Apply Likelihood-Ratio Tests to Nested Pricing Models

Why Test Nested Pricing Models?

A pricing analyst may need to decide whether two portfolios can share one severity assumption or require separate assumptions. A model with separate severity parameters can fit at least as well as a model with a common parameter, but the improvement may reflect random variation rather than a meaningful difference. A likelihood-ratio test (LRT) quantifies that improvement while accounting for the number of additional parameters.

The observations used in such a test must be represented on the correct policy basis. A payment capped by a policy limit is not an exact observation of the underlying loss: it tells us that the loss reached or exceeded the limit. The likelihood must use that information rather than treat the limit as an ordinary exact loss.

Definition: Two statistical models are nested when one is obtained from the other by imposing restrictions on its parameters. For example, a model with separate severity rates \(\lambda_1\) and \(\lambda_2\) contains the common-rate model as the special case \(\lambda_1=\lambda_2\). The restricted model is the null model, and the less restricted model is the alternative model.

Likelihood Contributions for Payments Capped by a Limit

Suppose ground-up claim severity \(X\) has an exponential distribution with rate \(\lambda>0\), so \(f_X(x)=\lambda e^{-\lambda x}\) for \(x>0\). Assume each claim has no deductible and has a known payment limit \(c\). The recorded payment is \(Y=\min(X,c)\). An exact payment \(y<c\) contributes the density \(f_X(y)\); a payment recorded at the limit contributes the probability \(P(X\geq c)\). The latter is a right-censored observation.

These contributions follow directly from the observation process. If the payment is below the limit, the underlying loss is observed exactly, giving likelihood contribution \(\lambda e^{-\lambda y}\). If the payment reaches the limit, all we know is \(X\geq c\), giving contribution \(S_X(c)=e^{-\lambda c}\). Thus, for \(r\) exact payments \(y_1,\ldots,y_r\) and \(s\) capped payments with limits \(c_1,\ldots,c_s\), the likelihood and log likelihood are:

L(\lambda)=\prod_{j=1}^{r}\lambda e^{-\lambda y_j} \prod_{k=1}^{s}e^{-\lambda c_k}, \qquad \ell(\lambda)=r\log\lambda-\lambda T, \qquad T=\sum_{j=1}^{r}y_j+\sum_{k=1}^{s}c_k.

Here \(T\) is total observed severity exposure in the likelihood: exact amounts contribute their observed values and capped amounts contribute their limits. Differentiating gives \(\ell'(\lambda)=r/\lambda-T\). If \(r>0\), the maximum-likelihood estimate is \(\widehat{\lambda}=r/T\); the second derivative \(\ell''(\lambda)=-r/\lambda^2<0\) confirms it is a maximum. A capped observation counts as information that a claim lasted at least to its limit, not as an exact claim of that size.

This likelihood assumes independent claims from the stated severity model, known limits, and accurate identification of exact versus capped payments. A deductible or a different reporting mechanism changes the observation likelihood; the formula above should not be reused without deriving the appropriate contributions.

Worked Example: Comparing Two Candidate Rates After Applying a Limit

A hypothetical equipment policy pays each claim up to \(8\) thousand dollars. Five payments are observed, in thousands: exact payments of \(2\), \(5\), and \(1\), and two payments capped at \(8\). Compare the likelihoods at candidate rates \(\lambda=0.10\) and \(\lambda=0.20\) per thousand dollars.

Method. Use the exact-payment density for the three uncapped claims and the survival probability at \(8\) for each capped claim. The total likelihood exposure is \(T=2+5+1+8+8=24\), and the number of exact observations is \(r=3\).

At \(\lambda=0.10\), the log likelihood is \(\ell(0.10)=3\log(0.10)-(0.10)(24)=-6.907755-2.4=-9.307755\).

At \(\lambda=0.20\), it is \(\ell(0.20)=3\log(0.20)-(0.20)(24)=-4.828314-4.8=-9.628314\). The log-likelihood difference is \(-9.307755-(-9.628314)=0.320559\), so the likelihood ratio is \(e^{0.320559}\approx1.3777\). These data give the larger likelihood to \(0.10\) among the two candidate rates; this comparison alone is not a likelihood-ratio test between fitted nested models.

