Likelihood Contributions For Interval-Censored Losses

Why Interval Reports Require a Probability Contribution

A loss record does not always give the exact amount. A hypothetical reporting system might record that a claim is between two monetary thresholds, rather than preserving its precise value. The information in such a record is that the loss fell in an interval. Its likelihood contribution must therefore be the probability of that interval under the assumed loss model—not the density at a guessed point inside it.

This distinction matters when estimating or comparing loss models. An exact observation and an interval-censored observation contain different information, and they enter a likelihood in different ways. The formulas below assume continuous losses, independent observations, and reporting limits that are known and do not depend on the unobserved loss beyond determining the reported interval. There is no selection or truncation of the sampled losses.

Definition: Suppose $X$ is a continuous loss with cumulative distribution function $F_X(x;\theta)$ and density $f_X(x;\theta)$, where $\theta$ denotes the model parameter or parameter vector. If the record says only that $X$ lies between known limits $a$ and $b$, with $a<b$, its likelihood contribution is $$ P(a<X\le b)=F_X(b;\theta)-F_X(a;\theta). $$ For a continuous distribution, the choice of including or excluding either endpoint does not change this probability.

Deriving the Interval Contribution

The derivation follows from the definition of a cumulative distribution function. The event that a loss is greater than $a$ and no greater than $b$ is the part of the event $X\le b$ that remains after removing $X\le a$. Therefore,

P(a<X\le b)=P(X\le b)-P(X\le a)=F_X(b;\theta)-F_X(a;\theta).

This is a probability, so the contribution is between zero and one. If $S_X(x;\theta)=1-F_X(x;\theta)$, the same contribution can also be written $S_X(a;\theta)-S_X(b;\theta)$. Either expression is correct; choose the one that is easiest to evaluate accurately.

An exact observation $X=x$ is handled differently. For a continuous distribution, $P(X=x)=0$, so it would be incorrect to use that probability as a likelihood contribution. Instead, the contribution is the density $f_X(x;\theta)$. One way to see why is to consider a small interval of width $\Delta$ around $x$: its probability divided by its width approaches the density as the width shrinks.

f_X(x;\theta)=\lim_{\Delta\to 0^+} \frac{P(x<X\le x+\Delta)}{\Delta}.

Consequently, a density value is not itself a probability and can exceed one. It has units reciprocal to those used for the loss, while an interval probability is dimensionless. Likelihoods involving exact observations inherit these density units; that does not prevent comparing parameter values within the same model and measurement scale.

Likelihood rule: For independent losses with known reporting forms, multiply one contribution per record. Use $F_X(b;\theta)-F_X(a;\theta)$ for an interval record $(a,b]$, and $f_X(x;\theta)$ for an exact record $x$. The resulting likelihood is a function of $\theta$.

Worked Applications

Worked Example: An Interval Under an Exponential Loss Model

A hypothetical portfolio has individual losses $X$, measured in thousands of dollars, with an exponential distribution of rate $\lambda=0.2$ per thousand dollars. One loss is reported only as greater than 1 and no greater than 4 thousand dollars. Find its likelihood contribution.

Method. For an exponential loss with rate $\lambda$, the cumulative distribution function for $x\ge 0$ is $F_X(x)=1-e^{-\lambda x}$. The interval contribution is the difference between the cumulative probabilities at the two limits:

F_X(4)-F_X(1) =(1-e^{-0.2(4)})-(1-e^{-0.2(1)}) =e^{-0.2}-e^{-0.8}.

Evaluate the exponentials: $e^{-0.2}=0.8187308$ and $e^{-0.8}=0.4493290$. Thus the contribution is $0.8187308-0.4493290=0.3694018$, or 0.3694 to four decimal places. The result is a probability and is between zero and one.

Common mistake avoided. The interval midpoint is 2.5 thousand dollars, but the record does not say that the loss equalled 2.5. Substituting that midpoint into the density would give $0.2e^{-0.2(2.5)}=0.1213$, a density rather than the probability of the reported interval.

Worked Example: Combining an Interval and an Exact Loss

Two independent hypothetical losses, measured in thousands of dollars, follow a Uniform distribution on $(0,12)$. The first is recorded in the interval $[3,9]$; the second is observed exactly as 5. Find the joint likelihood contribution for these records.

Method. Under this model, the density is $1/12$ throughout $(0,12)$, and the cumulative distribution function is $F_X(x)=x/12$ on that range. The first record contributes

F_X(9)-F_X(3)=\frac{9}{12}-\frac{3}{12} =\frac{6}{12}=0.5.

The second record is exact, so it contributes the density at 5, not the probability of an exact value:

f_X(5)=\frac{1}{12}=0.0833333.

Independence means the joint likelihood contribution is the product: $0.5(1/12)=1/24=0.0416667$. The joint contribution is therefore 0.0417 to four decimal places. It has the units of a density per thousand dollars because one of its factors is a density.

