Censoring Thresholds When Loss Distributions Have Atoms

Why Equality at a Threshold Matters

A censoring threshold divides possible losses into categories that are observed differently. If a loss equals the threshold and the loss distribution is continuous, that equality has probability zero. With a discrete or mixed distribution, however, the threshold can have positive probability. A loss exactly at the threshold can then be a real, non-negligible outcome, and the reporting rule determines whether it is recorded as an exact value or grouped with censored losses.

The threshold rule is part of the observation process, not a convention that can be inferred from the distribution alone. For example, one system might report an exact loss equal to $c$ and censor only losses greater than $c$. Another might record every loss at least $c$ as “$c$ or more.” Those rules produce different likelihood contributions whenever $P(X=c)>0$.

Definition: For a loss $X$ with CDF $F_X$, define the left limit $F_X(c-)=P(X<c)$. The probability of an atom at $c$ is $P(X=c)=F_X(c)-F_X(c-)$. Consequently, $$ P(X\ge c)=1-F_X(c-), \qquad P(X>c)=1-F_X(c). $$ A report that groups losses $X\ge c$ contributes $P(X\ge c)$; a report that groups losses $X>c$ contributes $P(X>c)$.

The two expressions differ by exactly the threshold atom: $$ P(X\ge c)-P(X>c)=P(X=c). $$ For an absolutely continuous distribution, $P(X=c)=0$, so the expressions agree. That familiar continuous-model simplification must not be carried over automatically to a discrete or mixed model.

Translate the Reporting Rule Before Writing the Likelihood

For each record, first identify what the observer actually reports. An exact observation $X=x$ contributes a probability mass if $x$ is a discrete atom. If the model has a continuous component, an exact observation in that component contributes its density, not the probability of the singleton event $X=x$. A censored record contributes the probability of the entire set of losses consistent with the report.

For example, if the report says only “at least $c$,” the compatible set is $[c,\infty)$, including equality. If it says “greater than $c$,” the compatible set is $(c,\infty)$, excluding equality. The likelihood uses the probability of the specified set under the loss model.

For independent losses $X_1,\ldots,X_n$, the likelihood is the product of the individual contributions, provided the observation mechanisms are fixed or independent of the loss-model parameters. If censoring thresholds are random, the usual likelihood for the loss parameters also requires an appropriate independent, noninformative censoring assumption, or an explicit model for the censoring mechanism.

Write the equality rule.
Determine whether the censoring category is $X\ge c$ or $X>c$, and whether an exact report at $c$ is possible.

Identify the distributional type.
Check whether $c$ has a point mass. For a mixed model, distinguish the atom from the continuous part.

Assign a contribution to each record.
Use a point probability for a discrete exact value, a density for a continuous exact value, and a set probability for a censored report.

Multiply the contributions.
Form the likelihood using the specified parameter values or leave it as a function of the model parameters.

Worked Likelihoods at a Threshold

Worked Example: A Discrete Loss Censored at or Above the Threshold

A hypothetical warranty loss $X$ has three possible values: $40$, $100$, and $180$ dollars, with probabilities $\theta$, $0.30$, and $0.70-\theta$, respectively. The reporting system records exact losses below $100$ and records “$100$ or more” for every loss $X\ge100$. Two independent claims are observed: one exact loss of $40$, and one “$100$ or more” report. Find the likelihood and evaluate it at $\theta=0.20$.

Method and boundary rule. The threshold category includes $X=100$. Its probability is the sum of the probabilities at $100$ and $180$: $$ P(X\ge100)=0.30+(0.70-\theta)=1-\theta. $$ The exact observation $X=40$ contributes its point probability $\theta$. Thus the likelihood is $$ L(\theta)=\theta(1-\theta). $$

At $\theta=0.20$, the exact-loss contribution is $0.20$, and the threshold-category contribution is $1-0.20=0.80$. Therefore, $$ L(0.20)=0.20(0.80)=0.16. $$ The probabilities sum to $0.20+0.30+0.50=1.00$, and the threshold category is $0.30+0.50=0.80$, as required.

Common mistake avoided: Using only $P(X>100)=0.50$ would omit the mass $0.30$ at $100$. That would describe a report of “greater than $100$,” not the stated report of “$100$ or more.”

Worked Example: An Exact Threshold Atom and Strict Right Censoring

Consider a hypothetical mixed loss model with probability $p$ of a loss exactly equal to $50$ dollars; with probability $1-p$, the loss follows an exponential distribution with rate $\lambda$ per dollar. The reporting rule records exact losses at or below $50$ and reports only “greater than $50$” for larger losses. The independent records are an exact loss of $20$, an exact loss of $50$, and a censored loss greater than $50$. Find the likelihood and evaluate it at $p=0.25$, $\lambda=0.02$.

Exact loss of $20$. This value comes from the exponential component, whose density is $\lambda e^{-\lambda x}$ for $x>0$. Its contribution is $$ (1-p)\lambda e^{-20\lambda}. $$

Exact loss of $50$. The exponential component assigns probability zero to any single exact value. The observed threshold value therefore contributes the atom $p$, not the exponential density evaluated at $50$.

