COMPUTER SESSION 3: ESTIMATION AND FORECASTING.

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1 UPPSALA UNIVERSITY Department of Mathematics JR Analysis of Time Series 1MS014 Spring 2010 COMPUTER SESSION 3: ESTIMATION AND FORECASTING. 1 Introduction The purpose of this exercise is two-fold: (i) By studying simulated data, hopefully a better understanding of notions and methodology may be gained from the student s perspective of the course; (ii) By examining real data, you practice on some of the important tools for an analyst working with time-series data after the course. However, model building is a difficult issue; experience and knowledge of the particular field of application is important. 2 Estimation in ARIMA models 2.1 The AR(1)-process Simulate 100 observations of an AR(1) process with φ = 0.8. The method-of-moment estimate can be found as follows: x <- arima.sim(n=100,list(ar=0.8)); acf(x)$acf[1] We may also, since we deal specifically with an AR process, use the function ar for estimation: ar(x) We now turn to ML estimation for the AR(1) process. The routine arima is useful for estimation of AR, MA, ARMA and ARIMA process. In this case, estmod <- arima(x,order=c(1,0,0),method= ML ) phi <- estmod$coef[1] You do not need to save the estimated model, as we did in the object estmod, but may access the parameter estimate directly by arima(x,order=c(1,0,0),method= ML )$coef[1] Now, study the sampling distribution of the parameter φ by simulating 1000 time series, each of 100 observations. The distribution is visualized in a histogram (by hist). 1

2 a <- 0 for (k in 1:1000) { x <- arima.sim(n=100,list(ar=0.8)); a[k] <- arima(x,order=c(1,0,0),method= ML )$coef[1] } hist(a); grid() Would you claim that the distribution seems normal? Let us check unbiasedness by comparing the true value (φ = 0.8) to the value obtained by mean(a). Also, compare the estimate from data, sd(a), with the large-sample result: V[ ˆφ] 1 = ˆφ 2 n Conclusion? 2.2 The AR(2) process Now, consider an AR(2) process. Experiment with parameters from various regions (cf. the figure from computer session 2). Estimation is performed by Yule Walker or Maximum Likelihood. Compare the results! x <- arima.sim(n=100,list(ar=c(0.6,0.3))) mest.yw <- ar(x,order.max=2,method= yw ) mest.ml <- ar(x,order.max=2,method= mle ) 2.3 MA, ARMA and ARIMA processes Use of arima for estimation in the more complex models is straight-forward; here are some examples: x1 <- arima.sim(n=200,list(ar=0.7,ma=-0.4)) # An ARMA(1,1) mest.ml <- arima(x1,order=c(1,0,1),method= ML ) x2 <- arima.sim(n=200,list(order=c(0,1,1),ma=0.1)) mest.ml <- arima(x2,order=c(0,1,1),method= ML ) # An IMA(1,1) 2.4 Model choice In all situations treated so far, we knew what type of process to fit (the input parameter order in the call of arima). In reality, however, the sample ACF and PACF can give us hints what model to fit. When in doubt and comparing several models, measures like AIC (returned by arima) might be helpful, as well as checking the variance of the estimated residuals. Consider the following sequence of commands: x1 <- arima.sim(n=200,list(ar=0.7,ma=-0.4)) m1 <- arima(x1,order=c(1,0,0),method= ML ) m2 <- arima(x1,order=c(0,0,1),method= ML ) m3 <- arima(x1,order=c(1,0,1),method= ML ) m1$aic; m2$aic; m3$aic m1$sigma^2; m2$sigma^2; m3$sigma^3 # An ARMA(1,1) 2

