# ARMA$$(p, q)$$ Processes

### ARMA$$(p, q)$$ process: definitions

• $$\{X_t\}$$ is an ARMA$$(p, q)$$ process if it is stationary, and for all $$t$$, $X_t - \phi_1 X_{t-1} - \cdots - \phi_p X_{t-p} = Z_t + \theta_1 Z_{t-1} + \cdots + \theta_q Z_{t-q}$ where $$\{Z_t\} \sim \textrm{WN}(0, \sigma^2)$$ and the polynomials $\phi(z) = 1 - \phi_1 z - \cdots - \phi_p z^p, \quad \theta(z) = 1 + \theta_1 z + \cdots + \theta_q z^q$ have no common factors

• Equivalent formula using the backward shift operator $\phi(B) X_t = \theta(B) Z_t$

• An ARMA$$(p,q)$$ process with mean $$\mu$$: we can study $$\{X_t - \mu\}$$ $(X_t-\mu) - \phi_1 (X_{t-1}-\mu) - \cdots - \phi_p (X_{t-p}-\mu) = Z_t + \theta_1 Z_{t-1} + \cdots + \theta_q Z_{t-q}$

## Stationary solution

### Stationary solution: existence and uniqueness

• A stationary solution exists and is unique if and only if $\phi(z) \neq 0, \quad \text{for all complex } z \text{ with } |z| = 1$

• The unit circle: the region in $$z \in \mathbb{C}$$ defined by $$|z|=1$$

• Stationary solution: $X_t = \theta(B) / \phi(B) Z_t = \psi(B) Z_t = \sum_{j=-\infty}^{\infty} \psi_j Z_{t-j}$

## Causality

### Causality: $$\phi(z)$$ has no zeros inside the unit circle

• An ARMA$$(p, q)$$ process is causal: if there exist $$\psi_0, \psi_1, \ldots$$ $\sum_{j=0}^{\infty} |\psi_j| < \infty, \quad \text{and}$ $X_t = \sum_{j=0}^{\infty}\psi_j Z_{t-j}, \quad \text{for all } t$

• Theorem (equivalent condition of causaility): $\phi(z) = 1 - \phi_1 z - \cdots - \phi_p z^p \neq 0, \quad \text{for all } |z| \leq 1$

• Example: ARMA$$(1, 1)$$ $$X_t - \phi X_{t-1} = Z_t + \theta Z_{t-1}$$ $1-\phi z = 0 \Longrightarrow \text{ only zero } z = 1/\phi$ So $$|z| = 1/|\phi| > 1$$, i.e., $$|\phi| < 1$$ is equivalent of causality

### How do we get $$\psi_j$$’s?

• Letting $$\theta_0 = 1$$ and matching coefficients of $$z^j$$ based on $1 + \theta_1 z + \cdots \theta_q z^q = (1 - \phi_1 z - \cdots \phi_p z^p)(\psi_0 + \psi_1 z + \cdots),$ gives $\theta_j \mathbf{1}_{[j \leq q]} = \psi_j - \sum_{j=1}^p \phi_k \psi_{j-k}, \quad j = 0, 1, \ldots$

• Example: causal ARMA$$(1, 1)$$
\begin{align*} 1 & = \psi_0\\ \theta & = \psi_1 - \phi \psi_0 \Longrightarrow \psi_1 = \theta + \psi\\ 0 & = \psi_j - \phi \psi_{j-1} \text{ for } j \geq 2 \Longrightarrow \psi_j = \phi \psi_{j-1} \end{align*}

Therefore, $\psi_0 = 1, \quad \psi_j = \phi^{j-1}(\theta + \psi) \text{ for } j \geq 1$

## Invertibility

### Invertibility: $$\theta(z)$$ has no zeros inside the unit circle

• An ARMA$$(p, q)$$ process is invertible: if there exist $$\pi_0, \pi_1, \ldots$$ $\sum_{j=0}^{\infty} |\pi_j| < \infty, \quad \text{and}$ $Z_t = \sum_{j=0}^{\infty}\pi_j X_{t-j}, \quad \text{for all } t$

• Theorem (equivalent condition of invertibility): $\theta(z) = 1 + \theta_1 z + \cdots + \theta_q z^q \neq 0, \quad \text{for all } |z| \leq 1$

• Example: ARMA$$(1, 1)$$ $$X_t - \phi X_{t-1} = Z_t + \theta Z_{t-1}$$ $1 + \theta z = 0 \Longrightarrow \text{ only zero } z = -1/\theta$ So $$|z| = 1/|\theta| > 1$$, i.e., $$|\theta| < 1$$ is equivalent of invertibility

### How do we get $$\pi_j$$’s?

• Letting $$\phi_0 = -1$$ and matching coefficients of $$z^j$$ based on $1 - \phi_1 z - \cdots \phi_p z^p = (1 + \theta_1 z + \cdots \theta_q z^q)(\pi_0 + \pi_1 z + \cdots),$ gives $-\phi_j \mathbf{1}_{[j \leq p]} = \pi_j + \sum_{j=1}^q \theta_k \pi_{j-k}, \quad j = 0, 1, \ldots$

