Why Parameter Counting Matters
Information criteria balance how well a time-series model fits against how many parameters it uses. For seasonal ARIMA models, the fit may include nonseasonal and seasonal autoregressive and moving-average terms, a mean or other deterministic term, and an innovation variance. Counting these consistently is essential: an incorrect parameter count changes the penalty, and can change which candidate model appears preferable.
The key distinction is between a fixed model order and a parameter estimated from the data. The orders say which coefficients may be estimated; the coefficients themselves, and any other freely estimated quantities, determine the parameter count used in the criterion.
A Practical Counting Rule
For a standard seasonal ARIMA fit with freely estimated AR and MA coefficients, the baseline count is \(p+q+P+Q\). Add one for each additional, separately estimated deterministic coefficient, such as a mean or a drift term when included. Add one for the innovation variance when it is estimated. Do not add \(d\), \(D\), or \(s\) merely because they are part of the model notation.
In this bookkeeping formula, \(c\) is the number of separately estimated deterministic coefficients, \(m\) is the number of other separately estimated model coefficients not already included (often zero), and \(v=1\) when the innovation variance is estimated and \(v=0\) when it is fixed or otherwise excluded under the stated criterion convention. In the common case with one estimated mean or drift and an estimated variance, \(c=1\) and \(v=1\). Count the parameters in the likelihood actually fitted, rather than assuming every model uses the same deterministic terms.
A mean and an intercept are not automatically two distinct parameters. A model may express its location using a mean parameter or an intercept parameter; if only one is freely estimated, count one. Likewise, count a drift only when it is actually included as an estimated coefficient. Whether a deterministic term is appropriate after differencing is a modeling question; for parameter counting, the relevant question is whether it is freely estimated in the fitted model.
Why Fixed Differencing Orders Are Not Parameters
AIC, AICc, and BIC penalize the dimension of the estimated parameter vector. For a likelihood \(L\), the conventional forms are \( -2\log L+2k\) for AIC and \( -2\log L+k\log(n)\) for BIC; AICc adds a finite-sample correction when applicable. The \(k\) in these expressions counts estimated parameters, not every choice used to define the candidate model.
This gives a direct justification for excluding fixed differencing orders. If the analyst selects \(d=1\) and \(D=1\) before fitting, the differencing operations are specified transformations. They do not introduce freely estimated coefficients into the likelihood parameter vector. By contrast, an AR coefficient such as \(\phi_1\) is estimated from the observed series and is one component of that vector. The same reasoning includes an estimated variance: it is an unknown likelihood parameter even when the likelihood calculation profiles it out during computation.
Here \(\ell\) is the maximized log likelihood, \(k\) is the number of estimated parameters under the criterion's convention, and \(n\) is the sample size used for that likelihood. For AICc, use the appropriate finite-sample correction with the same \(k\) convention and the likelihood sample size specified for the fit. When comparing results from different software, check whether its reported count includes the innovation variance and deterministic terms; do not mix conventions.
Worked Example: A Seasonal Model With One Seasonal AR Term
A hypothetical monthly portfolio series is fitted with \(\operatorname{ARIMA}(1,0,1)(1,0,0)_{12}\). The model estimates one mean, the innovation variance, and all indicated AR and MA coefficients. Its maximized log likelihood is \(-84.6\), based on \(n=60\) observations. Count \(k\), then calculate AIC and AICc using \( \text{AICc}=\text{AIC}+\frac{2k(k+1)}{n-k-1}\).
Method. Count each free coefficient once. The nonseasonal AR order contributes \(p=1\), the nonseasonal MA order contributes \(q=1\), and the seasonal AR order contributes \(P=1\). There is no seasonal MA coefficient because \(Q=0\). The estimated mean and estimated variance contribute one each. Thus
The differencing orders \(d=0\) and \(D=0\), and the period \(s=12\), do not add to \(k\). First, \(-2\ell=-2(-84.6)=169.2\). Therefore
For AICc, the correction is \(2(5)(6)/(60-5-1)=60/54=1.1111\) to four decimal places. Hence AICc is \(179.2+1.1111=180.3111\), to four decimal places. The calculation is checked by noting that the correction is positive, so AICc exceeds AIC, as expected when \(n>k+1\). The common mistake avoided here is counting the seasonal period 12 as twelve estimated parameters.
Multiplicative Seasonal Terms and Cross-Products
Seasonal ARIMA models commonly use multiplicative seasonal and nonseasonal polynomials. Multiplication creates terms at combined lags, but it does not make every resulting product a separate free coefficient. A cross-product is determined by the coefficients that were already estimated.
Worked Example: Counting a Multiplicative Seasonal Model
A hypothetical quarterly claims index is modeled by \(\operatorname{ARIMA}(2,0,1)(1,0,1)_4\), with an estimated mean and an estimated innovation variance. Its maximized log likelihood is \(-110\), based on \(n=50\) observations. Find \(k\) and AIC.
Method. The nonseasonal AR part has two free coefficients, the nonseasonal MA part has one, the seasonal AR part has one, and the seasonal MA part has one. Thus there are \(2+1+1+1=5\) AR and MA coefficients. Adding one estimated mean and one estimated variance gives \(k=7\).
