[SOLVED] ETW3420 Principles of Forecasting and Applications

ETW3420 Principles of Forecasting and Applications

Principles of Forecasting and Applications

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Topic 5 Exercises Part 2

Question 1

(a) Show that the forecast variance for an ETS(A,N,N) model is given by

1 + 2(h 1)

(b) Write down the corresponding 95% prediction interval as a function of `T , , h and ,

assuming Gaussian errors.

Question 2

For this question, use the quarterly UK passenger vehicle production data from 1977Q1

2005Q1 (data set ukcars).

(a) Plot the data and describe the main features of the series.

(b) Use ets() to choose a seasonal model for the data.

(c) Check the residuals of the ETS model.

(d) Produce and plot the forecasts for h = 24 from the fitted ets model. Comment on why

the forecasts show no trend.

Question 3

For this question, use the monthly Australian short-term overseas visitors data, May 1985

April 2005. (Data set: visitors.)

(a) Make a time plot of your data and describe the main features of the series.

(b) Split your data into a training set and a test set comprising the last two years of

available data. Forecast the test set using Holt-Winters multiplicative method.

#Splitting data

train <- window(visitors, end=end(visitors)-c(2,0))test <- window(visitors, start = end(visitors) – c(2,-1))#Forecast test set and plot forecastsfcast <- hw(train, h=24, seasonal=”multiplicative”)autoplot(fcast) +autolayer(visitors)(c) Why is multiplicative seasonality necessary here?(d) Forecast the two-year test set using each of the following methods: an ETS model; an ETS model applied to a Box-Cox transformed series; a seasonal naive method; an STL decomposition applied to the Box-Cox transformed data followed by an ETSmodel applied to the seasonally adjusted (transformed) data.#Forecastsf1 <- forecast(ets(train), h = 24)#Alternatively, ets(train) %>% forecast(h = 24)

f2 <- forecast(ets(train, lambda=0), h = 24)f3 <- snaive(train, h = 24)f4 <- stlf(train, lambda = 0, etsmodel = “ZZN”, h = 24)#Why ‘N’ for Seasonal component?#Print output#Plot forecastsautoplot(visitors) +autolayer(f1, PI=FALSE, series=”ETS”) +autolayer(f2, PI=FALSE, series=”ETS with Box-Cox”) +autolayer(f3, PI=FALSE, series=”Seasonal naive”) +autolayer(f4, PI=FALSE, series=”STL+ETS with Box-Cox”)(e) Which method gives the best forecasts? Does it pass the residual tests?(f) Compare the same four methods using time series cross-validation with the tsCV func-tion instead of using a training and test set. Do you come to the same conclusions? Recall the usage of tsCV: tsCV(y, forecastfunction, h = 1, window =NULL,…). The second argument requires us to specify the forecast function. Since snaive and stlf are inbuilt forecast functions, we can specify them directlyfor the forecastfunction argument. That is, tsCV(visitors, forecastfunction= snaive, …) and tsCV(visitors, forecastfunction = stlf,…). However, we will have to write the forecast function for the ETS models that will thenenter as the argument for forecastfunction. ets() does not produce forecasts forus – it only selects the ETS components and estimates the corresponding parametersof the models. So lets write a forecast function that produces forecasts for the ETS model:#Lets call the function f1f1 <- function(y, h) {forecast(ets(y), h = h)Subsequently, we can use the tsCV() function, save the subsequent forecast residuals andcalculate the mean squared error (Recall that the tsCV() function returns a vector of forecaste1 <- tsCV(y = visitors, forecastfunction = f1, h = 1)mean(e1^2, na.rm = T) Lets now write the forecast function that produces forecasts for the ETS model appliedto a Box-Cox transformed series:#Lets call the function f2f2 <- function(y, h) {forecast(ets(y, lambda = 0), h = h)Subsequently, we can use the tsCV() function, save the subsequent forecast residuals andcalculate the mean squared error (Recall that the tsCV() function returns a vector of forecaste2 <- tsCV(y = visitors, forecastfunction = f2, h = 1)mean(e2^2, na.rm = T)We can now proceed with time series cross-validation with the seasonal naive and STLFfunctions and calculate their respective mean squared error:e3 <- tsCV(visitors, forecastfunction = snaive)e4 <- tsCV(visitors, forecastfunction = stlf, lambda=0, h = 1)mean(e3^2, na.rm = T)mean(e4^2, na.rm = T)Now the STLF method appears better (based on 1-step forecasts), even though it was worston the test set earlier.Question 4The fets() function below returns ETS forecasts.fets <- function(y, h) {forecast(ets(y), h = h)(a) Apply tsCV() for a forecast horizon of h = 4, for both ETS and seasonal naive meth-ods to the qcement data. Do so by using the newly created fets() and the existingsnaive() functions as your forecast function arguments. Recall that the tsCV() func-tion returns a vector of forecast errors.e1 <- tsCV(qcement, fets, h=4)e2 <- tsCV(qcement, snaive, h=4)(b) Compute the MSE of the resulting 4-step-ahead errors. (Hint: make sure you removemissing values.) Comment on which forecasts are more accurate. Is this what youcolMeans(e1^2, na.rm=TRUE)colMeans(e2^2, na.rm=TRUE)Question 1Question 2Question 3Question 4 CS: assignmentchef QQ: 1823890830 Email: [email protected]