Centering ARMA Processes and Interpreting the Mean

Why the Intercept Is Not Usually the Mean

An ARMA model combines lagged values of a series with current and lagged innovations. Its intercept is a constant in the model equation, but it is generally not the long-run mean of the series. The autoregressive terms carry part of the mean forward, so the intercept must be interpreted together with the AR coefficients. Moving-average terms affect dependence over time, but zero-mean innovations contribute no constant to the unconditional mean.

This distinction is important when interpreting a fitted model or converting it to a centered form. The results here are aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Definition: In this tutorial, an ARMA($p,q$) model with intercept $c$ is written $$ \phi(B)X_t=c+\theta(B)\varepsilon_t, $$ where $B$ is the lag operator, $BX_t=X_{t-1}$, $$ \phi(B)=1-\phi_1B-\cdots-\phi_pB^p,\qquad \theta(B)=1+\theta_1B+\cdots+\theta_qB^q. $$ The innovations $\varepsilon_t$ are white noise with $E[\varepsilon_t]=0$ and constant finite variance $\sigma^2$. With this sign convention, the intercept is $c$, and $\phi(1)=1-\phi_1-\cdots-\phi_p$.

A white-noise innovation is zero-mean by definition here. It need not equal zero in any particular period; rather, its expected value is zero. The standard formula for the process mean assumes a stationary process with a finite, time-independent expectation and $\phi(1)\ne 0$. Stationarity conditions on the AR polynomial are a separate issue; the formula below does not make a nonstationary model stationary.

Deriving the Unconditional Mean

Let $\mu=E[X_t]$ for a stationary process. Taking expectations of the model gives the mean equation. Since every innovation has mean zero, the expected value of the moving-average part is zero.

E[\phi(B)X_t] =E[X_t]-\phi_1E[X_{t-1}]-\cdots-\phi_pE[X_{t-p}] =\mu\left(1-\sum_{j=1}^{p}\phi_j\right).

The right-hand side of the original model has expectation $c$. Equating the two expectations yields the mean formula.

Result: For a stationary ARMA($p,q$) model with zero-mean innovations and $\phi(1)\ne 0$, the unconditional mean is $$ \mu=E[X_t]=\frac{c}{\phi(1)} =\frac{c}{1-\phi_1-\cdots-\phi_p}. $$ The moving-average coefficients do not appear in this formula because the innovation terms have expectation zero.

This derivation also explains why $c$ is generally not the mean: the AR coefficients multiply the mean on the left side of the expectation equation. If $c=0$ and $\phi(1)\ne0$, the unconditional mean is zero. If $c=\mu\phi(1)$, the model can be expressed directly in terms of its mean.

Worked Example: Finding the Mean and Centering an ARMA(1,1)

A hypothetical monthly claim-index series follows $$ X_t=0.8+0.6X_{t-1}+\varepsilon_t+0.25\varepsilon_{t-1}, $$ where $\varepsilon_t$ is zero-mean white noise with variance $1.44$. Assume the model is stationary. Find the unconditional mean and write the centered equation. State why the intercept is not the mean.

Method and assumptions. This is an ARMA(1,1) model in the stated plus-sign convention. The innovation mean is zero, so the moving-average terms make no contribution to the expectation.

Step 1: Identify the AR coefficient and intercept. Here $c=0.8$, $\phi_1=0.6$, and $\theta_1=0.25$. Thus $\phi(1)=1-0.6=0.4$.

Step 2: Calculate the mean. $$ \mu=\frac{0.8}{0.4}=2. $$

Step 3: Check by substitution in the expectation equation. Taking expectations in the original equation gives $E[X_t]=0.8+0.6E[X_{t-1}]$. At a stationary mean of $2$, the right side is $0.8+0.6(2)=2$, as required.

Step 4: Center the process. Define $Z_t=X_t-2$. Subtracting the mean from the model gives $$ Z_t=0.6Z_{t-1}+\varepsilon_t+0.25\varepsilon_{t-1}, $$ or $(1-0.6B)Z_t=(1+0.25B)\varepsilon_t$. The intercept $0.8$ is not the process mean; it is the constant that, together with the AR coefficient, produces mean $2$.

Common mistake avoided. Setting $\mu=c=0.8$ ignores the autoregressive contribution. The expectation check exposes that error: $0.8+0.6(0.8)=1.28\ne0.8$.

