Std. Error

z-Statistic

Prob.

-0.045702 0.013304 -0.046538 0.014090 0.044716 0.013812

-3.435162 -3.303005 3.237367

0.0006 0.0010 0.0012

41.55146 9.307385

0.0000 0.0000

Variance Equation C RESID(-1)^2

0.000194 0.243443

R-squared Adjusted R-squared S.E. of regression Sum squared resid

0.004705 0.003102 0.015958 0.632539

4.67E-06 0.026156

Mean dependent var S.D. dependent var Akaike info criterion(σ) Schwarz criterion(σ)

0.000244 0.015982 -5.493524 -5.481833

The coefficients differ from the earlier estimated ones because here they are estimated simultaneously with the ARCH term. The newly estimated model is: rt = -0.0457 rt-5 - 0.0465 rt-5 + 0.0447 rt-8 + et , with the error process et2 = 0.000194 + 0.2434 et-12 + at where at should be a pure random series. The AR representation of the error process gives the opportunity to forecast the volatility.

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xii.

GARCH(1,1) model on the base of the simple model r=const.

Here you should model the return rate as a real GARCH process. The generalized autoregressive conditional heteroscedasticity model (GARCH (p,q)) describes a process where the conditional error variance on all information Ωt-1 available at time t ht2 = var (u t Ωt −1 ) is assumed to obey an ARMA(p,q) model: ht2 = α 0 + α 1ht2−1 + ... + α p ht2− p + β 1ut2−1 + β 2 ut2−2 + ... + β q ut2−q with u t being the error process . Table 17: Dependent Variable: r Method: ML - ARCH (Marquardt) - Normal distribution Sample (adjusted): 10 2498; observations: 2489 Convergence achieved after 12 iterations GARCH = C(5) + C(6)*RESID(-1)^2 + C(7)*GARCH(-1)

C

rt-5 rt-5 rt-8

Coefficient Std. Error

z-Statistic

Prob.

0.000710 -0.029571 -0.036495 0.012575

2.882234 -1.473904 -1.792478 0.611309

0.0039 0.1405 0.0731 0.5410

4.980196 8.564982 96.17172

0.0000 0.0000 0.0000

0.000246 0.020063 0.020360 0.020570

Variance Equation C RESID(-1)^2 GARCH(-1)

2.01E-06 0.082160 0.911339

R-squared Adjusted R-squared S.E. of regression Sum squared resid

0.003340 0.000930 0.015975 0.633406

4.03E-07 0.009593 0.009476

Mean dependent var S.D. dependent var Akaike info criterion(σ) Schwarz criterion (σ)

0.000244 0.015982 -5.752055 -5.735688

While in the original AR(8) model the constant could be omitted here the constant is the only significant term in the regression part of the model. Therefore all lagged r terms can be omitted now:

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Table 18: Dependent Variable: r Method: ML – ARCH Sample (adjusted): 2 2498; observations: 2497 Convergence achieved after 12 iterations GARCH = C(2) + C(3)*RESID(-1)^2 + C(4)*GARCH(-1)

C

Coefficient Std. Error

z-Statistic

Prob.

0.000662

2.753134

0.0059

5.116231 8.664398 96.35969

0.0000 0.0000 0.0000

0.000240

Variance Equation C RESID(-1)^2 GARCH(-1)

2.08E-06 0.083192 0.910103

R-squared Adjusted R-squared S.E. of regression Sum squared resid

-0.000681 -0.001886 0.015977 0.636337

4.06E-07 0.009602 0.009445

Mean dependent var S.D. dependent var Akaike info criterion(σ) Schwarz criterion(σ)

0.000245 0.015961 -5.756824 -5.747496

The model proves very simple: DAX return equals the constant 0,00066 (i.e. 0,06 % per day on an average) with sort of an ARMA(1,1) conditional variance that describes the development of volatility or risk: ht2 = 2,08 · !0-06 + 0,9101 ht-12 + 0.08319 et-12 + at 2

with ht being the conditional variance of rt on base of the information by time t, and et being the deviation of r from its mean in this model. In the meaning of Akaike and Schwarz criteria, the model fits better than all considered before: The values of these criteria are the lower ones despite the model is extremely simple. The model shows that conditional variance as a measure of volatility and investment risk is highly determined by the variance of the last day, i.e. rather by the more theoretical conditional variance ht-12 than by the directly measurable deviation et-1.

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