This document summarizes a session on financial econometric models. It discusses time series analysis principles including auto regressive processes, moving average processes, and ARMA models. The session aims to remind participants of key concepts from the previous session and conclude by reviewing time series analysis steps of identifying trends and seasonality, fitting appropriate models, and forecasting.

1. ESGF 5IFM Q1 2012
Financial Econometric Models
Vincent JEANNIN – ESGF 5IFM
Q1 2012
vinzjeannin@hotmail.com
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2. ESGF 5IFM Q1 2012
Summary of the session (Est. 3h)
• Reminder of Last Session
• Time Series Analysis Principles
• Auto Regressive Process
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• Moving Average Process
• ARMA
• Conclusion
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3. Be logic!
Reminder of Last Session
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4. Differentiation possible
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������������������������������ = ln⁡ ������)
(
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5. Time can be a factor of a regression
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6. Differentiation can add value
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7. Check ACF/PACF for autocorrelation
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8. Time Series Analysis Principles
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Reminders of the 3 steps
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Identify
Fit
Forecast
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9. ESGF 4IFM Q1 2012
Reminders of the 3 components
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Trend
Seasonality
Residual
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10. Lag
������������������ = ������������−1
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Difference
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∆������������ = ������������ − ������������−1
Seasonality Difference
∆30 ������������ = ������������ − ������������−30 10

11. ESGF 4IFM Q1 2012
Differentiate series to obtain stationary series
Time series analysis and forecast simpler with stationary series
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Different models involved with stationary or heteroscedasticity
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12. Properties of stationary series
Same distribution of the following
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(������1 , ������2 , ������3 , … , ������������ )
(������2 , ������3 , ������4 , … , ������������+1 )
Distribution not time dependent
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Rare occurrence
Stationarity accepted if
������(������������ ) = ������ Constant in the time
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������������������(������������ , ������������−������ ) Depends only on n

13. Acceptable Shortcut
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A series is stationary if the mean and the variance are stable
Which one is more likely to be stationary?
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14. About the residuals…
White noise!
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Normality test
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Have an idea with
Skewness
Kurtosis
Proper tests: KS, Durbin Watson, Portmanteau,…
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15. Auto Regressive Process
Main principle
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There is a correlation between current data and previous data
������������ = ������ + ������1 ������������−1 + ������2 ������������−2 + ⋯ + ������������ ������������−������ + ������������
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������������ Parameters of the model
������������ White noise
AR(n)
If the parameters are identified, the prediction will be easy 15

16. Let’s upload some data
DATA<-read.csv(file="C:/Users/vin/Desktop/Series1.csv",header=T)
plot(DATA$Val, type="l")
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17. Is this a white noise?
hist(DATA$Val, breaks=20)
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18. Probably not…
Portmanteau test
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Test the autocorrelation of a series
If there is autocorrelation, data aren’t independently distributed
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Let’s use Ljung–Box statistics
H0: Data are independently distributed
H1: Data aren’t independently distributed
������
������2 ������ With α confidence interval rejection
������ = ������(������ + 2) following a Chi Square distribution
������ − ������
������=1
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������������ Autocorrelation at the lag k ������ > Χ 21−������,ℎ

19. > Box.test(DATA$Val)
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Box-Pierce test
data: DATA$Val
X-squared = 188.3263, df = 1, p-value < 2.2e-16
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H0 is rejected, the data aren’t independently distributed
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20. Let’s try a regression and analyse residuals
TReg<-lm(DATA$Val~DATA$t)
plot(DATA$Val, type="l")
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abline(TReg, col="blue")
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21. eps<-resid(TReg)
ks.test(eps, "pnorm")
layout(matrix(1:4,2,2))
plot(TReg)
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22. Box-Pierce test
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data: eps
X-squared = 187.6299, df = 1, p-value < 2.2e-16
Residuals aren’t a white noise
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Regression rejected
Not a surprise, did the series look stationary?
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What next then?

23. lag.plot(DATA$Val, 9, do.lines=FALSE)
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Differentiation seems to be interesting

24. Does the differentiation create a stationary series?
plot(diff(DATA$Val), type="l")
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25. ACF & PACF
par(mfrow=c(2,1))
acf(diff(DATA$Val),20)
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pacf(diff(DATA$Val),20)
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ACF decreasing
PACF cancelling after order 1
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26. Decreasing ACF
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PACF cancel after order 1
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Typically an Autoregressive Process
AR(1) ?
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27. Let’s try to fit an AR(1) model
Modl<-ar(diff(DATA$Val),order.max=20)
plot(Modl$aic)
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The likelihood for the order 1 is significant