• Example: invertible ARMA$$(1, 1)$$
\begin{align*} 1 & = \pi_0\\ -\phi & = \pi_1 + \theta \psi_0 \Longrightarrow \pi_1 = -(\psi +\theta)\\ 0 & = \pi_j + \theta \pi_{j-1} \text{ for } j \geq 2 \Longrightarrow \pi_j = -\theta \pi_{j-1} \end{align*}

Therefore, $\pi_0 = 1, \quad \pi_j = (-1)^j \theta^{j-1}(\psi + \theta) \text{ for } j \geq 1$

# ACF and PACF of an ARMA$$(p, q)$$ Process

## Calculation of the ACVF

### Calculation of the ACVF

• Assume the ARMA$$(p, q)$$ process $$\{X_t\}$$ is causal and invertible

• Method 1: If $$X_t = \sum_{j=0}^{\infty} \psi_j Z_{t-j}$$, then $\gamma(h) = E(X_{t+h} E_t) = \sigma^2 \sum_{j=0}^{\infty} \psi_j \psi_{j + |h|}$

• Method 2 (difference equation method): multiple the ARMA formula with $$X_t, X_{t-1}, \ldots$$ and take expectation

### Example: ARMA$$(1, 1)$$

• Recall that for a causal ARMA$$(1, 1)$$, in $$X_t = \sum_{j=0}^{\infty} \psi_j Z_{t-j}$$, $\psi_0 = 1, \quad \psi_j = \phi^{j-1}(\theta + \psi) \text{ for } j \geq 1$

• Lag-0 autocorrelation $\gamma(0) = \sigma^2 \sum_{j=0}^{\infty} \psi_j^2 = \sigma^2\left[ 1 + (\theta+\phi)^2 \sum_{j=0}^{\infty}\phi^{2j}\right] = \sigma^2\left[ 1 + \frac{(\theta+\phi)^2}{1-\phi^2} \right]$

• Lag-1 autocorrelation $\gamma(1) = \sigma^2 \sum_{j=0}^{\infty} \psi_j \psi_{j+1} % = \sigma^2\left[ \theta+\phi + (\theta+\phi)^2\phi % \sum_{j=0}^{\infty}\phi^{2j}\right] = \sigma^2\left[ \theta+\phi + \frac{(\theta+\phi)^2 \phi}{1-\phi^2} \right]$

• Lag-$$k$$ autocorrelation ($$k \geq 2$$) $\gamma(k) = \phi^{k-1} \gamma(1), \quad k \geq 2$

### Use the difference equation method on ARMA$$(1, 1)$$

1. Multiple $$X_t - \phi X_{t-1} = Z_t + \theta Z_{t-1}$$ by $$X_t$$, then take expectation $E(X_t^2) - \phi E(X_t X_{t-1}) = E(X_t Z_t) + \theta E(X_t Z_{t-1})$ Since $$E(X_t Z_k) = E[(\sum_{j=0}^{\infty} \psi_j Z_{t-j})Z_k] = \psi_{t-k} \sigma^2$$, we have $\gamma(0) - \phi \gamma(1) = \sigma^2 + \theta (\theta + \phi) \sigma^2$

2. Multiply by $$X_{t-1}$$ $E(X_{t-1} X_t) - \phi E(X_{t-1}^2) = E(X_{t-1} Z_t) + \theta E(X_{t-1} Z_{t-1})$ $\gamma(1) - \phi \gamma(0) = 0 + \theta \sigma^2 \psi_0 = \theta \sigma^2$

Using the two equations from 1 and 2, we can solve $$\gamma(0), \gamma(1)$$

1. Multiply by $$X_{t-k}$$, for $$k \geq 2$$ $E(X_{t-k} X_t) - \phi E(X_{t-k} X_{t-1}) = E(X_{t-k} Z_t) + \theta E(X_{t-k} Z_{t-1})$ $\gamma(k) - \phi \gamma(k-1) = 0 \Longrightarrow \gamma(k) = \phi \gamma(k-1)$

## Test for MAs and ARs from the ACF and PACF

### ACF of an MA$$(q)$$ process

• Suppose $$\{X_t\}$$ is an MA$$(q)$$, then $$\rho(h) = 0$$ for all $$h > q$$
• By asymptotic normality $\hat{\rho}(q + 1) \stackrel{\cdot}{\sim} \textrm{N}\left(0, \frac{w_{q+1, q+1}}{n}\right)$ and Bartlett \begin{align*} w_{q+1, q+1} & = \sum_{k=1}^{\infty} \left[ \rho(k+q+1) + \rho(k-q-1) - 2 \rho(k+1) \rho(q) \right]^2 \\ & = \sum_{k=1}^{\infty}\rho(k-q-1)^2\\ & = 1 + 2 \sum_{j=1}^q \rho(j)^2 \end{align*}

