A Complete-Data Maximum Likelihood Workflow

From Data to a Fitted Model

A complete-data maximum likelihood analysis is a sequence of decisions, not just a formula applied to a dataset. The analyst must identify what was observed, specify a probability model and its parameter space, write the likelihood for those observations, and then check that the proposed estimate is an admissible maximum. In a claim study, a frequent source of error is to use the number of policies instead of their total exposure, or to interpret a severity rate as an average loss.

This tutorial uses a hypothetical portfolio with claim counts observed over known exposure and exact, fully observed claim amounts. The frequency model is Poisson, and the severity model is exponential. These choices make the workflow transparent; the same order of work applies to other parametric models, but their likelihoods and score equations can differ.

Model and assumptions: For independent exposure units \(j=1,\ldots,m\), let \(N_j\) be the observed claim count and \(e_j>0\) the exposure, measured in a consistent unit such as policy-years. Assume \(N_j\sim\operatorname{Poisson}(\lambda e_j)\), independently, with \(\lambda>0\). Conditional on the observed claims, let the exact positive severities \(X_1,\ldots,X_n\) be independent and identically distributed exponential random variables with rate \(\beta>0\), independent of the counts. Thus \(f_X(x)=\beta e^{-\beta x}\) for \(x>0\). All claims in the severity sample are assumed observed exactly: there is no censoring, truncation, or missing severity information.

The frequency parameter \(\lambda\) is the expected number of claims per exposure unit. The severity parameter \(\beta\) is a rate, with units inverse to the units used for \(X\). For this exponential model, the mean severity is \(E[X]=1/\beta\). If losses are recorded in thousands of dollars, for example, \(\beta\) is measured per thousand dollars and \(1/\beta\) is in thousands of dollars.

Constructing the Likelihood

A likelihood is the joint probability mass or density of the observed data, viewed as a function of the unknown parameters. It is not a probability distribution over parameter values. Under the stated independent models, the frequency and severity likelihood contributions multiply. Write \(N=\sum_{j=1}^m N_j\) for the total claim count and \(E=\sum_{j=1}^m e_j\) for total exposure.

L(\lambda,\beta) = \prod_{j=1}^{m} \frac{e^{-\lambda e_j}(\lambda e_j)^{N_j}}{N_j!} \prod_{k=1}^{n}\beta e^{-\beta X_k} = \left[ e^{-\lambda E}\lambda^N \prod_{j=1}^{m}\frac{e_j^{N_j}}{N_j!} \right] \left[ \beta^n e^{-\beta\sum_{k=1}^{n}X_k} \right].

The exposure/count term and the severity term involve separate parameters, so their maximizations can be carried out separately. Taking logarithms preserves the maximizer and turns products into sums. Terms independent of the parameter being estimated may be retained for a full likelihood calculation or omitted when solving a score equation; they must not be mistaken for parameter-dependent terms.

\ell(\lambda,\beta) = -\lambda E+N\log\lambda +\sum_{j=1}^{m}\left(N_j\log e_j-\log(N_j!)\right) +n\log\beta-\beta\sum_{k=1}^{n}X_k.

Estimating Frequency and Severity

For \(N>0\), differentiate the frequency log-likelihood with respect to \(\lambda\). Its score is \(N/\lambda-E\), which is zero at \(N/E\). The second derivative is negative throughout the positive parameter space, so this stationary point is the unique maximum. For severity, differentiating gives the score \(n/\beta-\sum X_k\). When \(n>0\), it is zero at \(n/\sum X_k\), and the second derivative is negative for every \(\beta>0\).

\widehat{\lambda}=\frac{N}{E}\quad(N>0), \qquad \widehat{\beta}=\frac{n}{\sum_{k=1}^{n}X_k}\quad(n>0), \qquad \widehat{E[X]}=\frac{1}{\widehat{\beta}}=\frac{\sum_{k=1}^{n}X_k}{n}.

