Score Contributions for Censored Models

Why Censoring Changes the Score

In maximum likelihood estimation, an observation contributes information through the likelihood factor that matches what was actually observed. An exact loss supplies a density value. A censored loss supplies a probability for a range of possible values. Since a score is the derivative of the log-likelihood, these different likelihood factors produce different score contributions.

This distinction matters even when the same parametric severity model is used for every record. A censored record does not tell us the exact loss; substituting the censoring threshold as though it were the loss generally gives the wrong likelihood and score. The methods below concern censoring, not truncation: censoring records that a value lies in a range, whereas truncation conditions on whether a value entered the observed sample.

Definition: For a continuous loss $X$ with density $f_X(x;\theta)$, distribution function $F_X(x;\theta)$, and survival function $S_X(x;\theta)=1-F_X(x;\theta)$, a record contributes a likelihood factor for the exact value or range it reveals. Its score contribution is the derivative of the log of that factor with respect to the model parameter $\theta$. For a vector parameter, the score is the gradient of the log factor.

For the formulas in this tutorial, assume losses are independent across records, the severity model is continuous, censoring thresholds are known, and the censoring mechanism is independent of the loss and has no unknown severity parameter. Under these conditions, the severity likelihood can be formed from the observed values and censoring indicators. If censoring is informative, or its mechanism depends on the severity parameter, that mechanism may need to be included in a joint likelihood.

Likelihood Factors and Score Formulas

For one record, let $\ell(\theta)$ denote its log-likelihood contribution and $U(\theta)=\partial\ell(\theta)/\partial\theta$ its scalar score contribution. For a vector parameter, replace the derivative by the gradient $\nabla_\theta$. The following factors apply to continuous losses with fixed censoring bounds.

Record	Likelihood factor	Scalar score contribution
Exact value $x$	$f_X(x;\theta)$	$\partial_\theta\log f_X(x;\theta)$
Left-censored at $c$, so $X\le c$	$F_X(c;\theta)$	$\partial_\theta\log F_X(c;\theta)$
Right-censored at $c$, so $X>c$	$S_X(c;\theta)$	$\partial_\theta\log S_X(c;\theta)$
Interval-censored in $(a,b]$	$F_X(b;\theta)-F_X(a;\theta)$	$\partial_\theta\log\{F_X(b;\theta)-F_X(a;\theta)\}$

For an interval-censored record, the event is that the loss lies between the two bounds. In a continuous model, the choice of including either endpoint does not change its probability. If $F_X$ is differentiable with respect to $\theta$, the interval score can also be written as

U_{(a,b]}(\theta)= \frac{\partial_\theta F_X(b;\theta)-\partial_\theta F_X(a;\theta)} {F_X(b;\theta)-F_X(a;\theta)}.

This expression is a probability-weighted derivative: the numerator is the derivative of the interval probability, and division by that probability converts it to a derivative of the log probability. The denominator must be positive; otherwise the model assigns zero probability to the reported interval and its log-likelihood is not finite.

Why the Contributions Add

Let independent records have likelihood factors $L_1(\theta),\ldots,L_n(\theta)$, where each factor is selected according to that record’s observation type. Independence gives the sample likelihood as their product. Taking a logarithm turns the product into a sum, and differentiating gives the total score:

L(\theta)=\prod_{j=1}^{n}L_j(\theta), \qquad \ell(\theta)=\sum_{j=1}^{n}\log L_j(\theta), \qquad U(\theta)=\sum_{j=1}^{n}\frac{\partial}{\partial\theta}\log L_j(\theta).

This justifies calculating each record’s score separately and adding the results. The score equation sets the total score to zero when an interior maximum-likelihood estimate is sought. That equation alone does not guarantee a maximum or an interior solution; those questions depend on the model and parameter space.

Worked Examples With a Lognormal Severity Model

For the examples, use a hypothetical claim severity $X=B\exp(W)$, where $B=\$10{,}000$ and $W\sim N(\mu,\sigma^2)$. Thus $X$ is lognormal. Treat $\sigma=1$ as known and calculate scores with respect to $\mu$, evaluated at $\mu=0$. Working with $W=\ln(X/B)$ makes the derivatives straightforward. For an exact claim, the density on the dollar scale includes the Jacobian factor $1/x$; it does not depend on $\mu$, so it does not change the score. The normal values used below are rounded for display, and final scores are rounded to four decimal places.

Worked Example: Combining an Exact Claim and a Right-Censored Claim

A hypothetical portfolio has two independent records under the stated lognormal model. The first claim is observed exactly at $X=B\exp(0.5)$. The second is right-censored at $X=B\exp(0.25)$, so its loss is known to exceed that threshold. Find each score contribution and the total score at $\mu=0$.

Method. For $W\sim N(\mu,\sigma^2)$, the exact-record density on the $W$ scale has log-likelihood, apart from constants, $-\ln(\sigma)-(w-\mu)^2/(2\sigma^2)$. Its score is $(w-\mu)/\sigma^2$. With $\sigma=1$, the exact record contributes $(0.5-0)/1=0.5000$.

For the right-censored record, define $z=(0.25-\mu)/1=0.25$. Its likelihood factor is $S_W(0.25)=1-\Phi(0.25)$, where $\Phi$ is the standard normal distribution function. The score is $\phi(z)/S_W(0.25)$, with $\phi$ the standard normal density. Using $\phi(0.25)=0.386668$ and $\Phi(0.25)=0.598706$, the survival probability is $1-0.598706=0.401294$. The contribution is $0.386668/0.401294=0.9636$.