Censored loss greater than $50$. The atom at $50$ is excluded by the strict inequality. The exponential component must exceed $50$, giving contribution $$ (1-p)e^{-50\lambda}. $$

Multiplying the three contributions gives $$ L(p,\lambda)=\big((1-p)\lambda e^{-20\lambda}\big)p\big((1-p)e^{-50\lambda}\big) =p(1-p)^2\lambda e^{-70\lambda}. $$ At the stated parameter values, $1-p=0.75$, $70\lambda=1.4$, and $e^{-1.4}\approx0.246597$. Hence $$ L(0.25,0.02)=0.25(0.75)^2(0.02)e^{-1.4} \approx0.00069355. $$ Rounded to six decimal places, the likelihood is $0.000694$. The exponent $70\lambda$ is the sum of the two exponential loss amounts, $20\lambda+50\lambda$. The threshold atom appears only in the exact-$50$ contribution.

Common mistake avoided: Treating the censored report as $X\ge50$ would add the atom $p$, even though the stated report is strictly greater than $50$ and the exact-$50$ outcome is recorded separately.

Worked Example: A Mixed Threshold Category That Includes an Atom

A hypothetical loss $X$ has probability $q$ of being exactly $100$ dollars. With probability $1-q$, it is uniformly distributed on $(0,200)$. A system records exact losses below $100$, but reports “$100$ or more” for $X\ge100$. The records are an exact loss of $60$ and one threshold report. Find the likelihood and evaluate it at $q=0.20$.

Exact loss of $60$. This observation must come from the uniform component. Its density on $(0,200)$ is $1/200$, so the contribution is $$ (1-q)\frac{1}{200}. $$

Threshold report. The atom at $100$ is included. The uniform component has probability $100/200=0.5$ of lying in $[100,200)$; the endpoint $200$ has probability zero under the uniform model. The total contribution is therefore $$ P(X\ge100)=q+(1-q)(0.5). $$

The likelihood is $$ L(q)=\frac{1-q}{200}\left(q+\frac{1-q}{2}\right). $$ At $q=0.20$, the exact-loss contribution is $0.80/200=0.004$. The threshold contribution is $0.20+0.80(0.5)=0.60$. Thus $L(0.20)=0.004(0.60)=0.0024$. Each component probability is valid, and the threshold report includes the full $0.20$ atom at $100$.

Common mistake avoided: Replacing the threshold contribution by the uniform tail probability $0.5$ ignores that the model also places probability $q$ at the threshold.

Why the CDF Endpoint Must Be Chosen Carefully

The formulas for $P(X\ge c)$ and $P(X>c)$ follow directly from complements. The complement of $\{X\ge c\}$ is $\{X<c\}$, whose probability is $F_X(c-)$. The complement of $\{X>c\}$ is $\{X\le c\}$, whose probability is $F_X(c)$. Therefore, $$ P(X\ge c)=1-F_X(c-), \qquad P(X>c)=1-F_X(c). $$ Subtracting the second expression from the first gives $F_X(c)-F_X(c-)=P(X=c)$. This shows exactly why the endpoint choice matters in the presence of an atom.

When the data record only a threshold category, do not count that record as an exact observation at $c$ merely because the recorded number displayed is $c$. A displayed threshold can be a label for a whole interval of possible losses. Conversely, if the protocol reports exact values at $c$, the atom may appear as an exact-value contribution, while a separate strict-censoring category excludes it.

The distinction is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus: likelihoods must reflect both the loss model and the way incomplete observations are reported. Before simplifying a survival probability or assigning a density, make the support of each observation explicit.

Common Exam Traps

Using $1-F_X(c)$ for every threshold report. This is correct for $X>c$, but not for $X\ge c$ when there is an atom at $c$.
Using $F_X(c)$ to calculate the mass strictly below $c$. The correct quantity is $F_X(c-)$; $F_X(c)$ includes the threshold atom.
Using a density for a discrete exact observation. A discrete exact value contributes its probability mass.
Using a point probability for an exact observation from a continuous component. A continuous exact-value likelihood contribution is a density, with units determined by the measurement scale.
Ignoring the observation rule. Whether equality is exact or censored must come from the stated protocol. It cannot be decided from the threshold number alone.

Key takeaway: At a threshold $c$ with positive mass, $P(X\ge c)$ and $P(X>c)$ are different. Determine whether equality is included in the reported censoring category, then use the matching probability—or the matching exact-value contribution—for every record.

Check Your Understanding

For each problem, identify the relevant threshold category before forming or evaluating its contribution.

A discrete loss has probabilities $0.25$ at $30$, $0.35$ at $80$, and $0.40$ at $150$. A record says “$80$ or more.” What probability contributes? (Answer: $0.75$)
A discrete model has $P(X=25)=0.10$, $P(X=60)=0.50$, and $P(X=120)=0.40$. A record says “greater than $60$.” What probability contributes? (Answer: $0.40$)
A mixed model has mass $0.30$ at $40$ and, otherwise, an exponential loss with rate $0.05$ per dollar. A report says $X\ge40$. What probability contributes? (Answer: $0.30+0.70e^{-2}\approx0.3947$)
A model assigns mass $q$ at $90$ and probability $1-q$ to a uniform distribution on $(0,180)$. It records “$90$ or more.” Write the likelihood contribution for one such report and evaluate at $q=0.40$. (Answer: $q+(1-q)/2=0.7000$)
A loss has mass $0.15$ at $25$; its remaining probability follows an exponential distribution with rate $0.04$ per dollar. Exact values at or below $25$ are recorded, and larger losses are reported as “greater than $25$.” What contribution corresponds to that censored report? (Answer: $0.85e^{-1}\approx0.3127$)

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