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# Financial Econometric Models III

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Third Session, MSc 5th Year

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### Financial Econometric Models III

1. 1. ESGF 5IFM Q1 2012Financial Econometric Models Vincent JEANNIN – ESGF 5IFM Q1 2012 vinzjeannin@hotmail.com 1
2. 2. ESGF 5IFM Q1 2012Summary of the session (Est. 3h)• Reminder of Last Session• Time Series Analysis Principles• Auto Regressive Process vinzjeannin@hotmail.com• Moving Average Process• ARMA• Conclusion 2
3. 3. Be logic! Reminder of Last Session vinzjeannin@hotmail.com ESGF 5IFM Q1 2012 3
4. 4. Differentiation possible ESGF 5IFM Q1 2012 vinzjeannin@hotmail.com = ln⁡ ) ( 4
5. 5. Time can be a factor of a regression vinzjeannin@hotmail.com ESGF 5IFM Q1 2012 5
6. 6. Differentiation can add value vinzjeannin@hotmail.com ESGF 5IFM Q1 2012 6
7. 7. Check ACF/PACF for autocorrelation vinzjeannin@hotmail.com ESGF 5IFM Q1 2012 7
8. 8. Time Series Analysis Principles ESGF 4IFM Q1 2012 Reminders of the 3 steps vinzjeannin@hotmail.com Identify Fit Forecast 8
9. 9. ESGF 4IFM Q1 2012Reminders of the 3 components vinzjeannin@hotmail.com Trend Seasonality Residual 9
10. 10. Lag = −1 ESGF 4IFM Q1 2012 Difference vinzjeannin@hotmail.com∆ = − −1 Seasonality Difference∆30 = − −30 10
11. 11. ESGF 4IFM Q1 2012Differentiate series to obtain stationary seriesTime series analysis and forecast simpler with stationary series vinzjeannin@hotmail.comDifferent models involved with stationary or heteroscedasticity 11
12. 12. Properties of stationary series Same distribution of the following ESGF 4IFM Q1 2012 (1 , 2 , 3 , … , ) (2 , 3 , 4 , … , +1 ) Distribution not time dependent vinzjeannin@hotmail.com Rare occurrence Stationarity accepted if ( ) = Constant in the time 12 ( , − ) Depends only on n
13. 13. Acceptable Shortcut ESGF 4IFM Q1 2012A series is stationary if the mean and the variance are stableWhich one is more likely to be stationary? vinzjeannin@hotmail.com 13
14. 14. About the residuals… White noise! ESGF 4IFM Q1 2012Normality test vinzjeannin@hotmail.com Have an idea with Skewness Kurtosis Proper tests: KS, Durbin Watson, Portmanteau,… 14
15. 15. Auto Regressive Process Main principle ESGF 4IFM Q1 2012 There is a correlation between current data and previous data = + 1 −1 + 2 −2 + ⋯ + − + vinzjeannin@hotmail.com Parameters of the model White noise AR(n) If the parameters are identified, the prediction will be easy 15
17. 17. Is this a white noise? hist(DATA\$Val, breaks=20) ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 17
18. 18. Probably not… Portmanteau test ESGF 4IFM Q1 2012 Test the autocorrelation of a series If there is autocorrelation, data aren’t independently distributed vinzjeannin@hotmail.com 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 18 Autocorrelation at the lag k Χ 21−,ℎ
19. 19. Box.test(DATA\$Val) ESGF 4IFM Q1 2012 Box-Pierce testdata: DATA\$ValX-squared = 188.3263, df = 1, p-value 2.2e-16 vinzjeannin@hotmail.com H0 is rejected, the data aren’t independently distributed 19
20. 20. Let’s try a regression and analyse residuals TReg-lm(DATA\$Val~DATA\$t) plot(DATA\$Val, type=l) ESGF 4IFM Q1 2012 abline(TReg, col=blue) vinzjeannin@hotmail.com 20
21. 21. eps-resid(TReg)ks.test(eps, pnorm)layout(matrix(1:4,2,2))plot(TReg) ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 21
22. 22. Box-Pierce test ESGF 4IFM Q1 2012data: epsX-squared = 187.6299, df = 1, p-value 2.2e-16Residuals aren’t a white noise vinzjeannin@hotmail.comRegression rejected Not a surprise, did the series look stationary? 22 What next then?
