Why a Larger Model Cannot Fit Worse
A more flexible model can usually adapt more closely to the observed data. That is a mathematical advantage in maximizing likelihood, but it is not automatically evidence that the more flexible model describes the underlying risk process better. The distinction matters whenever a model comparison uses maximized likelihoods: first determine whether one model contains the other, then interpret the likelihood gain in light of that flexibility.
Let \(x\) denote the same observed data for each candidate model. A model specifies a likelihood \(L(\theta;x)\), the probability of the observed data for discrete data or its joint density for continuous data, as a function of the model parameter \(\theta\). Its maximized likelihood is the largest value attainable over that model’s parameter space.
The models must be fitted to the same observations using compatible likelihoods. A larger likelihood then means a larger probability or density assigned to those observations at the fitted parameter values. It does not mean that the model has a higher probability of being true.
The Likelihood Ordering
Suppose \(\Theta_0\) is the parameter space for the restricted model and \(\Theta_1\) is the space for the larger model, with \(\Theta_0\subseteq\Theta_1\). Maximizing over a set that includes every value in the smaller set cannot yield a lower maximum:
This argument is the complete reason for the ordering; it does not depend on the form of the likelihood. It also applies when the maximum is expressed as a supremum rather than attained at a finite parameter value. Taking logarithms preserves the ordering because the logarithm is an increasing function. The likelihood-ratio statistic for comparing the restricted model with the larger model is therefore nonnegative:
The inequality can be strict, but it need not be. If the larger model’s maximizing fit also satisfies the restrictions defining the smaller model, the two maximized likelihoods are equal. Thus, adding a parameter gives the larger model an opportunity to improve the fit, not a guarantee of a strictly larger fitted likelihood.
Worked Example: Separate Claim Probabilities for Two Groups
Method: maximize the Bernoulli likelihood under the common-probability restriction and under the group-specific model. A hypothetical insurer observes claim indicators for two groups of five policies. Group A has three claims and two nonclaims; Group B has one claim and four nonclaims. Treat the indicators as independent Bernoulli observations. The restricted model has one common claim probability \(p\); the larger model has probabilities \(p_A\) and \(p_B\). Find both maximum likelihood estimates and compare the maximized likelihoods. Use the likelihood of the observed, labeled indicators, without binomial coefficient factors.
Restricted fit. There are four claims and six nonclaims among the ten policies. The common-probability log-likelihood is \(4\log p+6\log(1-p)\). Its maximum occurs at \(\widehat p=4/10=0.4\). The likelihood is
Group-specific fit. The Bernoulli maximum likelihood estimate in each group is its observed claim fraction. Thus \(\widehat p_A=3/5=0.6\) and \(\widehat p_B=1/5=0.2\), giving
The ratio is \(\widehat L_1/\widehat L_0=64/27\approx2.3704\), and the likelihood-ratio statistic is \(2\log(64/27)\approx1.7261\). The larger model fits better in the narrow sense that its maximized likelihood is higher. That fact alone does not establish that the two groups truly have different claim probabilities; the improvement is the flexibility reward that must be assessed, not an automatic reason to select the larger model.
Common mistake avoided: do not compare the fitted probabilities themselves as if their distance measured evidence. The comparison here is between likelihoods for the same ten observations, maximized under each model.
A Second Example: Severity Rates
The same principle applies when the likelihood is built from continuous severity observations. For an exponential severity model with rate \(\lambda>0\), the density is \(f(x;\lambda)=\lambda e^{-\lambda x}\) for \(x\geq0\). For independent observations \(x_1,\ldots,x_n\), the likelihood is \(\lambda^n e^{-\lambda\sum x_j}\), and the maximum likelihood estimate is \(n/\sum x_j\). The examples below use uncensored observations and this ordinary exponential likelihood.
Worked Example: A Common Exponential Rate Versus Two Rates
Method: maximize the exponential likelihood first with a common rate, then with one rate per group. In a hypothetical portfolio, logged claim severities are not being considered; the observed claim severities themselves, in thousands of dollars, are \(1,3\) in Group A and \(2,6\) in Group B. Suppose the exponential model is appropriate for the purpose of this comparison. The restricted model uses one rate, while the larger model uses \(\lambda_A\) and \(\lambda_B\).
Restricted fit. The total of the four observations is \(1+3+2+6=12\). Hence \(\widehat\lambda=4/12=1/3\) per thousand dollars. The maximized likelihood is
Group-specific fit. Group A has sum \(1+3=4\), so \(\widehat\lambda_A=2/4=1/2\). Group B has sum \(2+6=8\), so \(\widehat\lambda_B=2/8=1/4\). Therefore
The ratio is \(\widehat L_1/\widehat L_0=81/64=1.265625\), which is greater than one, as nesting requires. The separate-rate fit has a higher likelihood, although the sample is small and that ordering alone does not decide whether separate rates are justified. The numerical likelihoods are densities, not probabilities of the exact observed values.
Common mistake avoided: the total exponential likelihood uses the product of all individual densities. The two group likelihood contributions must both be included; omitting one group would compare models on different data.
When Extra Flexibility Gives No Gain
More parameters do not always improve the maximum. If the data already fit the restricted model’s constraints at the larger model’s optimum, the extra freedom remains unused. This can happen even when the larger model has an additional parameter.
