Read and tidy up the data using the code provided.
What is your data about (no more than 50 words)? Produce appropriate plots in order to become familiar with your data. Make sure you label your axes and plots appropriately. Comment on these. What do you see? (no more than 50 words per plot). (14 marks)
Marks breakdown:
Expectation:
Common errors:
Suggested coding:
myts |> autoplot(y) +
labs(title="Add you title here",
y= "add y label")
myts |> gg_season(y) +
labs(title="Add you title here",
y= "add y label")
myts |> gg_subseries(y)+
labs(title="Add you title here",
y= "add y label")
# Add other features to plotsWould transforming your data be useful (no more than 50 words)? Compare different transformations graphically. Choose the best transformation if you think a transformation is needed. Justify your choice (no more than 50 words). (5 marks)
Marks breakdown:
Expectation:
Common errors:
If the data show variation that increases or decreases with the level of the series, then a transformation can be useful. Have a read of this section on the online textbook for a clear understanding of transformations. https://otexts.com/fpp3/transformations.html
Transformations simplify the patterns in the historical data by removing known sources of variation or by making the pattern more consistent across the whole data set. Simpler patterns usually lead to more accurate forecasts.
At least a couple of transformations were expected, but some students were presenting only the plot of their finally chosen one.
Also in some assignments the BoxCox with the Guerrero feature was chosen, but the value of the lambda was not mentioned.
BoxCox transformed data with lambda other than 1 could not have “$m” as units in the y label of the plots.
Suggested coding:
library(patchwork)
p1 <- myts |> autoplot(sqrt(y))
p2 <- myts |> autoplot(log(y))
lambda <- myts |> features(y, features = guerrero) |>
pull(lambda_guerrero)
lambda[1] 0.1393678p3 <- myts |> autoplot(box_cox(y, lambda))
p1/p2/p3
I would probably go with the Box Cox transformation in the case with \(\lambda\)= 0.1393678.
Split your data into training and test sets. Leave the last two years’ worth of observations as the test set. Plot these on the same graph to make sure you have done this properly. (3 marks)
Marks breakdown:
Common errors:
Suggested coding:
Split the data into training and test sets. The data can be correctly split into training and test sets as:
myts_train <- myts |>
slice(1:(n() - 24))
myts_train |> tail()# A tsibble: 6 x 2 [1M]
Month y
<mth> <dbl>
1 2020 Jun 28577.
2 2020 Jul 30021.
3 2020 Aug 28580.
4 2020 Sep 28350.
5 2020 Oct 30340
6 2020 Nov 33119.myts_test <- myts |>
slice((n() - 23):n())
myts |>
autoplot(y) +
geom_line(aes(colour = "Test"), data = myts_test) +
labs(title= "Add you title here", y="add y label") +
guides(colour=guide_legend(title = "Test"))
Splitting your transformed data was not a mistake, if it was explained in the labels or heading. But remember at the end of the day your task is to generated forecasts for the original series not the transformed one.
Apply the two most appropriate benchmark methods on the training set. Generate forecasts for the test set and plot them on the same graph. Compare their forecasting performance on the test set. Which method would you choose as the appropriate benchmark? Justify your answer (no more 100 words). (Hint: it will be useful to tabulate your results.) (12 marks)
Marks Breakdown:
Expectation:
Common errors:
fit_train <- myts_train |>
model(
snaive = SNAIVE(log(y)),
drift = RW(log(y) ~ drift())
)
fit_train |> glance()# A tibble: 2 × 2
.model sigma2
<chr> <dbl>
1 snaive 0.000971
2 drift 0.0156 fc_test <- fit_train |>
forecast(h = "2 years")
fc_test |> autoplot(myts, alpha=0.6)
fc_test |> autoplot(myts, level=NULL)
# Train
fit_train |> accuracy() |> select(.model, MAE, RMSE, MAPE, MASE, RMSSE)# A tibble: 2 × 6
.model MAE RMSE MAPE MASE RMSSE
<chr> <dbl> <dbl> <dbl> <dbl> <dbl>
1 snaive 686. 815. 5.39 1 1
2 drift 1165. 1996. 8.40 1.70 2.45# Test
fc_test |> accuracy(myts) |> select(.model, MAE, RMSE, MAPE, MASE, RMSSE)# A tibble: 2 × 6
.model MAE RMSE MAPE MASE RMSSE
<chr> <dbl> <dbl> <dbl> <dbl> <dbl>
1 drift 7081. 7384. 22.3 10.3 9.07
2 snaive 3403. 4001. 10.3 4.96 4.91The error values should have been presented. However, some students did not put the labels to identify the corresponding methods/errors.
For the best method, do a residual analysis. Do the residuals appear to be white noise? What did your forecasting method miss? (no more than 150 words) (8 marks)
Marks Breakdown:
Expectation:
Common errors:
Suggested Coding:
fit <- myts |>
model(
snaive = SNAIVE(log(y)),
)
fit |>
gg_tsresiduals()Warning: Removed 12 rows containing missing values or values outside the scale range
(`geom_line()`).Warning: Removed 12 rows containing missing values or values outside the scale range
(`geom_point()`).Warning: Removed 12 rows containing non-finite outside the scale range
(`stat_bin()`).
Lots of autocorrelation remaining in the residuals - the seasonal naive fails to capture this. The residuals do not look normal.
Generate forecasts for the next 2 years (future) from the benchmark method you considered best and plot them. Comment on these (no more than 50 words).(4 marks)
Marks Breakdown:
Expectation:
Common errors:
Suggested coding:
Suppose the best method chosen in Question 4 was the seasonal naïve method. Then generate and plot point forecasts and prediction intervals. In this case I would comment that the point forecasts look reasonable while the prediction intervals look very wide (recall that lots of autocorrelation was left over in the residuals hence these will affect my predictions intervals).
fit |>
forecast(h = "2 years") |>
autoplot(myts)