Likelihood Contributions For Right-Censored Losses

Why Right-Censored Records Need a Different Likelihood Contribution

A loss record is right-censored when the exact loss is not observed, but the record establishes that the loss exceeded a known threshold. For example, a hypothetical claim file might record that a loss exceeded a reporting cap, without revealing its full amount. The information is not an exact value at the cap: it is evidence that the underlying loss lies above it.

That distinction changes the likelihood. An exact continuous observation contributes a density evaluated at the observed loss. A right-censored observation contributes the probability that the loss exceeds the censoring threshold. Treating the threshold as though it were the exact loss discards the information that the loss was larger.

Definition: Let $X$ be a nonnegative continuous loss with density $f_X(x;\theta)$, distribution function $F_X(x;\theta)$, and survival function $S_X(x;\theta)=1-F_X(x;\theta)$, where $\theta$ is a model parameter or parameter vector. An exact observation $X=x$ contributes $f_X(x;\theta)$ to the likelihood. A record right-censored at $c$, meaning that $X>c$ is known but the exact value is not, contributes $S_X(c;\theta)$.

This formulation assumes the loss distribution is continuous at the censoring threshold. Thus, using $X\ge c$ rather than $X>c$ makes no difference to the probability. For a discrete model or a model with an atom at $c$, the recorded inequality and the observation protocol must be used precisely; $P(X\ge c)$ and $P(X>c)$ can differ.

Deriving the Censored Contribution

Suppose the record tells us only that $X>c$. The probability of that information under the model is $P(X>c)=S_X(c;\theta)$, so this is the likelihood contribution. One way to see why it is not a density at $c$ is to compare the two types of record directly.

Exact value.
If the observed loss lies in a small interval $[x,x+\Delta x)$, its probability is approximately $f_X(x;\theta)\Delta x$. The density is the factor that describes an exact continuous observation.

Only known to exceed $c$.
The observed event is the whole tail $X>c$, not a small interval around $c$. Its probability is $P(X>c)=S_X(c;\theta)$.

Form the likelihood.
For independent loss records, multiply the contribution for every exact value by the contribution for every censored record. Each factor is evaluated using the model being fitted.

More formally, if $X$ has a continuous distribution, then $P(x\le X<x+\Delta x)$ is approximately $f_X(x;\theta)\Delta x$ for small $\Delta x$. By contrast, observing only $X>c$ specifies an event with probability $1-F_X(c;\theta)$. Therefore, an exact observation calls for a density, whereas a right-censored observation calls for a survival probability.

Likelihood for independent records: Write $\delta_j=1$ if record $j$ is exact and $\delta_j=0$ if it is right-censored. Let $z_j$ be the observed exact loss when $\delta_j=1$, or the censoring threshold when $\delta_j=0$. The loss-distribution part of the likelihood is $$ L(\theta)=\prod_{j=1}^{n}\left[f_X(z_j;\theta)\right]^{\delta_j}\left[S_X(z_j;\theta)\right]^{1-\delta_j}. $$ For an exact record, the factor is $f_X(z_j;\theta)$; for a censored record, it is $S_X(z_j;\theta)$. This expression assumes independent losses and censoring that is fixed by design or independent of the loss, so that the censoring mechanism adds no parameter-dependent factor for the loss parameters being estimated.

If censoring thresholds are generated by a random variable $C$, independent of $X$, a full joint model for $(X,C)$ can include factors describing the distribution of $C$. When that distribution has no parameters of interest in the loss model, those factors do not change the likelihood maximization for $\theta$. If censoring depends on the unobserved loss in a way not accounted for, the simple contribution above need not yield a valid likelihood for the loss distribution.

Worked Example: Exponential Losses With Exact and Censored Records

Worked Example: Estimating an Exponential Rate From Mixed Records

A hypothetical portfolio has independent losses $X_j\sim\operatorname{Exponential}(\lambda)$, with density $f_X(x;\lambda)=\lambda e^{-\lambda x}$ and survival function $S_X(x;\lambda)=e^{-\lambda x}$ for $x\ge0$. Loss amounts are measured in thousands of dollars, and $\lambda>0$ is measured per thousand dollars. Two exact losses are $2$ and $5$; two other records are right-censored at $3$ and $4$. Find the likelihood and maximum likelihood estimate (MLE) of $\lambda$.

