Testing Shared Parameters Across Risk Groups

Why Test Whether Groups Share a Parameter?

Actuarial data are often divided into risk groups: policy types, regions, or classes defined by underwriting characteristics. A common model may be convenient and stable, but it assumes the groups have the same value of a parameter. For example, two classes might be assigned a common claim frequency, even when their observed claim counts differ. A likelihood-ratio test provides a systematic way to assess whether the data support that restriction.

The comparison is between a restricted model, in which the parameter is shared, and an unrestricted model, in which each group has its own parameter. Both models describe the same observations. The unrestricted model can fit at least as well because it includes the restricted model as a special case; the question is whether its improvement in fit is large relative to the number of restrictions removed.

Definition: Suppose \(G\) independent risk groups follow the same parametric family, with a \(q\)-dimensional parameter \(\boldsymbol{\theta}_g\) for group \(g\). The shared-parameter hypothesis is \(H_0:\boldsymbol{\theta}_1=\cdots=\boldsymbol{\theta}_G\). The alternative allows the group parameters to differ. If the group-specific model is regular and the null restrictions are independent, the difference in parameter counts is \((G-1)q\). Under the usual large-sample conditions, the likelihood-ratio statistic is compared with a chi-square distribution having \((G-1)q\) degrees of freedom.

Constructing the Comparison

Let \(L_0\) be the largest likelihood attainable under the shared-parameter hypothesis, and let \(L_1\) be the largest likelihood attainable when the parameters may differ by group. The likelihood-ratio statistic is

T=2\left(\log L_1-\log L_0\right).

The unrestricted maximum cannot be smaller than the restricted maximum, so \(T\geq 0\). A large value means that allowing separate parameters improves the fit substantially. The degrees of freedom come from the number of independent parameter restrictions, not from the number of observations or the numerical size of the parameter estimates.

For independent groups, the joint likelihood is the product of the group likelihoods, so its log is the sum of their log-likelihoods. Under the alternative, each group can be maximized separately. Under the null, the common parameter must be estimated using all groups together, with each observation contributing according to the specified model. “Pooling” therefore means maximizing the restricted likelihood; it does not necessarily mean taking an unweighted average of group estimates.

Why the Pooled Estimate Is Exposure-Weighted for Poisson Counts

Consider independent Poisson claim counts \(N_g\) observed over known exposures \(E_g\). Assume \(N_g\sim\operatorname{Poisson}(E_g\lambda_g)\), where \(\lambda_g\) is the claim rate per exposure unit. Under the shared-rate hypothesis, every group has the same rate \(\lambda\). Ignoring terms that do not depend on \(\lambda\), the null log-likelihood is

\ell_0(\lambda)=\sum_{g=1}^G N_g\log\lambda-\lambda\sum_{g=1}^G E_g+\text{constant}.

Differentiating and setting the score equal to zero gives

\frac{d\ell_0}{d\lambda} =\frac{\sum_g N_g}{\lambda}-\sum_g E_g=0 \quad\Longrightarrow\quad \widehat{\lambda}_0=\frac{\sum_g N_g}{\sum_g E_g}.

Thus the common estimate is total claims divided by total exposure. The exposure matters because a group observed for more policy-years contributes more information about the shared rate. Under the alternative, \(\widehat{\lambda}_g=N_g/E_g\) for each group. Substituting these estimates into the log-likelihoods gives a convenient form of the test statistic:

T=2\sum_{g=1}^G N_g\log\left(\frac{\widehat{\lambda}_g}{\widehat{\lambda}_0}\right).

This expression follows because the exposure terms cancel: at the separate and common maximum-likelihood estimates, \(\sum_g E_g\widehat{\lambda}_g=\sum_gN_g\) and \(\widehat{\lambda}_0\sum_gE_g=\sum_gN_g\). The formula applies to independent Poisson counts with known exposures and positive fitted rates. It is not a general formula for arbitrary distributions or for models whose likelihood includes other group-specific terms.

Worked Likelihood-Ratio Tests

Worked Example: Comparing Poisson Claim Rates with Unequal Exposures

Method: Compare a common Poisson rate with two group-specific rates. A hypothetical insurer observes 18 claims over 600 policy-years in Group A and 32 claims over 400 policy-years in Group B. Assume independent Poisson counts with constant rates within groups. Test whether the rates are equal at the 1% significance level.