28. > ar(diff(DATA$Val),order.max=20)
Call:
ar(x = diff(DATA$Val), order.max = 20)
Coefficients:
1 2 3
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0.5925 -0.1669 0.1385
Order selected 3 sigma^2 estimated as 0.8514
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We have our coefficient and standard deviation
> ARDif<-diff(DATA$Val)
> ARDif[1]
[1] 0.3757723
We know the first term of our series
������������ = 0.3757723 + 0.5925. ������������−1 + ������������
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Here is our model

29. Need to test the residuals
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Box.test(Modl$resid)
Box-Pierce test
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data: Modl$resid
X-squared = 7e-04, df = 1, p-value = 0.9789
H0 accepted, residuals are independently distributed (white noise)
The differentiated series is a AR(1)
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30. > predict(arima(diff(DATA$Val), order = c(1,0,0)), n.ahead = 7)
$pred
Time Series:
Start = 193
End = 199
Frequency = 1
[1] -0.81359048 -0.43300609 -0.22850452 -0.11861853 -0.05957287 -
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0.02784553 -0.01079729
$se
Time Series:
Start = 193
End = 199
Frequency = 1
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[1] 0.923352 1.048210 1.081582 1.091027 1.093739 1.094521 1.094747
120
115
110
105
100
95
90
30
85
80
1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101106111116121126131136141146151156161166171176181186191196

31. ESGF 4IFM Q1 2012
Another typical example?
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You make the comments!
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32. DATA<-read.csv(file="C:/Users/vin/Desktop/Series2.csv",header=T)
plot(DATA$Ser2, type="l")
hist(DATA$Ser2, breaks=20)
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33. > Box.test(DATA$Ser2)
Box-Pierce test
data: DATA$Ser2
X-squared = 149.9227, df = 1, p-value < 2.2e-16
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TReg<-lm(DATA$Ser2~DATA$t)
plot(DATA$Ser2, type="l")
abline(TReg, col="blue")
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34. > eps<-resid(TReg)
> Box.test(eps)
Box-Pierce test
data: eps
X-squared = 148.5669, df = 1, p-value < 2.2e-16
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> layout(matrix(1:4,2,2))
> plot(TReg)
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35. > lag.plot(DATA$Ser2, 9, do.lines=FALSE)
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Much less obvious but clues of autoregression

36. par(mfrow=c(2,1))
plot(diff(DATA$Ser2), type="l")
plot(diff(DATA$Ser2, lag=2), type="l")
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37. par(mfrow=c(2,1))
plot(diff(DATA$Ser2), type="l")
plot(diff(DATA$Ser2, lag=2), type="l")
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ACF decreases 2 by 2
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PACF cancelling after
order 2
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38. First order differentiation, strong AR(2) clues
par(mfrow=c(1,1))
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Modl<-ar(diff(DATA$Ser2),order.max=20)
plot(Modl$aic)
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39. Parameters estimation
> ar(diff(DATA$Ser2),order.max=20)
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Call:
ar(x = diff(DATA$Ser2), order.max = 20)
Coefficients:
1 2 3
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0.5919 -0.8326 0.1086
Order selected 3 sigma^2 estimated as 0.877
> ARDif<-diff(DATA$Ser2)
> ARDif[1]
[1] 0.3757723
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40. > predict(arima(diff(DATA$Ser2), order = c(2,0,0)), n.ahead = 7)
$pred
Time Series:
Start = 193
End = 199
Frequency = 1
[1] 0.4505213 2.0075741 0.6639701 -1.2321156 -1.1409989 0.3866745
1.0879588
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$se
Time Series:
Start = 193
End = 199
Frequency = 1
[1] 0.9220713 1.0332515 1.1413067 1.2938326 1.2957576 1.3932158 1.4080266
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115
110
105
100
95
90
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85
80
101
106
111
116
121
126
131
136
141
146
151
156
161
166
171
176
181
186
191
196
1
6
11
16
21
26
31
36
41
46
51
56
61
66
71
76
81
86
91
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41. > Box.test(Modl$resid)
Box-Pierce test
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data: Modl$resid
X-squared = 0.0023, df = 1, p-value = 0.9619
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Model accepted
The more factors the harder the prediction is
The more factors there are the more stationary need to be
the series for a good prediction
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42. Moving Average Process
Main principle
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Stationary series with auto correlation of errors
������������ = ������ + ������������ + ������1 ������������−1 + ������2 ������������−2 + ⋯ + ������������ ������������−������
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������������ Parameters of the model
������������ White noise
MA(n)
More difficult to estimate than a AR(n) 42

43. plot(Data, type="l")
hist(Data, breaks=20)
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44. acf(Data,20)
pacf(Data,20)
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ACF cancels
after order 1
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PACF decays to 0
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ACF & PACF suggest MA(1)

