Introduction
A likelihood-ratio calculation is short, but a small sign or bookkeeping error can change its meaning completely. Subtracting likelihoods rather than log-likelihoods does not give the test statistic; reversing the ratio can produce a negative value; and comparing fits based on different observations does not give a valid nested-model likelihood-ratio test.
The central check is straightforward: for a nested comparison, the unrestricted model must be at least as well fitted as the restricted model when both are maximized on the same observations under the same likelihood setup. This tutorial develops that check, shows how to diagnose the three errors, and distinguishes an invalid calculation from a correctly calculated statistic whose usual reference distribution may not apply. The treatment is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.
Three Checks Before Interpreting a Statistic
A valid calculation should pass three checks. First, use log-likelihoods or the logarithm of a likelihood ratio, not a difference of raw likelihoods. Second, put the unrestricted model in the numerator, so the statistic is nonnegative for nested models. Third, ensure that both maximized likelihoods describe the same observed data and sampling setup. These checks diagnose arithmetic and setup errors; they do not by themselves establish that a chi-square approximation is appropriate.
Under the usual regularity conditions for a nested likelihood-ratio test, \(T\) is compared with a chi-square distribution whose degrees of freedom equal the difference in the number of free parameters. Boundary restrictions and other failures of regularity can change that reference distribution. A negative statistic, however, is not an alternative calibration issue: for a genuinely nested comparison of maximized likelihoods on the same data, it signals an error in the calculation or model setup.
Why the Nested-Model Statistic Is Nonnegative
The nonnegativity check follows directly from the maximization domains. Every parameter value allowed by \(H_0\) is also allowed by \(H_1\). Thus, the maximum likelihood attainable under \(H_1\) cannot be smaller than the maximum attainable under \(H_0\), provided both likelihoods use the same observations and likelihood definition.
The logarithm also explains why likelihoods are subtracted on the log scale: \(\log(L_1/L_0)=\log L_1-\log L_0=\ell_1-\ell_0\). By contrast, \(L_1-L_0\) is neither a log-likelihood difference nor a likelihood ratio. It depends on the scale of the likelihood and does not have the standard likelihood-ratio test interpretation.
Check that the restricted parameter values are contained in the larger model.
Verify that both likelihoods use the same observations and the same sampling or conditioning setup.
Calculate \(T=2(\ell_1-\ell_0)\), with \(\ell_1\) from the larger model.
For a valid nested calculation, confirm \(T\geq0\), then use a reference distribution only if its assumptions are justified.
Worked Examples
Worked Example: Subtracting Log-Likelihoods, Not Likelihoods
A hypothetical portfolio has four independent claim counts, \(0,1,2,3\). Suppose the counts follow a Poisson distribution. Test the restriction \(H_0:\lambda=1\) against the larger model \(H_1:\lambda\) is unrestricted. Calculate the likelihood-ratio statistic and show why subtracting the likelihoods is the wrong operation.
Method. Maximize the Poisson log-likelihood under each model using the same four counts. For observations \(x_1,\ldots,x_n\), the log-likelihood is \(\ell(\lambda)=\sum_i x_i\log\lambda-n\lambda-\sum_i\log(x_i!)\). Here \(n=4\), \(\sum_i x_i=6\), and \(\sum_i\log(x_i!)=\log(0!1!2!3!)=\log 12=2.4849\).
Under \(H_1\), the maximum-likelihood estimate is \(\hat{\lambda}=\bar{x}=6/4=1.5\). Under \(H_0\), \(\lambda=1\). Therefore,
The log-likelihood difference is \(\ell_1-\ell_0=-6.0521-(-6.4849)=0.4328\). Thus \(T=2(0.4328)=0.8656\), to four decimal places. As a check, \(L_1/L_0=\exp(0.4328)\), so \(2\log(L_1/L_0)=0.8656\) as well.
Common mistake avoided. The maximized likelihoods are approximately \(L_0=e^{-6.4849}=0.0015\) and \(L_1=e^{-6.0521}=0.0024\). Their raw difference is about \(0.0008\), not the test statistic. The correct calculation uses the logarithm of their ratio, or equivalently twice the difference of their log-likelihoods. The larger model has the higher maximized likelihood, as required by nesting.
Worked Example: Correcting a Reversed Ratio
A hypothetical severity sample consists of four independent positive losses: \(0.5,1,1.5,3\), measured in thousands of dollars. Assume an exponential distribution with mean parameter \(\theta\). Test \(H_0:\theta=1\) against an unrestricted positive \(\theta\). A calculation reports \(2\log(L_0/L_1)=-0.7563\). Diagnose the result and calculate the correctly oriented statistic.
Method. For an exponential distribution with mean \(\theta\), the density is \(f(x;\theta)=\theta^{-1}e^{-x/\theta}\) for \(x>0\). For these four observations, \(n=4\) and \(\sum_i x_i=6\), so \(\ell(\theta)=-4\log\theta-6/\theta\). The unrestricted estimate is \(\hat{\theta}=\bar{x}=6/4=1.5\).