The estimates follow from the likelihood, not from treating every data field as an equally weighted observation. In particular, the frequency denominator is total exposure. The severity rate estimate is the reciprocal of the sample mean, while the implied fitted mean is the sample mean itself.

Boundary condition: The stated frequency parameter space is \(\lambda>0\). If \(N=0\), the likelihood is proportional to \(e^{-\lambda E}\), which increases toward 1 as \(\lambda\to0^+\) but does not attain 1 at any allowed value. Therefore there is no MLE in the stated parameter space; the supremum is approached as \(\lambda\to0^+\). If the parameter space is extended to include \(\lambda=0\), the boundary MLE is 0. Similarly, if there are no observed claims, there are no severity observations from which to estimate \(\beta\).

Worked Examples

Worked Example: Estimating a Frequency Rate With Unequal Exposure

A hypothetical insurer observes three independent groups with exposures \(e_1=1\), \(e_2=2\), and \(e_3=3\) policy-years. Their fully observed claim counts are \(N_1=1\), \(N_2=0\), and \(N_3=3\). Assume independent Poisson counts with mean \(\lambda e_j\) and \(\lambda>0\). Find the MLE and verify that it is a maximum.

Method and sufficient totals. The total number of claims is \(N=1+0+3=4\), and the total exposure is \(E=1+2+3=6\) policy-years. Therefore the candidate estimate is \(\widehat{\lambda}=N/E=4/6=2/3\) claims per policy-year.

Likelihood check. Keeping the exposure-dependent constants, the likelihood is \[ L(\lambda)= \frac{e^{-\lambda}\lambda^1}{1!} \frac{e^{-2\lambda}(2\lambda)^0}{0!} \frac{e^{-3\lambda}(3\lambda)^3}{3!} = \frac{27}{6}e^{-6\lambda}\lambda^4 = 4.5e^{-6\lambda}\lambda^4. \] The log-likelihood is \(\ell(\lambda)=\log(4.5)-6\lambda+4\log\lambda\). Its score is \(\ell'(\lambda)=-6+4/\lambda\), which is zero at \(2/3\). Its second derivative is \(\ell''(\lambda)=-4/\lambda^2<0\) for every \(\lambda>0\), verifying a maximum. Also, the likelihood tends to zero as \(\lambda\to0^+\) and as \(\lambda\to\infty\), so this stationary point is the global maximum.

Numerical likelihood and interpretation. At the estimate, \(L(2/3)=4.5e^{-4}(2/3)^4=(8/9)e^{-4}\approx0.0163\). The fitted frequency is \(0.667\) claims per policy-year, to three significant figures. The likelihood’s numerical value is a likelihood for the observed count data, not the probability that \(\lambda=2/3\).

Common mistake avoided. Dividing four claims by the three groups gives \(4/3\), but that is claims per group, not claims per policy-year. The model’s Poisson means use each group’s exposure, so the correct denominator is six policy-years.

Worked Example: Fitting an Exponential Severity Model

A hypothetical complete sample contains four exact claim amounts, recorded in thousands of dollars: \(2,3,5,\) and \(10\). Assume independent exponential severities with density \(f_X(x)=\beta e^{-\beta x}\) for \(x>0\), where \(\beta>0\). Estimate the rate and mean, and check the likelihood maximum.

Compute the total and candidate estimates. The sample size is \(n=4\), and the total severity is \(2+3+5+10=20\) thousand dollars. Hence \(\widehat{\beta}=4/20=0.2\) per thousand dollars. The fitted mean is \(1/\widehat{\beta}=5\) thousand dollars.

Verify through the likelihood. The likelihood is \[ L(\beta)=\prod_{k=1}^{4}\beta e^{-\beta X_k} =\beta^4e^{-20\beta}. \] Thus \(\ell(\beta)=4\log\beta-20\beta\), with score \(4/\beta-20\), equal to zero at \(0.2\). The second derivative is \(-4/\beta^2<0\), so the stationary point is a maximum; the log-likelihood tends to \(-\infty\) at either parameter-space limit. At the estimate, \(L(0.2)=(0.2)^4e^{-4}=0.0016e^{-4}\approx0.0000293\), in units corresponding to the four severity densities.