Check. The capped claims contribute \(8+8\) to \(T\), not two exact observed losses whose ground-up amounts are known to equal \(8\). The result also has the expected direction: the maximum-likelihood estimate from these data is \(3/24=0.125\), between the two candidates and nearer \(0.10\).

The Likelihood-Ratio Statistic

Let \(\ell_0\) be the maximized log likelihood under the null restrictions and \(\ell_1\) the maximized log likelihood under the alternative. Because the alternative contains the null, \(\ell_1\geq\ell_0\). The likelihood-ratio statistic is:

G^2=2(\ell_1-\ell_0).

A larger value indicates that relaxing the restrictions improves fit more substantially. Under regularity conditions, when the null model is correct, \(G^2\) has approximately a chi-square distribution with degrees of freedom equal to the difference in the numbers of free parameters, \(q=p_1-p_0\). The approximate \(p\)-value is \(P(\chi^2_q\geq G^2)\).

Specify the nesting.
State the null restrictions and the alternative parameters. Confirm that the null is genuinely a special case of the alternative.

Build the policy-aware likelihood.
Use a density for each exact payment and a survival probability for each payment capped at its limit. Use the same observations and observation rules under both models.

Fit both models.
Maximize each log likelihood subject to its own restrictions. The alternative maximum cannot be lower than the null maximum if both are optimized correctly.

Calculate and interpret the test.
Compute \(G^2=2(\ell_1-\ell_0)\), use \(q=p_1-p_0\), and compare the \(p\)-value with the chosen significance level. Then assess the pricing implications separately.

The chi-square approximation requires more than nesting. Common conditions include independent observations, an identifiable and correctly specified likelihood, regular differentiability, and a true parameter lying in the interior of the parameter space. The sample must also be large enough for the asymptotic approximation to be reasonable. If a tested parameter lies on a boundary, or if parameters are not identifiable under the null, the usual chi-square reference distribution may fail; a specialized result or simulation-based calibration may be needed.

Worked Example: Testing a Common Rate Against Separate Rates

Two hypothetical portfolios have exponential ground-up severities, but claims are observed subject to different policy limits. In Portfolio A, four exact payments, in thousands of dollars, are \(0.50, 0.75, 1.00,\) and \(1.75\), all below its \(2\)-thousand-dollar limit. In Portfolio B, two claims are both capped at its \(6\)-thousand-dollar limit. Test whether one exponential rate can be used for both portfolios against the alternative of separate rates.

Method and fitted rates. For Portfolio A, the exact payments sum to \(4\), so \(r_A=4\), \(T_A=4\), and \(\widehat{\lambda}_A=4/4=1\). For Portfolio B, there are \(r_B=0\) exact payments and \(T_B=12\) from the two limits. The log likelihood in that group is \(-12\lambda_B\), which increases as \(\lambda_B\) decreases and has no finite maximizer over \(\lambda_B>0\). This data set therefore cannot furnish the intended regular two-rate LRT: it has no exact events in Portfolio B.

This is an important fitting check, not a reason to force a finite estimate. To make a valid comparison, suppose instead that Portfolio B has two exact payments of \(3\) and \(4\) thousand dollars and two payments capped at \(6\). Then \(r_B=2\) and \(T_B=3+4+6+6=19\), giving \(\widehat{\lambda}_B=2/19\). The combined fit has \(r=6\), \(T=23\), and \(\widehat{\lambda}_0=6/23\).

The maximized alternative log likelihood is \(\ell_1=[4\log(1)-4]+[2\log(2/19)-2]\). Since \(\log(2/19)=-2.251292\), this is \(-4+2(-2.251292)-2=-10.502584\). The common-rate log likelihood is \(\ell_0=6\log(6/23)-6\). Since \(\log(6/23)=-1.343735\), \(\ell_0=-8.062410-6=-14.062410\).

There is one added parameter, so \(q=1\). The statistic is \(G^2=2[-10.502584-(-14.062410)]=7.119652\), approximately \(7.1197\). Its chi-square-one \(p\)-value is about \(0.0076\), so at the \(5\%\) level the common-rate null is rejected, subject to the regularity and sample-size cautions. The correction in the data description matters: with no exact claims in Portfolio B, its rate estimate is not an ordinary finite MLE, and the standard test calculation is not justified.

