###### Projections Beyond the Short-term Fore ...

Seasonality is a time series feature in which data shows regular and predictable patterns that recur every year. For example, retail sales tend to peak for the Christmas season and then decline after the holidays.

A seasonal lag is the value of a time series one period before the current period, incorporated as an additional term in an autoregressive model. For example, if quarterly data are used, the seasonal lag is 4; if monthly data are used, the seasonal lag is 12; and so on. Seasonality is detected by autocorrelation of error terms when it differs significantly from 0. Using seasonal lags eliminates autocorrelation in the error terms.

Consider quarterly sales data. The 4^{th} autocorrelation will not be statistically different from zero if there is quarterly seasonality.

Seasonality can be corrected by incorporating an additional lagged term. For quarterly data, we add a prior year quarterly seasonal lag as follows:

$$\text{x}_{\text{t}}=\text{b}_{0}+\text{b}_{1}\text{x}_{\text{t}-1}+\text{b}_{2}\text{x}_{\text{t}-4}+\epsilon_{\text{t}}$$

In the above expression, the seasonal lag is \(\text{b}_{2}\text{x}_{\text{t}-4}\). We then run a regression analysis on the time series data to test whether the seasonal lag will eliminate statistically significant autocorrelation of error terms.

Consider an AR(1) model used to forecast quarterly retail sales of a certain company based on 100 observations.

$$\text{x}_{\text{t}}=\text{b}_{0}+\text{b}_{1}+\epsilon_{\text{t}}$$

The residual autocorrelations relating to a certain year are as presented in the following table:

$$\small{\begin{array}{c|c} \textbf{Lag} & \textbf{Autocorrelation}\\ \hline1 & -0.0584 \\ \hline2 & -0.0492 \\ \hline3 & 0.0625 \\ \hline4 & 0.6580 \\ \end{array}}$$

Test for seasonality in the time series and suggest how to correct it.

The first step is to calculate the t-statistic for each autocorrelation using the formula:

$$\text{t}_{\text{Statistic}}=\frac{\text{Residual autocorrelation}}{\frac{1}{\sqrt{\text{T}}}}$$

The t-statistic for lag one is calculated as:

$$\text{t}_{\text{Statistic}}=\frac{-0.0584}{\sqrt{100}}=-0.584$$

$$\small{\begin{array}{c|c|c} \textbf{Lag} & \textbf{Autocorrelation} & \textbf{t-statistic} \\\hline1 & -0.0584 & -0.584 \\\hline2 & -0.0492 & -0.492 \\\hline3 & 0.0625 & 0.625 \\\hline4 & 0.658 & 6.580\\ \end{array}}$$

There are 100 observations and two parameters, \(\text{b}_{0}\) and \(\text{b}_{1}\) to be estimated. Thus 98 (100-2) degrees of freedom. The critical t-value at the 5% significance level with 98 degrees of freedom is 1.98. The table above shows that the 4^{th }lag autocorrelation and the t-statistic are the highest.

The t-statistics of the first three lagged autocorrelations are less than the critical value at the 5% significance level. We can therefore conclude that none of the first three lagged autocorrelations is significantly different from zero.

However, notice that the t-statistic for the 4^{th} lag autocorrelation is greater than the critical value. We thus reject the null hypothesis that the 4^{th} lag autocorrelation is zero. The conclusion is that there is seasonality in the time series. This implies that the model is misspecified and not appropriate for use.

Seasonality can be corrected by incorporating an additional lag of the dependent variable that corresponds to the same period in the previous year to the original AR(1) model as another independent variable to make the model more correctly specified.

In the previous example, we have seen that there is seasonality in the quarterly time series. This implies that retail sales in each quarter are related to both the previous quarter and the corresponding quarter in the previous year.

We can incorporate a seasonality lag to the original AR(1) model as follows:

$$\text{x}_{\text{t}}=\text{b}_{0}+\text{b}_{1}\text{x}_{\text{t}-1}+\text{b}_{2}\text{x}_{\text{t}-4}+\epsilon_{\text{t}}$$

We have added the seasonal lag \(\text{b}_{2}\text{x}_{\text{t}-4}\) to eliminate the regular quarterly pattern and seasonal non-stationarity if they exist.

Assume that the regression coefficients after incorporating the seasonal lag are estimated as:

\(\text{b}_{0}=0.0080, \text{b}_{1}=-0.0650\) and \(\text{b}_2=0.8068\).

The estimated equation is:

$$\text{x}_{\text{t}}=0.0080-0.0650\text{x}_{\text{t}-1}+0.8068\text{x}_{\text{t}-4}+\epsilon_{\text{t}}$$

Where \(\text{x}_{\text{t}}\) is the retail sales for the i^{th} quarter.

Given the following quarterly retail sales:

$$\small{\begin{array}{c|c|c|c|c} \textbf{Quarter} & 2020.1 & 2020.2 & 2020.3 & 2020.4 \\\hline\textbf{Retail sales (USD Millions)} & 100 & 300 & 150 & 200\\ \end{array}}$$

We can forecast the retail sales for the first quarter of 2021 as follows:

$$\begin{align*}\text{y}_{2021.1}&=0.0080-0.0650(\text{y}_{2020.4})+0.8068(\text{y}_{2020.1})\\&=0.0080-0.0650(200)+0.8068(100)\\&=67.69\end{align*}$$

The forecasted value of the retail sales for the first quarter of 2021 is 67.69 million.

## Question

Consider the following monthly AR model with an additional lag incorporated to eliminate seasonality.

$$\text{x}_{\text{t}}=\text{b}_{0}+\text{b}_{1}\text{x}_{\text{t}-1}+\beta_{2}\text{x}_{\text{t}-12}+\epsilon_{\text{t}}$$

Given the following information:

$$\small{\begin{array}{l|c} {}& \text{Coefficients} \\ \hline\text{Intercept} & 0.0005 \\ \hline\text{lag 1} & –0.12 \\ \hline\text{lag 12} & 0.87\\ \end{array}}$$

$$\small{\begin{array}{c|c|c} 2019 & \text{Jan} & 2.8 \\\hline2019 & \text{Feb} & 3.0 \\\hline2019 & \text{Mar} & 3.5 \\\hline2019 & \text{Apr} & 4.0 \\\hline2019 & \text{May} & 4.6 \\\hline2019 & \text{Jun} & 5.0 \\\hline2019 & \text{Jul} & 5.4 \\\hline2019 & \text{Aug} & 6.0 \\\hline2019 & \text{Sep} & 7.0 \\\hline2019 & \text{Oct} & 5.4 \\\hline2019 & \text{Nov} & 6.0 \\\hline2019 & \text{Dec} & 7.0\\ \end{array}}$$

The forecasted value for January 2020 is

closest to:

- 1.6.
- 3.3.
- 5.8.
## Solution

The correct answer is A.$$\text{x}_{\text{t}}=\text{b}_{0}+\text{b}_{1}\text{x}_{\text{t}-1}+\text{b}_{2}\text{x}_{\text{t}-12}+\epsilon_{\text{t}}$$

$$\begin{align*}\text{x}_{2020.1}&=0.0005-0.12(\text{x}_{2019.12})+0.87(\text{x}_{2019.1})\\&=0.0005-0.12\times7+0.87\times2.8\\&=1.6\end{align*}$$

Reading 5: Time Series Analysis

*LOS 5 (l) Explain how to test and correct for seasonality in a time-series model and calculate and interpret a forecasted value using an AR model with a seasonal lag.*