Mixed Exact and Right-Censored Samples

Why Exact and Censored Records Belong in One Likelihood

A severity data set may contain some losses measured exactly and others known only to exceed a threshold. For example, a hypothetical claims study might record the full amount for claims settled during the observation period, while an unsettled claim is known only to exceed the amount currently established. The second record is incomplete, but it still provides information: it tells us the loss was at least that large.

The likelihood must preserve that distinction. An exact loss contributes a density; a right-censored loss contributes a survival probability. This construction is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: For a continuous severity \(X\) with density \(f_X(x;\boldsymbol{\theta})\), distribution function \(F_X(x;\boldsymbol{\theta})\), and survival function \(S_X(x;\boldsymbol{\theta})=1-F_X(x;\boldsymbol{\theta})\), an exact observation at \(x\) contributes \(f_X(x;\boldsymbol{\theta})\). A right-censored observation at threshold \(c\), which tells us \(X>c\), contributes \(S_X(c;\boldsymbol{\theta})\). Here \(\boldsymbol{\theta}\) is the common parameter vector governing every record.

Constructing the Mixed-Sample Likelihood

For record \(i\), let \(Y_i\) be the observed amount and let \(\delta_i\) equal 1 when the amount is exact and 0 when it is right-censored. Thus, \(Y_i=X_i\) if \(\delta_i=1\); if \(\delta_i=0\), then \(Y_i=c_i\) and the information is \(X_i>c_i\). Assuming the severities are independent and share the same parameter vector, the severity likelihood is

L(\boldsymbol{\theta}) =\prod_{i=1}^{n} \left[f_X(Y_i;\boldsymbol{\theta})\right]^{\delta_i} \left[S_X(Y_i;\boldsymbol{\theta})\right]^{1-\delta_i}.

The corresponding log-likelihood is

\ell(\boldsymbol{\theta}) =\sum_{i=1}^{n} \left\{ \delta_i\log f_X(Y_i;\boldsymbol{\theta}) +(1-\delta_i)\log S_X(Y_i;\boldsymbol{\theta}) \right\}.

The density contribution is not the probability that a continuous loss equals \(Y_i\), which is zero. It is the density evaluated at that exact observation, as required for a continuous-data likelihood. A censored record, by contrast, identifies an interval extending from its threshold to infinity, so its probability contribution is the probability of exceeding that threshold.

Assumption: The censoring mechanism is noninformative for the severity parameters: conditional on any modeled covariates, censoring does not convey additional information about the unobserved severity. If censoring is fixed by design, or is independent of severity and governed by separate parameters, its factor can be omitted when estimating \(\boldsymbol{\theta}\). If censoring depends on severity or shares parameters with the severity model, a joint model may be needed; the severity-only likelihood above is then not sufficient.

The key justification is that each independent record contributes the probability or density of the observation it actually reveals. For an exact value \(x\), the likelihood contribution over a small interval of width \(h\) is approximately \(f_X(x;\boldsymbol{\theta})h\). For a censored threshold \(c\), the revealed event is \(X>c\), with probability \(S_X(c;\boldsymbol{\theta})\). Multiplying independent contributions gives the likelihood; terms such as \(h\) that do not depend on the parameters do not affect likelihood-based inference. Taking logarithms converts the product into a sum without changing the parameter value that maximizes it.

Classify each record.
Record whether the value is exact or right-censored, and retain the observed amount or censoring threshold.

Write its contribution.
Use \(f_X(y_i;\boldsymbol{\theta})\) for an exact record and \(S_X(y_i;\boldsymbol{\theta})\) for a censored record.

Combine and simplify.
Multiply the contributions, then take logarithms if a sum is easier to differentiate or evaluate.

Check the model and assumptions.
Confirm that every record uses the same stated parameter vector and that censoring can be treated as noninformative.