Centering the Model

Centering means subtracting the process mean from each observation. It expresses the model in terms of deviations from the long-run level rather than deviations from zero. For the lag-polynomial form above, set $Z_t=X_t-\mu$. Because a constant remains unchanged under a lag, $\phi(B)\mu=\phi(1)\mu$. Consequently, if $c=\phi(1)\mu$, the constant cancels.

\phi(B)X_t=c+\theta(B)\varepsilon_t,\qquad X_t=\mu+Z_t $$ $$ \phi(B)Z_t+\phi(1)\mu=c+\theta(B)\varepsilon_t.

Since $c=\phi(1)\mu$, this reduces to

\phi(B)Z_t=\theta(B)\varepsilon_t,\qquad Z_t=X_t-\mu.

Thus a model written around its mean has no separate intercept: $$ \phi(B)(X_t-\mu)=\theta(B)\varepsilon_t. $$ This is an equivalent parameterization, not a different stochastic model. It also makes clear that the centered series has mean zero, while the original series has mean $\mu$. The innovation mean is a third quantity: under the standard model it is zero, regardless of whether the process mean is zero.

Worked Example: An ARMA(2,1) Model with a Negative AR Coefficient

A hypothetical quarterly reserve indicator is modeled by $$ X_t=1.5+0.3X_{t-1}-0.2X_{t-2}+\varepsilon_t+0.4\varepsilon_{t-1}, $$ where $\varepsilon_t$ is zero-mean white noise with variance $0.64$. Assume a stationary solution exists. Find the unconditional mean, check it, and write the centered model.

Method and assumptions. The model has $c=1.5$, $\phi_1=0.3$, and $\phi_2=-0.2$. The negative sign is part of the coefficient: the lag-two term is $-0.2X_{t-2}$. The moving-average term has expectation zero.

Step 1: Evaluate the AR polynomial at one. $$ \phi(1)=1-\phi_1-\phi_2 =1-0.3-(-0.2)=0.9. $$

Step 2: Calculate the mean. $$ \mu=\frac{1.5}{0.9}=\frac{5}{3}=1.6667 $$ to four decimal places.

Step 3: Check the expectation equation. Taking expectations gives $\mu=1.5+0.3\mu-0.2\mu$. Substituting $\mu=5/3$, the right side is $1.5+0.1(5/3)=1.5+1/6=5/3$. The equation is satisfied.

Step 4: Write the centered model. With $Z_t=X_t-5/3$, the model becomes $$ Z_t=0.3Z_{t-1}-0.2Z_{t-2}+\varepsilon_t+0.4\varepsilon_{t-1}. $$ Equivalently, $(1-0.3B+0.2B^2)Z_t=(1+0.4B)\varepsilon_t$.

Common mistake avoided. Replacing $\phi_2=-0.2$ by $+0.2$ when evaluating $1-\phi_1-\phi_2$ would give the wrong denominator. Keeping the signed coefficient visible gives $1-0.3-(-0.2)=0.9$, not $0.5$.

When the Disturbance Has a Nonzero Mean

Sometimes an equation is written with a white-noise disturbance that has mean $m\ne0$. Under the convention used here, that disturbance is not a zero-mean innovation. Its mean must be included in the unconditional mean calculation. For clarity, write the disturbance as $u_t=m+\varepsilon_t$, where $\varepsilon_t$ is zero-mean white noise. If the model is $\phi(B)X_t=c+\theta(B)u_t$, then $E[\theta(B)u_t]=\theta(1)m$. Therefore, provided a stationary finite mean exists and $\phi(1)\ne0$,

Result: For a model written with a disturbance $u_t$ whose mean is $m$, $$ \mu=\frac{c+\theta(1)m}{\phi(1)}. $$ Equivalently, replacing $u_t$ by the zero-mean innovation $\varepsilon_t=u_t-m$ changes the intercept to $c+\theta(1)m$.

The change in intercept includes all the moving-average weights: the constant part $m$ appears once for the current disturbance and once for each lagged disturbance. This adjustment is only needed when the disturbance in the written equation truly has nonzero mean. Do not add it to the standard zero-mean innovation formula.

Worked Example: Reparameterizing a Nonzero-Mean Disturbance

Consider the hypothetical model $$ X_t=0.5+0.6X_{t-1}+u_t+0.25u_{t-1}, $$ where $u_t$ is white noise around mean $m=0.2$, with variance $0.25$. Assume the series has a stationary finite mean. Find that mean and rewrite the model using zero-mean innovations.

