Why Discrete Losses Need Different Likelihood Contributions
A continuous loss observed exactly at \(x\) contributes a density to a likelihood. A discrete loss observed exactly at \(x\), by contrast, contributes the probability mass at \(x\). This distinction matters whenever losses take values on a grid, such as fixed benefit amounts or integer-valued claim counts. Censoring and truncation also require probabilities of sets of possible values, not continuous tail integrals.
The central task is to translate each record into the event it tells us about the loss. An exact record identifies one value. A censored record identifies a range or set of values. A truncated sample includes only losses meeting a selection rule, so its likelihood contributions must account for that selection.
A point mass is not a density evaluated at a point. In particular, a discrete model has no density contribution \(f_X(x)\) for an exact observation. For a censored record, first identify the compatible set \(B\), then add the masses for all values in that set. The record's wording determines whether the threshold itself belongs to \(B\).
Turning Records Into Probabilities
For a sample of independent losses under a specified observation mechanism, multiply the contribution from each record. An exact observation \(X=x\) contributes \(p_\theta(x)\). A left-censored report \(X\leq c\) contributes \(F_\theta(c)\); a right-censored report \(X>c\) contributes \(1-F_\theta(c)\). An interval-censored report \(a\leq X\leq b\) contributes the sum of the masses at all support points in that interval.
In discrete data, the event at a threshold must be read literally. “Greater than 100” excludes 100; “at least 100” includes it. These events have different probabilities whenever there is positive mass at 100. A reported censoring threshold alone does not settle the issue: use the event described by the recording rule.
| Record says the loss is... | Compatible event | Likelihood contribution |
|---|---|---|
| Exactly \(x\) | \(X=x\) | \(p_\theta(x)\) |
| At most \(c\) | \(X\leq c\) | \(F_\theta(c)\) |
| Strictly greater than \(c\) | \(X>c\) | \(1-F_\theta(c)\) |
| At least \(c\) | \(X\geq c\) | \(1-F_\theta(c-)\) |
| Between \(a\) and \(b\), inclusive | \(a\leq X\leq b\) | \(\sum_{a\leq x\leq b}p_\theta(x)\) |
Worked Example: Exact and Strictly Right-Censored Losses
A hypothetical product has losses supported at \(\$100\), \(\$200\), and \(\$400\), with probabilities \(\theta\), \(2\theta\), and \(1-3\theta\), respectively, where \(0<\theta<1/3\). In an independent sample, one loss is observed exactly as \(\$100\), one is known to be strictly greater than \(\$100\), and one is observed exactly as \(\$200\). Find the likelihood at \(\theta=0.20\).
Method. Use the mass for each exact loss and sum the masses for the values compatible with the censored record. The wording “strictly greater” excludes \(\$100\).
The exact \(\$100\) record contributes \(p_\theta(100)=\theta\). The strictly greater-than-\(\$100\) record is compatible with \(\$200\) and \(\$400\), so it contributes \(2\theta+(1-3\theta)=1-\theta\). The exact \(\$200\) record contributes \(2\theta\). Therefore,
At \(\theta=0.20\), the three contributions are \(0.20\), \(0.80\), and \(0.40\). Their product is \(0.20(0.80)(0.40)=0.064\). The likelihood value is \(0.064\).
Check. The model probabilities sum to \(0.20+0.40+0.40=1\), and the censored contribution \(0.80\) is the sum of the two masses above \(\$100\). A common mistake is to use \(P(X\geq100)=1\) for the strictly greater-than-\(\$100\) record; that would give \(0.20(1)(0.40)=0.080\), which includes the excluded \(\$100\) outcome.
Truncation: Condition on Being Included
Censoring means that a sampled loss is only partly observed. Truncation means that losses outside a selection region are not included in the sample at all. For a fixed-size sample drawn from losses satisfying \(X\in A\), the probability of an observed event \(X\in B\), where \(B\subseteq A\), is conditional on inclusion:
For an exact observed value \(x\in A\), take \(B=\{x\}\), giving \(p_\theta(x)/P_\theta(X\in A)\). For a censored record, sum the masses over the values compatible with that record, then divide by the selection probability. The denominator is parameter-dependent whenever the model probabilities of selected losses depend on \(\theta\).
Worked Example: A Truncated Sample With a Censored Record
A hypothetical severity model has support \(\$1{,}000,\$2{,}000,\$3{,}000,\$4{,}000\), with respective probabilities \(\theta,\theta,2\theta,1-4\theta\), where \(0<\theta<1/4\). Losses are included only when \(X\geq\$2{,}000\). In a fixed-size sample from this selected population, one loss is observed exactly as \(\$2{,}000\), and another is reported as strictly greater than \(\$2{,}000\). Calculate the joint likelihood contribution at \(\theta=0.10\).
Step 1: Find the inclusion probability. The included values are \(\$2{,}000,\$3{,}000,\$4{,}000\), so \(P_\theta(X\geq\$2{,}000)=\theta+2\theta+(1-4\theta)=1-\theta\). At \(\theta=0.10\), this is \(0.90\).
Step 2: Condition the exact record on inclusion. Its contribution is \(\theta/(1-\theta)=0.10/0.90=1/9\).
Step 3: Condition the censored record on inclusion. Strictly greater than \(\$2{,}000\) means \(\$3{,}000\) or \(\$4{,}000\), with total probability \(2\theta+1-4\theta=1-2\theta\). Its conditional contribution is \((1-2\theta)/(1-\theta)=0.80/0.90=8/9\).
