Likelihood Constants And Consistent Criterion Calculations

Why Likelihood Conventions Matter

A likelihood calculation can contain terms that do not depend on the model parameters. Those terms can often be dropped while estimating parameters, because they do not change the maximizing parameter value. Information criteria require more care: a term that is harmless for estimation may affect a criterion value, and it is harmless for a ranking only when its effect is the same for every candidate model.

This distinction matters in actuarial work because likelihoods may be reported as full likelihoods, as kernels with data-only terms omitted, or as densities expressed in different units. The AIC and BIC formulas use the maximized log likelihood, so candidate models must be evaluated using a consistent convention.

Definition: For candidate model $M_j$, let $\ell_j$ be its maximized full log likelihood, and let $k_j$ be its number of free parameters. Then $$ \operatorname{AIC}_j=-2\ell_j+2k_j, \qquad \operatorname{BIC}_j=-2\ell_j+k_j\ln n, $$ where $n$ is the number of independent observations used in the likelihood. A likelihood kernel is a log likelihood with terms that do not depend on the model parameters omitted.

If a calculation uses a reduced log likelihood $\tilde{\ell}_j=\ell_j+C_j$, the resulting criterion is the full criterion minus $2C_j$. If the same constant $C$ was omitted for every candidate, all criterion values shift by the same amount and their ranking is preserved. If the omitted constants differ by candidate, the shift is not common and the ranking can change.

\widetilde{\operatorname{AIC}}_j =\operatorname{AIC}_j-2C_j, \qquad \widetilde{\operatorname{BIC}}_j =\operatorname{BIC}_j-2C_j.

The parameter count and the penalty term do not disappear when a likelihood constant is omitted. Nor does a constant become common merely because it is independent of the parameters within each model. It must also be the same across the candidates being compared.

When a Common Constant Preserves the Ranking

Suppose all candidate log likelihoods are calculated from the same observations and convention, and each reduced log likelihood omits the same data-only term $C$. Then the difference between two reduced AIC values equals the difference between their full AIC values:

\widetilde{\operatorname{AIC}}_a-\widetilde{\operatorname{AIC}}_b =(-2\ell_a+2C+2k_a)-(-2\ell_b+2C+2k_b) =\operatorname{AIC}_a-\operatorname{AIC}_b.

The same cancellation holds for BIC, since its penalty terms are unchanged and the same $2C$ is added to both criteria. This is why likelihood kernels are useful for optimization and sometimes for comparing models within a carefully matched calculation. But a reported criterion based on the kernel is not the full AIC or BIC: its numerical value has been shifted.

Worked Example: A Shared Poisson Factorial Term

A hypothetical portfolio is summarized by four independent counts $0,1,2,3$. Compare a one-rate Poisson model $M_0$, with one estimated rate, against a two-rate Poisson model $M_1$, with separate rates for the first two and last two observations. Calculate AIC and BIC both with the full likelihood and with the factorial terms omitted. Use natural logarithms.

Model and assumptions. Under each model, the observations are independent Poisson random variables. For $M_0$, the common mean is $\lambda$. For $M_1$, the first pair has mean $\lambda_1$ and the second pair has mean $\lambda_2$. The parameter counts are $k_0=1$ and $k_1=2$; $n=4$. No interest rate or mortality assumption applies to this count-model comparison.

For independent Poisson observations, the full log likelihood is $\sum_i(x_i\ln\lambda_i-\lambda_i-\ln(x_i!))$. The term omitted in the kernel is common to these two models: $\sum_i\ln(x_i!)=\ln(0!1!2!3!)=\ln 12=2.4849066498$.

Fit $M_0$. The maximum-likelihood estimate is $\hat{\lambda}=(0+1+2+3)/4=1.5$. Its kernel value is $6\ln(1.5)-6=2.4327906486-6=-3.5672093514$. The full maximized log likelihood is $-3.5672093514-2.4849066498=-6.0521160012$.

Fit $M_1$. The pair totals are $1$ and $5$, so $\hat{\lambda}_1=1/2=0.5$ and $\hat{\lambda}_2=5/2=2.5$. The two kernel contributions are $1\ln(0.5)-1=-1.6931471806$ and $5\ln(2.5)-5=-0.4185463406$. Thus the kernel log likelihood is $-2.1116935212$, and the full value is $-2.1116935212-2.4849066498=-4.5966001710$.

Full AIC. For $M_0$, $-2(-6.0521160012)+2(1)=14.1042320$. For $M_1$, $-2(-4.5966001710)+2(2)=13.1932003$. Full AIC favors $M_1$.