Worked Example: An Additional Group Mean That Does Not Help
Method: compare the maximum likelihoods for nested normal models and check whether the fitted group means differ. A hypothetical severity analysis uses four observed log-severity values, in a consistent log scale: Group A has \(1,3\), and Group B has \(2,2\). Assume independent normal observations with known variance \(\sigma^2=1\). The restricted model has one unknown common mean; the larger model has separate unknown group means, with the same known variance.
The sample mean in Group A is \((1+3)/2=2\), and the sample mean in Group B is \((2+2)/2=2\). The combined mean is \((1+3+2+2)/4=2\). Thus the maximum likelihood estimates in both models assign mean 2 to every observation, even though the larger model permits two means.
For independent normal observations with known variance 1, the likelihood at the fitted means is \((2\pi)^{-2}\exp(-\tfrac12\sum_j(x_j-\widehat\mu_j)^2)\). The sum of squared residuals is \((1-2)^2+(3-2)^2+(2-2)^2+(2-2)^2=2\). Consequently,
The larger model has one more mean parameter, but the likelihood-ratio statistic is \(2\log(\widehat L_1/\widehat L_0)=0\). The result is not a violation of the nesting rule: equality is permitted. The added parameter simply has no likelihood advantage for these observations.
Common mistake avoided: do not assume that each additional parameter must strictly raise the maximized likelihood. The general result guarantees “no lower,” not “strictly higher.”
Why Raw Likelihood Is Not a Model-Selection Rule
The likelihood ordering is a property of the parameter spaces, not a general score of model quality. In a nested comparison, the larger model receives at least the smaller model’s best fit by construction. If one chose whichever model has the greatest raw maximized likelihood, an increasingly flexible model could keep gaining an advantage by adapting to accidental features of the observed sample. In the extreme, a model could devote enough parameters to reproducing individual observations while representing the broader risk process poorly.
A likelihood gain can still be informative. The likelihood-ratio framework asks whether the observed gain is unusually large under a specified restricted model, with its calibration depending on the applicable assumptions and conditions. This is different from declaring the model with the numerically larger likelihood to be preferable. A likelihood-ratio calculation also requires correctly specified nested models and the same observations; it does not turn every pair of likelihood values into a valid test.
Raw likelihood magnitudes also require care because they depend on what is being observed and how it is represented. For continuous data, a likelihood is a density and changes when the measurement unit changes. If \(n\) observations are converted from dollars to thousands of dollars, the joint density is multiplied by the corresponding unit-conversion factor for all \(n\) observations. For candidate models fitted to the same data in the same units, that common factor does not change their likelihood ordering or ratio. But an absolute density value should not be read as a probability or compared casually across differently scaled data.
A defensible comparison therefore separates three questions:
- Are the models nested? If so, the larger model’s maximized likelihood cannot be smaller.
- How large is the improvement? Compare likelihoods on the same data, often through the likelihood ratio or log-likelihood difference.
- Does that improvement justify added flexibility? This requires an appropriate inferential or predictive criterion, alongside consideration of assumptions and actuarial purpose. Raw likelihood alone does not supply the answer.
This distinction is useful in claim-frequency, claim-severity, and group-rating analyses. A more detailed segmentation can improve in-sample fit simply because it estimates more parameters. A candidate should not be selected solely for that mechanical improvement; the analyst must also consider whether the additional structure is supported and useful.
Check Your Understanding
For each item, identify the relevant nesting or likelihood principle before computing or interpreting the result.
- A hypothetical two-group Bernoulli model has eight policies in each group. Group A has four claims; Group B has two. Compare the maximum likelihood estimates under a common claim probability and under separate group probabilities. State the likelihood ordering. (Answer: \(\widehat p=6/16=0.375\) under the common model; \(\widehat p_A=4/8=0.5\), \(\widehat p_B=2/8=0.25\) under the larger model; its maximized likelihood is at least the common-model maximum.)
- Independent exponential severities in two hypothetical groups are \(2,4,6\) and \(5,5\), respectively. Find the pooled-rate estimate and each group-rate estimate, with rates per unit severity. (Answer: \(\widehat\lambda=5/22\approx0.2273\), \(\widehat\lambda_A=3/12=0.25\), and \(\widehat\lambda_B=2/10=0.2\).)
- For a nested comparison, the restricted maximum likelihood is \(0.012\) and the larger model’s maximum is \(0.015\). Calculate \(\widehat L_1/\widehat L_0\) and \(-2\log(\widehat L_0/\widehat L_1)\), rounding the statistic to four decimals. Does the ordering contradict nesting? (Answer: the ratio is \(1.25\); the statistic is \(2\log(1.25)\approx0.4463\); no.)
- A common-mean normal model is nested in a separate-means model. The fitted means in the larger model happen to be equal. What relationship must hold between the two maximized likelihoods, assuming the same variance treatment and data? (Answer: they are equal, since the larger model’s maximizing fit is also allowed by the common-mean restriction.)
- A severity data set is converted from dollars to thousands of dollars before fitting two candidate models to the same \(n\) observations. How does this affect the absolute continuous-data likelihoods, and does it change their ordering if the same conversion is applied to both? (Answer: each joint density is multiplied by the same unit-conversion factor \(1000^n\); the likelihood ordering and ratio are unchanged.)