Why Incomplete Records Can Leave Parameters Unresolved
A likelihood calculation can be algebraically correct and still fail to produce a useful estimate of every model parameter. Censoring may leave too little information about the loss distribution, while truncation may remove precisely the part of the distribution that distinguishes parameter values. The result can be a likelihood that is flat in one direction, a maximum at the edge of the parameter space, or a supremum that no admissible finite parameter value attains.
The key question is not simply whether an optimizer returns numbers. It is which features of the model are determined by the observed records? The likelihood must be examined for flat directions, equivalent parameter combinations, and boundary behavior before interpreting an estimate.
For example, suppose an observation design records only whether a loss exceeds one fixed threshold. The observed distribution is then determined by a single exceedance probability. A model with two parameters may not allow those two parameters to be separated using that probability alone. By contrast, if exact loss values are also observed, the observable distribution may contain more information. The records and the likelihood—not the number of parameter symbols in the model—determine what can be estimated.
Three Likelihood Behaviors to Distinguish
A likelihood may be uninformative in several different ways. It helps to state exactly which behavior occurs before reporting an estimate.
| Behavior | What the likelihood shows | Interpretation |
|---|---|---|
| Structural nonidentifiability | Different parameter values give the same distribution of observable records. | The data design cannot distinguish those parameter values, even in principle. |
| Weak identification | The likelihood has a broad, shallow region rather than a sharp peak. | The sample supplies little precision about some parameter or parameter combination. |
| Boundary estimate | The largest likelihood occurs at an allowed edge of the parameter space. | A maximum may exist, but standard interior-estimate interpretations need care. |
| Unattained supremum | The likelihood increases toward a limiting value as a parameter approaches a boundary excluded from the parameter space. | No finite admissible maximum likelihood estimate exists. |
A boundary estimate and an unattained supremum are not interchangeable. If the boundary is included in the parameter space and the likelihood is defined there, the maximum can be attained. If the likelihood only approaches its supremum as a parameter tends to infinity or to an excluded value, there is no maximum at an admissible parameter value.
Worked Example: A Truncated Pareto Sample Cannot Identify Its Scale
Worked Example: A Truncated Pareto Sample Cannot Identify Its Scale
A hypothetical portfolio study includes only losses exceeding \(a=\$100\). Losses follow a Pareto Type I distribution with scale parameter \(\theta\) and shape parameter \(\alpha\), where \(0<\theta\leq 100\) and \(\alpha>0\). Three exact observed losses are \(\$125,\$200,\$400\). Find the conditional maximum likelihood estimate of \(\alpha\), and determine whether \(\theta\) is estimable from these records.
Method and observation model. For a Pareto Type I loss \(X\), the density is \(f_X(x)=\alpha\theta^\alpha x^{-(\alpha+1)}\) for \(x\geq\theta\), and \(S_X(a)=(\theta/a)^\alpha\) for \(a\geq\theta\). Because records are included only when \(X>a\), the likelihood uses the conditional density:
The factor \(\theta^\alpha\) cancels. Thus, for all three observed losses, the conditional likelihood does not depend on \(\theta\), provided \(0<\theta\leq100\). No value of \(\theta\) in that range is favored by these exact truncated records.
Estimate \(\alpha\). The likelihood for the three losses \(x_1,x_2,x_3\) is
Set \(z_j=\ln(x_j/100)\). The three ratios are \(1.25,2,4\), so
Ignoring terms constant in \(\alpha\), the log likelihood is \(3\ln\alpha-\alpha\sum_j z_j\). Its derivative is \(3/\alpha-\sum_j z_j\), which equals zero at
The second derivative is \(-3/\alpha^2<0\) for every \(\alpha>0\), so this stationary point is the unique maximum for \(\alpha\). The estimate is \(\widehat{\alpha}=1.3029\) to four decimal places. In contrast, \(\theta\) is structurally unidentifiable from this conditional likelihood: every \(\theta\) satisfying \(0<\theta\leq100\) gives the same likelihood. The calculation checks by substitution: \(3/\widehat{\alpha}=\ln(10)\), so the score is zero.