23. 23. lag.plot(DATA\$Val, 9, do.lines=FALSE) ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 23Differentiation seems to be interesting
24. 24. Does the differentiation create a stationary series? plot(diff(DATA\$Val), type=l) ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 24
25. 25. ACF PACF par(mfrow=c(2,1)) acf(diff(DATA\$Val),20) ESGF 4IFM Q1 2012 pacf(diff(DATA\$Val),20) vinzjeannin@hotmail.com ACF decreasing PACF cancelling after order 1 25
26. 26. Decreasing ACF ESGF 4IFM Q1 2012PACF cancel after order 1 vinzjeannin@hotmail.com Typically an Autoregressive Process AR(1) ? 26
27. 27. Let’s try to fit an AR(1) model Modl-ar(diff(DATA\$Val),order.max=20) plot(Modl\$aic) ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 27The likelihood for the order 1 is significant
28. 28. ar(diff(DATA\$Val),order.max=20) Call: ar(x = diff(DATA\$Val), order.max = 20) Coefficients: 1 2 3 ESGF 4IFM Q1 2012 0.5925 -0.1669 0.1385 Order selected 3 sigma^2 estimated as 0.8514 vinzjeannin@hotmail.comWe have our coefficient and standard deviation ARDif-diff(DATA\$Val) ARDif[1] [1] 0.3757723We know the first term of our series = 0.3757723 + 0.5925. −1 + 28Here is our model
29. 29. Need to test the residuals ESGF 4IFM Q1 2012Box.test(Modl\$resid) Box-Pierce test vinzjeannin@hotmail.comdata: Modl\$residX-squared = 7e-04, df = 1, p-value = 0.9789 H0 accepted, residuals are independently distributed (white noise) The differentiated series is a AR(1) 29
30. 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 - ESGF 4IFM Q1 2012 0.02784553 -0.01079729 \$se Time Series: Start = 193 End = 199 Frequency = 1 vinzjeannin@hotmail.com [1] 0.923352 1.048210 1.081582 1.091027 1.093739 1.094521 1.0947471201151101051009590 308580 1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101106111116121126131136141146151156161166171176181186191196
31. 31. ESGF 4IFM Q1 2012Another typical example? vinzjeannin@hotmail.com You make the comments! 31
33. 33. Box.test(DATA\$Ser2) Box-Pierce testdata: DATA\$Ser2X-squared = 149.9227, df = 1, p-value 2.2e-16 ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com TReg-lm(DATA\$Ser2~DATA\$t) plot(DATA\$Ser2, type=l) abline(TReg, col=blue) 33
34. 34. eps-resid(TReg) Box.test(eps) Box-Pierce testdata: epsX-squared = 148.5669, df = 1, p-value 2.2e-16 ESGF 4IFM Q1 2012 layout(matrix(1:4,2,2)) plot(TReg) vinzjeannin@hotmail.com 34
35. 35. lag.plot(DATA\$Ser2, 9, do.lines=FALSE) ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 35Much less obvious but clues of autoregression
36. 36. par(mfrow=c(2,1))plot(diff(DATA\$Ser2), type=l)plot(diff(DATA\$Ser2, lag=2), type=l) ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 36
37. 37. par(mfrow=c(2,1))plot(diff(DATA\$Ser2), type=l)plot(diff(DATA\$Ser2, lag=2), type=l) ESGF 4IFM Q1 2012 ACF decreases 2 by 2 vinzjeannin@hotmail.com PACF cancelling after order 2 37
38. 38. First order differentiation, strong AR(2) clues par(mfrow=c(1,1)) ESGF 4IFM Q1 2012 Modl-ar(diff(DATA\$Ser2),order.max=20) plot(Modl\$aic) vinzjeannin@hotmail.com 38
39. 39. Parameters estimation ar(diff(DATA\$Ser2),order.max=20) ESGF 4IFM Q1 2012Call:ar(x = diff(DATA\$Ser2), order.max = 20)Coefficients: 1 2 3 vinzjeannin@hotmail.com 0.5919 -0.8326 0.1086Order selected 3 sigma^2 estimated as 0.877 ARDif-diff(DATA\$Ser2) ARDif[1][1] 0.3757723 39
40. 40. predict(arima(diff(DATA\$Ser2), order = c(2,0,0)), n.ahead = 7)\$predTime Series:Start = 193End = 199Frequency = 1[1] 0.4505213 2.0075741 0.6639701 -1.2321156 -1.1409989 0.38667451.0879588 ESGF 4IFM Q1 2012\$seTime Series:Start = 193End = 199Frequency = 1[1] 0.9220713 1.0332515 1.1413067 1.2938326 1.2957576 1.3932158 1.4080266 vinzjeannin@hotmail.com1151101051009590 408580 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 96
41. 41. Box.test(Modl\$resid) Box-Pierce test ESGF 4IFM Q1 2012data: Modl\$residX-squared = 0.0023, df = 1, p-value = 0.9619 vinzjeannin@hotmail.com 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 41
