Below are the solutions to these exercises on forecasting with the extended ARIMA model.
#################### # # # Exercise 1 # # # #################### require(ggplot2) require(gridExtra) df <- read.csv("Icecream.csv") p1 <- ggplot(df, aes(x = X, y = cons)) + ylab("Consumption") + xlab("") + geom_line() + expand_limits(x = 0, y = 0) p2 <- ggplot(df, aes(x = X, y = temp)) + ylab("Temperature") + xlab("") + geom_line() + expand_limits(x = 0, y = 0) p3 <- ggplot(df, aes(x = X, y = income)) + ylab("Income") + xlab("Period") + geom_line() + expand_limits(x = 0, y = 0) grid.arrange(p1, p2, p3, ncol=1, nrow=3)
#################### # # # Exercise 2 # # # #################### require(forecast) fit_cons <- auto.arima(df$cons) fcast_cons <- forecast(fit_cons, h = 6) #################### # # # Exercise 3 # # # #################### require(forecast) autoplot.forecast(fcast_cons)
#################### # # # Exercise 4 # # # #################### require(forecast) accuracy(fit_cons)
## ME RMSE MAE MPE MAPE ## Training set 0.0001020514 0.03525274 0.02692065 -0.9289035 7.203075 ## MASE ACF1 ## Training set 0.8200619 -0.1002901
# The MASE is equal to 0.8200619 #################### # # # Exercise 5 # # # #################### require(forecast) fit_cons_temp <- auto.arima(df$cons, xreg = df$temp) fcast_temp <- c(70.5, 66, 60.5, 45.5, 36, 28) fcast_cons_temp <- forecast(fit_cons_temp, xreg = fcast_temp, h = 6) autoplot.forecast(fcast_cons_temp)
#################### # # # Exercise 6 # # # #################### summary(fcast_cons_temp)
## ## Forecast method: Regression with ARIMA(0,1,0) errors ## ## Model Information: ## Series: df$cons ## Regression with ARIMA(0,1,0) errors ## ## Coefficients: ## df$temp ## 0.0028 ## s.e. 0.0007 ## ## sigma^2 estimated as 0.001108: log likelihood=56.03 ## AIC=-108.06 AICc=-107.59 BIC=-105.32 ## ## Error measures: ## ME RMSE MAE MPE MAPE MASE ## Training set 0.002563685 0.03216453 0.02414157 0.564013 6.478971 0.7354048 ## ACF1 ## Training set -0.1457977 ## ## Forecasts: ## Point Forecast Lo 80 Hi 80 Lo 95 Hi 95 ## 31 0.5465774 0.5039101 0.5892446 0.4813234 0.6118313 ## 32 0.5337735 0.4734329 0.5941142 0.4414905 0.6260566 ## 33 0.5181244 0.4442225 0.5920263 0.4051012 0.6311476 ## 34 0.4754450 0.3901105 0.5607796 0.3449371 0.6059529 ## 35 0.4484147 0.3530078 0.5438217 0.3025024 0.5943270 ## 36 0.4256524 0.3211393 0.5301654 0.2658135 0.5854913
# the coefficient for the temperature variable is 0.0028 # the standard error of the coefficient is 0.0007 # the mean absolute scaled error is 0.7354048, which is smaller than # the error for the initial model (0.8200619) #################### # # # Exercise 7 # # # #################### require(lmtest) coeftest(fit_cons_temp)
## ## z test of coefficients: ## ## Estimate Std. Error z value Pr(>|z|) ## df$temp 0.00284529 0.00074313 3.8288 0.0001288 *** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
# The p-value is equal to 0.0001288, it implies that the coefficient is significant at # 5% level. #################### # # # Exercise 8 # # # #################### temp_column <- matrix(df$temp, ncol = 1) income <- c(NA, NA, df$income) income_matrix <- embed(income, 3) vars_matrix <- cbind(temp_column, income_matrix) print(vars_matrix)
## [,1] [,2] [,3] [,4] ## [1,] 41 78 NA NA ## [2,] 56 79 78 NA ## [3,] 63 81 79 78 ## [4,] 68 80 81 79 ## [5,] 69 76 80 81 ## [6,] 65 78 76 80 ## [7,] 61 82 78 76 ## [8,] 47 79 82 78 ## [9,] 32 76 79 82 ## [10,] 24 79 76 79 ## [11,] 28 82 79 76 ## [12,] 26 85 82 79 ## [13,] 32 86 85 82 ## [14,] 40 83 86 85 ## [15,] 55 84 83 86 ## [16,] 63 82 84 83 ## [17,] 72 80 82 84 ## [18,] 72 78 80 82 ## [19,] 67 84 78 80 ## [20,] 60 86 84 78 ## [21,] 44 85 86 84 ## [22,] 40 87 85 86 ## [23,] 32 94 87 85 ## [24,] 27 92 94 87 ## [25,] 28 95 92 94 ## [26,] 33 96 95 92 ## [27,] 41 94 96 95 ## [28,] 52 96 94 96 ## [29,] 64 91 96 94 ## [30,] 71 90 91 96
#################### # # # Exercise 9 # # # #################### require(forecast) fit_vars_0 <- auto.arima(df$cons, xreg = vars_matrix[, 1:2]) fit_vars_1 <- auto.arima(df$cons, xreg = vars_matrix[, 1:3]) fit_vars_2 <- auto.arima(df$cons, xreg = vars_matrix[, 1:4]) print(fit_vars_0$aic)
## [1] -113.3357
print(fit_vars_1$aic)
## [1] -111.9228
print(fit_vars_2$aic)
## [1] -110.2497
# The AIC can be used because the models have the same order of integration (0). # The model with the lowest value of the AIC is the first model. # Its AIC is equal to -113.3357. #################### # # # Exercise 10 # # # #################### require(forecast) expected_temp_income <- matrix(c(fcast_temp, 91, 91, 93, 96, 96, 96), ncol = 2, nrow = 6) fcast_cons_temp_income <- forecast(fit_vars_0, xreg = expected_temp_income, h = 6) autoplot.forecast(fcast_cons_temp_income)
accuracy(fit_cons)[, "MASE"]
## [1] 0.8200619
accuracy(fit_cons_temp)[, "MASE"]
## [1] 0.7354048
accuracy(fit_vars_0)[, "MASE"]
## [1] 0.7290753
# the model with two external regressors has the lowest # mean absolute scaled error (0.7290753)
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Nice exercise, but i have a question.
isn`t there any problem of comparing MASE from E6 to E4 cause with the new ARIMA has different parameters E4.(2,0,1) and E6. (0,1,0) ?
Thank you for the question.
The mean absolute scaled error (MASE), as well as other error measures (such as RMSE, MAE, MAPE), is a measure of the difference between the forecast values and the actual (observed) values. It does not matter what models (ARIMA, exponential smoothing, the naive method or something else) were used to obtain forecasts. Moreover, one may compare MASE, RMSE, MAE, and so on, from forecasting of different time series.
What can matter is (1) a forecast period (it should be the same), and (2) whether the units, in which different time series are expressed, are different or the same. Some error measures (for example, RMSE, MAE) make sense only if the units are the same, other (such as MASE) are unit-independent.
For more details, you may want to look at this text:
http://otexts.org/fpp2/accuracy.html
Very good exercise.
I just want to ask a question.
The dataset is not stationary. Why you sill apply auto.arima into it.
Thanks