Method and assumptions. Here $u_t$ is not a zero-mean innovation. Set $\varepsilon_t=u_t-0.2$, so $u_t=0.2+\varepsilon_t$ and $E[\varepsilon_t]=0$.

Step 1: Compute the mean directly. The AR polynomial has $\phi(1)=1-0.6=0.4$, while $\theta(1)=1+0.25=1.25$. The numerator is $$ c+\theta(1)m=0.5+1.25(0.2)=0.5+0.25=0.75. $$ Thus $$ \mu=\frac{0.75}{0.4}=1.875. $$

Step 2: Check the expectation equation. Taking expectations in the original equation gives $$ E[X_t]=0.5+0.6E[X_{t-1}]+0.2+0.25(0.2). $$ At $\mu=1.875$, the right side is $0.5+0.6(1.875)+0.2+0.05=0.5+1.125+0.25=1.875$. The result checks.

Step 3: Rewrite with zero-mean innovations. Substituting $u_t=0.2+\varepsilon_t$ into both disturbance terms gives a total added constant $0.2+0.25(0.2)=0.25$. The equivalent model is $$ X_t=0.75+0.6X_{t-1}+\varepsilon_t+0.25\varepsilon_{t-1}. $$ Its mean is $0.75/(1-0.6)=1.875$, the same as before.

Common mistake avoided. Using $c/(1-\phi_1)=0.5/0.4=1.25$ would incorrectly treat $u_t$ as zero-mean. The missing $0.25$ in the numerator is the mean contribution from the current and lagged disturbances.

Common Interpretive Traps

Calling the intercept the mean. For a stationary ARMA model, calculate $c/\phi(1)$; do not equate $c$ with $\mu$ unless the parameterization explicitly puts the process around its mean.
Including MA coefficients when innovations are zero-mean. Their coefficients affect dependence and variance, but their expected contribution is zero under the stated model.
Forgetting a disturbance mean. If the equation uses a nonzero-mean disturbance, account for the entire moving-average polynomial through $\theta(1)m$, or recenter the disturbance and adjust the intercept.
Using the formula without its conditions. The derivation assumes a stationary finite mean and $\phi(1)\ne0$. If $\phi(1)=0$, the expectation equation does not determine a unique finite mean by division; the formula cannot be used as written.

Key takeaway: With zero-mean innovations, the ARMA unconditional mean is $c/\phi(1)$, not generally $c$. Centering by that mean removes the intercept. A nonzero disturbance mean must first be accounted for, either directly or by converting to zero-mean innovations and adjusting the constant.

Check Your Understanding

For each problem, use the stated lag-polynomial sign convention and assume a stationary finite-mean solution where needed.

A hypothetical ARMA(1,1) series has intercept $1.2$, AR coefficient $0.5$, and zero-mean innovations. Its MA coefficient is $-0.3$. Find its unconditional mean and say whether the MA coefficient changes that mean. (Answer: $\mu=2.4$; no.)
A hypothetical AR(2) model has intercept $0.6$, $\phi_1=0.2$, and $\phi_2=0.3$. Find the unconditional mean and verify it in the expectation equation. (Answer: $\mu=1.2$; $0.6+0.2(1.2)+0.3(1.2)=1.2$.)
A hypothetical ARMA(1,2) model has intercept $-0.4$, $\phi_1=-0.25$, and zero-mean innovations. The MA coefficients are $0.5$ and $-0.2$. Find the mean and the centered equation in lag-polynomial form. (Answer: $\mu=-0.32$; $(1+0.25B)(X_t+0.32)=(1+0.5B-0.2B^2)\varepsilon_t$.)
A hypothetical ARMA(1,1) equation uses disturbances $u_t$ with mean $0.1$: $X_t=0.3+0.4X_{t-1}+u_t+0.2u_{t-1}$. Find the mean and the intercept after converting to zero-mean innovations. (Answer: $\mu=0.7$; the zero-mean innovation intercept is $0.42$.)
A centered ARMA(2,1) model has process mean $4$, $\phi_1=0.1$, and $\phi_2=-0.3$. Find the intercept in the equivalent uncentered form. (Answer: $c=4[1-0.1-(-0.3)]=4.8$.)

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