Kernel-based AIC. For $M_0$, $-2(-3.5672093514)+2=9.1344187$. For $M_1$, $-2(-2.1116935212)+4=8.2233870$. These values are not the full AIC values: each is lower by $2\ln12=4.9698133$. Their ranking is nevertheless the same because the omitted factorial term is common.

Full BIC. Since $\ln4=1.3862943611$, $M_0$ has BIC $12.1042320+1.3862944=13.4905264$, and $M_1$ has BIC $9.1932003+2(1.3862944)=11.9657891$. The kernel-based BIC values are $8.5207131$ and $6.9959758$, respectively; again each is lower by $4.9698133$, with the ranking unchanged.

Check. Both models use the same four counts, and both omit exactly $\ln12$. The full and kernel calculations therefore agree that $M_1$ has the lower AIC and BIC. The example also shows why criterion values from the two conventions should not be reported as if they were on the same numerical scale.

Density Units and the Likelihood Jacobian

A continuous density is expressed per unit of its measurement scale. Changing from dollars to thousands of dollars changes the numerical density, even though it does not change the underlying observations. For $z=x/1000$, the density transformation is $f_Z(z)=1000f_X(1000z)$. For $n$ observations, this multiplies the likelihood by $1000^n$ and adds $n\ln1000$ to the log likelihood.

Key result: If the same one-to-one change of measurement units is applied to every candidate model for the same observations, the log likelihoods all gain the same Jacobian term. The absolute AIC and BIC values shift, but the ranking is unchanged. A density per dollar and a density per thousand dollars must not be treated as numerically identical.

Worked Example: Comparing Severity Densities in Two Units

A hypothetical analyst observes two positive losses, $\$1{,}000$ and $\$2{,}000$, and compares an exponential severity model with a gamma model whose shape is fixed at $2$. Calculate the maximized log likelihoods and AIC values first in dollars and then in thousands of dollars. Use natural logarithms.

Model and assumptions. The losses are independent and have continuous severity distributions. For the exponential model, the rate is estimated; for the gamma model, the shape is fixed at $2$ and the scale is estimated. Each model has one free parameter, so $k=1$, and $n=2$. No interest rate or mortality assumption applies.

In dollars: exponential model. The fitted exponential mean is $\bar{x}=(1000+2000)/2=1500$, so the fitted rate is $1/1500$ per dollar. Its log likelihood is $-2\ln(1500)-(1000+2000)/1500=-14.6264407742-2=-16.6264407742$. Thus AIC is $-2(-16.6264407742)+2=35.2528815$.

In dollars: gamma model. With shape $2$, the fitted scale is $\hat{\theta}=\bar{x}/2=750$ dollars. Since $\Gamma(2)=1$, the log likelihood is $\ln(1000)+\ln(2000)-3000/750-4\ln(750)$. The terms are $14.5086577385$, $-4$, and $-26.4802928261$, respectively, giving $-15.9716350876$. Its AIC is $-2(-15.9716350876)+2=33.9432702$. In dollars, the gamma model has the lower AIC.

Convert to thousands of dollars. The observations are now $1$ and $2$. The exponential fitted mean is $1.5$ thousand dollars, with rate $2/3$ per thousand dollars. Its log likelihood is $-2\ln(1.5)-3/1.5=-2.8109302162$, giving AIC $7.6218604$.

The gamma fitted scale is $0.75$ thousand dollars. Its log likelihood is $\ln(1)+\ln(2)-3/0.75-4\ln(0.75)$ $=0+0.6931471806-4+1.1507282898=-2.1561245296$. Its AIC is $4.3122490592+2=6.3122491$.

Check. For each model, the log likelihood in thousands exceeds the log likelihood in dollars by $2\ln1000=13.8155106$. Consequently, each AIC decreases by $4\ln1000=27.6310211$. The gamma model remains preferred in either unit system. The exact AIC numbers differ because the density units differ; the ranking does not.

When Mixed Conventions Change the Answer

The common-constant condition can fail when one candidate is evaluated with its full likelihood and another with a reduced likelihood. A factorial term may be data-only for a Poisson model, but that does not make it safe to omit only from the Poisson candidate while retaining the full likelihood for another model. The resulting criterion values are on inconsistent scales.

Worked Example: A Mixed Poisson and Geometric Comparison

Use the counts $0,1,2,3$ from the earlier hypothetical portfolio. Compare an unknown-rate Poisson model with a geometric model on $0,1,2,\ldots$ whose probability $p=0.4$ was fixed in advance. Calculate AIC using full log likelihoods, then show what goes wrong if the Poisson factorial term alone is omitted.

