Likelihood Contributions For Left-Censored Losses

Why Left-Censored Records Need a Different Contribution

A loss record may give an exact amount, or it may say only that the loss was below a reporting threshold. For example, a hypothetical system might record small losses as “at most $500” while retaining more precise amounts for larger losses. The threshold is not itself an observed loss: the information in that record is that the loss fell in a range.

This distinction determines the likelihood contribution. An exact observation contributes a density; a left-censored observation contributes the probability of the range consistent with what was reported. Confusing these contributions can change both the numerical likelihood and any parameter estimate derived from it.

Definition: Suppose $X$ is a continuous loss with distribution function $F_X(x;\theta)$ and density $f_X(x;\theta)$, where $\theta$ denotes the model parameter or parameters. An exact observation $X=x$ contributes $f_X(x;\theta)$ to the likelihood. A left-censored observation known only to satisfy $X\le c$, for a known reporting threshold $c$, contributes $F_X(c;\theta)=P(X\le c)$.

For the formulas in this tutorial, assume the losses are independent, the model for $X$ is specified, and the reporting threshold is known and fixed. The censoring rule is treated as noninformative about the model parameters apart from the recorded statement $X\le c$. If the mechanism that sets thresholds depends on unobserved losses or on unknown parameters, it may need to be modeled as part of the likelihood.

For a continuous model, $P(X=x)=0$ at any single value $x$. Thus, an exact observation is represented by a density, not by a point probability. By contrast, a left-censored record identifies an event with positive probability: the loss lies at or below the threshold.

Constructing the Likelihood

Suppose a sample has exact observations $x_i$ for indices in a set $E$, and left-censored observations with thresholds $c_j$ for indices in a set $C$. Under independence, the likelihood is the product of the contributions for all records:

L(\theta) = \prod_{i\in E} f_X(x_i;\theta) \prod_{j\in C} F_X(c_j;\theta).

Each exact record supplies the density at the observed amount. Each left-censored record supplies the probability of falling in the set of values compatible with the report. When maximizing the likelihood, it is often convenient to use the log-likelihood:

\ell(\theta) = \sum_{i\in E}\log f_X(x_i;\theta) + \sum_{j\in C}\log F_X(c_j;\theta).

This expression presumes the observations are independent under the loss model. If a record is exact at a value that has positive probability mass under a mixed or discrete model, its contribution is a probability mass rather than a density. A left-censored record still contributes $P(X\le c)=F_X(c)$, including any mass at or below $c$.

Why the Censored Contribution Is a Cumulative Probability

The likelihood contribution follows from the information the record actually supplies. If an observation is exact at $x$, a density describes the probability of observing a value in a very small interval around $x$. For a small interval width $\Delta$, that probability is approximately $f_X(x;\theta)\Delta$. The factor $\Delta$ does not depend on $\theta$, so the likelihood contribution for the exact record is proportional to $f_X(x;\theta)$.

A left-censored record is different: it identifies the whole event $X\le c$, not a small interval around $c$. Its probability is the integral of the density over all values up to the threshold:

P(X\le c) = \int_{-\infty}^{c} f_X(x;\theta)\,dx = F_X(c;\theta).

The lower limit can be replaced by the lower endpoint of the loss support when the distribution has a restricted support. The key point is unchanged: include the probability of the entire set consistent with the report, rather than the density at its boundary.

Worked Example: Two Exact Exponential Losses and One Left-Censored Loss

In a hypothetical portfolio, independent losses follow an exponential distribution with rate $\theta=0.2$ per dollar unit. Two losses are exact at $x_1=1$ and $x_2=3$. A third is known only to be at most $c=2$. Write and evaluate the likelihood to four decimal places.

Method. For an exponential loss with rate $\theta$, the density and distribution function are $f_X(x;\theta)=\theta e^{-\theta x}$ for $x\ge 0$ and $F_X(c;\theta)=1-e^{-\theta c}$ for $c\ge 0$. The two exact records contribute densities, and the left-censored record contributes a cumulative probability.

The exact contributions are $f_X(1;0.2)=0.2e^{-0.2}$ and $f_X(3;0.2)=0.2e^{-0.6}$. The censored contribution is $F_X(2;0.2)=1-e^{-0.4}$. Therefore,

L = (0.2e^{-0.2})(0.2e^{-0.6})(1-e^{-0.4}) = 0.2^2 e^{-0.8}(1-e^{-0.4}).

Using $e^{-0.8}\approx0.4493290$ and $1-e^{-0.4}\approx0.3296800$,

L \approx 0.04(0.4493290)(0.3296800) \approx 0.0059254.

Rounded to four decimal places, the likelihood is 0.0059. The check is that both exact records use density factors and the censored record uses a probability no greater than 1. A common mistake would be to use $f_X(2;0.2)$ for the third record, as though its loss were exactly 2.

Different Thresholds, Different CDF Factors

The reporting threshold can vary from one censored record to another. Each record contributes the probability below its own threshold. Do not replace several thresholds by a single average threshold: that would describe a different set of observations.

Worked Example: One Exact Loss and Two Reporting Thresholds

Suppose independent hypothetical losses follow an exponential distribution with rate $\theta=0.5$ per dollar unit. One loss is exact at $x=2$. The other two records say, respectively, that the losses are at most 1 and at most 2. Write the likelihood and evaluate it to four decimal places.