45. > arima(Data, order = c(0, 0, 1),include.mean = FALSE)
Call:
arima(x = Data, order = c(0, 0, 1), include.mean = FALSE)
Coefficients:
ma1
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-0.4621
s.e. 0.0903
sigma^2 estimated as 0.937: log likelihood = -138.76, aic = 281.52
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> Box.test(Rslt$residuals)
Box-Pierce test
data: Rslt$residuals
X-squared = 0, df = 1, p-value = 0.9967
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It works, MA(1), 0 mean, parameter -0.4621

46. Fore<-predict(Rslt, n.ahead=5)
U = Fore$pred + 2*Fore$se
L = Fore$pred - 2*Fore$se
minx=min(Data,L)
maxx=max(Data,U)
ts.plot(Data,Fore$pred,col=1:2,
ylim=c(minx,maxx))
lines(U, col="blue", lty="dashed")
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lines(L, col="blue", lty="dashed")
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47. ESGF 4IFM Q1 2012
Another typical example?
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You make the comments!
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48. plot(Data, type="l")
hist(Data, breaks=20)
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49. vinzjeannin@hotmail.com ESGF 4IFM Q1 2012
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50. > arima(Data, order = c(0, 0, 2),include.mean = FALSE)
Call:
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arima(x = Data, order = c(0, 0, 2), include.mean = FALSE)
Coefficients:
ma1 ma2
-0.5365 0.6489
s.e. 0.0701 0.1044
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sigma^2 estimated as 1.005: log likelihood = -142.74, aic = 291.48
> Box.test(Rslt$residuals)
Box-Pierce test
data: Rslt$residuals
X-squared = 0.0283, df = 1, p-value = 0.8664
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MA(2)

51. Fore<-predict(Rslt, n.ahead=5)
U = Fore$pred + 2*Fore$se
L = Fore$pred - 2*Fore$se
minx=min(Data,L)
maxx=max(Data,U)
ts.plot(Data,Fore$pred,col=1:2,
ylim=c(minx,maxx))
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lines(U, col="blue", lty="dashed")
lines(L, col="blue", lty="dashed")
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52. ARMA
Main principle
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The series is a function of past values plus current and past values of the noise
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ARMA(p,q)
Combines AR(p) & MA(q)
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53. plot(Data, type="l")
hist(Data, breaks=20)
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54. ESGF 4IFM Q1 2012
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Both ACF and PACF decreases exponentially after order 1

55. > Rslt<-arima(Data, order = c(1, 0, 1),include.mean = FALSE)
> Rslt
Call:
arima(x = Data, order = c(1, 0, 1), include.mean = FALSE)
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Coefficients:
ar1 ma1
0.7214 0.7563
s.e. 0.0716 0.0721
sigma^2 estimated as 0.961: log likelihood = -141.13, aic = 288.27
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> Box.test(Rslt$residuals)
Box-Pierce test
data: Rslt$residuals
X-squared = 0.0098, df = 1, p-value = 0.9213
ARMA(1,1) fits
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56. > par(mfrow=c(1,1))
> Fore<-predict(Rslt, n.ahead=5)
> U = Fore$pred + 2*Fore$se
> L = Fore$pred - 2*Fore$se
> minx=min(Data,L)
> maxx=max(Data,U)
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57. Identification can get tricky at this stage
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58. What do you think?
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59. > Rslt<-arima(Data, order = c(4, 0, 3),include.mean = FALSE)
> Rslt
Call:
arima(x = Data, order = c(4, 0, 3), include.mean = FALSE)
Coefficients:
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ar1 ar2 ar3 ar4 ma1 ma2 ma3
0.2722 -0.5276 0.0202 -0.2663 0.8765 -0.4672 -0.5248
s.e. 0.2018 0.2308 0.1968 0.1546 0.1992 0.1690 0.1882
sigma^2 estimated as 1.140: log likelihood = -151.19, aic = 318.38
> Box.test(Rslt$residuals)
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Box-Pierce test
data: Rslt$residuals
X-squared = 0.2953, df = 1, p-value = 0.5869
Was supposed to fit pretty wel….
Data<-arima.sim(model=list(ar=c(0.5,-0.5,0.3,-
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0.3),ma=c(0.75,-0.5,-0.5)),n=100)

60. Identification can be difficult
ESGF 4IFM Q1 2012
Easiest model is AR
Imagine when the series is not stationary…
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Step by step approach, exploration, tries,…
Sometimes you find a satisfying model
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Sometimes you don’t!

61. Conclusion
AR
MA
ARMA
Times series
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