Worked Likelihoods and Estimation

Worked Example: Exponential Severity With Two Censored Claims

A hypothetical severity sample, measured in thousands of dollars, contains exact losses of \(1.2\) and \(2.0\), and right-censoring thresholds of \(0.8\) and \(1.5\). Assume independent observations from an exponential severity model with rate \(\lambda>0\) per thousand dollars, and noninformative censoring. Construct the likelihood and find the maximum likelihood estimate of \(\lambda\).

Method. For the exponential model, \(f_X(x;\lambda)=\lambda e^{-\lambda x}\) and \(S_X(x;\lambda)=e^{-\lambda x}\). The exact observations contribute \(\lambda e^{-1.2\lambda}\) and \(\lambda e^{-2.0\lambda}\). The censored observations contribute \(e^{-0.8\lambda}\) and \(e^{-1.5\lambda}\).

Multiplying gives

L(\lambda)=\lambda^2 e^{-\lambda(1.2+2.0+0.8+1.5)} =\lambda^2e^{-5.5\lambda}, \qquad \ell(\lambda)=2\log\lambda-5.5\lambda.

Differentiate the log-likelihood and set the score equal to zero:

\frac{d\ell}{d\lambda}=\frac{2}{\lambda}-5.5=0 \quad\Longrightarrow\quad \widehat{\lambda}=\frac{2}{5.5}=\frac{4}{11} =0.3636.

The second derivative is \(-2/\lambda^2<0\) for every \(\lambda>0\), so this stationary point is a maximum. The rate is \(0.3636\) per thousand dollars. As a check, the estimate equals the number of exact events, 2, divided by the sum of all four observed amounts and thresholds, 5.5.

Common mistake avoided: Counting all four records as exact would give \(4/5.5=0.7273\). The censored records contribute exposure through their thresholds, but they do not contribute exact-event factors of \(\lambda\).

Worked Example: Evaluating a Weibull Mixed-Sample Likelihood

A hypothetical sample of loss amounts in thousands of dollars has exact values \(2\) and \(4\), and a right-censoring threshold of \(3\). Assume independent observations from a Weibull model with shape \(k=2\) and scale \(\eta=4\), both common to all records, and noninformative censoring. Evaluate the likelihood and log-likelihood at these parameter values.

Method. Use the Weibull density and survival function

f_X(x;k,\eta)=\frac{k}{\eta} \left(\frac{x}{\eta}\right)^{k-1} e^{-(x/\eta)^k}, \qquad S_X(x;k,\eta)=e^{-(x/\eta)^k}.

At \(x=2\), the exact density is \((2/4)(2/4)e^{-1/4}=\frac14e^{-1/4}\). At \(x=4\), it is \((2/4)(4/4)e^{-1}=\frac12e^{-1}\). The censored threshold contributes \(S_X(3;2,4)=e^{-9/16}\). Therefore,

L(2,4)=\left(\frac14e^{-1/4}\right) \left(\frac12e^{-1}\right)e^{-9/16} =\frac18e^{-29/16} \approx 0.0204.

The log-likelihood is

\ell(2,4)=\log(1/8)-\frac{29}{16} \approx -3.8919.

The exponent check is \(1/4+1+9/16=29/16\), and there are two density factors and one survival factor. This verifies that the censored observation has not been treated as an exact loss. The calculation evaluates the likelihood at specified parameters; it does not claim that these values maximize it.

Common mistake avoided: Applying the Weibull density at \(3\) would describe an exact loss of \(3\), not the information \(X>3\). The correct contribution is the survival probability \(e^{-9/16}\).

Worked Example: A Gamma Model With a Non-Exponential Survival Function

A hypothetical loss sample, in thousands of dollars, has exact observations \(1\) and \(2\), and right-censored thresholds \(1\) and \(3\). Assume independent observations from a gamma model with fixed shape 2 and scale \(\theta=2\), with noninformative censoring. Find the likelihood and log-likelihood at the specified model.

Method. For a gamma distribution with shape 2 and scale \(\theta\), the density and survival function are

f_X(x;2,\theta)=\frac{x}{\theta^2}e^{-x/\theta}, \qquad S_X(x;2,\theta)=e^{-x/\theta}\left(1+\frac{x}{\theta}\right).

