Activities: Week 10

Australian tobacco expenditure

In this exercise, we will compare an ETS model, an ARIMA model and an STL+ARIMA model fitted to Australian tobacco expenditure data.

library(fpp3)
tobacco <- aus_tobacco |>
  summarise(Expenditure = sum(Expenditure))
tobacco |> autoplot(Expenditure)

First we will find an appropriate STL decomposition for this data set.

tobacco |>
  model(
    stl = STL(log(Expenditure) ~ season(window = ???) + trend(window = ???))
  ) |>
  components() |>
  autoplot()

Now fit an ETS, ARIMA and STL+ARIMA model with the same window values. For the ARIMA and STL+ARIMA models, use a log transformation.

fit <- tobacco |>
  model(
    ets = ETS(Expenditure),
    arima = ARIMA(log(Expenditure)),
    stl_arima = decomposition_model(
      STL(log(Expenditure) ~ season(window = ???) + trend(window = ???)),
      ARIMA(season_adjust),
      SNAIVE(season_year)
    )
  )

Write down the equations for each model.

fit |> select(ets) |> report()
fit |> select(arima) |> report()
fit |> select(stl_arima) |> report()

Check if the residuals from each model look like white noise

fit |> select(ets) |> gg_tsresiduals()
fit |> select(arima) |> gg_tsresiduals()
fit |> select(stl_arima) |> gg_tsresiduals()

Which model fits the data best?
```
fit |> accuracy()
```

Do the forecasts look similar?

fit |>
  forecast(h = 24) |>
  autoplot(tobacco |> filter(year(Quarter) >= 2010), level = 80, alpha = 0.5)

Find the one-step median forecast and compute the 95% prediction interval from first principles. Check that you get the same result as obtained with the following code.

fit |>
  select(arima) |>
  forecast(h = 1) |>
  mutate(
    .median = median(Expenditure),
    PI = hilo(Expenditure, level = 95)
  )
fit |> select(arima) |> report()