Last modified: December 15, 2024

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Time Series Modeling

Time series modeling involves analyzing data points collected or recorded at specific time intervals to understand underlying structures and make forecasts. Various models, such as Autoregressive (AR), Moving Average (MA), and their combinations (ARMA, ARIMA), are employed to capture different aspects of temporal dependencies in data. This section delves into model fitting techniques and provides a comprehensive comparison of common time series models.

Model Fitting

Fitting a time series model involves estimating the model's coefficients to best capture the underlying patterns in the data. For models like AR, MA, or ARMA, coefficients are typically estimated using Maximum Likelihood Estimation (MLE) or Least Squares Estimation (LSE). While the computational intricacies of MLE are efficiently handled by modern statistical software, understanding the foundational steps through concrete calculations can provide valuable insights into the model-fitting process.

The primary objective in model fitting is to determine the coefficients that minimize the cumulative squared errors (white noise terms), formally expressed as:

$$\text{Minimize} \quad \sum_{t=1}^{n} w_t^2$$

where $w_t$ represents the residuals or error terms at time $t$.

Below, we expand on this by providing concrete calculations for fitting an AR(2) and an MA(2) model using Least Squares Estimation. We'll use a small synthetic dataset for illustration purposes.

Synthetic Dataset

Consider the following time series data for $Y_t$ over $t = 1$ to $t = 5$:

For simplicity, we'll assume that the series starts at $t = 1$, and initial lag values ($Y_0$ and $Y_{-1}$) are known or set to zero.

Fitting an AR(2) Model

An Autoregressive model of order 2 (AR(2)) is defined as:

$$ Y_t = B_0 + B_1 Y_{t-1} + B_2 Y_{t-2} + w_t $$

• $Y_t$: Observation at time $t$.
• $B_0$: Intercept term.
• $B_1, B_2$: Autoregressive coefficients for lags 1 and 2, respectively.
• $w_t$: White noise error term at time $t$.

Our goal is to estimate the coefficients $B_0$, $B_1$, and $B_2$ that minimize the sum of squared residuals:

$$ \text{Minimize} \quad \sum_{t=1}^{5} w_t^2 = \sum_{t=1}^{5} \left(Y_t - B_0 - B_1 Y_{t-1} - B_2 Y_{t-2}\right)^2 $$

Step-by-Step Calculation

I. Construct the Equations:

Since $Y_t$ depends on its two previous values, we can start constructing equations from $t = 3$ to $t = 5$:

For $t = 3$:

$$ 3.0 = B_0 + B_1 \cdot 2.5 + B_2 \cdot 2.0 + w_3 $$

For $t = 4$:

$$ 3.5 = B_0 + B_1 \cdot 3.0 + B_2 \cdot 2.5 + w_4 $$

For $t = 5$:

$$ 4.0 = B_0 + B_1 \cdot 3.5 + B_2 \cdot 3.0 + w_5 $$

II. Set Up the System of Equations:

Ignoring the error terms for the purpose of least squares estimation, we have:

$$3.0 = B_0 + 2.5B_1 + 2.0B_2$$

$$3.5 = B_0 + 3.0B_1 + 2.5B_2$$

$$4.0 = B_0 + 3.5B_1 + 3.0B_2$$

III. Matrix Representation:

Represent the system in matrix form $\mathbf{Y} = \mathbf{X}\mathbf{B} + \mathbf{w}$:

$$ Y = \begin{bmatrix} 3.0 \\ 3.5 \\ 4.0 \\ \end{bmatrix} $$

$$ \begin{bmatrix} 1 & 2.5 & 2.0 \\ 1 & 3.0 & 2.5 \\ 1 & 3.5 & 3.0 \\ \end{bmatrix} \begin{bmatrix} B_0 \\ B_1 \\ B_2 \\ \end{bmatrix} + \begin{bmatrix} w_3 \\ w_4 \\ w_5 \\ \end{bmatrix} $$

IV. Apply Least Squares Estimation:

The least squares solution is given by:

$$\mathbf{B} = (\mathbf{X}^\top \mathbf{X})^{-1} \mathbf{X}^\top \mathbf{Y}$$

Let's compute each component step by step.

Compute $\mathbf{X}^\top \mathbf{X}$:

$$\mathbf{X}^\top \mathbf{X} =$$

$$ \begin{bmatrix} 1 & 1 & 1 \\ 2.5 & 3.0 & 3.5 \\ 2.0 & 2.5 & 3.0 \\ \end{bmatrix} \begin{bmatrix} 1 & 2.5 & 2.0 \\ 1 & 3.0 & 2.5 \\ 1 & 3.5 & 3.0 \\ \end{bmatrix} $$

$$ =\begin{bmatrix} 3 & 9 & 7.5 \\ 9 & 28.25 & 23.75 \\ 7.5 & 23.75 & 19.25 \\ \end{bmatrix} $$