Step 1: Estimate the rates under the alternative. The group-specific maximum-likelihood estimates are

\widehat{\lambda}_A=\frac{18}{600}=0.03,\qquad \widehat{\lambda}_B=\frac{32}{400}=0.08

claims per policy-year.

Step 2: Estimate the common rate under the null. Pool claims and exposure, rather than averaging the two rates:

\widehat{\lambda}_0=\frac{18+32}{600+400} =\frac{50}{1000}=0.05.

Step 3: Calculate the likelihood-ratio statistic.

\begin{aligned} T &=2\left[18\log\left(\frac{0.03}{0.05}\right) +32\log\left(\frac{0.08}{0.05}\right)\right]\\ &=2\left[18\log(0.6)+32\log(1.6)\right]\\ &=2\left[-9.1949+15.0401\right]\\ &=11.6905. \end{aligned}

Step 4: Determine degrees of freedom and decide. The alternative has two rates and the null has one, so there is one restriction and the reference distribution is approximately \(\chi^2_1\). The p-value is approximately 0.0006, below 0.01. Reject the common-rate hypothesis; these data provide evidence that the group rates differ.

Check and trap avoided: The pooled estimate of 0.05 lies between 0.03 and 0.08, and multiplying it by the total exposure gives 50 fitted claims, matching the observed total. An unweighted average of the group rates would be 0.055, which is not the null maximum-likelihood estimate and would produce an incorrect test statistic.

Worked Example: Testing a Shared Exponential Severity Rate

Method: Compare a common exponential rate with separate rates. A hypothetical portfolio has 12 independent observed losses in Group A with total loss 720 dollars, and 8 in Group B with total loss 240 dollars. Assume each loss is exponentially distributed with density \(f(x)=\lambda e^{-\lambda x}\) for \(x\geq 0\), and test a common rate at the 5% level.

Step 1: Estimate the rates under the alternative. For an exponential sample, the maximum-likelihood rate is sample size divided by total loss:

\widehat{\lambda}_A=\frac{12}{720}=\frac{1}{60}, \qquad \widehat{\lambda}_B=\frac{8}{240}=\frac{1}{30}.

The corresponding fitted mean severities are 60 and 30 dollars.

Step 2: Estimate the common rate. Under the null, the pooled sample has 20 losses and total loss \(720+240=960\):

\widehat{\lambda}_0=\frac{20}{960}=\frac{1}{48}.

Step 3: Compute the statistic. The exponential log-likelihood for a group with \(n\) observations and total loss \(s\) is \(n\log\lambda-\lambda s\). At the maximum-likelihood estimate, \(\lambda s=n\). Consequently, the linear terms cancel when the separate and pooled maxima are compared:

\begin{aligned} T &=2\left[12\log\left(\frac{1/60}{1/48}\right) +8\log\left(\frac{1/30}{1/48}\right)\right]\\ &=2\left[12\log(0.8)+8\log(1.6)\right]\\ &=2\left[-2.6777+3.7600\right]\\ &=2.1646. \end{aligned}

Step 4: Decide. There is one restriction, since two group-specific rates are replaced by one common rate. Against \(\chi^2_1\), the p-value is approximately 0.141. Do not reject at 5%; the observed difference in fitted mean severities is not strong enough, by this test, to establish different rates.

Check and trap avoided: The pooled rate is \(20/960\), not the unweighted mean of \(1/60\) and \(1/30\). In addition, a larger exponential rate means a smaller mean severity; comparing rates and comparing means reverse the direction.

Worked Example: Testing a Shared Binomial Probability

Method: Compare a common success probability with group-specific probabilities. In a hypothetical service study, Group A has 14 favorable outcomes among 20 independent cases, while Group B has 6 among 20. Assume independent binomial observations with fixed group sample sizes, and test equality of probabilities at the 5% level.

Step 1: Fit the alternative. The group-specific estimates are

\widehat{p}_A=\frac{14}{20}=0.7,\qquad \widehat{p}_B=\frac{6}{20}=0.3.

Step 2: Fit the null. Under a common probability, the maximum-likelihood estimate is the total number of favorable outcomes divided by the total number of cases:

\widehat{p}_0=\frac{14+6}{20+20}=\frac{20}{40}=0.5.