### Test for an MA$$(q)$$: from the ACF

1. Hypotheses $H_0: \{X_t\} \sim \textrm{MA}(q) \quad \longleftrightarrow \quad H_A: \text{Not } H_0$

2. Test statistic $Z = \frac{\hat{\rho}(q + 1) - 0}{\sqrt{\frac{1 + 2 \sum_{j=1}^q \hat{\rho}(j)^2}{n}}}$

3. Reject $$H_0$$ if $$|Z| \geq z_{\alpha/2}$$

• Note: under the null hypothesis, we use the sample ACF plot with bounds $$\pm 1.96 \times \sqrt{\frac{1 + 2 \sum_{j=1}^q \hat{\rho}(j)^2}{n}}$$ to check if $$\hat{\rho}(h)$$ for all $$h \geq q+1$$ are inside the bounds. But this may have some multiple testing problems.

### Partial autocorrelation function (PACF)

• We define the partial autocorrelation function (PACF) of an ARMA process as the function $$\alpha(\cdot)$$ $\alpha(0) = 1, \quad \alpha(h) = \phi_{hh}, \text{ for } h \geq 1$ Here, $$\phi_{hh}$$ is the last entry of $\boldsymbol\phi_h = \boldsymbol\Gamma_h^{-1} \boldsymbol\gamma_h, \quad \text{where } \boldsymbol\Gamma_h = [\gamma(i-j)]_{i,j=1}^h, \ \boldsymbol\gamma_h = [\gamma(1), \ldots, \gamma(h)]'$

• Sample PACF $$\hat{\alpha}(\cdot)$$: change all $$\gamma(\cdot)$$ above to $$\hat{\gamma}(\cdot)$$

• Recall: in DL algorithm $$\hat{X}_{n+1} = \phi_{n1}X_n + \cdots + \phi_{nn} X_1$$, $\phi_{nn} = \alpha(n), \quad \text{PACF at lag }n$

### PACF property

• $$\phi_{nn}$$ is the correlation between the prediction errors $\alpha(n) = \text{Corr} \left( X_n - P(X_n | X_1, \ldots, X_{n-1}), X_0 - P(X_0 | X_1, \ldots, X_{n-1})\right)$

• Theorem: A stationary series is AR$$(p)$$ if and only if $\alpha(h) = 0 \text{ for all } h > p$

• If $$\{X_t\}$$ is an AR$$(p)$$, then we have asymptotic normality $\hat{\alpha}(h) \stackrel{\cdot}{\sim} \textrm{N}\left(0, \frac{1}{n}\right), \quad \text{for all } h > p$

### Test for an AR$$(p)$$: from the PACF

1. Hypotheses $H_0: \{X_t\} \sim \textrm{AR}(p) \quad \longleftrightarrow \quad H_A: \text{Not } H_0$

2. Test statistic $Z = \frac{\hat{\alpha}(p + 1) - 0}{\sqrt{\frac{1}{n}}}$

3. Reject $$H_0$$ if $$|Z| \geq z_{\alpha/2}$$

• Note: under the null hypothesis, we use the sample PACF plot with bounds $$\pm 1.96 / \sqrt{n}$$ to check if $$\hat{\alpha}(h)$$ for all $$h \geq p+1$$ are inside the bounds. But this may have some multiple testing problems.

# Forecast ARMA Processes

### Forecast ARMA$$(p, q)$$ using the innovation algorithm

• Let $$m = \max(p, q)$$

• One-step prediction $\hat{X}_{n+1} = \begin{cases} \sum_{j=1}^n \theta_{nj}\left( X_{n+1-j} - \hat{X}_{n+1-j} \right), & n < m\\ \sum_{i=1}^p \phi_i X_{n+1-i} + \sum_{j=1}^q \theta_{nj}\left( X_{n+1-j} - \hat{X}_{n+1-j} \right), & n \geq m\\ \end{cases}$

• Special case: AR$$(p)$$ process $\hat{X}_{n+1} = \sum_{i=1}^p \phi_k X_{n+1-i}, \quad n\geq p$
• $$h$$-step prediction: for $$n>m$$ and all $$h \geq 1$$, $P_n X_{n+h} = \sum_{i=1}^p \phi_i P_n X_{n+h-i} + \sum_{j=h}^q \theta_{n+h-1,j}\left( X_{n+1-j} - \hat{X}_{n+1-j} \right)$

### Innovation algorithm parameters vs MA parameters

• Innovation algorithm parameters converges to the MA parameters: If $$\{X_t\}$$ is invertible, then as $$n\rightarrow \infty$$, $\theta_{nj} \longrightarrow \theta_j, \quad j = 1, 2, \ldots, q$

• Prediction MSE converges to $$\sigma^2$$: Let $v_n = E(X_{n+1} - \hat{X}_{n+1})^2, \quad \text{and } v_n = r_n \sigma^2$ If $$\{X_t\}$$ is invertible, then as $$n\rightarrow \infty$$, $r_n \longrightarrow 1$