Worked Test: Keep the Sign and Pricing Interpretation Straight

Worked Example: A Pooled Rate Versus Two Portfolio Rates

For a separate hypothetical two-portfolio analysis, the exponential likelihood summaries are \(r_A=4,T_A=4\) and \(r_B=2,T_B=12\), where \(r\) counts exact claims and \(T\) includes exact amounts and applicable policy-limit exposures. Test a common rate against separate rates.

Fit the null. Under the common-rate model, \(\widehat{\lambda}_0=(4+2)/(4+12)=6/16=3/8\). Its maximized log likelihood is \(\ell_0=6\log(3/8)-6\). Using \(\log(3/8)=-0.980829253\), \(\ell_0=-5.884975518-6=-11.884975518\).

Fit the alternative. The separate estimates are \(\widehat{\lambda}_A=4/4=1\) and \(\widehat{\lambda}_B=2/12=1/6\). Thus \(\ell_1=[4\log(1)-4]+[2\log(1/6)-2]\). Since \(\log(1)=0\) and \(\log(1/6)=-1.791759469\), \(\ell_1=-4+2(-1.791759469)-2=-9.583518938\).

The signed difference in the order “null minus alternative” is \(\ell_0-\ell_1=-11.884975518-(-9.583518938)=-2.301456580\). Equivalently, the improvement “alternative minus null” is \(+2.301456580\). The LRT uses the latter: \(G^2=2(\ell_1-\ell_0)=2(2.301456580)=4.602913160\), or \(4.6029\) to four decimals. The degrees of freedom are \(2-1=1\), and the approximate \(p\)-value is \(0.0319\) to four decimals.

At a \(5\%\) significance level, this is evidence against a common severity rate. It may support considering portfolio-specific severity assumptions in pricing. It does not, by itself, establish that different indicated premiums are material or appropriate: exposure, credibility, expenses, coverage terms, and the consequences of model misspecification still matter.

Common Mistakes and Practical Meaning

Reversing the log-likelihood difference. The null-minus-alternative difference in the example is \(-2.3015\), not \(+2.3015\). The test statistic uses alternative-minus-null, which is \(+2.3015\), before multiplying by two.
Using the wrong degrees of freedom. Count the added free parameters, not the number of portfolios or observations. A common rate versus two separate rates adds one parameter.
Treating a capped payment as exact. The likelihood contribution at a cap is a survival probability. Substituting the density at the limit discards the fact that the claim could have been larger.
Using different data under the two models. A valid nested comparison evaluates both likelihoods on the same observed claims, with the same policy and censoring rules.
Turning statistical significance directly into a price change. The test assesses fit under specified models. Pricing also requires assessing expected covered payments and other business assumptions.

Key takeaway: Fit nested models to the likelihood of what the policy actually records. Subtract the maximized null log likelihood from the maximized alternative log likelihood, multiply by two, and use the parameter-count difference only when the regularity conditions support the chi-square approximation.

Check Your Understanding

For each problem, use policy-aware likelihood contributions and state the requested result.

A hypothetical policy caps payments at \(3\) thousand dollars. The observations, in thousands, are exact payments \(1\) and \(2\), plus one capped payment. At rate \(\lambda=0.5\), calculate the log likelihood. (Answer: \(2\log(0.5)-0.5(1+2+3)=-4.3863\).)
Under the exponential exact/censored likelihood, a portfolio has \(7\) exact claims and total likelihood exposure \(21\). Find its maximum-likelihood rate. (Answer: \(\widehat{\lambda}=7/21=0.3333\) per exposure unit.)
Three portfolios are modeled with separate exponential rates under the alternative and one common rate under the null. How many chi-square degrees of freedom apply under regular conditions? (Answer: \(3-1=2\).)
Two groups have \((r_1,T_1)=(3,12)\) and \((r_2,T_2)=(2,8)\). Compare their fitted rates and give the LRT statistic for a common-rate null. (Answer: both rates are \(0.25\), so the common and separate fits coincide and \(G^2=0\).)
A nested model’s null restriction sets a parameter at the boundary of its permitted range. Can the usual chi-square degrees-of-freedom rule automatically be applied? (Answer: No; boundary parameters can invalidate the usual chi-square reference distribution, so an appropriate alternative calibration is needed.)

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