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Third Session, MSc 5th Year
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Financial Econometric Models Vincent JEANNIN – ESGF 5IFM
Q1 2012
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Summary of the session (Est. 3h) • Reminder of Last Session • Time Series Analysis Principles • Auto Regressive Process • Moving Average Process • ARMA • Conclusion
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Reminder of Last Session
Be logic!
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𝑌𝐷𝑖𝑓𝑓 = ln(𝑌)
Differentiation possible
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Time can be a factor of a regression
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Differentiation can add value
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Check ACF/PACF for autocorrelation
Time Series Analysis Principles
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Reminders of the 3 steps
Identify
Fit
Forecast
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Reminders of the 3 components
Trend
Seasonality
Residual
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Lag
𝐵𝑥𝑡 = 𝑥𝑡−1
Difference
∆𝑥𝑡= 𝑥𝑡 − 𝑥𝑡−1
Seasonality Difference
∆30𝑥𝑡 = 𝑥𝑡 − 𝑥𝑡−30
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Differentiate series to obtain stationary series
Time series analysis and forecast simpler with stationary series
Different models involved with stationary or heteroscedasticity
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Properties of stationary series
(𝑌1, 𝑌2, 𝑌3, … , 𝑌𝑛)
(𝑌2, 𝑌3, 𝑌4, … , 𝑌𝑛+1)
Same distribution of the following
Distribution not time dependent
Rare occurrence
Stationarity accepted if
𝐸(𝑌𝑡) = 𝜇 Constant in the time
𝐶𝑜𝑣(𝑌𝑡 , 𝑌𝑡−𝑛) Depends only on n
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Acceptable Shortcut
A series is stationary if the mean and the variance are stable
Which one is more likely to be stationary?
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About the residuals…
White noise!
Normality test
Have an idea with
Skewness
Kurtosis
Proper tests: KS, Durbin Watson, Portmanteau,…
Auto Regressive Process
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There is a correlation between current data and previous data
Main principle
𝑋𝑡 = 𝑐 + 𝜑1𝑋𝑡−1 + 𝜑2𝑋𝑡−2 + ⋯+ 𝜑𝑛𝑋𝑡−𝑛 + 𝜀𝑡
𝜑𝑛 Parameters of the model
𝜀𝑛 White noise
If the parameters are identified, the prediction will be easy
AR(n)
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DATA<-read.csv(file="C:/Users/vin/Desktop/Series1.csv",header=T)
plot(DATA$Val, type="l")
Let’s upload some data
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Is this a white noise?
hist(DATA$Val, breaks=20)
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Probably not…
Portmanteau test
Test the autocorrelation of a series
If there is autocorrelation, data aren’t independently distributed
Let’s use Ljung–Box statistics
𝑄 = 𝑛(𝑛 + 2) 𝜌 2𝑘
𝑛 − 𝑘
𝑛
𝑘=1
𝜌 𝑘 Autocorrelation at the lag k
H0: Data are independently distributed H1: Data aren’t independently distributed
𝑄 > Χ21−𝛼,ℎ
With α confidence interval rejection following a Chi Square distribution
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> Box.test(DATA$Val)
Box-Pierce test
data: DATA$Val
X-squared = 188.3263, df = 1, p-value < 2.2e-16
H0 is rejected, the data aren’t independently distributed
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Let’s try a regression and analyse residuals
TReg<-lm(DATA$Val~DATA$t)
plot(DATA$Val, type="l")
abline(TReg, col="blue")
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eps<-resid(TReg)
ks.test(eps, "pnorm")
layout(matrix(1:4,2,2))
plot(TReg)
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Box-Pierce test
data: eps
X-squared = 187.6299, df = 1, p-value < 2.2e-16
Residuals aren’t a white noise
Regression rejected
Not a surprise, did the series look stationary?
What next then?