The exact observation contributes $f_X(2;0.5)=0.5e^{-1}$. The first censored record contributes $F_X(1;0.5)=1-e^{-0.5}$, and the second contributes $F_X(2;0.5)=1-e^{-1}$. Independence gives

L = 0.5e^{-1}(1-e^{-0.5})(1-e^{-1}).

Substituting the exponential values, $0.5e^{-1}\approx0.1839397$, $1-e^{-0.5}\approx0.3934693$, and $1-e^{-1}\approx0.6321206$. Thus,

L \approx (0.1839397)(0.3934693)(0.6321206) \approx 0.0457495.

To four decimal places, $L=\mathbf{0.0457}$. In particular, the censored record with threshold 2 contributes $F_X(2;0.5)$; it does not duplicate the exact-loss contribution $f_X(2;0.5)$. The two expressions refer to different information, even though the threshold and the exact amount have the same numerical value.

Using Left-Censored Records in a Score Equation

The likelihood can also be used to estimate unknown model parameters. The following example illustrates how a censored observation changes the score equation. The score is the derivative of the log-likelihood with respect to the parameter; a maximum-likelihood estimate commonly solves the score equation, subject to checking that it is a maximum.

Worked Example: Estimating an Exponential Rate

A hypothetical sample contains two independent exponential losses with unknown rate $\theta>0$. One is exact at $x=1$, and the other is known only to be at most $c=1$. Find the likelihood and the positive solution to the score equation, giving the estimate to four decimal places.

The exact observation contributes $\theta e^{-\theta}$. The left-censored observation contributes $F_X(1;\theta)=1-e^{-\theta}$. Therefore,

L(\theta) = \theta e^{-\theta}(1-e^{-\theta}).

Taking logarithms and differentiating gives

\ell(\theta) = \log\theta-\theta+\log(1-e^{-\theta}), \qquad \ell'(\theta) = \frac{1}{\theta}-1+\frac{1}{e^\theta-1}.

The score equation is therefore $\frac{1}{\theta}-1+\frac{1}{e^\theta-1}=0$. For example, its left-hand side is positive at $\theta=1.4$ (approximately $0.0416$) and negative at $\theta=1.5$ (approximately $-0.0461$), so the root lies between those values. Numerically solving the equation gives $\hat{\theta}\approx1.4456$.

This root is the unique positive solution: the derivative of the score is $-1/\theta^2-e^\theta/(e^\theta-1)^2$, which is negative for every $\theta>0$. Hence the score decreases strictly, and its sign change identifies its sole root. The important modeling step is the term $\log(1-e^{-\theta})$, which comes from the censored record. Treating that record as exact at 1 would produce a different likelihood and a different score equation.

Common Mistakes and a Practical Checklist

A frequent error is to treat the censoring threshold as the observed loss amount. For instance, with an exponential model of rate $\theta=1$, suppose one loss is exact at 1 and another is known only to be at most 1. The incorrect calculation uses the density for both:

\text{Wrong:}\qquad f_X(1;1)f_X(1;1)=e^{-1}e^{-1}=e^{-2}\approx0.1353.

The censored observation is not known to equal 1. Its correct contribution is $F_X(1;1)=1-e^{-1}$, so

\text{Correct:}\qquad f_X(1;1)F_X(1;1) = e^{-1}(1-e^{-1}) \approx0.2325.

The difference is not a rounding issue: the two calculations encode different observed information. A useful checklist is:

For each exact record, identify the observed amount and use its density under the stated model.
For each left-censored record, identify its threshold and use the probability $P(X\le c)$.
Use each record’s own threshold, and multiply contributions only under the stated independence assumption.
Keep the distinction between censoring and observing a loss at the threshold: a threshold is a bound, not an exact value.
If the reporting mechanism contains additional parameter information, do not omit that mechanism from the likelihood without justification.

These likelihood contributions are useful whenever a data set preserves exact larger losses but records only that some smaller losses fell below specified thresholds. Getting each contribution right is essential before maximizing the likelihood or comparing parameter values.

Check Your Understanding

For each problem, identify the contribution type before simplifying the likelihood.

A hypothetical exponential loss model has rate $\theta=0.4$. One loss is exact at 2, and another is known to be at most 1. Write and evaluate the likelihood to four decimal places. (Answer: $0.4e^{-0.8}(1-e^{-0.4})\approx0.0593$)
Independent hypothetical exponential losses have rate $\theta=0.25$. One is exact at 4; two are left-censored at thresholds 2 and 1. Write the likelihood. (Answer: $0.25e^{-1}(1-e^{-0.5})(1-e^{-0.25})$)
A hypothetical loss is uniform on $[0,8]$. One independent observation is exact at 5, and another is known to be at most 3. Find their joint likelihood contribution. (Answer: $(1/8)(3/8)=0.046875$)
For independent exponential losses with rate $\theta=0.6$, two observations are exact at 0.5 and 1.5, and a third is at most 0.75. Write the likelihood, leaving exponentials unevaluated. (Answer: $0.6^2e^{-0.6(0.5+1.5)}(1-e^{-0.6(0.75)})$)
An exponential model has unknown rate $\theta>0$. One loss is exact at 2 and another is known only to be at most 1. Write the log-likelihood score equation. (Answer: $1/\theta-2+1/(e^\theta-1)=0$)

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