Why Likelihood Errors Change the Estimate
A maximum likelihood estimate is obtained by maximizing the likelihood of the observed data under a specified model. In a complete-data problem, the observation record is fully available: for example, each loss amount is known exactly, or each period’s claim count is observed. A likelihood error can change the estimate, create a false boundary solution, or make the estimate correspond to a different model.
This tutorial focuses on auditing four common errors: omitting an observation, using a survival probability in place of a density for an exact loss, dropping a parameter-dependent term, and combining frequency and severity contributions incorrectly. The examples use hypothetical data and familiar distributions. They are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.
For independent, identically distributed, exactly observed continuous losses \(X_1,\ldots,X_n\) with density \(f_X(x;\theta)\), the likelihood is the product of the individual densities. For discrete observations, use the probability mass function instead. Independence justifies the product; exact observation determines that the contribution is a density or mass, rather than a tail probability.
Taking logarithms is convenient because it turns a product into a sum. It does not change the maximizer, since the logarithm is a strictly increasing function on positive values. Terms may be removed from a log-likelihood only when they do not depend on any parameter being estimated. A term can be numerically constant for the observed data and still be essential if it contains an unknown parameter.
Audit the Data Contributions First
Before differentiating, match each observation in the problem statement to a contribution in the likelihood. For an independent complete sample of size \(n\), there should be \(n\) contributions. If the data are grouped, each observed group count must be represented, usually through a probability raised to that count. A missing factor is not repaired by solving the resulting score equation accurately: that equation belongs to the wrong likelihood.
Worked Example: Detecting an Omitted Exact Loss
A hypothetical portfolio has four independent, exactly observed losses \(X_1=1\), \(X_2=2\), \(X_3=3\), and \(X_4=4\), in thousands of dollars. Assume each loss has an exponential distribution with rate \(\lambda>0\), so \(f_X(x;\lambda)=\lambda e^{-\lambda x}\) for \(x>0\). Find the MLE and quantify the effect of omitting the fourth observation.
Method. Write the density contribution for all four exact observations, then compare it with the likelihood formed from only the first three. The common mistake is to use the reduced sample while still treating the problem as if all four losses had been observed.
For all four observations, the likelihood and log-likelihood are
The score is \(4/\lambda-10\). Setting it equal to zero gives \(\widehat{\lambda}=4/10=0.4\) per thousand dollars. The second derivative is \(-4/\lambda^2<0\), so this stationary point is a maximum. The estimate also agrees with the exponential complete-sample formula \(n/\sum_j x_j\).
If the fourth loss is accidentally omitted, the likelihood becomes \(\lambda^3e^{-6\lambda}\). Its score equation is \(3/\lambda-6=0\), giving the incorrect estimate \(\widehat{\lambda}_{\mathrm{omit}}=0.5\). The omitted loss was relatively large, so leaving it out lowers the sample mean and raises the estimated rate. The check is direct: the complete data contain four losses summing to 10, not three losses summing to 6.
Exact Losses Require Densities, Not Tail Probabilities
For a continuous loss observed exactly at \(x\), its likelihood contribution is \(f_X(x;\theta)\). The survival function \(S_X(x)=P(X>x)\) describes the probability of exceeding a threshold; it is not the likelihood contribution for an exact value. A survival contribution can be appropriate when the record says only that a loss exceeds a threshold, but that is different information from an exact observation.
Worked Example: Correcting a Density-for-Survival Substitution
Suppose three independent, exact hypothetical losses are \(2\), \(5\), and \(8\), measured in thousands of dollars. Assume an exponential model with rate \(\lambda>0\). Find the MLE, then determine what happens if the survival function is mistakenly used for every exact loss.
Method. Use the exponential density for each exact loss. The trap is replacing \(f_X(x;\lambda)=\lambda e^{-\lambda x}\) with \(S_X(x)=e^{-\lambda x}\), which records exceedance information rather than an exact amount.
The correct likelihood is
Thus \(\ell(\lambda)=3\log\lambda-15\lambda\), and the score is \(3/\lambda-15\). Solving gives \(\widehat{\lambda}=3/15=0.2\) per thousand dollars. Since \(\ell''(\lambda)=-3/\lambda^2<0\), this is a maximum.