Interpretation and trap. The estimate \(0.2\) is not the mean claim amount: it is a rate per thousand dollars. Its reciprocal, \(5\), is the fitted mean in thousands of dollars. Reporting the mean as \(0.2\) thousand dollars confuses the rate with its reciprocal.

Why the Estimates Have These Forms

The score equations explain the structure of the results. For frequency, the parameter-dependent log-likelihood is \(N\log\lambda-\lambda E\). Increasing \(\lambda\) raises the contribution \(N\log\lambda\), reflecting the observed claims, but lowers the contribution \(-\lambda E\), reflecting the exposure at risk. Balancing these terms gives \(N/\lambda=E\), or \(\lambda=N/E\). For exponential severity, the corresponding terms are \(n\log\beta-\beta\sum X_k\). Their balance gives \(n/\beta=\sum X_k\), so the estimated rate is the number of losses divided by their total.

State the data and model.
Identify the count observations and exposures, the exact severity observations and units, and the assumed distributions, independence, and parameter spaces.

Write the joint likelihood.
Use a Poisson probability mass for each count and a severity density for each exact loss. Keep the exposure and factorial terms if evaluating the full likelihood.

Maximize and check.
Take logarithms, solve the score equations, check curvature and parameter constraints, and inspect boundary cases such as zero claims.

Interpret the fitted values.
Report the frequency per exposure unit and translate a severity rate into the corresponding fitted mean using its reciprocal.

Worked Example: Completing a Joint Frequency-Severity Fit

A hypothetical portfolio has two independent groups: group 1 has exposure \(e_1=2\) policy-years and \(N_1=2\) claims; group 2 has exposure \(e_2=3\) policy-years and \(N_2=3\) claims. The five exact severities, in thousands of dollars, are \(1,2,2,4,\) and \(6\). Assume the Poisson frequency and exponential severity models stated above, with frequency independent of severity. Find both MLEs, construct the joint likelihood, and verify the estimates.

Frequency fit. Here \(N=2+3=5\) claims and \(E=2+3=5\) policy-years, so \(\widehat{\lambda}=5/5=1\) claim per policy-year. The count likelihood is \[ L_F(\lambda) = \frac{e^{-2\lambda}(2\lambda)^2}{2!} \frac{e^{-3\lambda}(3\lambda)^3}{3!} = 9e^{-5\lambda}\lambda^5. \] Its log-likelihood has score \(5/\lambda-5\), zero at 1, and second derivative \(-5/\lambda^2<0\).

Severity fit. There are \(n=5\) losses, with total \(1+2+2+4+6=15\) thousand dollars. Thus \(\widehat{\beta}=5/15=1/3\) per thousand dollars, and the fitted mean severity is \(1/\widehat{\beta}=3\) thousand dollars. The severity likelihood is \(L_S(\beta)=\beta^5e^{-15\beta}\); its score \(5/\beta-15\) vanishes at \(1/3\), and its second derivative \(-5/\beta^2\) is negative.

Joint likelihood and numerical check. Independence gives \[ L(\lambda,\beta)=9e^{-5\lambda}\lambda^5\beta^5e^{-15\beta}. \] At \((\widehat{\lambda},\widehat{\beta})=(1,1/3)\), this is \(9e^{-5}(1/3)^5e^{-5}=(1/27)e^{-10}\approx0.00000168\). The joint log-likelihood is the sum of the two log-likelihoods; its second derivatives in each parameter are negative and its mixed derivative is zero. Therefore the separate maxima together give the joint maximum. The fitted expected aggregate loss per policy-year, under these assumptions, is \(1\) claim per policy-year times \(3\) thousand dollars per claim, or \(3\) thousand dollars per policy-year.

