library(fpp2)

First, we need to load the library fpp2(). This will automatically load all the data used in the book.

library(fpp2)

a)

Use the ses() function in R to find the optimal values of \(\alpha\) and \(l_0\), and generate forecasts for the next four months.

Before, we continue, I would like to have an idea of how this data set looks like. Please note that this data set describe the monthly total number of pigs slaughtered in Victoria, Australia (Jan 1980 – Aug 1995).

Now, let’s have a visualization of the regular decomposition of the data.

Now, we can proceed to use the ses() function in order to estimate the optimal parameters values of \(\alpha\) and \(l_0\).

In order to do so, I will employ h = 4, since the frequency of the data set is 12; which represents monthly data.

fc <- ses(pigs, h=4)

From the above, we can determine that our optimal parameter values for the Simple Exponential Smothing model are \(\alpha\) = 0.2971 and \(l_0\) = 77260.06.

## Simple exponential smoothing 
## 
## Call:
##  ses(y = pigs, h = 4) 
## 
##   Smoothing parameters:
##     alpha = 0.2971 
## 
##   Initial states:
##     l = 77260.0561 
## 
##   sigma:  10308.58
## 
##      AIC     AICc      BIC 
## 4462.955 4463.086 4472.665

The below table, will represent the forecast for the next four months.

The below graph will represent a visualization of the data with the fitted values.

b)

Compute a 95% prediction interval for the first forecast using \(\widehat{y} \pm 1.96s\) where \(s\) is the standard deviation of the residuals. Compare your interval with the interval produced by R.

Let’s estimate the standard deviation from the residuals with some visualizations.

## 
##  Ljung-Box test
## 
## data:  Residuals from Simple exponential smoothing
## Q* = 55.356, df = 22, p-value = 0.0001057
## 
## Model df: 2.   Total lags used: 24

From the above, the standard deviation of the residuals is: \(s\) = 10273.69.

Let’s take a look at the following table:

And by comparing, we notice how the values differ, however is very interesting to note that the difference in between the Lo.95 and Hi.95 are very similar but with opposite signs.

a)

Plot the series and discuss the main features of the data.

First, let’s have a look at the following table.

Now, let’s visualize it.

From the above graph, we can identify two different series; one for Paperback and one for Hardcover. There seems to be some seasonality that is in opposite trends, that is, when one trend goes up, the other group’s trend seems to go down, over all the trend seems to be increasing for the whole period of time.

b)

Use the ses() function to forecast each series, and plot the forecasts.

In order to do so, I will employ h = 4, since the frequency of the data set is 1; which represents yearly data in a daily fashion.

fc.p <- ses(books[,1], h = 4) # Paperback
fc.h <- ses(books[,2], h = 4) # Hardcover

The below table, will represent the Paper forecast for the next four days.

The below table, will represent the Hardcover forecast for the next four days.

Let’s have a visualization of the forecasts.

c)

Compute the RMSE values for the training data in each case.

Since the RMSE stands for Root-Mean-Square-Error; and the formula is given by:

\[RMSE = \sqrt{\frac{1}{N} \sum_{i=1}^{N}{(y_i - \widehat{y_i})^2}}\] We have as follows:

RMSE.p <- sqrt( sum( residuals(fc.p)^2 ) / length(books[,1]) ) # Paperback
RMSE.h <- sqrt( sum( residuals(fc.h)^2 ) / length(books[,2]) ) # Hardcover

From above, we conclude as follows:

From above, we obtain the following RMSE values.

Values returned from the Paperback model.

Values returned from the Hardcover model.

And, as you might notice, the manually calculated values match the model returned values.

a)

Now apply Holt’s linear method to the paperback and hardback series and compute four-day forecasts in each case.

The below table, will represent the Paper forecast for the next four days.

The below table, will represent the Hardcover forecast for the next four days.