The mistaken survival-based likelihood is \(L_{\mathrm{wrong}}(\lambda)=e^{-15\lambda}\). Its derivative is \(-15e^{-15\lambda}<0\), so it decreases throughout \(\lambda>0\). It has no maximizer in that parameter space; its supremum occurs as \(\lambda\) approaches zero. That boundary behavior is a warning that the wrong data contribution has removed all three factors of \(\lambda\). The correction is to use the density for each amount known exactly.
Keep Every Parameter-Dependent Term
A log-likelihood may contain terms that look like constants when the data are fixed but still depend on an unknown parameter. Such terms must remain when differentiating. By contrast, factors that depend only on the observations and fixed model constants can be omitted if the goal is to find the MLE. The safe audit is to mark each term by what it depends on, not by whether it looks complicated.
Worked Example: A Parameter Term Hidden in the Weibull Density
Three independent exact hypothetical losses are \(1\), \(2\), and \(3\), in thousands of dollars. Assume a Weibull distribution with known shape \(k=2\) and unknown scale \(\theta>0\), with density \(f_X(x;\theta)=\frac{k}{\theta}(x/\theta)^{k-1}\exp[-(x/\theta)^k]\) for \(x>0\). Find the MLE of \(\theta\) and diagnose the result of dropping a parameter-dependent term.
Method. Substitute \(k=2\) into the density before forming the log-likelihood. The common mistake is treating the scale factor in the density as constant because the observations are fixed.
For one observation, the density is \(2x\theta^{-2}\exp(-x^2/\theta^2)\). The three-observation likelihood is therefore
Its log-likelihood is \(\ell(\theta)=\log 48-6\log\theta-14/\theta^2\). The term \(\log48\) is independent of \(\theta\) and may be dropped; the term \(-6\log\theta\) depends on \(\theta\) and must remain. Differentiating gives
At this point, \(\ell''(\theta)=6/\theta^2-84/\theta^4\). Substituting \(\theta^2=14/3\) gives \(9/7-27/7=-18/7<0\), confirming a local maximum. Also, the likelihood tends to zero as \(\theta\) tends to zero or infinity, so this interior maximum is global.
If the parameter-dependent term \(-6\log\theta\) is dropped, the remaining expression, apart from a constant, is \(-14/\theta^2\). Its derivative is \(28/\theta^3>0\) for every \(\theta>0\); it increases toward zero as \(\theta\) grows and has no finite maximizer. The incorrect result is not an alternative estimate: it comes from altering the likelihood.
Keep Frequency and Severity Contributions Separate
When claim counts and claim amounts are both observed, the joint likelihood contains a frequency component and a severity component. Under a model in which counts follow a Poisson distribution and severities are independent of counts, with independent and identically distributed severities conditional on the count, the joint likelihood factors into the count likelihood and the conditional severity likelihood. They describe different random quantities and generally have different parameters and units.
Here \(t_r\) is the exposure in period \(r\), \(n_r\) is its observed count, \(\lambda\) is the Poisson rate per exposure unit, and \(x_1,\ldots,x_N\) are the exact severities. The severity product has \(N\) factors because there is one amount for each observed claim. The count likelihood uses the period counts and exposures; severity amounts do not replace exposure.
Worked Example: Separating a Poisson Rate From a Severity Rate
A hypothetical insurer observes two one-year periods with one exposure unit in each. The claim counts are \(n_1=2\) and \(n_2=1\). The three exact losses, in thousands of dollars, are \(2\), \(3\), and \(4\). Assume independent Poisson counts with common rate \(\lambda\) per exposure unit and independent exponential severities with rate \(\beta\) per thousand dollars. Estimate both parameters.
Method. Form the Poisson likelihood from counts and exposures, then multiply by the exponential density for each exact loss. The trap is using the total loss as the frequency exposure or using the count likelihood to estimate the severity parameter.