3 Repeat the simulations several times. The lowest ˆσ 2 and AIC is to prefer; in most cases, this should occur for fitted model m3. Experiment with the influence of sample size, for instance n = 20, n = Forecasting In this section, we first perform forecasting on simulated data, whereafter analysis of real datasets is made. Actually, the routine arima is the basis also for forecasting. We illustrate with a simulated AR(2) process (with a mean µ = 100). In all, 52 values are simulated, but the last 12 values are set aside to compare forecast to the actual values. library(tsa) x <- arima.sim(n=52,list(ar=c(1.5,-0.75)))+100 actual <- window(x,start=41) xdat <- window(x,end=40) model <- arima(xdat,order=c(2,0,0),method= ML ); model Forecasting, 12 time steps ahead: result=plot(model,n.ahead=12,ylab= Series & Forecasts,col=NULL,pch=19) abline(h=coef(model)[names(coef(model))== intercept ]) Note how the forecasts mimic the pseudo-periodic nature of the series but also decay toward the series mean µ as they go further into the future. We now compare the forecasts to the values set aside: forecast=result$pred; cbind(actual,forecast) Finally, we plot the forecasts together with 95% forecast limits. Do the actual values fall within the forecast limits? plot(model,n1=25,n.ahead=12,ylab= Series, Forecasts, Actuals & Limits, pch=19) points(x=(41:52),y=actual,pch=3) # Future actual values as + signs abline(h=coef(model)[names(coef(model))== intercept ]) 4 Case study: Quarterly earnings of Johnson and Johnson 4.1 Initial studies. Stationarity, transformation The quarterly earnings per share for of the U.S. company Johnson & Johnson are saved in the file JJ. We first plot the time series as well as the logarithm of the series. Do log transformation seem to be advantageous in this case? data(jj); win.graph(width=7,height=3,pointsize=8); oldpar=par; par(mfrow=c(1,2)) plot(jj,ylab= Earnings,type= o ); plot(log(jj),ylab= Log(Earnings),type= o ) par=oldpar 3

4 In the plot at the left, note that at the higher values of the series there is also much more variation. The plot of the logs, at the right, shows much more equal variation at all levels of the series. Hence, use logs for all of the remaining modelling. The series is clearly not stationary. Let us take first differences and plot that series. Does stationarity now seem reasonable? plot(diff(log(jj)),ylab= Difference of Log(Earnings),type= o ) We do not expect stationary series to have less variability in the middle of the series as this one does, but we might assume a stationary model and see where this leads us. 4.2 Modelling seasonality Calculate and graph the sample ACF of the first differences. Interpret the results. acf(diff(log(as.vector(jj))),ci.type= ma ) In this quarterly series, the strongest autocorrelations are at the seasonal lags of 4, 8, 12, and 16. Clearly, we need to address the seasonality in this series. Display the plot of seasonal differences and the first differences. Interpret the plot. (Recall that for quarterly data, a season is of length 4.) series=diff(diff(log(jj),lag=4)) plot(series,ylab= Seasonal & First Difference,type= l ) points(y=series,x=time(series),pch=as.vector(season(series))) The various quarters seem to be quite randomly distributed among high, middle, and low values, so that most of the seasonality is accounted for in the seasonal difference. Graph and interpret the sample ACF of seasonal differences with the first differences: acf(as.vector(series),ci.type= ma ) The only significant autocorrelations are at lags 1 and 7. Lag 4 (the quarterly lag) is nearly significant. Fit the model ARIMA(0,1,1) (0,1,1) 4, and assess the significance of the estimated coefficients. model=arima(log(jj),order=c(0,1,1),seasonal=list(order=c(0,1,1),period=4)); 4.3 Diagnostics We know perform diagnostic tests on the residuals, e.g. by using tsdiag: win.graph(width=5,height=5) tsdiag(model) These diagnostic plots do not show any inadequacies with the model. No outliers are detected and there is little autocorrelation in the residuals. Check of normality, by simple visual checks or Shapiro Wilk test: hist(residuals(model),xlab= Residuals ) qqnorm(residuals(model)); qqline(residuals(model)); shapiro.test(residuals(model)) Do normality of the error terms seems reasonable? 4

5 4.4 Forecasting Calculate and plot forecasts for the next two years of the series. Be sure to include forecast limits. win.graph(width=7,height=3,pointsize=8) plot(model,n1=c(1978,1),n.ahead=8,pch=19,ylab= Log(Earnings) ) The forecasts follow the general pattern of seasonality and trend in the earnings series and the forecast limits give a good indication of the confidence in these forecasts. Lastly, we display the forecasts in orginal terms. plot(model,n1=c(1978,1),n.ahead=8,pch=19,ylab= Earnings,transform=exp) In original terms, the uncertainty in the forecasts is easier to understand. 5