Let’s have a visualization of the forecasts.

b)

Compare the RMSE measures of Holt’s method for the two series to those of simple exponential smoothing in the previous question. (Remember that Holt’s method is using one more parameter than SES.) Discuss the merits of the two forecasting methods for these data sets.

First, let’s compare the respective RMSEs. From above, the previous Simple Exponential Smoothing model, we obtained the following RMSE values.

Holt’s values returned from the Paperback model.

Holt’s values returned from the Hardcover model.

By comparing, the Simple Exponential Smoothing and Holt’s RMSE, we can notice a difference in between them; that is, Holt’s model shows a lower RMSE value compared to the Simple Exponential Smoothing model.

c)

Compare the forecasts for the two series using both methods. Which do you think is best?

First, let’s perform a subtraction of the values in the following fashion:

[Forecast Simple Exponential Smoothing] - [Forecast Holt’s method]

Let’s check the Paperback first.

Let’s check the Hardcover now.

If we compare the two forecasts, by comparing the RMSE values, definitely, the Holt’s linear method seems to predict the behavior of the trends, in a better fashion compared to the Simple Exponential Smoothing method.

Also, by looking at the above differences, something interesting to note is that the values forecast by the Holt’s method are bigger and mostly wider intervals compared to the Simple Exponential Smoothing method; thus, definitely helps to capture a better confidence interval in the forecast.

d)

Calculate a 95% prediction interval for the first forecast for each series, using the RMSE values and assuming normal errors. Compare your intervals with those produced using ses and holt.

First, let’s estimate the standard deviation from the Paperback residuals with some visualizations.

## 
##  Ljung-Box test
## 
## data:  Residuals from Holt's method
## Q* = 15.081, df = 3, p-value = 0.001749
## 
## Model df: 4.   Total lags used: 7

From the above, the standard deviation of the residuals is: \(s\) = 31.44.

Now, let’s estimate the standard deviation from the Hardcover residuals with some visualizations.

## 
##  Ljung-Box test
## 
## data:  Residuals from Holt's method
## Q* = 9.416, df = 3, p-value = 0.02424
## 
## Model df: 4.   Total lags used: 7

From the above, the standard deviation of the residuals is: \(s\) = 27.66.

Let’s take a look at the following Paperback table:

Let’s take a look at the following Hardcover table:

Previously, we calculated the values for SES; for simplicity, let’s recap the tables:

Here is the SES Paperback.

And here is the SES’ Hardcover table.

a)

Why is multiplicative seasonality necessary for this series?

Let’s recall that series:

read_excel(retail.xlsx)

My previously selected time series was A3349627V; and it represents the Turnover in New South Wales about Liquor retailing.

b)

Apply Holt-Winters’ multiplicative method to the data. Experiment with making the trend damped.

c)

Compare the RMSE of the one-step forecasts from the two methods. Which do you prefer?

From the above table, seems that the HW multiplicative forecasts Damped has better results.

e)

Check that the residuals from the best method look like white noise.

For white noise series, we expect each auto correlation to be close to zero. Of course, they will not be exactly equal to zero as there is some random variation. For a white noise series, we expect 95% of the spikes in the ACF to lie within \(\pm \frac{2}{\sqrt{T}}\) where \(T\) is the length of the time series.

Let’s calculate this number:

From the calculations; it is noted that about 2.6 % of values lie outside the bound of \(\pm\) 0.1. Hence, even though they look like white noise is not considered white noise based on the definition of white noise.

e)

Now find the test set RMSE, while training the model to the end of 2010. Can you beat the seasonal naïve approach from Exercise 8 in Section 3.7?

Now, lets find the values from the seasonal naïve approach from Exercise 8 in Section 3.7.

Here are the results for the RMSE accuracy results:

HW multiplicative forecasts

HW multiplicative forecasts Damped

Seasonal naïve approach

From the above tables, we can conclude that in effect, we can beat the Seasonal naïve approach with the HW multiplicative forecasts approach.