By independence, the total score is $0.5000+0.9636=\mathbf{1.4636}$. The positive right-censoring score indicates that, at the specified parameter value, increasing $\mu$ increases the likelihood of a loss exceeding the threshold. The result is a score, not an estimate of $\mu$.

Worked Example: Adding a Left-Censored Record

A second hypothetical sample contains one claim known only to satisfy $W\le -1$, and one exact claim with $W=0.25$. Under the same model and parameter values, calculate both contributions and their sum.

For the left-censored record, the likelihood factor is $\Phi(z)$, where $z=(-1-\mu)/1=-1$ at $\mu=0$. Differentiating its log with respect to $\mu$ gives $-\phi(z)/\Phi(z)$. Using $\phi(-1)=0.241971$ and $\Phi(-1)=0.158655$, its score is $-0.241971/0.158655=-1.5251$.

The exact record contributes $(0.25-0)/1=0.2500$. Therefore the total is $-1.5251+0.2500=\mathbf{-1.2751}$. The left-censored record pulls the score downward because it reports a value below the threshold, rather than an exact loss equal to that threshold.

Worked Example: Scoring an Interval-Censored Claim

A hypothetical claim is known to have $W\in(-0.5,1]$, but its exact value is unavailable. Calculate its score contribution at $\mu=0$, with $\sigma=1$.

The likelihood factor is $\Phi(1-\mu)-\Phi(-0.5-\mu)$. At $\mu=0$, the interval probability is $\Phi(1)-\Phi(-0.5)=0.841345-0.308538=0.532807$.

Differentiating the interval probability with respect to $\mu$ gives $-\phi(1-\mu)+\phi(-0.5-\mu)$. At $\mu=0$, this is $-0.241971+0.352065=0.110094$. Dividing by the interval probability gives the score $0.110094/0.532807=\mathbf{0.2066}$. The positive sign reflects that a small increase in $\mu$ increases the probability of this particular interval at the specified parameter value.

Common Exam Trap: Using the Wrong Tail

The censoring label describes what is known about the unobserved loss. A right-censored record at $c$ means $X>c$, so its likelihood factor is $S_X(c;\theta)$, not $F_X(c;\theta)$. In Worked Example 1, using the left-tail probability instead would produce the incorrect score

\frac{\partial}{\partial\mu}\log\Phi(0.25-\mu) = -\frac{\phi(0.25)}{\Phi(0.25)} = -\frac{0.386668}{0.598706} \approx -0.6457,

rather than the correct right-censored score $+0.9636$. The error changes both the magnitude and the sign. A reliable check is to write the observed event in words first: “below the threshold” calls for a distribution probability, “above the threshold” calls for a survival probability, and “between two bounds” calls for a difference of distribution probabilities.

Key takeaway: Different observation types require different likelihood factors, and their scores are derivatives of the log of those factors. For independent records, add the record-level scores. Do not replace a censored value by its threshold or condition on inclusion as though censoring were truncation.

Check Your Understanding

For each question, use a lognormal severity model $X=B\exp(W)$ with $W\sim N(\mu,\sigma^2)$, known $\sigma$, and evaluate the score with respect to $\mu$ at the stated parameter value. Round final scores to four decimal places.

An exact claim has $W=0.2$, with $\mu=0$ and $\sigma=0.5$. Find its score contribution. (Answer: $0.8000$)
A record is right-censored at $W=0.5$, with $\mu=0$ and $\sigma=1$. Use $\phi(0.5)=0.352065$ and $\Phi(0.5)=0.691462$. Find its score contribution. (Answer: $1.1411$)
A record is left-censored at $W=0$, with $\mu=0.25$ and $\sigma=0.5$. Use $\phi(-0.5)=0.352065$ and $\Phi(-0.5)=0.308538$. Find its score contribution. (Answer: $-2.2822$)
A claim is interval-censored in $(0,1]$, with $\mu=0$ and $\sigma=1$. Use $\phi(0)=0.398942$, $\phi(1)=0.241971$, $\Phi(0)=0.500000$, and $\Phi(1)=0.841345$. Find its score contribution. (Answer: $0.4599$)
An independent sample has an exact record $W=0.3$ and a right-censored record at $W=0.1$, with $\mu=0$ and $\sigma=1$. Use $\phi(0.1)=0.396953$ and $\Phi(0.1)=0.539828$. Find the total score. (Answer: $1.1626$)

Record	Likelihood factor	Scalar score contribution
Exact value \(x\)	\(f_X(x;\theta)\)	\(\partial_\theta\log f_X(x;\theta)\)
Left-censored at \(c\), so \(X\le c\)	\(F_X(c;\theta)\)	\(\partial_\theta\log F_X(c;\theta)\)
Right-censored at \(c\), so \(X>c\)	\(S_X(c;\theta)\)	\(\partial_\theta\log S_X(c;\theta)\)
Interval-censored in \((a,b]\)	\(F_X(b;\theta)-F_X(a;\theta)\)	\(\partial_\theta\log\{F_X(b;\theta)-F_X(a;\theta)\}\)

What You'll Learn