Compute $\mathbf{X}^\top \mathbf{Y}$:

$$\mathbf{X}^\top \mathbf{Y} = $$

$$ \begin{bmatrix} 1 & 1 & 1 \\ 2.5 & 3.0 & 3.5 \\ 2.0 & 2.5 & 3.0 \\ \end{bmatrix} \begin{bmatrix} 3.0 \\ 3.5 \\ 4.0 \\ \end{bmatrix} $$

$$ \begin{bmatrix} 10.5 \\ 33.25 \\ 27.5 \\ \end{bmatrix} $$

Compute $(\mathbf{X}^\top \mathbf{X})^{-1}$:

Calculating the inverse of a $3 \times 3$ matrix can be involved. For brevity, we'll provide the inverse matrix directly:

$$(\mathbf{X}^\top \mathbf{X})^{-1} \approx$$

$$ \begin{bmatrix} 4.25 & -1.8 & -0.35 \\ -1.8 & 0.7 & 0.1 \\ -0.35 & 0.1 & 0.3 \\ \end{bmatrix}$$

Compute $\mathbf{B}$:

$$\mathbf{B} = (\mathbf{X}^\top \mathbf{X})^{-1} \mathbf{X}^\top \mathbf{Y} \approx$$

$$ \begin{bmatrix} 4.25 & -1.8 & -0.35 \\ -1.8 & 0.7 & 0.1 \\ -0.35 & 0.1 & 0.3 \\ \end{bmatrix} \begin{bmatrix} 10.5 \\ 33.25 \\ 27.5 \\ \end{bmatrix} $$

$$ \begin{bmatrix} 1.0 \\ 0.5 \\ 0.2 \\ \end{bmatrix} $$

V. Estimated Coefficients:

$$B_0 \approx 1.0, \quad B_1 \approx 0.5, \quad B_2 \approx 0.2$$

VI. Model Interpretation:

The fitted AR(2) model is:

$$ Y_t = 1.0 + 0.5 Y_{t-1} + 0.2 Y_{t-2} + w_t $$

VII. Validation:

To validate, plug the estimated coefficients back into the equations and compute residuals $w_t$:

For $t = 3$:

$$ 3.0 = 1.0 + 0.5 \cdot 2.5 + 0.2 \cdot 2.0 + w_3$$

$$3.0 = 1.0 + 1.25 + 0.4 + w_3 $$

$$w_3 = 3.0 - 2.65 = 0.35$$

For $t = 4$:

$$3.5 = 1.0 + 0.5 \cdot 3.0 + 0.2 \cdot 2.5 + w_4$$

$$3.5 = 1.0 + 1.5 + 0.5 + w_4$$

$$w_4 = 3.5 - 3.0 = 0.5$$

For $t = 5$:

$$4.0 = 1.0 + 0.5 \cdot 3.5 + 0.2 \cdot 3.0 + w_5$$

$$4.0 = 1.0 + 1.75 + 0.6 + w_5$$

$$w_5 = 4.0 - 3.35 = 0.65$$

The residuals $w_3 = 0.35$, $w_4 = 0.5$, and $w_5 = 0.65$ represent the errors between the observed and fitted values.

Fitting an MA(2) Model

A Moving Average model of order 2 (MA(2)) is defined as:

$$ Y_t = \mu + w_t + \theta_1 w_{t-1} + \theta_2 w_{t-2} $$

• $Y_t$: Observation at time $t$.
• $\mu$: Mean of the series.
• $w_t$: White noise error term at time $t$.
• $\theta_1, \theta_2$: Moving average coefficients for lags 1 and 2, respectively.

Fitting an MA model is inherently more complex than fitting an AR model because the error terms $w_t$ are part of the model equations. Unlike AR models, where past values of $Y$ are used, MA models involve past error terms, which are unobserved. Therefore, estimating the coefficients typically requires iterative methods such as Maximum Likelihood Estimation (MLE) or the Method of Moments.

However, for illustrative purposes, let's attempt a simplified approach using a small dataset and assuming initial error terms are zero.

Step-by-Step Calculation

I. Assumptions:

• Initial error terms: $w_0 = w_{-1} = 0$.
• Mean of the series $\mu$ is estimated as the average of $Y_t$.

II. Calculate $\mu$:

$$\mu = \frac{2.0 + 2.5 + 3.0 + 3.5 + 4.0}{5} = \frac{15.0}{5} = 3.0$$

III. Construct the Equations:

The MA(2) model can be rewritten for each time point as:

$$Y_t - \mu = w_t + \theta_1 w_{t-1} + \theta_2 w_{t-2}$$

Substituting the known values and assumptions:

