Homework 5, Problem 1 Andrii Baryshpolets 6 April 2017

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1 Homework 5, Problem 1 Andrii Baryshpolets 6 April 2017 Total Private Residential Construction Spending library(quandl) Warning: package 'Quandl' was built under R version Loading required package: xts Loading required package: zoo Attaching package: 'zoo' The following objects are masked from 'package:base': as.date, as.date.numeric library(zoo) library(xts) library(dygraphs) library(knitr) library(forecast) Loading required package: timedate This is forecast 7.3 library(urca) y <- Quandl("FRED/PRRESCON", type="ts") We plot the original and log-transformed Total Private Residential Construction Spending. ly <- log(y) plot(y, xlab=" Years ", ylab="", main="total Private Residential Construction Spending" ) 1

2 Total Private Residential Construction Spending Years plot(ly, xlab="years ", ylab="", main="log Total Private Residential Construction Spending" ) 2

3 Log Total Private Residential Construction Spending Years The time series shows an exponential trend (with a possible structural break) as well as a seasonal pattern is present.for further analysis we will use log-transformed data untill the end of 2013 to build an estimation model. endof2013 < (11/12) ly.est <- window(ly, end = endof2013) ly.pred <- window(ly, start = endof (1/12)) dl.y <- diff(ly.est,lag=1) dl.y12 <- diff(ly.est,12) dl.y12_1 <- diff(diff(ly.est,12), 1) par(mfrow=c(2,2)) plot(ly.est, xlab="", ylab="", main=expression(log(y))) plot(dl.y, xlab="", ylab="", main=expression(paste(delta, "log(y)"))) plot(dl.y12, xlab="", ylab="", main=expression(paste(delta[12], "log(y)"))) plot(dl.y12_1, xlab="", ylab="", main=expression(paste(delta, Delta[12], "log(y)"))) 3

4 log(y) log(y) log(y) 12 log(y) The time series in differences and seasonal differences is the only that looks weakly stationary. To verify this is true, we look at ACFs and PACFs. library(forecast) maxlag <-24 par(mfrow=c(2,4)) Acf(ly.est, type='correlation', lag=maxlag, ylab="", main=expression(paste("acf for log(y)"))) Acf(dl.y, type='correlation', lag=maxlag, ylab="", main=expression(paste("acf for ", Delta,"log(y)"))) Acf(dl.y12, type='correlation', lag=maxlag, ylab="", main=expression(paste("acf for ", Delta[12], "log(y Acf(dl.y12_1, type='correlation', lag=maxlag, ylab="", main=expression(paste("acf for ", Delta, Delta[12 Acf(ly, type='partial', lag=maxlag, ylab="", main=expression(paste("pacf for log(y)"))) Acf(dl.y, type='partial', lag=maxlag, ylab="", main=expression(paste("pacf for ", Delta, "log(y)"))) Acf(dl.y12, type='partial', lag=maxlag, ylab="", main=expression(paste("pacf for ", Delta[12], "log(y)") Acf(dl.y12_1, type='partial', lag=maxlag, ylab="", main=expression(paste("pacf for ", Delta,Delta[12], " 4

5 ACF for log(y) ACF for log(y) ACF for 12 log(y) ACF for 12 log(y) PACF for log(y) PACF for log(y) PACF for 12 log(y) PACF for 12 log(y) So the last time series is the only weekly stationary according to ACFs and PACFs. Tests for presence of a unit root ADF test library(tseries) adf.test(ly) Augmented Dickey-Fuller Test data: ly Dickey-Fuller = , order = 6, p-value = alternative hypothesis: stationary By ADF test we cannot reject the H 0 hypothesis that the time series has a unit root. KPSS test library(tseries) kpss.test(ly, null="trend") Warning in kpss.test(ly, null = "Trend"): p-value smaller than printed p- value KPSS Test for Trend Stationarity 5

6 data: ly KPSS Trend = , Truncation lag parameter = 3, p-value = 0.01 library(urca) ly.urkpss <- ur.kpss(ly, type="tau", lags="short") summary(ly.urkpss) # # KPSS Unit Root Test # # Test is of type: tau with 5 lags. Value of test-statistic is: Critical value for a significance level of: 10pct 5pct 2.5pct 1pct critical values By KPSS test we reject the H 0 hypothesis that the time series is stationary. Building the model We use autoarima function to find the best specification for our model. M12.bic <- auto.arima(ly.est, ic="bic", seasonal=true, stationary=false, stepwise=false) M12.bic Series: ly.est ARIMA(1,1,2)(0,0,2)[12] Coefficients: ar1 ma1 ma2 sma1 sma s.e sigma^2 estimated as : log likelihood= AIC= AICc= BIC= M12.aic <- auto.arima(ly.est, ic="aic", seasonal=true, stationary=false, stepwise=false) M12.aic Series: ly.est ARIMA(1,1,2)(0,0,2)[12] with drift Coefficients: ar1 ma1 ma2 sma1 sma2 drift s.e sigma^2 estimated as : log likelihood= AIC= AICc= BIC= M12.aicc <- auto.arima(ly.est, ic="aicc", seasonal=true, stationary=false, stepwise=false) M12.aicc 6