Model and assumptions. The observations are independent counts. The Poisson rate is estimated, so $k_{\mathrm{Pois}}=1$. The geometric probability is fixed, so $k_{\mathrm{Geom}}=0$. Both models use $n=4$. No interest rate or mortality assumption applies.

Poisson full likelihood. From the calculation above, $\hat{\lambda}=1.5$ and $\ell_{\mathrm{Pois}}=-6.0521160012$. Therefore, $\operatorname{AIC}_{\mathrm{Pois}}=-2(-6.0521160012)+2=14.1042320$.

Geometric full likelihood. The probability mass function is $P(X=x)=p(1-p)^x$. The log likelihood at the prespecified $p=0.4$ is $4\ln(0.4)+6\ln(0.6)=-3.6651629276-3.0649537428=-6.7301166704$. With no estimated parameters, its AIC is $-2(-6.7301166704)+0=13.4602333$. The consistent full-likelihood comparison favors the fixed geometric model.

The inconsistent calculation. If the Poisson factorial term is dropped but the geometric likelihood remains full, the Poisson kernel is $-3.5672093514$, so its apparent AIC is $-2(-3.5672093514)+2=9.1344187$. This incorrectly makes the Poisson model appear preferable to the geometric AIC of $13.4602333$.

Correction and check. Restore the Poisson term: $\ell_{\mathrm{Pois}}=-3.5672093514-\ln12=-6.0521160012$. Its full AIC is $14.1042320$, which is greater than $13.4602333$; the fixed geometric model is preferred. The incorrect Poisson AIC was reduced by $2\ln12=4.9698133$, while the geometric AIC was not shifted. This is precisely the situation in which the rankings need not agree.

A Practical Consistency Check

Before comparing candidate criteria, identify which likelihood each number represents. A compact audit prevents a frequent exam error: using a convenient likelihood kernel for one model and a full likelihood for another.

Write the likelihood convention.
State whether each maximized log likelihood includes all probability or density terms, including factorials, normalizing constants, and any density-unit Jacobian.

Identify omitted terms.
For each candidate, record the value of every omitted term and check whether it is identical across the candidates.

Use the full values or one consistent reduction.
Compute criteria from full maximized log likelihoods, or use the same common reduction for every candidate and label the resulting values as shifted criteria.

Check the scale and penalty.
Use natural logarithms, the appropriate parameter count, and the same stated sample-size convention in the BIC penalty.

A related trap is to use base-10 logarithms in the AIC or BIC formula. Those formulas use natural log likelihoods. If a log likelihood is given as $\log_{10}L$, convert it using $\ln L=(\ln10)\log_{10}L$ before applying the criteria; multiplying a base-10 log likelihood by $-2$ directly does not give the standard AIC contribution.

Key takeaway: Omitting a parameter-independent term is safe for a criterion ranking only when the same term is omitted for every candidate. Unit changes add a common Jacobian term when applied consistently, but mixed full and reduced likelihoods can change the selected model. Use full likelihoods for interpretable AIC and BIC values.

Check Your Understanding

For each problem, determine the criterion consequence using the stated convention. Answers are given at the end of each question.

Two candidate models have full maximized log likelihoods $-12.0$ and $-13.2$, with $k=2$ and $k=3$, respectively. A common data-only term of $3$ is omitted from both log likelihoods. Find the full and reduced AIC values and identify the preferred model. (Answer: Full AICs $28.0$ and $32.4$; reduced AICs $22.0$ and $26.4$; the first model is preferred.)
Three positive observations are recorded in dollars and then converted to thousands of dollars. For a candidate density model, what is the change in maximized log likelihood and the change in AIC under the conversion? (Answer: The log likelihood increases by $3\ln1000\approx20.7233$; AIC decreases by $6\ln1000\approx41.4465$.)
For independent Poisson counts $1,1,2$, find the common factorial term in the full log likelihood. If it is omitted from every candidate, by how much does each resulting AIC differ from its full AIC? (Answer: The term is $\ln(1!1!2!)=\ln2$; each kernel-based AIC is lower by $2\ln2\approx1.3863$, with rankings unchanged if the omission is common.)
A candidate has $k=1$ and reported maximized $\log_{10}L=-2$. Calculate its standard AIC. (Answer: $\ell=-2\ln10$, so AIC $=4\ln10+2\approx11.2103$.)
Candidate A has full AIC $14.0$, and candidate B has full AIC $13.0$. A data-only term of $2$ is omitted from A alone. What reduced AICs result, and why is their ranking not valid? (Answer: A’s reduced value is $14-2(2)=10.0$, while B remains $13.0$; the apparent preference for A is invalid because the omitted term was not applied commonly.)

What You'll Learn