The polynomial expansion explains why no extra parameter is needed for cross-products. The AR polynomial is
The coefficients at lags 5 and 6 are products of existing coefficients, not new independent parameters. Similarly, the MA polynomial expands as
The lag-5 coefficient is also a product, so the MA part has two free coefficients, not three. Finally, \(-2\ell=220\), and
The check is that the count comprises five free AR and MA coefficients, one mean, and one variance. A common wrong calculation counts the two AR cross-products and the MA cross-product as additional parameters, giving \(k=10\) and AIC \(=240\). That overcounts because those products are fixed functions of the already estimated coefficients.
Models With Differencing and an Estimated Constant
When differencing is present, the orders \(d\) and \(D\) still do not count as estimated parameters. But a constant, mean, or drift term can contribute to \(k\) if the fitted model estimates it. The exact name and interpretation of the deterministic term depend on the model specification. For counting, record how many independent deterministic coefficients the likelihood estimates; do not infer that count from the differencing orders alone.
Worked Example: Fixed Differencing Orders With an Estimated Drift
A hypothetical monthly demand series is fitted with \(\operatorname{ARIMA}(0,1,1)(0,1,1)_{12}\). The specified model includes one estimated deterministic constant in the differenced equation and estimates the innovation variance. The maximized log likelihood is \(-96\), with \(n=40\) observations in the likelihood. Calculate \(k\), AIC, and AICc.
Method. There is one nonseasonal MA coefficient and one seasonal MA coefficient. There are no AR coefficients. The fitted deterministic constant contributes one parameter, and the estimated variance contributes one. Therefore
Neither \(d=1\) nor \(D=1\) is added. The period 12 is also not a parameter. Since \(-2\ell=192\), AIC is
The AICc correction is \(2(4)(5)/(40-4-1)=40/35=1.1429\), to four decimal places. Thus AICc is \(200+1.1429=201.1429\), to four decimal places. The calculation is checked by confirming that the model has four estimated quantities and that the correction increases AIC. The trap is to count \(d\) and \(D\) as two more parameters, or to assume a deterministic term is absent merely because differencing is used.
A Consistent Counting Checklist
Before calculating a criterion, write down the fitted model's estimated quantities. This short audit is especially useful when comparing candidates with different seasonal orders or deterministic specifications.
Count the freely estimated nonseasonal AR and MA coefficients, then the freely estimated seasonal AR and MA coefficients.
Count each independently estimated mean, intercept, drift, or other deterministic coefficient that is part of the fitted likelihood. Do not count two labels for the same single parameter.
Include the innovation variance if it is estimated and included in the criterion's parameter count. Verify the convention when using software output.
Do not add the differencing orders \(d,D\), the seasonal period \(s\), or polynomial cross-products determined by other coefficients.
Use the resulting \(k\) with the maximized likelihood and the sample-size convention for that fitted model. Keep the convention consistent across candidate models.
Common Counting Errors
- Counting differencing orders. A model with \(d=1,D=1\) does not acquire two estimated parameters solely from those fixed orders.
- Counting seasonal lags instead of coefficients. A seasonal AR(1) term contributes one coefficient, even though it acts at lag \(s\).
- Counting products as free coefficients. In a multiplicative specification, a product such as \(\phi_1\Phi_1\) is determined by its factors and is not counted again.
- Omitting the variance by habit. For the usual likelihood-based information-criterion count, include an estimated innovation variance. Check the specified convention rather than assuming it is omitted.
- Double-counting a location term. If a model uses one estimated intercept to represent its location, do not also add a separate mean unless the fitted model genuinely estimates both as distinct parameters.
Accurate parameter counting does not decide whether a candidate is adequate: residual diagnostics and the comparability of the likelihoods remain important. It ensures that the information criterion applies its complexity penalty to the estimated quantities the model actually uses. This bookkeeping discipline is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.
Check Your Understanding
For each question, count only the estimated quantities stated and use the usual convention that an estimated innovation variance is included.
- A fitted \(\operatorname{ARIMA}(2,1,0)(0,1,1)_{12}\) has no mean or drift term. Its innovation variance is estimated. What is \(k\)? (Answer: \(k=4\).)
- A fitted \(\operatorname{ARIMA}(0,0,0)(1,0,1)_4\) estimates one mean and the innovation variance. What is \(k\)? (Answer: \(k=4\).)
- A fitted \(\operatorname{ARIMA}(1,1,1)(1,1,0)_4\) includes one estimated drift term and an estimated variance. What is \(k\)? (Answer: \(k=5\).)
- A candidate model has \(k=5\) and maximized log likelihood \(-72.5\). Calculate its AIC. (Answer: \(\text{AIC}=155\).)
- Two candidates both estimate a mean and an innovation variance. Model A is \(\operatorname{ARIMA}(1,0,1)(0,0,1)_4\); Model B is \(\operatorname{ARIMA}(1,0,1)(1,0,1)_4\). What are their respective parameter counts? (Answer: Model A has \(k=5\); Model B has \(k=6\).)