For the two count periods, the frequency likelihood is
The frequency log-likelihood, omitting only the parameter-independent term \(-\log2\), is \(-2\lambda+3\log\lambda\). Its score is \(-2+3/\lambda\), so \(\widehat{\lambda}=3/2=1.5\) claims per exposure unit. The total exposure is \(1+1=2\), and the result checks as total claims divided by total exposure: \(3/2\).
For the three severities, the conditional severity likelihood is
The severity log-likelihood is \(3\log\beta-9\beta\). Its score is \(3/\beta-9\), giving \(\widehat{\beta}=3/9=1/3\) per thousand dollars. Its reciprocal, \(3\) thousand dollars, equals the observed mean severity.
The joint likelihood is \(L_F(\lambda)L_S(\beta)\); the parameters separate, so each estimate maximizes its own factor. A mistaken calculation \(\lambda=3/9=1/3\) uses total loss in place of total exposure. It has the wrong units and does not maximize the count likelihood. The estimates are therefore \(\widehat{\lambda}=1.5\) per exposure unit and \(\widehat{\beta}=1/3\) per thousand dollars.
A Practical Likelihood-Error Checklist
A short audit before solving can catch errors more efficiently than trying to diagnose an implausible estimate after the calculation. The checklist below applies to complete-data models; if the record is censored or truncated, the likelihood contribution requires additional information about how the record arose.
Count the exact values, counts, periods, and exposures in the problem. Confirm that every required datum appears in the likelihood.
Use a density for an exact continuous value, a mass for an exact discrete value, and do not substitute a tail probability for either.
Keep every factor and log-likelihood term involving an unknown parameter. Remove only terms constant with respect to all parameters being estimated.
For a frequency-severity model, identify which data inform the count parameters and which exact amounts inform the severity parameters.
Verify the score equation, parameter-space restrictions, and whether the candidate is a maximum. Compare with a simple model-based check when available.
These checks matter because a plausible-looking numerical estimate does not establish that the likelihood was constructed correctly. In particular, omitting a large observation may still produce a valid-looking interior solution, while using survival probabilities can remove the density’s parameter factors and produce a boundary result. Dimensional checks are useful too: a frequency rate is measured per exposure unit, whereas a severity rate is measured per unit of loss.
Check Your Understanding
For each problem, first identify the data contribution that determines the likelihood, then calculate or diagnose the result.
- A hypothetical sample of exact losses is \(1, 2, 5\), in thousands of dollars, from an exponential model with rate \(\lambda>0\). Find the MLE. (Answer: \(\widehat{\lambda}=3/8=0.375\) per thousand dollars.)
- Two exact independent losses are \(3\) and \(6\), in thousands of dollars, with exponential rate \(\lambda>0\). A candidate uses \(e^{-9\lambda}\) as the likelihood. What key factor is missing, and what is the correct MLE? (Answer: The two density factors \(\lambda\) are missing; \(L(\lambda)=\lambda^2e^{-9\lambda}\) and \(\widehat{\lambda}=2/9\) per thousand dollars.)
- Two exact losses, \(1\) and \(3\), in thousands of dollars, follow a Weibull model with known shape \(k=2\) and unknown scale \(\theta>0\). Find the MLE of \(\theta\). (Answer: \(\widehat{\theta}=\sqrt{(1^2+3^2)/2}=\sqrt{5}\approx2.2361\) thousand dollars.)
- A hypothetical portfolio has four claims over two exposure units. The exact severities are \(1, 2, 5, 8\), in thousands of dollars. Assume a Poisson frequency rate \(\lambda\) per exposure unit and exponential severity rate \(\beta\) per thousand dollars. Find both MLEs. (Answer: \(\widehat{\lambda}=4/2=2\) per exposure unit and \(\widehat{\beta}=4/16=0.25\) per thousand dollars.)
- For \(n\) independent exact exponential losses \(x_1,\ldots,x_n\), a proposed log-likelihood is \(-\lambda\sum_j x_j\), with no \(n\log\lambda\) term. Diagnose the error and describe what happens to this proposed function for \(\lambda>0\). (Answer: It omits the parameter-dependent density term \(n\log\lambda\); the proposed function decreases toward its supremum as \(\lambda\) approaches zero and gives no valid interior MLE.)