Why the Constant Needs Interpretation
A constant in an ARIMA equation can describe different things depending on whether the model is written for the original series or for a differenced series. In a stationary AR model, the constant is generally not the process mean. In a first-differenced ARIMA model, the constant is generally not the long-run mean change either. Autoregressive terms alter both quantities.
The distinction matters when interpreting a fitted model or translating its equation into a forecast. A positive constant in a model for changes may imply a positive long-run trend, but its size need not equal that trend. The relevant adjustment depends on the autoregressive coefficients.
For the results below, assume the ARMA process after differencing is stationary and has a finite mean. In particular, its autoregressive polynomial must permit a stationary solution, and \(\phi(1)=1-\phi_1-\cdots-\phi_p\) must be nonzero. The moving-average terms do not change the unconditional mean when the innovations have mean zero.
Constant and Mean in a Stationary ARMA Model
First consider \(d=0\), so the equation is for the original series: \(\phi(B)y_t=c+\theta(B)\varepsilon_t\). Let \(\mu=E[y_t]\). Stationarity makes the mean the same at every time, and \(E[\varepsilon_t]=0\). Taking expectations gives \(\phi(1)\mu=c\), hence \(\mu=c/\phi(1)\). Therefore, the equation constant equals the mean only in special cases, such as a model with no autoregressive terms.
For an AR(1) model, \(\phi(B)=1-\phi B\), so the equation is \(y_t=c+\phi y_{t-1}+\varepsilon_t\). If \(|\phi|<1\), its mean is \(c/(1-\phi)\). If the desired mean is \(\mu\), the constant must instead be set to \(c=(1-\phi)\mu\).
Worked Example: Finding a Stationary Mean and Its Constant
A stationary series follows \(y_t=2+0.5y_{t-1}+\varepsilon_t\), where \(\varepsilon_t\) are independent \(N(0,4)\) innovations. Find the mean of \(y_t\), then find the constant required if the intended mean were 6 instead. State why the stationary-mean calculation applies.
Method and assumptions. This is an AR(1) model with \(\phi=0.5\). Since \(|0.5|<1\), it has a stationary solution. The innovations have mean zero; their variance, 4, affects variability but not the mean.
Mean under the given equation. Here \(c=2\), so \(\mu=c/(1-\phi)=2/(1-0.5)=2/0.5=4\). The equation constant 2 is not the series mean.
Constant for a target mean of 6. Rearranging \(\mu=c/(1-\phi)\) gives \(c=(1-\phi)\mu\). Thus \(c=(1-0.5)(6)=0.5(6)=3\). The equation with mean 6 would be \(y_t=3+0.5y_{t-1}+\varepsilon_t\).
Check and common mistake. Substitution gives \(3+0.5(6)=6\), as required. Treating the original constant 2 as the mean would be incorrect; the autoregressive feedback changes the mean.
Constant and Drift After First Differencing
For \(d=1\), write \(x_t=\Delta y_t=(1-B)y_t=y_t-y_{t-1}\). The ARIMA equation becomes an ARMA equation for \(x_t\): \(\phi(B)x_t=c+\theta(B)\varepsilon_t\). If \(x_t\) is stationary, its mean is \(\mu_x=c/\phi(1)\). This mean is the long-run average change in \(y_t\), or the drift. Thus \(c\) and drift are generally different whenever the differenced equation has autoregressive terms.
The word “long-run” is important. A particular one-step change can depend on recent changes and recent innovations. The drift is the stationary mean of the change process, not necessarily the conditional expected change at the current time. Under the stated stationarity conditions, the expected changes settle around this mean.
When there are no autoregressive terms in the differenced equation, \(\phi(1)=1\) and the drift equals \(c\). With autoregressive terms, feedback accumulates or offsets part of the constant, producing the divisor \(\phi(1)\). Moving-average terms can affect short-term forecasts but do not appear in this unconditional-mean calculation.
Worked Example: Drift Versus a One-Step Expected Change
A hypothetical index is modeled by \(\Delta y_t=2+0.5\Delta y_{t-1}+\varepsilon_t\), with independent \(N(0,1)\) innovations. At time \(T\), the observed index is \(y_T=120\) and the observed change is \(\Delta y_T=6\). Find the long-run drift and the one-step forecast \(E[y_{T+1}\mid\text{information through }T]\).
Method and assumptions. This is an AR(1) model for the change series, with autoregressive coefficient \(0.5\). Since \(|0.5|<1\), the change process is stationary. The innovation at \(T+1\) has conditional expectation zero.
Long-run drift. The equation constant is \(c=2\) and \(\phi(1)=1-0.5=0.5\). Therefore the mean change is \(2/0.5=4\) index units per time period.
One-step forecast. The equation gives \(E[\Delta y_{T+1}\mid\text{information through }T]=2+0.5(6)+0=5\). Since \(y_{T+1}=y_T+\Delta y_{T+1}\), the forecast is \(120+5=125\).
Check and common mistake. The one-step expected change, 5, differs from the long-run drift, 4, because the current change of 6 is above its stationary mean. Using 2 as the drift would confuse the equation constant with the mean change; using 4 as the immediate forecast would ignore the observed lagged change.
Drift in a Differenced ARMA Model
The same mean relation applies when the differenced model includes moving-average terms. Those terms affect forecasts because past innovations can contribute to a conditional expectation, but they do not alter the stationary mean of the change series. The innovation convention must be clear: here the moving-average polynomial is \(1+\theta_1B+\cdots+\theta_qB^q\), so a positive \(\theta_1\) multiplies the previous innovation with a positive sign.