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lag.plot(DATA$Val, 9, do.lines=FALSE)
Differentiation seems to be interesting
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Does the differentiation create a stationary series?
plot(diff(DATA$Val), type="l")
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ACF & PACF
par(mfrow=c(2,1))
acf(diff(DATA$Val),20)
pacf(diff(DATA$Val),20)
ACF decreasing
PACF cancelling after order 1
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Decreasing ACF
PACF cancel after order 1
Typically an Autoregressive Process
AR(1) ?
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Modl<-ar(diff(DATA$Val),order.max=20)
plot(Modl$aic)
Let’s try to fit an AR(1) model
The likelihood for the order 1 is significant
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> ar(diff(DATA$Val),order.max=20)
Call:
ar(x = diff(DATA$Val), order.max = 20)
Coefficients:
1 2 3
0.5925 -0.1669 0.1385
Order selected 3 sigma^2 estimated as 0.8514
> ARDif<-diff(DATA$Val)
> ARDif[1]
[1] 0.3757723
We have our coefficient and standard deviation
We know the first term of our series
𝑦𝑡 = 0.3757723 + 0.5925. 𝑦𝑡−1 + 𝜀𝑡
Here is our model
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Need to test the residuals
Box.test(Modl$resid)
Box-Pierce test
data: Modl$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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> 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 -
0.02784553 -0.01079729
$se
Time Series:
Start = 193
End = 199
Frequency = 1
[1] 0.923352 1.048210 1.081582 1.091027 1.093739 1.094521 1.094747
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Another typical example?
You make the comments!
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DATA<-read.csv(file="C:/Users/vin/Desktop/Series2.csv",header=T)
plot(DATA$Ser2, type="l")
hist(DATA$Ser2, breaks=20)
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> Box.test(DATA$Ser2)
Box-Pierce test
data: DATA$Ser2
X-squared = 149.9227, df = 1, p-value < 2.2e-16
TReg<-lm(DATA$Ser2~DATA$t)
plot(DATA$Ser2, type="l")
abline(TReg, col="blue")
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> eps<-resid(TReg)
> Box.test(eps)
Box-Pierce test
data: eps
X-squared = 148.5669, df = 1, p-value < 2.2e-16
> layout(matrix(1:4,2,2))
> plot(TReg)
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> lag.plot(DATA$Ser2, 9, do.lines=FALSE)
Much less obvious but clues of autoregression
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par(mfrow=c(2,1))
plot(diff(DATA$Ser2), type="l")
plot(diff(DATA$Ser2, lag=2), type="l")
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par(mfrow=c(2,1))
plot(diff(DATA$Ser2), type="l")
plot(diff(DATA$Ser2, lag=2), type="l")
ACF decreases 2 by 2
PACF cancelling after order 2
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First order differentiation, strong AR(2) clues
par(mfrow=c(1,1))
Modl<-ar(diff(DATA$Ser2),order.max=20)
plot(Modl$aic)
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Parameters estimation
> ar(diff(DATA$Ser2),order.max=20)
Call:
ar(x = diff(DATA$Ser2), order.max = 20)
Coefficients:
1 2 3
0.5919 -0.8326 0.1086
Order selected 3 sigma^2 estimated as 0.877
> ARDif<-diff(DATA$Ser2)
> ARDif[1]
[1] 0.3757723
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> 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
$se
Time Series:
Start = 193
End = 199
Frequency = 1
[1] 0.9220713 1.0332515 1.1413067 1.2938326 1.2957576 1.3932158 1.4080266
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The more factors the harder the prediction is
> Box.test(Modl$resid)
Box-Pierce test
data: Modl$resid
X-squared = 0.0023, df = 1, p-value = 0.9619
Model accepted
The more factors there are the more stationary need to be the series for a good prediction
Moving Average Process
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Stationary series with auto correlation of errors
Main principle
𝑋𝑡 = 𝜇 + 𝑍𝑡 + 𝜑1𝑍𝑡−1 + 𝜑2𝑍𝑡−2 + ⋯+ 𝜑𝑛𝑍𝑡−𝑛
𝜑𝑛 Parameters of the model
𝑍𝑛 White noise
More difficult to estimate than a AR(n)
MA(n)
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plot(Data, type="l")
hist(Data, breaks=20)
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acf(Data,20)
pacf(Data,20)
ACF & PACF suggest MA(1)
ACF cancels after order 1
PACF decays to 0
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> Box.test(Rslt$residuals)
Box-Pierce test
data: Rslt$residuals
X-squared = 0, df = 1, p-value = 0.9967
It works, MA(1), 0 mean, parameter -0.4621
> arima(Data, order = c(0, 0, 1),include.mean = FALSE)
Call:
arima(x = Data, order = c(0, 0, 1), include.mean = FALSE)
Coefficients:
ma1
-0.4621
s.e. 0.0903
sigma^2 estimated as 0.937: log likelihood = -138.76, aic = 281.52
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Fore<-predict(Rslt, n.ahead=5)
U = Fore$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")
lines(L, col="blue", lty="dashed")
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Another typical example?
