Right-Censored Severity Data at Policy Limits

Why a Policy Limit Can Censor Severity Data

A policy limit caps an insurer’s payment, but it does not necessarily cap the underlying loss. If a claim reaches the limit and the amount above the limit is not recorded, the data may tell us only that the loss was at least as large as that limit. This is right-censoring: the loss is known to lie to the right of a recorded threshold, but its precise value is unknown.

This distinction matters when fitting a severity model. An exact observed loss contributes a density to the likelihood. A loss known only to exceed a policy limit contributes the probability of exceeding that limit. Treating the limit as though it were the exact loss discards the information correctly represented by the observation.

Definition: Let $X$ be the ground-up loss under a specified severity model, with density $f_X$, distribution function $F_X$, and survival function $S_X(x)=P(X>x)=1-F_X(x)$. For a policy with known limit $L$, an exact observation $x<L$ contributes $f_X(x)$ to the likelihood. If the record establishes only that $X\ge L$, the observation is right-censored at $L$ and contributes $S_X(L)$. For a continuous severity model, $P(X=L)=0$, so using $P(X\ge L)$ instead of $P(X>L)$ makes no difference.

The model here concerns ground-up losses, not insurer payments treated as exact severities. The censoring interpretation applies when a payment at the cap indicates that the underlying loss reached or exceeded the cap, while its amount above the cap is unavailable. The policy limit is a known observation threshold; it is not an additional severity parameter.

Likelihood Contributions for a Censored Sample

Suppose there are $n$ independent losses with the same parametric severity model, but possibly different known policy limits $L_i$. For each loss, record $y_i$ and an indicator $\delta_i$: $\delta_i=1$ means the loss was observed exactly at $y_i$; $\delta_i=0$ means it is known only that $X_i\ge y_i$, where $y_i=L_i$ is the censoring limit. Under independent observations and a noninformative censoring mechanism (the limits are known and do not convey additional information about the unobserved loss), the likelihood is

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

Here $\theta$ denotes the severity-model parameter or parameters. An exact observation contributes a density, not a probability at a single point; a continuous random variable has probability zero of equaling any one specified value. A censored observation contributes an event probability, namely the probability that the loss reaches the censoring threshold. The likelihood is a function of $\theta$, with the observed data held fixed.

This product assumes that the losses are independent conditional on the model parameters. If policy limits differ, use each observation’s own threshold in its survival contribution. The formula does not apply unchanged if the records are selected only when losses exceed a threshold: that sampling rule changes the likelihood. Nor does it apply if a capped payment does not establish that the ground-up loss reached the cap.

Classify each record.
Decide whether it gives an exact ground-up loss or only a lower bound at a policy limit.

Choose the correct contribution.
Use $f_X(x;\theta)$ for an exact loss and $S_X(L;\theta)$ for a loss censored at $L$.

Multiply contributions.
Under independence, multiply the contributions across all policies to form the likelihood.

Check the interpretation and units.
A density contribution has inverse-loss units; a survival probability is dimensionless. The likelihood can therefore have units, which do not prevent it from being used to compare parameter values for the same data.

Why the Contributions Have These Forms

For an exact observation at $x$, the probability of falling in a small interval $[x,x+\Delta x)$ is approximately $f_X(x;\theta)\Delta x$ when $\Delta x$ is small. The factor $\Delta x$ is fixed by the measurement resolution and does not depend on $\theta$, so the likelihood contribution for estimating $\theta$ is the density $f_X(x;\theta)$. For a censored observation at $L$, the information is not that the loss lies in a narrow interval: it is the event $X\ge L$. Its probability is $S_X(L;\theta)$. Independence makes the joint likelihood the product of these contributions.

P(x\le X<x+\Delta x)\approx f_X(x;\theta)\Delta x, \qquad P(X\ge L)=S_X(L;\theta).

The first expression explains the density contribution; the second is the exact probability contribution for right-censoring. In particular, replacing the survival probability with a density at the limit would describe a different observation: an exact loss at that value.

Worked Likelihoods

Worked Example: One Exact Loss and One Weibull-Censored Loss

A hypothetical portfolio has two independent ground-up losses. The severity model is Weibull with shape $k=2$ and scale $\eta=\$1{,}000$, so

f_X(x)=\frac{k}{\eta}\left(\frac{x}{\eta}\right)^{k-1} e^{-(x/\eta)^k}, \qquad S_X(x)=e^{-(x/\eta)^k},\quad x>0.

The first loss is observed exactly at $\$600$. The second is censored at a policy limit of $\$900$, meaning its ground-up loss is known to be at least $\$900$. Find the likelihood at the stated parameters.

Method. Use the Weibull density for the exact loss and the Weibull survival function at the censoring limit.

For the exact loss, $k/\eta=2/1000=0.002$ per dollar and $600/1000=0.6$. Therefore,

f_X(600)=0.002(0.6)^{1}e^{-(0.6)^2} =0.0012e^{-0.36}.

For the censored loss, $900/1000=0.9$, so

S_X(900)=e^{-(0.9)^2}=e^{-0.81}.

Multiplying gives

L=(0.0012e^{-0.36})(e^{-0.81}) =0.0012e^{-1.17} \approx 0.0003724.

The likelihood is approximately $0.0003724$ per dollar, since there is one density contribution. The result is checked by combining the exponents, $-0.36-0.81=-1.17$, and evaluating $0.0012e^{-1.17}\approx0.0003724403$, which rounds to $0.0003724$ at four significant figures. Common mistake avoided: the second record is not an exact loss of $\$900$, so it contributes $S_X(900)$, not $f_X(900)$.