At the restricted value \(\theta=1\), \(\ell_0=-4\log(1)-6/1=-6.0000\). At the unrestricted estimate, \(\ell_1=-4\log(1.5)-6/1.5=-4(0.4055)-4=-5.6219\). Hence \(\ell_1-\ell_0=-5.6219-(-6.0000)=0.3781\), and the correct statistic is \(T=2(0.3781)=0.7563\), to four decimal places.
Common mistake avoided. The reported value uses \(L_0/L_1\), the reverse of the required ratio. Its negative sign is a warning: \(H_1\) contains \(H_0\) and is maximized on the same observations, so its maximized likelihood cannot be smaller. Reversing the ratio negates the correct log-likelihood difference. The calculation can also be checked from \(2\log(L_1/L_0)=2(0.3781)=0.7563\). The value is below the 5% critical value \(3.841\) for a chi-square distribution with one degree of freedom, if the regularity conditions for that calibration hold.
Worked Example: Rejecting a Comparison Based on Different Observations
A hypothetical audit records six independent Bernoulli outcomes, of which four are successes and two are failures. The null model fixes the success probability at \(p=0.5\); the larger model estimates \(p\). An analyst uses all six outcomes for the null fit but accidentally drops one success before fitting the larger model. Diagnose the comparison, then calculate the valid statistic using all six outcomes for both fits.
Method. For a specified sequence of Bernoulli outcomes with \(s\) successes and \(f\) failures, the log-likelihood is \(\ell(p)=s\log p+f\log(1-p)\). With the full data, \(s=4\), \(f=2\), and the unrestricted estimate is \(\hat{p}=4/6=2/3\).
For the null model, \(\ell_0=6\log(0.5)=6(-0.6931)=-4.1589\). For the unrestricted model on the same six outcomes, \(\ell_1=4\log(2/3)+2\log(1/3)=4(-0.4055)+2(-1.0986)=-3.8191\). Thus \(\ell_1-\ell_0=-3.8191-(-4.1589)=0.3398\), and \(T=2(0.3398)=0.6796\), to four decimal places.
The analyst's reduced data contain three successes and two failures, giving \(\hat{p}=3/5=0.6\) and a reduced-sample log-likelihood of \(3\log(0.6)+2\log(0.4)=-1.5325-1.8326=-3.3651\). Subtracting the full-sample null log-likelihood from this gives \(-3.3651-(-4.1589)=0.7938\), or a purported statistic of \(1.5876\). That number is invalid: the two terms come from different data sets, so their ratio is not the likelihood ratio for a single experiment.
Common mistake avoided. A nonnegative result does not prove that a likelihood-ratio calculation is valid. Here, even though the mismatched calculation is positive, it compares a five-observation fit with a six-observation fit. Refit both models on the same six observations; the valid statistic is \(0.6796\), as calculated above.
Common Exam Traps
- Subtracting raw likelihoods. \(L_1-L_0\) is not \(T\). Use \(2\log(L_1/L_0)\) or \(2(\ell_1-\ell_0)\).
- Reversing numerator and denominator. The unrestricted fit belongs in the numerator. A negative statistic for a nested comparison on the same data points to a reversed ratio, an optimization error, or a setup problem.
- Checking only the sign. A positive result can still be invalid if the likelihoods use different observations, different exposure definitions, or incompatible conditioning.
- Confusing a calculation with its calibration. A correctly computed, nonnegative statistic does not guarantee a chi-square reference distribution. Verify nesting and the applicable regularity conditions, including whether a parameter lies on a boundary.
- Mixing likelihood conventions. Likelihoods must represent the same observations and sampling basis. Constants that are identical in both log-likelihoods cancel in the difference; dropping a term from only one fit does not preserve that cancellation.
Check Your Understanding
For each problem, identify the calculation issue or compute the requested statistic. Assume regularity conditions only where the problem explicitly says they hold.
- A nested comparison gives \(\ell_0=-12.4\) and \(\ell_1=-11.9\), both maximized on the same data. Calculate \(T\). (Answer: \(1.0000\))
- A calculation uses \(2\log(L_0/L_1)\) for nested models, with \(\ell_0=-8.2\) and \(\ell_1=-7.6\). State the correctly oriented statistic and explain the negative result from the reversed ratio. (Answer: \(T=1.2000\); the reversed ratio gives \(-1.2000\))
- A restricted model has maximized likelihood \(0.08\), and a nested unrestricted model on the same observations has maximized likelihood \(0.10\). A candidate reports \(T=2(0.10-0.08)\). Identify the error and calculate \(T\). (Answer: the candidate subtracted raw likelihoods; \(T=2\log(0.10/0.08)=0.4463\))
- A null log-likelihood of \(-20.0\) uses 50 observations. An unrestricted log-likelihood of \(-18.7\) uses 48 observations because two records were dropped. Is \(2(-18.7-(-20.0))\) a valid nested likelihood-ratio statistic? (Answer: No; the fits use different observations, so the likelihood ratio is not valid for that comparison)
- A nested comparison uses the same sample and gives \(\ell_0=-5.75\), \(\ell_1=-6.10\), both claimed to be maximized. What should be checked before reporting a test result? (Answer: This violates \(\ell_1\geq\ell_0\); check the model nesting, maximization, likelihood definitions, and ratio direction before interpreting a statistic)