You make the comments!
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plot(Data, type="l")
hist(Data, breaks=20)
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> arima(Data, order = c(0, 0, 2),include.mean = FALSE)
Call:
arima(x = Data, order = c(0, 0, 2), include.mean = FALSE)
Coefficients:
ma1 ma2
-0.5365 0.6489
s.e. 0.0701 0.1044
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
MA(2)
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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$pred,col=1:2,
ylim=c(minx,maxx))
lines(U, col="blue", lty="dashed")
lines(L, col="blue", lty="dashed")
ARMA
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The series is a function of past values plus current and past values of the noise
Main principle
ARMA(p,q)
Combines AR(p) & MA(q)
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plot(Data, type="l")
hist(Data, breaks=20)
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Both ACF and PACF decreases exponentially after order 1
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> Rslt<-arima(Data, order = c(1, 0, 1),include.mean = FALSE)
> Rslt
Call:
arima(x = Data, order = c(1, 0, 1), include.mean = FALSE)
Coefficients:
ar1 ma1
0.7214 0.7563
s.e. 0.0716 0.0721
sigma^2 estimated as 0.961: log likelihood = -141.13, aic = 288.27
> Box.test(Rslt$residuals)
Box-Pierce test
data: Rslt$residuals
X-squared = 0.0098, df = 1, p-value = 0.9213
ARMA(1,1) fits
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> par(mfrow=c(1,1))
> Fore<-predict(Rslt, n.ahead=5)
> U = Fore$pred + 2*Fore$se
> L = Fore$pred - 2*Fore$se
> minx=min(Data,L)
> maxx=max(Data,U)
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Identification can get tricky at this stage
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What do you think?
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> Rslt<-arima(Data, order = c(4, 0, 3),include.mean = FALSE)
> Rslt
Call:
arima(x = Data, order = c(4, 0, 3), include.mean = FALSE)
Coefficients:
ar1 ar2 ar3 ar4 ma1 ma2 ma3
0.2722 -0.5276 0.0202 -0.2663 0.8765 -0.4672 -0.5248
s.e. 0.2018 0.2308 0.1968 0.1546 0.1992 0.1690 0.1882
sigma^2 estimated as 1.140: log likelihood = -151.19, aic = 318.38
> Box.test(Rslt$residuals)
Box-Pierce test
data: Rslt$residuals
X-squared = 0.2953, df = 1, p-value = 0.5869
Data<-arima.sim(model=list(ar=c(0.5,-0.5,0.3,-
0.3),ma=c(0.75,-0.5,-0.5)),n=100)
Was supposed to fit pretty wel….
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Identification can be difficult
Easiest model is AR
Imagine when the series is not stationary…
Step by step approach, exploration, tries,…
Sometimes you find a satisfying model
Sometimes you don’t!
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Conclusion
AR
MA
ARMA
Times series