Worked Example: Estimating an Exponential Rate with Censoring

Four independent hypothetical losses follow an exponential severity model with rate $\lambda>0$ per dollar:

f_X(x;\lambda)=\lambda e^{-\lambda x}, \qquad S_X(x;\lambda)=e^{-\lambda x},\quad x\ge0.

Two losses are observed exactly, at $\$400$ and $\$800$. Two other losses are censored at $\$1{,}000$ and $\$2{,}000$. Find the likelihood and the maximum likelihood estimate of $\lambda$.

Method. Multiply the two densities and the two survival probabilities. The exact observations supply two factors of $\lambda$:

L(\lambda) =(\lambda e^{-400\lambda})(\lambda e^{-800\lambda}) (e^{-1000\lambda})(e^{-2000\lambda}) =\lambda^2e^{-4200\lambda}.

To maximize it, use the log-likelihood:

\ell(\lambda)=\log L(\lambda)=2\log\lambda-4200\lambda, \qquad \ell'(\lambda)=\frac{2}{\lambda}-4200.

Setting the derivative equal to zero gives $2/\lambda=4200$, hence

\widehat{\lambda}=\frac{2}{4200} =\frac{1}{2100} \approx 0.0004762\text{ per dollar}.

The second derivative is $\ell''(\lambda)=-2/\lambda^2<0$ for $\lambda>0$, so this stationary point is a maximum. The numerator is the number of exact observations, not the total number of policies: censored observations add exposure through their limits but do not contribute a factor of $\lambda$. Common mistake avoided: counting all four losses in the numerator would incorrectly treat both censored records as exact.

Worked Example: Combining Exact and Censored Weibull Observations

Consider three independent hypothetical losses under a Weibull model with shape $k=2$ and scale $\eta=\$1{,}000$. Two losses are observed exactly at $\$500$ and $\$1{,}000$; the third is censored at a limit of $\$1{,}500$. Write and evaluate the likelihood at these parameter values.

Method. Use the Weibull density for each exact loss and the survival probability for the censored loss. At $\$500$, the ratio $x/\eta=0.5$, so

f_X(500)=0.002(0.5)e^{-(0.5)^2} =0.001e^{-0.25}.

At $\$1{,}000$, the ratio is $1$, giving

f_X(1000)=0.002(1)e^{-1}=0.002e^{-1}.

At the censoring limit $\$1{,}500$, the ratio is $1.5$, so

S_X(1500)=e^{-(1.5)^2}=e^{-2.25}.

Thus,

L=(0.001e^{-0.25})(0.002e^{-1})(e^{-2.25}) =2\times10^{-6}e^{-3.5} \approx 6.0395\times10^{-8}.

The likelihood has units of inverse dollars squared because there are two density factors. The exponent check is $-0.25-1-2.25=-3.5$, and $e^{-3.5}\approx0.0301974$, giving $2\times10^{-6}(0.0301974)\approx6.0395\times10^{-8}$. Common mistake avoided: the $\$1{,}500$ limit contributes a probability of exceeding that limit, not a third density factor.

Interpretation and Exam Traps

A censored observation is informative: it says the loss was not below the limit. It is simply less informative than an exact severity. In an exponential model, for example, a censored loss at $L$ contributes $e^{-\lambda L}$, which affects the likelihood and the rate estimate even though its exact amount is unknown. Ignoring censored records would discard that information.

A frequent error is to substitute the limit into the density. In the Weibull example above, doing so for the $\$900$ record would use $f_X(900)=0.0018e^{-0.81}$, whereas the correct contribution is $S_X(900)=e^{-0.81}$. These terms describe different observations: the first describes an exact loss at $\$900$, and the second describes a loss known only to have reached $\$900$. Another error is to use the survival probability at the wrong threshold; always use the policy limit associated with that censored record.

Key takeaway: For independent continuous severities observed exactly or right-censored at known policy limits, multiply a density contribution for every exact loss by a survival-probability contribution for every censored loss. The exact/censored classification determines the likelihood term.

Check Your Understanding

For each problem, identify which records are exact and which are censored before constructing the likelihood.

A hypothetical exponential severity model has rate $\lambda=0.002$ per dollar. One loss is observed exactly at $\$300$, and another is censored at $\$700$. Write and evaluate the likelihood. (Answer: $0.002e^{-2}\approx0.0002707$ per dollar.)
A hypothetical Weibull model has shape $k=1.5$ and scale $\eta=\$2{,}000$. One loss is exact at $\$1{,}000$, and another is censored at $\$2{,}000$. Write the likelihood at these parameters. (Answer: $\left[\frac{1.5}{2000}\sqrt{0.5}\,e^{-(0.5)^{1.5}}\right]e^{-1}$.)
Three independent losses follow an exponential model with rate $\lambda=0.0005$ per dollar. The first two are exact at $\$1{,}200$ and $\$3{,}000$; the third is censored at $\$5{,}000$. Give the likelihood. (Answer: $2.5\times10^{-7}e^{-4.6}\approx2.5130\times10^{-9}$ per dollar squared.)
Under an exponential model, one loss is exact at $\$600$, while two losses are censored at $\$900$ and $\$1{,}500$. Find the maximum likelihood estimate of the rate. (Answer: $\widehat{\lambda}=1/(600+900+1500)=1/3000\approx0.0003333$ per dollar.)
Under an exponential model, two losses are exact at $\$400$ and $\$900$, while two are censored at $\$800$ and $\$1{,}200$. Find the maximum likelihood estimate of the rate. (Answer: $\widehat{\lambda}=2/(400+900+800+1200)=2/3300\approx0.0006061$ per dollar.)

What You'll Learn