Build Likelihoods for Censored Policy-Limit Payments

Why a Payment at the Limit Is Not an Exact Loss

A policy limit changes what the insurer observes. If a hypothetical policy pays the loss up to a limit $u$, the recorded payment is $Z=\min(X,u)$, where $X$ is the ground-up loss. A recorded payment below $u$ identifies the loss amount under the assumptions below. A payment equal to $u$, however, can arise from any loss at least as large as $u$. It is therefore a censored observation, not an exact observation that $X=u$.

This distinction changes the likelihood. An exact observation contributes a density value; a claim reported at the limit contributes the probability of reaching or exceeding that limit. Confusing the two gives the capped claims too much information about their loss amounts and can materially distort a severity estimate. The likelihood methods here are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Assume a continuous ground-up loss $X$ has density $f_X$, distribution function $F_X$, and survival function $S_X(x)=1-F_X(x)$. Under a known policy limit $u$, the recorded payment is $Z=\min(X,u)$. An observed payment $z<u$ is an exact observation of $X=z$ and contributes $f_X(z)$ to the likelihood. An observed payment $z=u$ indicates $X\ge u$ and contributes $S_X(u)$. The likelihood is the product of these contributions for conditionally independent claims.

For a dataset with exact losses $x_1,\ldots,x_r$, each below its applicable policy limit, and censored claims with known thresholds $c_1,\ldots,c_m$, the likelihood is

L(\boldsymbol{\theta})= \prod_{i=1}^{r} f_X(x_i;\boldsymbol{\theta}) \prod_{j=1}^{m} S_X(c_j;\boldsymbol{\theta}),

where $\boldsymbol{\theta}$ denotes the severity-model parameter or parameters. A threshold $c_j$ is the ground-up loss corresponding to the cap on claim $j$. If each claim has a different policy limit, use that claim's own limit. This setup assumes claims are independent conditional on the model parameters, limits are known, and the observation mechanism does not select claims according to unaccounted-for loss information. It also assumes the dataset contains the claims being modeled, rather than omitting claims below an unmodeled reporting threshold.

For a continuous severity model, $P(X=u)=0$. Thus the event corresponding to a payment at the cap can be written $X\ge u$ or $X>u$ without changing its probability. If a severity model has discrete mass at the threshold, the precise event and its probability must instead be specified for that model. Also, if a reported amount is rounded or grouped rather than exact, its contribution is a probability over the corresponding interval, not a density at one point.

Why the Censored Contribution Is a Survival Probability

A capped observation tells us that the loss reached the cap, but not how far beyond it. To form its likelihood contribution, consider the event the observation records: $X\ge c$. The probability of that event is $1-F_X(c)$, which is the survival function. No particular loss above $c$ is singled out, so there is no density factor evaluated at the cap.

Translate the payment into information about $X$.
If $z<c$, the observation identifies $X=z$. If $z=c$, it identifies only $X\ge c$.

Write the corresponding likelihood contribution.
An exact continuous observation contributes $f_X(z;\boldsymbol{\theta})$. A cap observation contributes $P(X\ge c)=S_X(c;\boldsymbol{\theta})$.

Multiply contributions and estimate parameters.
Under conditional independence, multiply the contributions, then maximize the resulting likelihood over the allowed parameter values.

This argument explains the essential difference between censoring and treating the payment as an exact loss. A density at $c$ assigns likelihood to losses near $c$; a survival probability at $c$ assigns likelihood to all losses at or above $c$, as the observed cap requires.

Exponential Severity: A Useful Closed-Form Estimator

Suppose $X$ has an exponential distribution with scale parameter $\theta>0$, so that $f_X(x;\theta)=\theta^{-1}e^{-x/\theta}$ for $x\ge0$ and $S_X(c;\theta)=e^{-c/\theta}$. Let there be $r$ exact observations $x_i$ and $m$ censored observations at thresholds $c_j$. Define the total observed exposure

T=\sum_{i=1}^{r}x_i+\sum_{j=1}^{m}c_j.