Method. Use a density for each exact loss and a survival probability for each censored threshold. The four individual contributions are

\begin{aligned} X=2 &: \lambda e^{-2\lambda},\\ X=5 &: \lambda e^{-5\lambda},\\ X>3 &: e^{-3\lambda},\\ X>4 &: e^{-4\lambda}. \end{aligned}

Multiplying them gives

L(\lambda)=\lambda^2 e^{-(2+5+3+4)\lambda} =\lambda^2e^{-14\lambda},\qquad \lambda>0.

The log-likelihood is $\ell(\lambda)=2\ln\lambda-14\lambda$. Its derivative is $2/\lambda-14$, so setting the derivative to zero gives $\widehat{\lambda}=2/14=1/7$. The second derivative is $-2/\lambda^2<0$ for every $\lambda>0$, confirming this stationary point is a maximum.

At the estimate, the likelihood is

L(1/7)=\frac{1}{49}e^{-14/7} =\frac{e^{-2}}{49}\approx 0.002762.

The MLE is $1/7$ per thousand dollars. A check is that the likelihood has two density factors, hence the term $\lambda^2$, and the exponential term uses the total $2+5+3+4=14$, including both censoring thresholds. Common mistake avoided: using a density at $3$ or $4$ would falsely treat a censored record as an exact loss.

Exponential MLE and an Important Boundary Case

The previous example illustrates a useful result. Suppose exact exponential losses are $x_1,\ldots,x_r$, and right-censored records have thresholds $c_1,\ldots,c_m$. Define the total observed exposure $T=\sum_{j=1}^{r}x_j+\sum_{k=1}^{m}c_k$. The likelihood is $\lambda^r e^{-\lambda T}$, because each exact record contributes one factor of $\lambda$, and every record contributes its observed loss or censoring threshold to the exponential term.

Result: If $T>0$ and $r>0$, the exponential MLE is $\widehat{\lambda}=r/T$. If $T>0$ and $r=0$, with parameter space $\lambda>0$, no MLE exists: the likelihood has supremum $1$ as $\lambda\to0^+$, but it does not attain that value.

To verify the result when $r>0$, the log-likelihood is $r\ln\lambda-\lambda T$. Its derivative is $r/\lambda-T$, which vanishes at $r/T$, and its second derivative is $-r/\lambda^2<0$. For $r=0$, the likelihood instead reduces to $e^{-\lambda T}$. It is strictly decreasing for $\lambda>0$, so it approaches $1$ as $\lambda\to0^+$ without attaining it in the stated parameter space. Reporting $0/T$ as an MLE in that case is incorrect. If a different model explicitly includes $\lambda=0$, the boundary conclusion must be reconsidered under that model's parameter space.

Worked Example: A Right-Censored Pareto Loss

Worked Example: Estimating a Pareto Shape Parameter

A hypothetical severity model has a Pareto Type I loss distribution with known scale $\theta=2$ thousand dollars and unknown shape parameter $\alpha>0$. Its density and survival function for $x\ge\theta$ are $f_X(x;\alpha)=\alpha\theta^\alpha x^{-(\alpha+1)}$ and $S_X(x;\alpha)=(\theta/x)^\alpha$. One loss is observed exactly as $4$, and another is right-censored at $5$. Find the likelihood and MLE of $\alpha$.

Method. Both records are in the model's support. The exact loss contributes

f_X(4;\alpha)=\frac{\alpha 2^\alpha}{4^{\alpha+1}} =\frac{\alpha}{4}\left(\frac12\right)^\alpha.

The censored record contributes the probability that the loss exceeds $5$:

S_X(5;\alpha)=\left(\frac25\right)^\alpha.

Thus,

L(\alpha)=\frac{\alpha}{4}\left(\frac12\right)^\alpha \left(\frac25\right)^\alpha =\frac{\alpha}{4}\left(\frac15\right)^\alpha.

The log-likelihood is $\ell(\alpha)=\ln\alpha-\ln4-\alpha\ln5$. Differentiating gives $1/\alpha-\ln5=0$, so

\widehat{\alpha}=\frac{1}{\ln5}\approx 0.6213.