42. 42. Moving Average Process Main principle ESGF 4IFM Q1 2012 Stationary series with auto correlation of errors = + + 1 −1 + 2 −2 + ⋯ + − vinzjeannin@hotmail.com Parameters of the model White noise MA(n) More difficult to estimate than a AR(n) 42
43. 43. plot(Data, type=l)hist(Data, breaks=20) ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 43
44. 44. acf(Data,20)pacf(Data,20) ESGF 4IFM Q1 2012 ACF cancels after order 1 vinzjeannin@hotmail.com PACF decays to 0 44 ACF PACF suggest MA(1)
45. 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 ESGF 4IFM Q1 2012 -0.4621s.e. 0.0903sigma^2 estimated as 0.937: log likelihood = -138.76, aic = 281.52 vinzjeannin@hotmail.com Box.test(Rslt\$residuals) Box-Pierce test data: Rslt\$residuals X-squared = 0, df = 1, p-value = 0.9967 45 It works, MA(1), 0 mean, parameter -0.4621
46. 46. Fore-predict(Rslt, n.ahead=5)U = Fore\$pred + 2*Fore\$seL = Fore\$pred - 2*Fore\$seminx=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) ESGF 4IFM Q1 2012lines(L, col=blue, lty=dashed) vinzjeannin@hotmail.com 46
47. 47. ESGF 4IFM Q1 2012Another typical example? vinzjeannin@hotmail.com You make the comments! 47
48. 48. plot(Data, type=l)hist(Data, breaks=20) ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 48
49. 49. vinzjeannin@hotmail.com ESGF 4IFM Q1 201249
50. 50. arima(Data, order = c(0, 0, 2),include.mean = FALSE)Call: ESGF 4IFM Q1 2012arima(x = Data, order = c(0, 0, 2), include.mean = FALSE)Coefficients: ma1 ma2 -0.5365 0.6489s.e. 0.0701 0.1044 vinzjeannin@hotmail.comsigma^2 estimated as 1.005: log likelihood = -142.74, aic = 291.48 Box.test(Rslt\$residuals) Box-Pierce testdata: Rslt\$residualsX-squared = 0.0283, df = 1, p-value = 0.8664 50 MA(2)
51. 51. Fore-predict(Rslt, n.ahead=5)U = Fore\$pred + 2*Fore\$seL = Fore\$pred - 2*Fore\$seminx=min(Data,L)maxx=max(Data,U)ts.plot(Data,Fore\$pred,col=1:2,ylim=c(minx,maxx)) ESGF 4IFM Q1 2012lines(U, col=blue, lty=dashed)lines(L, col=blue, lty=dashed) vinzjeannin@hotmail.com 51
52. 52. ARMA Main principle ESGF 4IFM Q1 2012 The series is a function of past values plus current and past values of the noise vinzjeannin@hotmail.com ARMA(p,q) Combines AR(p) MA(q) 52
53. 53. plot(Data, type=l)hist(Data, breaks=20) ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 53
54. 54. ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 54Both ACF and PACF decreases exponentially after order 1
55. 55. Rslt-arima(Data, order = c(1, 0, 1),include.mean = FALSE) RsltCall:arima(x = Data, order = c(1, 0, 1), include.mean = FALSE) ESGF 4IFM Q1 2012Coefficients: ar1 ma1 0.7214 0.7563s.e. 0.0716 0.0721sigma^2 estimated as 0.961: log likelihood = -141.13, aic = 288.27 vinzjeannin@hotmail.com Box.test(Rslt\$residuals) Box-Pierce testdata: Rslt\$residualsX-squared = 0.0098, df = 1, p-value = 0.9213 ARMA(1,1) fits 55
56. 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) ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 56
57. 57. Identification can get tricky at this stage ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 57
58. 58. What do you think? vinzjeannin@hotmail.com ESGF 4IFM Q1 201258
59. 59. Rslt-arima(Data, order = c(4, 0, 3),include.mean = FALSE) RsltCall:arima(x = Data, order = c(4, 0, 3), include.mean = FALSE)Coefficients: ESGF 4IFM Q1 2012 ar1 ar2 ar3 ar4 ma1 ma2 ma3 0.2722 -0.5276 0.0202 -0.2663 0.8765 -0.4672 -0.5248s.e. 0.2018 0.2308 0.1968 0.1546 0.1992 0.1690 0.1882sigma^2 estimated as 1.140: log likelihood = -151.19, aic = 318.38 Box.test(Rslt\$residuals) vinzjeannin@hotmail.com Box-Pierce testdata: Rslt\$residualsX-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,- 59 0.3),ma=c(0.75,-0.5,-0.5)),n=100)
60. 60. Identification can be difficult ESGF 4IFM Q1 2012Easiest model is ARImagine when the series is not stationary… vinzjeannin@hotmail.comStep by step approach, exploration, tries,… Sometimes you find a satisfying model 60 Sometimes you don’t!
61. 61. Conclusion AR MA ARMA Times series vinzjeannin@hotmail.com ESGF 5IFM Q1 201261