The exact observations contribute one factor of $\theta^{-1}$ apiece, while both exact losses and censoring thresholds contribute to the exponential term. Consequently,

L(\theta)=\theta^{-r}\exp\left(-\frac{T}{\theta}\right), \qquad \ell(\theta)=-r\log\theta-\frac{T}{\theta}.

For $r>0$ and $T>0$, differentiating the log-likelihood gives

\frac{d\ell}{d\theta}=-\frac{r}{\theta}+\frac{T}{\theta^2} =\frac{T-r\theta}{\theta^2}.

The derivative is positive when $\theta<T/r$ and negative when $\theta>T/r$. The log-likelihood therefore increases up to $T/r$ and decreases after it, establishing the finite maximum at that value. This direction check also confirms that the stationary point is a maximum.

Result: For exponential severity with $r>0$ and $T>0$, the maximum likelihood estimate is $\widehat{\theta}=T/r$. The numerator includes exact losses and censoring thresholds; the denominator counts exact observations only. If $r>0$ and $T=0$, there is no finite MLE: $L(\theta)=\theta^{-r}$ is unbounded as $\theta\downarrow0$. If $r=0$ and $T>0$, $L(\theta)=e^{-T/\theta}$ increases toward 1 as $\theta$ tends to infinity, so there is no finite maximizing value of $\theta$.

Worked Example: Exponential Severity with Two Policy Limits

A hypothetical insurer observes four independent claims under policies with different limits. The ground-up losses are recorded exactly as $2$ and $4$ thousand dollars for two claims. The other two claims are reported at their limits of $5$ and $7$ thousand dollars. Assume exponential severity with scale $\theta$, no deductible, and known limits. Estimate $\theta$.

Method. Use a density contribution for each exact loss and a survival contribution for each claim at its limit. There are $r=2$ exact observations. The total $T$ includes the exact losses and the censoring thresholds:

T=2+4+5+7=18\text{ thousand dollars}.

The likelihood, omitting no parameter-dependent factors, is

L(\theta)=f_X(2;\theta)f_X(4;\theta)S_X(5;\theta)S_X(7;\theta) =\theta^{-2}\exp\left(-\frac{18}{\theta}\right).

Since $r=2>0$ and $T=18>0$, the estimator is

\widehat{\theta}=\frac{T}{r}=\frac{18}{2}=9\text{ thousand dollars}.

The common mistake is to call all four recorded payments exact and use $r=4$. That would produce the incorrect estimate $(2+4+5+7)/4=4.5$ thousand dollars. The sum happens to be unchanged, but the two capped claims do not contribute density factors, so they do not increase the exact-observation count. Check: substituting $\theta=9$ into the log-likelihood derivative gives $-2/9+18/81=0$, as required.

Known-Shape Weibull Severity

The same likelihood construction applies to other continuous severity families. For a Weibull model with known shape $k>0$ and scale $\lambda>0$, the density and survival function are

f_X(x;\lambda)=\frac{kx^{k-1}}{\lambda^k} \exp\left[-\left(\frac{x}{\lambda}\right)^k\right], \qquad S_X(c;\lambda)= \exp\left[-\left(\frac{c}{\lambda}\right)^k\right].

With $r$ exact positive losses $x_i$ and censored thresholds $c_j$, define $A=\sum_i x_i^k+\sum_j c_j^k$. Multiplying the contributions gives a likelihood proportional to $\lambda^{-kr}\exp(-A/\lambda^k)$. The exact claims produce the scale powers; the censored claims contribute survival probabilities but no density powers. When $r>0$ and $A>0$, differentiating its log gives $-kr/\lambda+kA/\lambda^{k+1}$. Setting this equal to zero yields $\widehat{\lambda}^{\,k}=A/r$, and the derivative changes from positive to negative there.

Worked Example: Weibull Scale with Censored Claims

A hypothetical portfolio is modeled with Weibull severity of known shape $k=2$ and unknown scale $\lambda$. Two losses are exact at $2$ and $4$ thousand dollars. A third claim is reported at a limit of $3$ thousand dollars, so its ground-up loss is right-censored at $3$. Estimate $\lambda$.

Method. Raise each exact loss and censoring threshold to the known shape $k=2$, then sum:

A=2^2+4^2+3^2=4+16+9=29, \qquad r=2.