Common mistake avoided: Using the unconditioned Pareto density would leave a factor involving \(\theta\) and would incorrectly suggest that these records provide information about it. The sampling rule \(X>100\) requires division by \(S_X(100)\); that normalizer is exactly what removes \(\theta\).
Why Truncation Can Remove a Parameter
The cancellation in the example follows from conditioning, not from a numerical coincidence. A left-truncated sample describes losses conditional on crossing the entry threshold. If the parameter changes the probability of crossing that threshold but leaves the conditional distribution above the threshold unchanged, the selected records cannot estimate that parameter when the number of eligible losses is not itself used as information.
For left truncation at \(a\), an observed exact loss has conditional density \(f_X(x)/P(X>a)\) for \(x>a\).
Include the truncation normalizer for each record when the likelihood is conditional on inclusion.
If a parameter cancels from every conditional contribution, the selected exact values do not identify that parameter under this likelihood.
This argument depends on the observation design. If the number of losses above the threshold and the total number exposed are also modeled, the rate of crossing the threshold may contain information about the parameter that canceled from the conditional likelihood. Do not add that information unless the sampling design and likelihood actually include it.
Worked Example: All Right-Censored Weibull Records Give a Boundary Supremum
Worked Example: All Right-Censored Weibull Records Give a Boundary Supremum
In a hypothetical study, five policies are followed for two years. All five losses are right-censored at time \(t=2\); no loss is observed before that time. Assume independent losses from a Weibull distribution with shape \(k>0\) and scale \(\eta>0\), with survival function \(S_X(t)=\exp[-(t/\eta)^k]\). Determine whether a finite maximum likelihood estimate exists.
Method. A right-censored record at \(t=2\) contributes \(S_X(2)\). Therefore the likelihood for five such records is
For every fixed \(k>0\), increasing \(\eta\) decreases \((2/\eta)^k\), so \(L(k,\eta)\) increases. Its limiting value as \(\eta\) tends to infinity is 1, but for every finite \(\eta\), the exponent is strictly negative and \(L(k,\eta)<1\). Thus the likelihood has supremum 1 but no finite maximizing \(\eta\). There is no finite maximum likelihood estimate.
The records also provide no finite estimate of \(k\). For a fixed likelihood value determined by \(q=(2/\eta)^k\), many shape-scale pairs satisfy the same equation. For example, let \(q=0.2\). If \(k=1\), then \(\eta=2/0.2=10\). If \(k=2\), then \(\eta=2/\sqrt{0.2}=4.4721\) to four decimal places. Both pairs give \(L=\exp(-5(0.2))=\exp(-1)=0.3679\) to four decimal places. This illustrates a likelihood ridge: the observed censored records do not separate the parameters along that contour.
Check. For either pair, \((2/\eta)^k=0.2\), so both likelihood calculations equal \(\exp(-1)\). As \(\eta\) grows without bound for fixed \(k\), the likelihood approaches 1 from below, confirming that the supremum is not attained.
Common mistake avoided: Setting a numerical score equation to zero and reporting a large finite scale estimate can disguise the fact that the likelihood keeps increasing toward infinity. The correct conclusion is that no finite MLE exists for these records.
Deriving a Boundary Estimate Under Truncation
A boundary maximum can occur even when the likelihood is not flat. The following derivation shows how a support constraint can place the estimate at the smallest parameter value consistent with the observations.
Worked Example: A Truncated Uniform Model Has an Attained Boundary Estimate
A hypothetical survey records exact losses from a Uniform\((0,\theta)\) model, but includes a loss only if it exceeds \(a=\$4\). The observed losses are \(\$5.20,\$6.10,\$7.00\). Find the conditional MLE of \(\theta\), assuming \(\theta>4\).
