Financial Econometric Models III

Financial Econometric Models Vincent JEANNIN – ESGF 5IFM

Q1 2012

Summary of the session (Est. 3h) • Reminder of Last Session • Time Series Analysis Principles • Auto Regressive Process • Moving Average Process • ARMA • Conclusion

Reminder of Last Session

Be logic!

𝑌𝐷𝑖𝑓𝑓 = ln(𝑌)

Differentiation possible

Time can be a factor of a regression

Differentiation can add value

Check ACF/PACF for autocorrelation

Time Series Analysis Principles

Reminders of the 3 steps

Identify

Forecast

Reminders of the 3 components

Seasonality

Residual

𝐵𝑥𝑡 = 𝑥𝑡−1

Difference

∆𝑥𝑡= 𝑥𝑡 − 𝑥𝑡−1

Seasonality Difference

∆30𝑥𝑡 = 𝑥𝑡 − 𝑥𝑡−30

Differentiate series to obtain stationary series

Time series analysis and forecast simpler with stationary series

Different models involved with stationary or heteroscedasticity

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

Acceptable Shortcut

A series is stationary if the mean and the variance are stable

Which one is more likely to be stationary?

About the residuals…

White noise!

Normality test

Have an idea with

Skewness

Kurtosis

Proper tests: KS, Durbin Watson, Portmanteau,…

Auto Regressive Process

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

DATA<-read.csv(file="C:/Users/vin/Desktop/Series1.csv",header=T)

plot(DATA$Val, type="l")

Let’s upload some data

Is this a white noise?

hist(DATA$Val, breaks=20)

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

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

> 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

Let’s try a regression and analyse residuals

TReg<-lm(DATA$Val~DATA$t)

plot(DATA$Val, type="l")

abline(TReg, col="blue")

eps<-resid(TReg)

ks.test(eps, "pnorm")

layout(matrix(1:4,2,2))

plot(TReg)

Box-Pierce test

data: eps

Residuals aren’t a white noise

Regression rejected

Not a surprise, did the series look stationary?

What next then?

lag.plot(DATA$Val, 9, do.lines=FALSE)

Differentiation seems to be interesting

Does the differentiation create a stationary series?

plot(diff(DATA$Val), type="l")

ACF & PACF

par(mfrow=c(2,1))

acf(diff(DATA$Val),20)

pacf(diff(DATA$Val),20)

ACF decreasing

PACF cancelling after order 1

Decreasing ACF

PACF cancel after order 1

Typically an Autoregressive Process

AR(1) ?

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

> ar(diff(DATA$Val),order.max=20)

ar(x = diff(DATA$Val), order.max = 20)

Coefficients:

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

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)

> predict(arima(diff(DATA$Val), order = c(1,0,0)), n.ahead = 7)

Time Series:

Start = 193

End = 199

Frequency = 1

[1] -0.81359048 -0.43300609 -0.22850452 -0.11861853 -0.05957287 -

0.02784553 -0.01079729

Time Series:

Start = 193

End = 199

Frequency = 1

[1] 0.923352 1.048210 1.081582 1.091027 1.093739 1.094521 1.094747

1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101106111116121126131136141146151156161166171176181186191196

Another typical example?

You make the comments!

DATA<-read.csv(file="C:/Users/vin/Desktop/Series2.csv",header=T)

plot(DATA$Ser2, type="l")

hist(DATA$Ser2, breaks=20)

> Box.test(DATA$Ser2)

Box-Pierce test

data: DATA$Ser2

TReg<-lm(DATA$Ser2~DATA$t)

plot(DATA$Ser2, type="l")

abline(TReg, col="blue")

> eps<-resid(TReg)

> Box.test(eps)

Box-Pierce test

data: eps

> layout(matrix(1:4,2,2))

> plot(TReg)

> lag.plot(DATA$Ser2, 9, do.lines=FALSE)

Much less obvious but clues of autoregression

par(mfrow=c(2,1))

plot(diff(DATA$Ser2), type="l")

plot(diff(DATA$Ser2, lag=2), type="l")

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

First order differentiation, strong AR(2) clues

par(mfrow=c(1,1))

Modl<-ar(diff(DATA$Ser2),order.max=20)

plot(Modl$aic)

Parameters estimation

> ar(diff(DATA$Ser2),order.max=20)

ar(x = diff(DATA$Ser2), order.max = 20)

Coefficients:

0.5919 -0.8326 0.1086

Order selected 3 sigma^2 estimated as 0.877

> ARDif<-diff(DATA$Ser2)

> ARDif[1]

[1] 0.3757723

> predict(arima(diff(DATA$Ser2), order = c(2,0,0)), n.ahead = 7)

Time Series:

Start = 193

End = 199

Frequency = 1

[1] 0.4505213 2.0075741 0.6639701 -1.2321156 -1.1409989 0.3866745

1.0879588

Time Series:

Start = 193

End = 199

Frequency = 1

[1] 0.9220713 1.0332515 1.1413067 1.2938326 1.2957576 1.3932158 1.4080266

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

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)

plot(Data, type="l")

hist(Data, breaks=20)

acf(Data,20)

pacf(Data,20)

ACF & PACF suggest MA(1)

ACF cancels after order 1

PACF decays to 0

> 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)

arima(x = Data, order = c(0, 0, 1), include.mean = FALSE)

Coefficients:

-0.4621

s.e. 0.0903

sigma^2 estimated as 0.937: log likelihood = -138.76, aic = 281.52

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")

Another typical example?

You make the comments!

> arima(Data, order = c(0, 0, 2),include.mean = FALSE)

Coefficients:

ma1 ma2

-0.5365 0.6489

s.e. 0.0701 0.1044

Box-Pierce test

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")

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)

Both ACF and PACF decreases exponentially after order 1

> Rslt<-arima(Data, order = c(1, 0, 1),include.mean = FALSE)

> Rslt

Coefficients:

ar1 ma1

0.7214 0.7563

s.e. 0.0716 0.0721

Box-Pierce test

ARMA(1,1) fits

> 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)

Identification can get tricky at this stage

What do you think?

> Rslt<-arima(Data, order = c(4, 0, 3),include.mean = FALSE)

> Rslt

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

Box-Pierce test

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….

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!

Conclusion

Times series

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