Why Compare Maximized Likelihoods?
When two non-nested models are plausible descriptions of the same actuarial data, neither model can be tested against the other by treating one as a restricted version of the other. A direct and useful first comparison is to fit each model separately and compare the likelihoods at their respective maximum-likelihood estimates. This identifies which model assigns the greater likelihood to the observations, given the candidate models and their assumptions.
The comparison is meaningful only when the models are fitted to the same observations under the same observation scheme. A likelihood is constructed from the probability model for the data actually recorded. If one calculation uses a different sample, discards observations, or represents the observed data differently, the likelihood values do not provide the intended like-for-like comparison.
For independent observations, the joint likelihood is the product of the individual probability masses or densities. It is often more convenient to calculate its logarithm, since products become sums. The logarithm preserves the ranking: if \(L_1>L_2>0\), then \(\log L_1>\log L_2\). A log-likelihood difference can also be converted to a likelihood ratio by exponentiation.
What the Ranking Means
A higher maximized likelihood says that, after each model’s parameters have been fitted, that model gives a higher probability mass or density to the particular observed data. It is a statement about relative fit to those data under those candidate models. It does not establish that the higher-ranking model is true, that its assumptions are adequate, or that the difference is statistically significant.
In particular, a likelihood ratio formed from two non-nested models does not automatically have the chi-square reference distribution used in standard likelihood-ratio tests. Such a test requires an appropriate null hypothesis and conditions justifying its calibration. Non-nested models do not have the usual restricted-versus-unrestricted relationship that supplies the standard likelihood-ratio test construction. A numerical ratio alone is therefore not a p-value and does not determine a rejection decision.
Likelihoods for continuous data are densities, not probabilities of observing an exact value; a density can exceed 1. Their numerical values depend on the measurement units, but a common change of units multiplies both models’ likelihoods by the same data-dependent Jacobian factor when both models use the same observations and transformation. The ranking is then unchanged. The essential discipline is to compare consistently constructed likelihoods, not to interpret their absolute values as probabilities.
A Short Justification for the Comparison
Suppose two models are evaluated at their separate maximum-likelihood estimates using the same observed data \(x\). Define the log-likelihood difference as \(\Delta=\ell_1(\hat{\theta}_1;x)-\ell_2(\hat{\theta}_2;x)\), where \(\ell_j=\log L_j\). Because the logarithm is strictly increasing, the sign of \(\Delta\) gives the same ranking as the likelihoods. In addition:
Thus, \(\Delta>0\) exactly when model 1 has the larger maximized likelihood, and \(\exp(\Delta)\) is the ratio of the two maximized likelihoods. This is a comparison identity, not a sampling-distribution result: it does not supply a null distribution for \(\Delta\), a standard error, or a significance level. Those require additional inferential methods and assumptions.
Write down exactly which observations are included and the probability model used for them.
Find the maximum-likelihood estimate within each model’s permitted parameter space.
Evaluate both likelihoods, or their log-likelihoods, using the same observations and conventions.
Identify which candidate gives the larger maximized likelihood, then clarify that this ranking by itself is not a significance test or a proof of model adequacy.
Worked Comparisons
Worked Example: Exponential Severity Versus a Bounded Uniform Model
A hypothetical portfolio has three fully observed claim severities \(x=(1,2,3)\), measured in thousands of dollars. Compare an exponential model with rate \(\lambda>0\) to a uniform model on \([0,\theta]\), with \(\theta>0\). Assume the observations are independent and identically distributed within each model. Both models are fitted to all three severities in the stated units.
Exponential fit. The density is \(f(x;\lambda)=\lambda e^{-\lambda x}\) for \(x\geq 0\). Its log-likelihood is \(3\log\lambda-6\lambda\), because the observed severities sum to 6. Differentiating gives \(3/\lambda-6=0\), so \(\hat{\lambda}=0.5\) per thousand dollars. The second derivative is \(-3/\lambda^2<0\), confirming a maximum.
The fitted likelihood is
The log-likelihood is \(3\log(0.5)-3\approx -5.0794\).
Uniform fit. The density is \(1/\theta\) for \(0\leq x\leq\theta\), and zero otherwise. A value \(\theta<3\) assigns zero likelihood because it excludes the observed maximum. For \(\theta\geq3\), the likelihood is \(\theta^{-3}\), which decreases as \(\theta\) increases. Therefore, the maximum occurs at \(\hat{\theta}=3\), and
The uniform model has the larger maximized likelihood: \(0.037037>0.006223\). The ratio is approximately \(0.037037/0.006223=5.95\). The appropriate conclusion is that this uniform model assigns greater fitted likelihood to these three observed severities, not that it has been shown to be the correct severity model or that the ratio is statistically significant.
Method and trap avoided. Each model was maximized over its own parameter, then evaluated on the same three observations. It would be incorrect to call 5.95 a p-value or to infer significance from its size alone. The uniform estimate also sits at the observed sample maximum, so its result depends strongly on the bounded-support model assumption.
Worked Example: Poisson Versus Geometric Claim Counts
In a hypothetical study, the numbers of claims in four equal exposure units are \(x=(0,1,2,3)\). Compare independent Poisson counts with mean \(\lambda\) against independent geometric counts with probability mass function \(P(X=k)=p(1-p)^k\), for \(k=0,1,2,\ldots\). Both models are fitted to all four counts.