The second derivative, $-1/\alpha^2$, is negative, so this is a maximum. As a numerical check, $L(\widehat{\alpha})=e^{-1}/(4\ln5)\approx0.05715$. Common mistake avoided: substituting a Pareto density at $5$ for the censored contribution would use information the record does not contain; the correct factor is $(2/5)^\alpha$.

Worked Example: When Every Record Is Censored

Worked Example: Checking Whether an MLE Exists

A hypothetical exponential loss model has $X\sim\operatorname{Exponential}(\lambda)$, $\lambda>0$. Three independent losses are right-censored at $1$, $2$, and $3$ thousand dollars; none is observed exactly. Determine the likelihood and whether an MLE exists.

Each record contributes a survival probability. The likelihood is

L(\lambda)=e^{-\lambda}e^{-2\lambda}e^{-3\lambda} =e^{-6\lambda},\qquad \lambda>0.

For every allowed $\lambda>0$, $L(\lambda)<1$, and it decreases as $\lambda$ increases. As $\lambda\to0^+$, $L(\lambda)\to1$. Therefore the likelihood has supremum $1$ but does not attain it in the parameter space $\lambda>0$; there is no MLE. The thresholds provide total exposure $T=6$, but there are no exact records, so $r=0$. Common mistake avoided: the formula $r/T$ cannot be called an MLE here. It would give zero, which is outside the specified parameter space.

Exam Checks and Common Traps

When writing a likelihood, first state what each record actually establishes. For an exact continuous loss, the record identifies a value and calls for a density. For a right-censored loss, it identifies a tail event and calls for a survival probability. Then multiply the contributions, keeping the model parameter in every factor.

Do not set a censored loss equal to its threshold. A threshold $c$ means the loss exceeded $c$, not that the loss was exactly $c$.
Do not use the CDF for a right-censored record. The information is $X>c$, so its probability is $S_X(c)=1-F_X(c)$, not $F_X(c)$.
Keep the observation mechanism in view. The simple likelihood contribution assumes censoring is fixed by design or independent of the loss for the loss parameters being estimated.
Check the parameter space before claiming an MLE. A stationary point or a limiting value at a boundary is not necessarily an attained maximum.

These distinctions matter because an incorrect contribution changes the likelihood's dependence on the parameters, and therefore can change an estimate as well as its interpretation. Correctly using censored information preserves what is known—exceedance—without inventing an exact loss value. These likelihood calculations are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify exact and censored contributions before simplifying. Loss amounts are in thousands of dollars unless stated otherwise.

A hypothetical exponential model has rate $\lambda=0.25$ per thousand dollars. One loss is exact at $3$, and another is censored at $6$. Write the likelihood contribution for these records and evaluate it. (Answer: $0.25e^{-2.25}\approx0.0264$.)
For independent exponential losses with unknown rate $\lambda>0$, exact observations are $1$ and $4$, and censoring thresholds are $2$ and $3$. Find $T$, the MLE, and the likelihood at the MLE. (Answer: $T=10$, $\widehat{\lambda}=0.2$, and $L(\widehat{\lambda})=0.04e^{-2}\approx0.0054$.)
A Pareto Type I model has known $\theta=1.5$ and unknown $\alpha>0$. One loss is exact at $3$, and one is censored at $6$. Write the likelihood and find the MLE of $\alpha$. (Answer: $L(\alpha)=\frac{\alpha}{3}(1/8)^\alpha$, and $\widehat{\alpha}=1/\ln8\approx0.4809$.)
In an exponential model with $\lambda>0$, four records are all censored at $1$, $2$, $4$, and $5$. State the likelihood and determine whether an MLE exists. (Answer: $L(\lambda)=e^{-12\lambda}$; no MLE exists in $\lambda>0$, and the likelihood has supremum $1$ as $\lambda\to0^+$.)
A continuous loss model has survival function $S_X(c;\theta)=e^{-\theta c}$, with $\theta>0$. One record is exact at $2$ with density $f_X(2;\theta)=\theta e^{-2\theta}$, and another is censored at $3$. Write the joint loss likelihood and identify the censored factor. (Answer: $L(\theta)=\theta e^{-5\theta}$; the censored factor is $S_X(3;\theta)=e^{-3\theta}$.)

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