Step 3: Calculate the statistic. Each group contributes successes and failures to its binomial likelihood:

\begin{aligned} T=2\Bigg[& 14\log\left(\frac{0.7}{0.5}\right) +6\log\left(\frac{0.3}{0.5}\right)\\ &+6\log\left(\frac{0.3}{0.5}\right) +14\log\left(\frac{0.7}{0.5}\right)\Bigg]\\ =2\left[28\log(1.4)+12\log(0.6)\right] &=2(9.4212-6.1299)\\ &=6.5826. \end{aligned}

Step 4: Decide. The unrestricted model has two probabilities and the null has one, giving one degree of freedom. The p-value is approximately 0.0103, below 0.05. Reject equality of the probabilities at the 5% level.

Check and trap avoided: The failure contributions must be included as well as the success contributions. Using only the two success terms would not be the binomial likelihood-ratio comparison.

Degrees of Freedom and Interpretation

For \(G\) groups and a \(q\)-parameter model in every group, the unrestricted model has \(Gq\) parameters and the shared model has \(q\). The difference is \(Gq-q=(G-1)q\), provided the restrictions are independent. For a single shared rate across three groups, for example, the number of restrictions is two. For a two-parameter severity model shared across four groups, it is six under the same assumptions.

A significant result is evidence against the shared-parameter assumption, not proof that every group differs from every other group. The test is an omnibus comparison: it indicates that allowing some group-specific variation improves the likelihood enough to matter, but does not identify which pairs differ. Conversely, failure to reject does not prove equality. It says that the data do not provide sufficient evidence against the restriction at the chosen significance level.

The chi-square calibration is an approximation for sufficiently informative samples under standard regularity conditions, with parameters in the interior of the parameter space and identifiable models. Sparse counts, boundary estimates, dependent groups, or a change in model structure beyond the shared-parameter restriction can make the usual reference distribution inappropriate. The test also requires the models to be fitted to the same observations under comparable likelihoods.

Common Mistakes

Averaging group estimates without justification. The restricted maximum-likelihood estimate depends on the likelihood. For Poisson rates it weights by exposure; for binomial probabilities it pools the successes and trial counts.
Counting the wrong restrictions. With two groups and one scalar parameter per group, there is one restriction, not two. The null parameter is estimated rather than fixed in advance.
Reversing the likelihood ratio. Use the unrestricted maximum in \(L_1\) and the restricted maximum in \(L_0\), so that \(T=2(\log L_1-\log L_0)\) is nonnegative.
Interpreting non-rejection as proof. A non-significant result may reflect limited data or weak evidence; it does not establish that the parameters are exactly equal.

Key takeaway: Fit the shared and group-specific models to the same data, calculate twice the difference in their maximized log-likelihoods, and use the number of independent equality restrictions for the degrees of freedom. Check that pooling respects the likelihood and the groups’ exposures or sample sizes.

Check Your Understanding

For each problem, identify the restriction count as well as the requested result. Assume the usual large-sample chi-square calibration when a decision is requested.

Two independent Poisson groups have 9 claims over 300 exposure units and 18 claims over 600 exposure units. Find the common-rate estimate under the null and the likelihood-ratio statistic. (Answer: \(\widehat{\lambda}_0=0.0300\); \(T=0.0000\), with 1 restriction.)
Two independent exponential samples have \(n_1=10\), total loss \(500\), and \(n_2=10\), total loss \(300\). Find the pooled rate estimate and the likelihood-ratio statistic for equal rates. (Answer: \(\widehat{\lambda}_0=20/800=0.0250\); \(T\approx1.2907\), with 1 restriction.)
Two binomial groups each have 15 cases. Group A has 9 successes and Group B has 3. Find the shared probability estimate and the likelihood-ratio statistic for equal probabilities. (Answer: \(\widehat{p}_0=12/30=0.4\); \(T\approx5.1784\), with 1 restriction.)
Three independent groups are modeled with a shared two-parameter severity distribution under the null and separate two-parameter distributions under the alternative. How many restrictions are tested? (Answer: \( (3-1)\times2=4\) restrictions.)
A two-group test of equal scalar Poisson rates gives \(T=4.00\). Using a \(\chi^2_1\) reference distribution, the 5% critical value is approximately 3.841. State the decision and a suitable interpretation. (Answer: Reject equality at 5%; the data provide evidence against a common rate, but do not by themselves identify which groups differ in other respects.)

What You'll Learn