Poisson fit. The sample total is \(0+1+2+3=6\), so the MLE is \(\hat{\lambda}=6/4=1.5\). The likelihood includes the factorial terms for each observed count:
Equivalently, \(\ell_{\mathrm{Pois}}=-6+6\log(1.5)-\log(12)\approx -6.0521\).
Geometric fit. The log-likelihood is \(4\log p+6\log(1-p)\). Differentiating gives \(4/p-6/(1-p)=0\), hence \(4(1-p)=6p\) and \(\hat{p}=0.4\). The log-likelihood is concave for \(0<p<1\), so this stationary point is its maximum. The fitted likelihood is
Its log-likelihood is \(4\log(0.4)+6\log(0.6)\approx -6.7301\). The Poisson model ranks higher, with likelihood ratio \(0.002353/0.001194\approx1.97\). This says only that its fitted likelihood is larger on these four counts. It does not establish that the difference is significant.
Method and trap avoided. The geometric distribution here counts failures before the first success and therefore has support starting at zero. Using a geometric convention whose support starts at one would define a different model and would not match these data as specified.
Worked Example: Comparing Two Continuous Models With Fixed Scales
Suppose a hypothetical set of four measurement residuals is \(x=(0,0,2,2)\). Compare a normal location model with known standard deviation \(\sigma=1\) to a Laplace location model with known scale \(b=1\). Assume independent observations. The scale values are fixed parts of the candidate models; only the location is estimated in each.
Normal fit. For known \(\sigma\), the normal MLE of the location is the sample mean, \(\hat{\mu}=(0+0+2+2)/4=1\). The sum of squared deviations is \(1+1+1+1=4\). Thus,
Laplace fit. The density is \(f(x;\mu)=\frac{1}{2b}\exp(-|x-\mu|/b)\). With \(b=1\), a location MLE is any sample median, so any \(\hat{\mu}\in[0,2]\) maximizes the likelihood; take \(\hat{\mu}=1\). The sum of absolute deviations is \(1+1+1+1=4\), giving
The normal model has the larger maximized likelihood, by a ratio of approximately \(0.003427/0.001145=2.99\). The calculation compares these two specified models, including their fixed scale assumptions. It does not show that a normal model with an estimated scale, or a Laplace model with an estimated scale, would have the same ranking.
Method and trap avoided. The location parameter is maximized separately under each model, while the scales remain fixed as stipulated. Silently estimating a scale in only one model would change its parameter space and invalidate this particular comparison.
Common Exam Traps and Practical Limits
- Comparing different data. A larger likelihood on a larger or otherwise different sample is not a like-for-like ranking. Check that both likelihoods describe precisely the same observations.
- Using a non-maximized value. Comparing one model’s likelihood at its MLE with another model evaluated at an arbitrary parameter value does not compare maximized likelihoods.
- Calling the likelihood ratio a test statistic with a known reference distribution. For non-nested models, the ratio alone does not provide the usual likelihood-ratio test calibration. Do not report a p-value unless a justified test procedure supplies one.
- Confusing density with probability. A continuous likelihood is built from densities at the observations; it is not the probability of observing those exact values.
- Ignoring model specification. A ranking is conditional on the chosen distributions, supports, fixed parameters, and treatment of the observations. A higher likelihood does not certify that these assumptions are sensible.
The comparison is useful as a transparent descriptive step: it shows which candidate, among those considered, fits the observed data more closely in the likelihood sense after fitting. A defensible model choice may require further assessment of assumptions and model adequacy. The key conclusion should remain proportionate: model A has the larger maximized likelihood on these data, not “model A is true” or “the difference is significant.”
Check Your Understanding
For each problem, compare the maximized likelihoods only as a ranking unless the question explicitly supplies a valid inferential procedure.
- A hypothetical sample of two positive severities is \(x=(1,3)\). Fit an exponential model with rate \(\lambda\) and a uniform model on \([0,\theta]\). Which has the higher maximized likelihood, and what is the ratio of the higher to the lower likelihood? (Answer: Uniform; ratio \( (1/9)/(0.25e^{-2})\approx 3.28\).)
- Claim counts are \(0\) and \(2\). Compare a Poisson model with mean \(\lambda\) to a geometric model with mass \(p(1-p)^k\). Which model ranks higher at its MLE? (Answer: Poisson; its maximized likelihood is \(e^{-2}/2\approx0.0677\), versus \(0.0625\) for the geometric model.)
- For data \((-1,1)\), compare a normal location model with known \(\sigma=2\) to a Laplace location model with known \(b=1\). Which has the larger maximized likelihood? (Answer: Laplace; \(0.25e^{-2}\approx0.0338\), compared with \(e^{-0.25}/(8\pi)\approx0.0310\) for the normal model.)
- Two non-nested models fitted to exactly the same observations have maximized log-likelihoods \(-12.4\) and \(-13.1\). Find the likelihood ratio for the higher-ranking model to the lower-ranking model, and state whether that result alone establishes statistical significance. (Answer: \(\exp(0.7)\approx2.01\), with the first model ranking higher; it does not establish statistical significance.)
- Model A is fitted to every observed claim, while Model B omits two claims before its likelihood is maximized. Can their maximized likelihoods be used for the direct comparison taught here? (Answer: No; the likelihoods are not based on the same observations.)