ACF and PACF of an ARMA(1,1) Process

Worked Solution

How to Think About It: An ARMA(1,1) is the minimal model that captures both momentum (AR) and mean-reversion-after-shock (MA) in a single process. The ACF decays geometrically starting from lag 1 -- that is the AR(1) signature -- but the lag-1 value itself is distorted by the MA term. The PACF, by contrast, does not cut off cleanly like a pure MA; it also decays geometrically. Neither the ACF nor PACF cuts off to zero at any finite lag, which is the diagnostic fingerprint of an ARMA model (versus a pure AR where the PACF cuts off, or a pure MA where the ACF cuts off).

Quick Sanity Checks: - If $\theta = 0$: reduces to AR(1), so $\rho(h) = \phi^h$ and PACF cuts off after lag 1. - If $\phi = 0$: reduces to MA(1), so $\rho(1) = \theta/(1+\theta^2)$ and $\rho(h) = 0$ for $h \ge 2$. - For $\phi > 0$, $\theta > 0$: both components add positive autocorrelation, so $\rho(1) > 0$. - For $\phi > 0$, $\theta < 0$: AR pushes $\rho(1)$ up, MA pushes it down -- sign depends on magnitudes.

Derivation:

Part 1 -- Autocovariance function:

Multiply both sides of $X_t = \phi X_{t-1} + \varepsilon_t + \theta \varepsilon_{t-1}$ by $X_{t-h}$ and take expectations.

First, compute $\gamma(0)$ and $\gamma(1)$ by multiplying both sides by $X_t$ and $X_{t-1}$ respectively:

$$\gamma(0) = \phi \gamma(1) + \sigma^2 + \theta \cdot E[\varepsilon_{t-1} X_t].$$

Note $E[\varepsilon_{t-1} X_t] = \theta\sigma^2$ and $E[\varepsilon_t X_t] = \sigma^2$. So: $$\gamma(0) = \phi \gamma(1) + \sigma^2(1 + \theta^2). \quad (*)$$

For $\gamma(1)$, multiply by $X_{t-1}$: $$\gamma(1) = \phi\gamma(0) + \theta\sigma^2.$$

From these two equations, solve simultaneously. Substitute $\gamma(1) = \phi\gamma(0) + \theta\sigma^2$ into $(*)$: $$\gamma(0) = \phi(\phi\gamma(0) + \theta\sigma^2) + \sigma^2(1+\theta^2)$$ $$\gamma(0)(1 - \phi^2) = \phi\theta\sigma^2 + \sigma^2(1+\theta^2) = \sigma^2(1 + 2\phi\theta + \theta^2).$$ $$\gamma(0) = \frac{(1 + 2\phi\theta + \theta^2)\sigma^2}{1 - \phi^2}.$$

And: $$\gamma(1) = \phi\gamma(0) + \theta\sigma^2 = \frac{(\phi + \theta)(1 + \phi\theta)\sigma^2}{1-\phi^2}.$$

For $h \ge 2$, the MA term no longer contributes (since $\varepsilon_{t-1}$ is independent of $X_{t-h}$ for $h \ge 2$), so the recursion becomes: $$\gamma(h) = \phi \gamma(h-1), \quad h \ge 2.$$

Hence $\gamma(h) = \phi^{h-1} \gamma(1)$ for $h \ge 1$.

The ACF for $h \ge 1$: $$\rho(h) = \rho(1) \cdot \phi^{h-1}, \quad \text{where} \quad \rho(1) = \frac{(\phi+\theta)(1+\phi\theta)}{1+2\phi\theta+\theta^2}.$$

Part 2 -- Sign of $\rho(1)$:

From the formula $\rho(1) = (\phi+\theta)(1+\phi\theta)/(1+2\phi\theta+\theta^2)$:

• The denominator $1 + 2\phi\theta + \theta^2 = (1+\phi\theta)^2 + \theta^2(1-\phi^2) > 0$ always.
• The factor $(1+\phi\theta) > 0$ when $|\phi\theta| < 1$ (always true since $|\phi|,|\theta| < 1$).
• So the sign of $\rho(1)$ equals the sign of $(\phi + \theta)$.

Conclusion: - $\rho(1) > 0$ when $\phi + \theta > 0$ - $\rho(1) < 0$ when $\phi + \theta < 0$ - $\rho(1) = 0$ when $\phi = -\theta$

Example: $\phi = 0.3$, $\theta = -0.8$: $\phi + \theta = -0.5 < 0$, so $\rho(1) < 0$ despite the positive AR coefficient.

Part 3 -- PACF behavior:

For an ARMA(1,1), the PACF at lag $h$ is: $$\alpha(h) = A \cdot (-\theta)^{h-1} \cdot \phi_1, \quad h \ge 2,$$ (approximately) where the exact form depends on the parameters, but the key qualitative behavior is: - $\alpha(1) = \rho(1)$ -- equal to the ACF at lag 1. - For $h \ge 2$, the PACF decays geometrically toward zero, at rate $\approx -\theta$ (or more precisely, as a mixture involving both $\phi$ and $\theta$).

Unlike a pure AR(1) (where PACF is zero for $h \ge 2$), the PACF of ARMA(1,1) tails off slowly. Neither the ACF nor PACF has a clean cutoff -- this is how you distinguish an ARMA from a pure AR or MA in practice.

Answer: $$\gamma(0) = \frac{(1+2\phi\theta+\theta^2)\sigma^2}{1-\phi^2}, \quad \gamma(h) = \phi^{h-1}\gamma(1)\text{ for }h\ge 1, \quad \rho(1) = \frac{(\phi+\theta)(1+\phi\theta)}{1+2\phi\theta+\theta^2}.$$ Sign of $\rho(1)$ is determined by the sign of $\phi + \theta$. Both ACF and PACF tail off geometrically -- no clean cutoff at any lag.