SDSC6012 - Assignment 1

#assignment #sdsc6012

Question 1

1. Trend Component Extraction (Moving Average Method)

   The trend component is extracted using the centered moving average method:

   $$\text{Trend}_t = \frac{1}{k} \sum_{i=t-m}^{t+m} x_i$$

   Where:

   • $k$ is the window size (here set to 12, corresponding to the annual cycle)

   • $m = \lfloor k/2 \rfloor$ (half-window width for centered moving average)

   • Boundary handling: when $t < m$ or $t > n-m$, calculate the mean using available data

2. Seasonal Component Extraction (Periodic Average Method)

   1. Calculate the detrended series: $d_t = x_t - \text{Trend}_t$

   2. For each periodic position $j$ ($j=0,1,\ldots,11$), compute the average:

      $$s_j = \frac{1}{N_j} \sum_{k=0}^{N_j-1} d_{j+12k}$$

      where $N_j$ is the number of occurrences of periodic position $j$

   3. Zero-mean adjustment:

      $$\text{Seasonal}_j = s_j - \bar{s}, \quad \bar{s} = \frac{1}{12}\sum_{j=0}^{11} s_j$$

   4. Construct the complete seasonal series: $\text{Seasonal}_t = \text{Seasonal}_{t \mod 12}$

3. Residual Calculation

   $$\varepsilon_t = x_t - \text{Trend}_t - \text{Seasonal}_t$$

4. Time Series Equation

   $$x_t = \text{Trend}_t + \text{Seasonal}_{t \mod 12} + \varepsilon_t$$

   where $\varepsilon_t$ is random noise with mean 0.

Question 2

Consider the time series

$$x_t = \beta_1 + \beta_2 t + w_t$$

where $\beta_1$ and $\beta_2$ are known constants and w_t is a white noise process with variance $\sigma_w^2$.
(a) Determine whether $x_t$ is stationary.
(b) Show that the process $y_t = x_t - x_{t-1}$ is stationary.
© Show that the mean of the moving average

$$v_t = \frac{1}{2q+1} \sum_{j=-q}^{q} x_{t-j}$$

is $\beta_1$ + $\beta_2$ t, and give a simplified expression for the autocovariance function.

(a) Determining the Stationarity of $x_t$

$x_t$ is not a stationary process.

• Mean function: $E[x_t] = \beta_1 + \beta_2 t$ (varies with time)

• Autocovariance function: $\gamma_x(h) = \begin{cases} \sigma_w^2 & h=0 \\ 0 & h \neq 0 \end{cases}$
  Although the autocovariance depends only on the time difference $h$, the mean is not constant, thus failing to satisfy the stationarity condition.

(b) Proving $y_t = x_t - x_{t-1}$ is Stationary

$y_t = \beta_2 + w_t - w_{t-1}$ is a stationary process.

1. Mean:
   $E[y_t] = \beta_2$ (constant)

2. Autocovariance Function:
   $\gamma_y(h) = \begin{cases} 2\sigma_w^2 & h=0 \\ -\sigma_w^2 & |h|=1 \\ 0 & |h|>1 \end{cases}$
   Depends only on the time difference $h$, satisfying the stationarity condition.

© Mean and Autocovariance of the Moving Average $v_t$

$$v_t = \frac{1}{2q+1} \sum_{j=-q}^{q} x_{t-j}$$

$$\sum_{j=-q}^{q} E[x_{t-j}] = \sum (\beta_1 + \beta_2(t-j)) = (2q+1)(\beta_1 + \beta_2 t)$$

Autocovariance Function:

• Number of non-zero terms $N(h) = 2q+1 - |h|$ (when $|h|\leq 2q$)

• Each term’s covariance is $\sigma_w^2$, denominator is $(2q+1)^2$

$$\gamma_v(h) = \begin{cases} \frac{2q+1-|h|}{(2q+1)^2} \sigma_w^2 & |h| \leq 2q \\ 0 & |h| > 2q \end{cases}$$