Method and derivation. Before truncation, the density is \(1/\theta\) on \(0<x<\theta\), and \(P(X>4)=(\theta-4)/\theta\). Hence the conditional density is \(1/(\theta-4)\) on \(4<x<\theta\). The largest observed loss is \(M=7\), so the likelihood is zero unless \(\theta\geq7\). For \(\theta\geq7\),
The derivative is \(\ell'(\theta)=-3/(\theta-4)<0\) throughout the feasible region. The log likelihood therefore decreases as \(\theta\) increases, so its maximum occurs at the smallest feasible value, \(\widehat{\theta}=7\). The parameter value 7 is allowed by the support convention and the likelihood is defined there; this is an attained boundary maximum.
At the estimate, each conditional density is \(1/(7-4)=1/3\) per dollar, and the likelihood is \(1/27=0.037037\ldots\) per dollar cubed. The score is \(-3/(7-4)=-1\), not zero; that is expected because this is a boundary estimate, not an interior stationary point. The result checks against the monotonicity of the likelihood.
Common mistake avoided: Requiring the score to equal zero in every MLE problem would miss this estimate. An interior score equation is necessary for a differentiable interior maximum, but a constrained maximum can occur at a boundary without a zero score.
What to Report and What Not to Infer
A flat or boundary likelihood changes how an estimate should be interpreted. Near an interior maximum, curvature is often used to summarize local precision. If the likelihood is nearly flat, curvature can be very small and the resulting uncertainty calculation unstable. If the estimate lies at a boundary, an interior approximation may be inappropriate because it treats nearby parameter values as if they were available on both sides. A singular curvature calculation is a warning about the information in the likelihood, not a reason to force a finite standard error.
When analyzing incomplete data, make the following checks part of the calculation:
- Check the parameter domain. Identify whether endpoints are included and whether parameters can tend to zero or infinity.
- Check the full likelihood, not only the score. A score equation may have no solution, may have multiple solutions, or may locate only an interior candidate.
- Look for cancellations and ridges. A parameter that disappears after conditioning cannot be estimated from that conditional likelihood.
- Evaluate boundary behavior. State whether the maximum is attained or whether only a limiting supremum exists.
- Separate model limitations from sample limitations. A parameter may be structurally identifiable but weakly estimated in a small sample; those are different conclusions.
These distinctions are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus. In an exam solution, a concise diagnosis should name the likelihood behavior, show the calculation that establishes it, and state precisely which parameter or parameter combination the observed records do and do not determine.
Check Your Understanding
For each problem, identify the likelihood behavior and give the requested estimate or conclusion.
- Three independent exponential losses with rate \(\lambda>0\) are each right-censored at times 1, 2, and 3, respectively. Write the likelihood and determine whether a finite MLE exists. (Answer: \(L(\lambda)=e^{-6\lambda}\); its supremum is 1 as \(\lambda\downarrow0\), so no finite MLE exists.)
- A Pareto Type I model is left-truncated at \(a=100\), with \(0<\theta\leq100\). Exact observed losses are 150, 200, and 300. Find the conditional MLE of \(\alpha\), and state whether \(\theta\) is identified. (Answer: \(\widehat{\alpha}=3/\ln(9)=1.3654\) to four decimal places; \(\theta\) cancels and is not identified.)
- Four Weibull losses with known shape \(k=2\) are all right-censored at \(t=3\). Their scale parameter is \(\eta>0\). Determine the likelihood's supremum and whether it is attained. (Answer: \(L(\eta)=\exp[-36/\eta^2]\); the supremum is 1 as \(\eta\to\infty\), and it is not attained for finite \(\eta\).)
- Losses follow Uniform\((0,\theta)\) and are observed only above \(a=2\). Exact observations are 3, 5, and 6. Find the conditional MLE of \(\theta\) and the maximized likelihood. (Answer: \(\widehat{\theta}=6\), an attained boundary estimate; \(L(6)=(1/4)^3=1/64\).)