At \(\theta=2\), the exact densities are \(f_X(1)=\frac14e^{-1/2}\) and \(f_X(2)=\frac12e^{-1}\). The censored contributions are \(S_X(1)=\frac32e^{-1/2}\) and \(S_X(3)=\frac52e^{-3/2}\). Thus,

L(2)= \left(\frac14e^{-1/2}\right) \left(\frac12e^{-1}\right) \left(\frac32e^{-1/2}\right) \left(\frac52e^{-3/2}\right) =\frac{15}{32}e^{-7/2} \approx 0.0142.

Taking logarithms gives

\ell(2)=\log(15/32)-\frac72\approx -4.2577.

The exponent is the sum \(1/2+1+1/2+3/2=7/2\), and the coefficient is \((1/4)(1/2)(3/2)(5/2)=15/32\). Both checks confirm the stated likelihood. Unlike the exponential survival function, the gamma survival function includes the factor \(1+x/\theta\); omitting it would change both the likelihood and log-likelihood.

Common mistake avoided: Replacing every survival contribution by a bare exponential tail can be wrong even when the density contains an exponential term. Use the survival function for the stated model, not a tail form borrowed from another distribution.

Common Errors and Interpretation

Using a density for a censored record. The record says \(X>c\), not \(X=c\); use \(S_X(c;\boldsymbol{\theta})\).
Omitting censored records altogether. A censoring threshold contributes information about the tail, even though it does not add an exact-event density.
Counting censored records as events. In the exponential model, only exact records contribute a factor \(\lambda\). Every record contributes its observed amount or threshold to the exponential survival factor.
Changing the parameter vector across records. In a common-parameter model, the same \(\boldsymbol{\theta}\) appears in every density and survival term.
Ignoring informative censoring. If high losses are more likely to be censored for reasons related to their severity, the independent-censoring assumption may fail; the censoring process may need to be modeled.

For an exponential model, the likelihood simplifies because both the density and survival contain \(e^{-\lambda x}\). With \(d\) exact observations and observed values \(y_i\) that include censoring thresholds, the likelihood is proportional to \(\lambda^d e^{-\lambda\sum_i y_i}\). This explains the rate estimate in the first example: censoring contributes to the total observed exposure, while only exact events contribute to the event count.

Key takeaway: A mixed exact and right-censored sample uses one factor per record: a density for an exact loss and a survival probability for a censored loss. The product, or equivalently its log, must use the same severity parameter vector throughout and must respect the censoring assumptions.

Check Your Understanding

For each problem, identify the appropriate mixed-sample contribution before using the stated model.

Under an exponential severity model with rate \(\lambda\), one loss is observed exactly at \(3\) and one is right-censored at \(5\). Write the likelihood. (Answer: \(L(\lambda)=\lambda e^{-8\lambda}\))
For an exponential sample, exact losses are \(3\) and \(5\), and censoring thresholds are \(2\) and \(4\), all in the same units. Find the maximum likelihood estimate of the rate. (Answer: \(\widehat{\lambda}=1/7\) per unit)
A Weibull model has \(k=1\) and \(\eta=5\). One loss is exact at \(2\), and another is censored at \(3\). Find the likelihood and log-likelihood. (Answer: \(L=(1/5)e^{-1}\approx0.0736,\ \ell=\log(1/5)-1\approx-2.6094\))
Under a gamma model with fixed shape 2 and scale \(\theta=1\), one loss is exact at \(1\) and one is censored at \(2\). Find the likelihood. (Answer: \(L=3e^{-3}\approx0.1494\))
An exponential sample has an exact loss of \(6\) months, a right-censoring threshold of \(18\) months, and another right-censoring threshold of \(1.5\) years. Find the maximum likelihood rate per year. (Answer: \(\widehat{\lambda}=2/7\approx0.2857\) per year)

What You'll Learn