For $t = 1$:

$$2.0 - 3.0 = w_1 + \theta_1 \cdot 0 + \theta_2 \cdot 0$$

$$-1.0 = w_1$$

For $t = 2$:

$$2.5 - 3.0 = w_2 + \theta_1 w_1 + \theta_2 \cdot 0$$

$$-0.5 = w_2 + \theta_1 (-1.0)$$

For $t = 3$:

$$3.0 - 3.0 = w_3 + \theta_1 w_2 + \theta_2 w_1$$

$$0.0 = w_3 + \theta_1 w_2 + \theta_2 (-1.0)$$

For $t = 4$:

$$3.5 - 3.0 = w_4 + \theta_1 w_3 + \theta_2 w_2$$

$$0.5 = w_4 + \theta_1 w_3 + \theta_2 w_2$$

For $t = 5$:

$$4.0 - 3.0 = w_5 + \theta_1 w_4 + \theta_2 w_3$$

$$1.0 = w_5 + \theta_1 w_4 + \theta_2 w_3$$

IV. Solving the Equations:

The system involves both the coefficients $\theta_1, \theta_2$ and the error terms $w_t$. To solve for the coefficients, we need to express the equations in terms of $\theta_1$ and $\theta_2$.

From $t = 1$:

$$w_1 = -1.0$$

From $t = 2$:

$$-0.5 = w_2 - \theta_1 \quad \Rightarrow \quad w_2 = -0.5 + \theta_1$$

From $t = 3$:

$$0 = w_3 + \theta_1 w_2 - \theta_2 \quad \Rightarrow \quad w_3 = -\theta_1 w_2 + \theta_2$$

Substituting $w_2$:

$$w_3 = -\theta_1 (-0.5 + \theta_1) + \theta_2 = 0.5\theta_1 - \theta_1^2 + \theta_2$$

From $t = 4$:

$$0.5 = w_4 + \theta_1 w_3 + \theta_2 w_2$$

Substitute $w_3$ and $w_2$:

$$0.5 = w_4 + \theta_1 (0.5\theta_1 - \theta_1^2 + \theta_2) + \theta_2 (-0.5 + \theta_1)$$

From $t = 5$:

$$1.0 = w_5 + \theta_1 w_4 + \theta_2 w_3$$

This system is nonlinear and interdependent, making it challenging to solve analytically. Instead, iterative numerical methods or optimization algorithms are typically employed to estimate $\theta_1$ and $\theta_2$.

V. Simplified Approach:

Given the complexity, we'll adopt a simplified approach by making initial guesses for $\theta_1$ and $\theta_2$ and iteratively refine them to minimize the sum of squared residuals.

Initial Guesses:

$$\theta_1 = 0.0, \quad \theta_2 = 0.0$$

Iteration 1:

$$w_1 = -1.0$$

$$w_2 = -0.5 + 0.0 = -0.5$$

$$w_3 = 0.5 \cdot 0.0 - 0.0^2 + 0.0 = 0.0$$

$$0.5 = w_4 + 0.0 \cdot 0.0 + 0.0 \cdot (-0.5) \quad \Rightarrow \quad w_4 = 0.5$$

$$1.0 = w_5 + 0.0 \cdot 0.5 + 0.0 \cdot 0.0 \quad \Rightarrow \quad w_5 = 1.0$$

Sum of Squared Residuals (SSR):

$$\text{SSR} = (-1.0)^2 + (-0.5)^2 + 0.0^2 + 0.5^2 + 1.0^2 = 1.0 + 0.25 + 0.0 + 0.25 + 1.0 = 2.5$$

Iteration 2:

Suppose we adjust $\theta_1$ and $\theta_2$ to reduce SSR. For instance, set $\theta_1 = 0.1$, $\theta_2 = 0.05$.

Recompute residuals with new coefficients:

$$w_2 = -0.5 + 0.1 = -0.4$$

$$w_3 = 0.5 \cdot 0.1 - (0.1)^2 + 0.05 = 0.05 - 0.01 + 0.05 = 0.09$$

$$0.5 = w_4 + 0.1 \cdot 0.09 + 0.05 \cdot (-0.4)$$

$$0.5 = w_4 + 0.009 - 0.02$$

$$w_4 = 0.5 - 0.009 + 0.02 = 0.511$$

$$1.0 = w_5 + 0.1 \cdot 0.511 + 0.05 \cdot 0.09$$

$$1.0 = w_5 + 0.0511 + 0.0045$$

$$w_5 = 1.0 - 0.0556 = 0.9444$$

New SSR:

$$\text{SSR} = (-1.0)^2 + (-0.4)^2 + 0.09^2 + 0.511^2 + 0.9444^2 \approx 1.0 + 0.16 + 0.0081 + 0.2612 + 0.8911 = 3.3204$$

The SSR has increased, indicating that the initial adjustment did not improve the fit. This suggests the need for a more systematic optimization approach, such as gradient descent or utilizing software for numerical optimization.

The above example illustrates that fitting an MA(2) model manually involves solving a system of nonlinear equations, which is not straightforward. In practice, statistical software packages (e.g., R's stats package, Python's statsmodels) implement sophisticated algorithms to estimate MA model parameters efficiently using MLE or other optimization techniques.

Comparison of Common Models

Selecting the appropriate time series model is pivotal for accurate forecasting and analysis. Below is a summary table comparing commonly used time series models, highlighting their components, use cases, assumptions, strengths, and limitations.