Step 4: Multiply the record contributions. Independence within the selected sample gives \((1/9)(8/9)=8/81\), or approximately \(0.098765\). The contributions are valid conditional probabilities: \(1/9\) for the exact record and \(8/9\) for the censored event, which together partition the selected sample space.
Method and trap avoided. This is a conditional likelihood for a sample drawn from the truncated population. Multiplying the unconditional masses \(0.10\) and \(0.80\) would ignore selection and produce \(0.080\), not \(8/81\).
Combining Discrete Record Types
A data set can contain exact and censored records together, and censoring thresholds can differ by record. The same rule applies to every record: identify its compatible set, calculate the probability of that set, and multiply the resulting contributions. If the sample is truncated, calculate each contribution conditional on the applicable inclusion rule as well.
Check that the model probabilities are nonnegative and sum to one for the permitted parameter values.
State whether the event is a singleton, a lower tail, an upper tail, or an interval, including its endpoints.
Use a point mass for an exact observation and a sum of point masses for a censored observation.
Divide by the probability of the inclusion region for each record sampled from the selected population.
Multiply record contributions for independent observations; parameter-free factors may be omitted when writing a likelihood proportional to the full likelihood.
Worked Example: Exact, Left-Censored, and Right-Censored Binomial Counts
A hypothetical claim count \(X\) has a binomial distribution with three trials and success probability \(q\), so \(X\sim\operatorname{Binomial}(3,q)\). Three independent records are: an exact count of 1, a report that the count is at most 1, and a report that the count is strictly greater than 1. Evaluate their joint likelihood contribution at \(q=0.40\).
Method. Calculate the exact mass and the two cumulative-event probabilities from the binomial masses. For \(k=0,1,2,3\), \(p_q(k)=\binom{3}{k}q^k(1-q)^{3-k}\). At \(q=0.40\), the masses are \(p(0)=0.6^3=0.216\), \(p(1)=3(0.4)(0.6)^2=0.432\), \(p(2)=3(0.4)^2(0.6)=0.288\), and \(p(3)=0.4^3=0.064\). They sum to \(1.000\).
The exact count of 1 contributes \(0.432\). The at-most-1 record contributes \(p(0)+p(1)=0.216+0.432=0.648\). The strictly-greater-than-1 record contributes \(p(2)+p(3)=0.288+0.064=0.352\). Thus the likelihood contribution is
The likelihood value is approximately \(0.098537\) (rounded to six decimal places). As a check, the two censored events partition the support: \(0.648+0.352=1\). The exact observation is a separate record, so its mass is multiplied by both event probabilities.
Common mistake avoided. A censored record is not an additional exact observation at the threshold. Treating “at most 1” as \(X=1\), for example, would replace \(0.648\) by \(0.432\) and discard the possible count of 0.
Common Errors and a Reliable Check
- Using a density for an exact discrete loss. Use \(p_\theta(x)\), the probability mass, not a density or an infinitesimal interval approximation.
- Using only one mass for a censored record. Add the masses of every support point consistent with the report.
- Including or excluding the threshold incorrectly. “Greater than \(c\)” and “at least \(c\)” differ by \(P_\theta(X=c)\). Translate the wording into an event before calculating.
- Forgetting truncation conditioning. When the sample consists of losses selected by \(X\in A\), divide the compatible-event probability by \(P_\theta(X\in A)\).
- Multiplying the wrong probabilities together. A likelihood is built from one contribution per record. Do not add record contributions, and do not treat a censoring statement as if it were an exact value.
A useful final check is to verify that every contribution is a probability under the stated observation rule and lies between 0 and 1. For truncated data, check that the numerator event is contained in the selection region and that the denominator is the probability of that region. For any censored event, sum its compatible masses directly at least once, even if a cumulative-probability expression is also available.
These likelihood constructions are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus. The main discipline is to model the observation, not just the underlying loss: exact values contribute masses, censored reports contribute sums of masses, and truncated records require conditioning on selection.
Check Your Understanding
For each problem, write the contribution for the stated record or records using the observation rule given.
- A loss has support \(\{0,1,2\}\) with probabilities \(\theta,1-2\theta,\theta\), where \(\theta=0.20\). One independent record is exactly 1 and another is strictly greater than 1. What is their joint likelihood contribution? (Answer: \(0.12\))
- A loss has support \(\{1,2,3\}\) with probabilities \(\theta,2\theta,1-3\theta\), where \(\theta=0.10\). The sample includes only losses \(X\geq2\), and a selected loss is observed exactly as 2. What is its conditional likelihood contribution? (Answer: \(2/9\), or approximately \(0.2222\))
- Let \(X\sim\operatorname{Binomial}(2,0.25)\). Two independent records are observed: one reports \(X\leq0\), and the other is exactly 1. Find their joint likelihood contribution. (Answer: \(0.2109375\))
- A discrete loss has probabilities \(0.10,0.20,0.30,0.40\) at support points \(0,1,2,3\), respectively. The sample includes only losses \(X\geq1\). A selected loss is interval-censored as \(1\leq X\leq2\). Find its conditional likelihood contribution. (Answer: \(5/9\), or approximately \(0.5556\))
- A loss has support \(\{1,2,3,4\}\) with probabilities \(\theta,\theta,\theta,1-3\theta\), where \(\theta=0.20\). Only losses \(X\geq2\) are sampled, and the record says strictly greater than 2. Find the conditional likelihood contribution. (Answer: \(0.75\))