The likelihood is

L(\lambda)= \left(\frac{2(2)}{\lambda^2}e^{-4/\lambda^2}\right) \left(\frac{2(4)}{\lambda^2}e^{-16/\lambda^2}\right) e^{-9/\lambda^2} =32\lambda^{-4}e^{-29/\lambda^2}.

Thus $\widehat{\lambda}^{\,2}=A/r=29/2=14.5$, and

\widehat{\lambda}=\sqrt{14.5}=3.807886\ldots \text{ thousand dollars}.

Rounded only at the final step, the estimate is $\$3{,}807.89$ when expressed in dollars, to the nearest cent. Treating the capped $3$ as an exact loss would add an extra density factor and change the power of $\lambda$; that would be the wrong likelihood. Check: the log-likelihood derivative is $-4/\lambda+58/\lambda^3$, which is zero when $\lambda^2=14.5$, the estimate obtained above.

Boundary Cases and Common Exam Traps

The exponential estimator requires attention to its conditions. In particular, $r>0$ alone is not enough to guarantee a finite estimate. If $r>0$ and $T=0$, every exact observation and censoring threshold is zero, and the likelihood is $\theta^{-r}$. As $\theta\downarrow0$, this likelihood grows without bound. Reporting $T/r=0$ as an MLE would be incorrect because the parameter space requires $\theta>0$, and no finite parameter value maximizes the likelihood.

This is a boundary pathology of the specified continuous model, not a reason to silently replace a zero observation with a positive amount. If the recorded zero represents rounding or a reporting interval, the likelihood should reflect that interval. More generally, verify the observation mechanism before treating recorded values as exact.

Counting capped claims as exact. A limit payment contributes $S_X(u)$, not $f_X(u)$. For the exponential model, it adds $u$ to $T$, but not one to $r$.
Using the wrong threshold. A cap at $u$ means the ground-up loss reached $u$. Use the threshold on the loss scale, not an unrelated recorded amount.
Ignoring policy-specific limits. If limits differ, use the applicable threshold in each survival contribution rather than one common threshold.
Assuming every recorded number is an exact loss. A payment at the cap and a rounded amount may each represent a range of possible ground-up losses.
Applying the closed-form estimate without checking conditions. For exponential severity, check $r$, $T$, and the parameter boundary before reporting $\widehat{\theta}=T/r$.

Key takeaway: A claim observed below its limit contributes a density at the exact loss; a claim observed at its limit contributes the survival probability at that threshold. In the exponential likelihood, exact losses and censoring thresholds both enter $T$, but only exact observations enter $r$. Check the conditions for a finite MLE before reporting an estimate.

Check Your Understanding

For each problem, identify exact observations and censoring thresholds before calculating. Assume independent claims and the stated severity model.

A hypothetical exponential portfolio has exact losses of $1$ and $5$ thousand dollars and claims censored at $2$ and $4$ thousand dollars. Find the MLE of $\theta$, in thousands of dollars. (Answer: $6$ thousand dollars.)
Under exponential severity, one loss is exact at $3$ thousand dollars and two claims are censored at $9$ and $12$ thousand dollars. Find the MLE of $\theta$, in thousands of dollars. (Answer: $24$ thousand dollars.)
A Weibull model has known shape $k=1$. There is one exact loss of $2$ thousand dollars and one claim censored at $1$ thousand dollars. Find the MLE of the scale $\lambda$, in thousands of dollars. (Answer: $3$ thousand dollars.)
For exponential severity, the observed data consist of one exact loss of $0$ and one claim censored at threshold $0$. Does a finite MLE exist? State the likelihood behavior that determines your answer. (Answer: No; $r=1$, $T=0$, and $L(\theta)=\theta^{-1}$ is unbounded as $\theta\downarrow0$.)
A hypothetical policy has a limit of $10$ thousand dollars. One claim is exactly $4$ thousand dollars and another is reported at the limit. Under exponential severity, find the MLE of $\theta$, in thousands of dollars, and state the censored claim's likelihood contribution. (Answer: $\widehat{\theta}=14$ thousand dollars; the contribution is $S_X(10;\theta)=e^{-10/\theta}$.)

What You'll Learn