Common mistake avoided. Treating both records as intervals, or both as exact values, would use the wrong contribution for at least one observation. The first record supplies a range probability; the second supplies a density.

Worked Example: A Likelihood with Two Intervals and One Exact Observation

Three independent hypothetical losses, measured in thousands of dollars, follow an exponential distribution with rate $\lambda=0.5$ per thousand dollars. The records are $(1,3]$, $(4,6]$, and an exact loss of 2.5. Calculate the likelihood at this specified rate.

Method. The first interval contributes $e^{-0.5(1)}-e^{-0.5(3)}=e^{-0.5}-e^{-1.5}$. Using $e^{-0.5}=0.6065307$ and $e^{-1.5}=0.2231302$, this is $0.3834005$.

The second interval contributes $e^{-0.5(4)}-e^{-0.5(6)}=e^{-2}-e^{-3}$. Using $e^{-2}=0.1353353$ and $e^{-3}=0.0497871$, this is $0.0855482$.

The exact observation contributes the exponential density at 2.5:

f_X(2.5)=0.5e^{-0.5(2.5)} =0.5e^{-1.25} =0.5(0.2865048)=0.1432524.

By independence, multiply the three contributions: $0.3834005(0.0855482)(0.1432524)=0.0046986$, approximately 0.004699 to six significant figures. Each interval factor is a probability; the exact-record factor is a density.

Common mistake avoided. Replacing either interval by its midpoint and evaluating a density there changes the observed data and the likelihood. For example, the first interval contributes $0.3834005$, not the density $0.5e^{-0.5(2)}=0.1839397$ at its midpoint.

Assumptions, Endpoints, and Interpretation

The interval formula is appropriate when the reporting process establishes that the loss lies within known limits and does not provide additional information about its value within those limits. A fixed reporting band is a familiar case. The formula does not by itself adjust for a sample that includes only losses meeting a selection threshold; that is a truncation issue, not merely an interval contribution.

For continuous loss models, endpoint conventions do not affect the result because an individual endpoint has probability zero. If losses instead have a discrete distribution, endpoint conventions do matter. For example, for an integer-valued loss and an interval $a\le X\le b$, the probability is $F_X(b)-F_X(a^-)$, where $F_X(a^-)=P(X<a)$. Using $F_X(b)-F_X(a)$ would omit the probability at $a$.

The likelihood is the product of contributions only when the observations are independent under the model. If independent records have different intervals, each interval can have its own limits but uses the same model parameter $\theta$. Taking logarithms changes the product into a sum, which is often convenient for estimation:

\ell(\theta)=\sum_{j\in I} \log\!\left(F_X(b_j;\theta)-F_X(a_j;\theta)\right) +\sum_{k\in E}\log f_X(x_k;\theta),

where $I$ indexes interval-censored records and $E$ indexes exact records. Each term must correspond to the record as observed. In particular, do not insert a midpoint as though it were exact, and do not replace an exact density with an interval probability unless the data were actually reported as an interval.

A useful numerical check is that every interval contribution is nonnegative and no greater than one. If $F_X(b)<F_X(a)$, the limits or subtraction order have been mishandled. For a valid density model, an exact contribution must be nonnegative, though it need not be at most one. These checks catch common calculation errors before the contributions are multiplied.

Key takeaway: An interval-censored loss contributes the probability mass between its reporting limits. An exact continuous loss contributes a density at its observed value. Keep those contributions distinct, then combine them according to the independence assumption.

Check Your Understanding

For each problem, identify the contribution that matches the record type. Round probabilities to four decimal places when needed.

A loss in thousands of dollars follows an exponential distribution with rate $\lambda=0.1$ per thousand dollars. It is reported as $(2,7]$. Find its contribution. (Answer: $e^{-0.2}-e^{-0.7}=0.3221$)
A loss follows a Uniform distribution on $(0,20)$. It is recorded as lying between 4 and 10. Find the interval contribution. (Answer: $6/20=0.3000$)
A loss follows an exponential distribution with rate $\lambda=0.4$ per unit. The exact observed value is 3 units. Find its density contribution. (Answer: $0.4e^{-1.2}=0.1205$ per unit)
Independent losses follow an exponential distribution with rate $\lambda=0.25$ per unit. One record is $(1,5]$, and the other is an exact value of 2. Find their joint likelihood contribution. (Answer: $(e^{-0.25}-e^{-1.25})(0.25e^{-0.5})\approx0.0746$)
Two independent observations follow $X\sim N(0,1)$. One is known to lie in $(-1,1]$, and the other is exactly 0. Find the joint contribution, using $P(-1<X\le1)=0.6827$ and $f_X(0)=0.3989$. (Answer: $0.6827(0.3989)\approx0.2724$)

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