7 Series: ly.est ARIMA(1,1,2)(0,0,2)[12] with drift Coefficients: ar1 ma1 ma2 sma1 sma2 drift s.e sigma^2 estimated as : log likelihood= AIC= AICc= BIC= ARIMA(1,1,0)(1,0,0)[12] is the best specification suggested by Autoarima using BIC, AIC, AICc criterias. Forecast At first, we construct the multiplestep forecast. library(forecast) M12.bic.f.h <- forecast(m12.bic, length(ly.pred)) plot(m12.bic.f.h, type="o", pch=16, xlim=c(2012,2017), ylim=c(9.5,11), main="arima Model Multistep Forecast - TPRCS") lines(m12.bic.f.h$mean, type="p", pch=16, lty="dashed", col="blue") lines(ly, type="o", pch=16, lty="dashed") ARIMA Model Multistep Forecast TPRCS Rolling scheme forecast 7

8 library(forecast) M12.bic.f.rol <- zoo() for(i in 1:length(ly.pred)) {M12.bic.rsc <- window( ly, start=1993+(i-1)/12, end=endof2013+(i-1)/12 ) M12.bic.updt <- arima(m12.bic.rsc, order=c(1,1,2), seasonal=list(order=c(0,0,2),period=na)) M12.bic.f.rol <- c(m12.bic.f.rol, forecast(m12.bic.updt, 1)$mean) } M12.bic.f.rol <- as.ts(m12.bic.f.rol) accuracy(m12.bic.f.rol, ly.pred) ME RMSE MAE MPE MAPE Test set e ACF1 Theil's U Test set plot(m12.bic.f.h, type="o", pch=16, xlim=c(2012,2017), ylim=c(9.5,11), main="multistep vs. Rolling Scheme forecasts - TPRCS") lines(m12.bic.f.h$mean, type="p", pch=16, lty="dashed", col="blue") lines(m12.bic.f.rol, type="p", pch=16, lty="dashed", col="red") lines(ly, type="o", pch=16, lty="dashed") Multistep vs. Rolling Scheme forecasts TPRCS Now we compare our model to the restricted Seasonal AR (9,1,0)(1,0,0)[12], which was offered in the class. ARIMA(1,1,0)(1,0,0)[12] class.arima <- arima(ly.est, order=c(9,1,0), seasonal=list(order=c(1,0,0),period=na)) class.arima 8

9 Call: arima(x = ly.est, order = c(9, 1, 0), seasonal = list(order = c(1, 0, 0), period = NA)) Coefficients: ar1 ar2 ar3 ar4 ar5 ar6 ar7 ar s.e ar9 sar s.e sigma^2 estimated as : log likelihood = , aic = Restricted ARIMA(1,1,0)(1,0,0)[12] class.restrarima <- arima(ly.est, order=c(9,1,0), seasonal=list(order=c(1,0,0),period=na), class.restrarima transform.pa Call: arima(x = ly.est, order = c(9, 1, 0), seasonal = list(order = c(1, 0, 0), period = NA), transform.pars = FALSE, fixed = c(na, NA, 0, 0, 0, 0, NA, NA, NA, NA)) Coefficients: ar1 ar2 ar3 ar4 ar5 ar6 ar7 ar8 ar9 sar s.e sigma^2 estimated as : log likelihood = , aic = We compare error accuracy between our model and restricted arima. library(forecast) par(mfrow=c(2,1)) accuracy(m12.bic.f.rol, ly.pred) ME RMSE MAE MPE MAPE Test set e ACF1 Theil's U Test set accuracy(class.restrarima) ME RMSE MAE MPE MAPE Training set MASE ACF1 Training set Restricted ARIMA rolling scheme library(forecast) class.restrarima.f.rol <- zoo() for(i in 1:length(ly.pred)) { class.restrarima.rsc <- window( ly, start=1993+(i-1)/12, end=endof2013+(i-1)/12 ) class.restrarima.updt <- arima(class.restrarima.rsc, order=c(9,1,0), seasonal=list(order=c(1,0,0),period class.restrarima.f.rol <- c(class.restrarima.f.rol, forecast(class.restrarima.updt, 1)$mean) 9

10 } class.restrarima.f.rol <- as.ts(class.restrarima.f.rol) accuracy(class.restrarima.f.rol, ly.pred) ME RMSE MAE MPE MAPE Test set ACF1 Theil's U Test set We compare our model and the restricted ARIMA plot(m12.bic.f.rol, type="o", pch=16, xlim=c(2013,2017), ylim=c(9.8,10.8), main="arima(9,1,0)(1,0,0) vs. ARIMA(1,1,2)(0,0,2) Rolling Scheme forecasts - TPRCS") lines(class.restrarima.f.rol, type="p", pch=16, lty="dashed", col="blue") lines(m12.bic.f.rol, type="p", pch=16, lty="dashed", col="red") lines(ly, type="o", pch=16, lty="dashed") ARIMA(9,1,0)(1,0,0) vs. ARIMA(1,1,2)(0,0,2) Rolling Scheme forecasts TPRCS M12.bic.f.rol Time Conclusion If we visually compare two rolling scheme forecasts for both ARIMA(9,1,0)(1,0,0) and ARIMA(1,1,2)(0,0,2), we can barely tell which one is a better forecast. However, discussed in class model seems to be preferable as it has lower ME, RMSE, MAE, Mape and only MPE is lower for our suggested model. 10

Transformations for variance stabilization

FORECASTING USING R Transformations for variance stabilization Rob Hyndman Author, forecast Variance stabilization If the data show increasing variation as the level of the series increases, then a transformation