Common mistake avoided. The severity total is \(15\) thousand dollars, not \(15\) dollars. The estimated rate is \(1/3\) per thousand dollars, and the mean is \(3\) thousand dollars. Keeping the loss unit attached throughout prevents a factor-of-one-thousand error.

Common Checks and Exam Traps

A complete-data fit should be checked both algebraically and in context. A score equation alone does not establish that a solution is an admissible maximum. Confirm the parameter is in its allowed space, inspect the second derivative or another appropriate maximum check, and consider whether the likelihood approaches a supremum at an excluded boundary. Then check that the reported units match the model.

Using the wrong exposure denominator. The Poisson rate estimate uses total exposure, not the number of groups, policies, or calendar periods unless that quantity is exactly the exposure defined by the model.
Dropping constants when a numerical likelihood is requested. Constants independent of the parameter can be omitted while maximizing, but they matter when reporting the full likelihood value.
Confusing a density with a point probability. For a continuous severity, \(f_X(x)\) is a density; the probability of observing one exact value is not obtained by treating that density as a point probability.
Ignoring parameter boundaries. With zero claims and \(\lambda>0\), there is no attained frequency maximum. State the supremum correctly rather than reporting zero as an interior estimate.
Mixing units or reciprocal parameters. If severity is in thousands of dollars, the exponential rate is per thousand dollars; the mean is the reciprocal rate in thousands of dollars.

Key takeaway: A defensible complete-data MLE workflow states the model and parameter space, writes the likelihood for the actual observations, maximizes it, checks curvature and boundaries, and interprets each estimate with its units. For independent Poisson frequency and exponential severity, the joint likelihood factors, giving \(\widehat{\lambda}=N/E\) and \(\widehat{\beta}=n/\sum X_k\) when the relevant totals are positive.

Check Your Understanding

For each problem, state estimates with units and use the supplied model assumptions. Claim counts are independent Poisson variables with mean \(\lambda e_j\); when severities are given, they are exact observations from an independent exponential model with rate \(\beta\).

A hypothetical portfolio has exposures \(1,2,1\) policy-years and counts \(0,2,1\). Find the frequency MLE and the full count likelihood evaluated at that estimate. (Answer: \(\widehat{\lambda}=0.75\) claims per policy-year; \(L(\widehat{\lambda})=(27/32)e^{-3}\approx0.0420\).)
Three exact severities, in thousands of dollars, are \(4,4,\) and \(8\). Find the exponential rate MLE and fitted mean. (Answer: \(\widehat{\beta}=3/16=0.1875\) per thousand dollars; fitted mean \(16/3\approx5.333\) thousand dollars.)
A portfolio has exposures \(2\) and \(1\) policy-years, with counts \(1\) and \(2\). Its exact severities are \(2,3,\) and \(7\) thousand dollars. Find the frequency and severity estimates and the fitted expected aggregate loss per policy-year. (Answer: \(\widehat{\lambda}=1\) claim per policy-year, \(\widehat{\beta}=0.25\) per thousand dollars, and fitted expected aggregate loss \(4\) thousand dollars per policy-year.)
Under the strict parameter space \(\lambda>0\), a dataset has zero claims over positive total exposure. Does the frequency likelihood have an MLE? What happens if \(\lambda=0\) is included in the parameter space? (Answer: No MLE exists for \(\lambda>0\); the likelihood’s supremum is approached as \(\lambda\to0^+\). If zero is included, the boundary MLE is \(0\).)
A complete exponential severity sample is \(500,1000,1500\) dollars. Give the rate estimate per dollar and the fitted mean in dollars, then express the rate per thousand dollars. (Answer: \(\widehat{\beta}=0.001\) per dollar, fitted mean \(1000\) dollars, and equivalently \(\widehat{\beta}=1\) per thousand dollars.)

What You'll Learn