Worked Example: ARIMA(1,1,1) with a Recent Innovation
A hypothetical claims index satisfies \(\Delta y_t=1.5+0.25\Delta y_{t-1}+\varepsilon_t+0.4\varepsilon_{t-1}\), where \(\varepsilon_t\) are independent \(N(0,1)\) innovations. At time \(T\), \(\Delta y_T=4\), \(\varepsilon_T=-0.5\), and \(y_T=80\). Find the long-run drift and the one-step conditional forecast of \(y_{T+1}\).
Method and assumptions. The differenced series has ARMA(1,1) form, with \(\phi=0.25\) and \(\theta_1=0.4\) under the stated plus-sign convention. Since \(|0.25|<1\), the differenced process is stationary. The next innovation has conditional expectation zero.
Long-run drift. The equation constant is \(c=1.5\), and \(\phi(1)=1-0.25=0.75\). Thus the mean change is \(1.5/0.75=2\) index units per period. The MA coefficient does not enter this calculation.
One-step forecast for the change. Substituting the known values and taking the conditional expectation of \(\varepsilon_{T+1}\) gives \(1.5+0.25(4)+0.4(-0.5)+0=1.5+1-0.2=2.3\).
Forecast for the index. Add the predicted change to the latest index: \(E[y_{T+1}\mid\text{information through }T]=80+2.3=82.3\).
Check and common mistake. The forecasted change is 2.3, while the long-run drift is 2. The extra 0.3 reflects the current change and recent innovation. Dropping the \(0.4\varepsilon_T\) term would give 2.5 and would be an incorrect one-step forecast.
Reading Constants by Difference Order
It is useful to identify exactly which series appears on the left side of an ARIMA equation before interpreting its constant. With \(d=0\), the constant is in the stationary equation for \(y_t\), and the stationary mean is \(c/\phi(1)\). With \(d=1\), it is in the equation for \(\Delta y_t\), and the stationary mean of that change—the drift—is \(c/\phi(1)\). In neither case should the constant automatically be labelled the mean.
If \(d=2\), the equation is for \(\Delta^2 y_t\). When this second difference has a stationary mean \(m=c/\phi(1)\), a nonzero \(m\) corresponds to a changing first difference and, in the original series, a quadratic rather than a linear deterministic trend. For example, the sequence with constant second difference \(m\) has a quadratic leading term \(m t^2/2\); initial levels and first differences contribute lower-order terms. The \(d=1\) interpretation of a constant as a steady average change should therefore not be carried over unchanged to \(d=2\).
| Difference order | Equation is for | Interpretation of \(c/\phi(1)\), when stationary |
|---|---|---|
| \(d=0\) | \(y_t\) | Mean level of the original series |
| \(d=1\) | \(\Delta y_t\) | Mean change, or drift, of the original series |
| \(d=2\) | \(\Delta^2 y_t\) | Mean second difference; nonzero mean implies quadratic trend behavior in the original series |
Common Exam Traps
- Calling the equation constant the mean. For a stationary AR(1), \(y_t=c+\phi y_{t-1}+\varepsilon_t\), the mean is \(c/(1-\phi)\), not \(c\).
- Calling the equation constant the drift. In a differenced AR model, calculate \(c/(1-\sum_j\phi_j)\). For example, with \(c=2\) and \(\phi_1=0.5\), the drift is \(2/(1-0.5)=4\), not 2.
- Using long-run drift as an immediate forecast. A one-step forecast may depend on observed lagged differences and past innovations. Substitute those values into the conditional forecast equation.
- Letting MA coefficients change the unconditional mean. With zero-mean innovations, the expected value of every MA innovation term is zero. MA terms may matter for conditional forecasts, as in the ARIMA(1,1,1) example, but not for the stationary mean formula.
- Ignoring the difference order. The constant is attached to the series on the left side of the equation. Confirm whether that is \(y_t\), \(\Delta y_t\), or a higher difference before describing its interpretation.
Check Your Understanding
For each problem, identify the relevant series before interpreting its constant. Assume zero-mean innovations and the stationarity conditions needed for the stated mean to exist.
- A differenced series follows \(\Delta y_t=1.2+0.4\Delta y_{t-1}+\varepsilon_t\). Find its long-run mean change. (Answer: \(1.2/(1-0.4)=2\) units per period.)
- A stationary AR(2) change model is \(\Delta y_t=0.5+0.3\Delta y_{t-1}+0.2\Delta y_{t-2}+\varepsilon_t\). Calculate the long-run drift and identify the denominator used. (Answer: \(0.5/(1-0.3-0.2)=1\) unit per period; the denominator is \(0.5\).)
- A stationary level model is \(y_t=2+0.5y_{t-1}+\varepsilon_t\). Find its mean. What constant would give the same autoregressive coefficient but a mean of 6? (Answer: The given mean is \(2/(1-0.5)=4\); the required constant is \((1-0.5)(6)=3\).)
- A differenced ARMA model is \(\Delta y_t=0.9-0.2\Delta y_{t-1}+\varepsilon_t+0.6\varepsilon_{t-1}\). Calculate its long-run drift. If the current change is 1 and the most recent innovation is 0.5, also calculate the next expected change. (Answer: Drift \(=0.9/[1-(-0.2)]=0.75\); next expected change \(=0.9-0.2(1)+0.6(0.5)=1.0\).)
- A differenced AR(2) model is \(\Delta y_t=1.8+0.4\Delta y_{t-1}-0.1\Delta y_{t-2}+\varepsilon_t\). Calculate the long-run drift and explain why it is not 1.8. (Answer: \(1.8/[1-0.4-(-0.1)]=1.8/0.7=18/7\approx2.5714\) units per period; autoregressive feedback changes the mean relative to the equation constant.)