DATA 624 Homework 2

Question 3.1

For the following series, find an appropriate Box-Cox transformation in order to stabilise the variance.

I will answer these questions with side by side visualizations of the data. The appropriate Box-Cox transformation will be on the right hand side with the lambda in the title.

side_by_side <- function(x, y){
  lambda <- BoxCox.lambda(x)
  plot1 <- autoplot(x) +
    ggtitle("Original") +
    ylab(y) + 
    theme(axis.title.x = element_blank())
  plot2 <- autoplot(BoxCox(x, lambda)) +
    ggtitle(paste0("Box-Cox Transformed (lambda=", round(lambda, 4),")")) +
    theme(axis.title.x = element_blank(),
        axis.title.y = element_blank()) 
  grid.arrange(plot1, plot2, ncol = 2)
}

side_by_side(usnetelec, "US Net Electricity Generation")

side_by_side(usgdp, "US GDP")

side_by_side(mcopper, "Monthly Copper Prices")

side_by_side(enplanements, "Monthly US Domestic Enplanements")

Question 3.2

Why is a Box-Cox transformation unhelpful for the cangas data?

side_by_side(cangas, "Monthly Canadian Gas Production")

The Box-Cox transformation does not help with this timeseries because the variation is initially small, then gets large, then gets small again. Box-Cox was not designed to handle this case. It was designed for cases where the variance increases or decreases over time.

Question 3.3

What Box-Cox transformation would you select for your retail data?

retaildata <- read_excel("retail.xlsx", skip = 1)
myts <- ts(retaildata[, "A3349873A"], frequency = 12, start = c(1982, 4))
side_by_side(myts, "Retail Sales")

The variation was increasing over time in the original data. It has become significantly more uniform once it is transformed with a lambda of 0.13. Because the variance was increasing over time this was an effective transformation.

Question 3.8

For your retail time series:

Split the data into two parts using

myts.train <- window(myts, end = c(2010, 12))
myts.test <- window(myts, start = 2011)

Check that your data have been split appropriately by producing the following plot.

autoplot(myts) +
  autolayer(myts.train, series = "Training") +
  autolayer(myts.test, series = "Test")

Calculate forecasts using snaive applied to myts.train.

fc <- snaive(myts.train)

Compare the accuracy of your forecasts against the actual values stored in myts.test.

accuracy(fc, myts.test)

                    ME     RMSE      MAE       MPE      MAPE     MASE      ACF1
Training set  7.772973 20.24576 15.95676  4.702754  8.109777 1.000000 0.7385090
Test set     55.300000 71.44309 55.78333 14.900996 15.082019 3.495907 0.5315239
             Theil's U
Training set        NA
Test set      1.297866

Check the residuals.

checkresiduals(fc)

    Ljung-Box test

data:  Residuals from Seasonal naive method
Q* = 624.45, df = 24, p-value < 2.2e-16

Model df: 0.   Total lags used: 24

Do the residuals appear to be uncorrelated and normally distributed?

They do appear to be normally distributed however with a sligh positve skew. The residuals do no appear to be uncorrelated. The Ljung-Box test has a p value that is less than 0.05. This suggests there is more information that can be discovered and that the seasonal naive model is not the best model.

How sensitive are the accuracy measures to the training/test split?

The accuracy measures are quite sensitive to the training/test split. The values are significantly different between the two. This would suggest that the model does not generalize well.