Why Check Likelihood Contributions?
An incomplete-data likelihood is built from what the observation process actually records, not necessarily from the underlying loss itself. A selected loss is observed only after it passes a truncation threshold; a censored loss is reported as an event such as “exceeds the study limit.” A useful check is to ask whether the proposed contributions account for all possible records and add or integrate to one under the stated model.
This check can expose a missing truncation normalizer, a density incorrectly used for a censored record, or a threshold atom assigned to the wrong record category. It also clarifies a basic consistency requirement: if truncation or censoring is removed, the contribution should reduce to the ordinary density or probability mass for a fully observed loss.
Conditional Contributions Under Left Truncation
Suppose a continuous loss is recorded only when \(X>a\), for a known threshold \(a\). If the likelihood is conditional on a loss having been retained, the density of a retained observation \(x>a\) is the original density divided by the probability of retention:
The denominator is not an optional adjustment. It makes the conditional density integrate to one and generally depends on the unknown parameters. For a discrete loss, the corresponding formula is \(P(X=k\mid X>a)=P(X=k)/P(X>a)\) for each allowable \(k>a\).
The strict inequality matters when there is an atom at the threshold. If inclusion is defined as \(X\geq a\), the denominator is \(P(X\geq a)\), which can differ from \(S_X(a)=P(X>a)\). The selection rule, including how equality is treated, determines the normalizer.
Write the event that makes a loss observable, such as \(X>a\), including its treatment of equality.
Use the model to calculate the probability of selection. Keep any parameter dependence in this probability.
Divide the exact density or mass by the selection probability, then verify that the conditional contributions integrate or sum to one.
Worked Example: Normalizing a Left-Truncated Exponential Density
A hypothetical study records only losses \(X>a\), where \(X\) has an exponential distribution with rate \(\lambda=2\) per loss unit and \(a=(\ln 2)/2\). Find the conditional density of a retained loss, verify its normalization, and calculate its value at \(x=\ln 2\).
Method and model. For \(x\geq 0\), the exponential density and survival function are \(f_X(x)=2e^{-2x}\) and \(S_X(x)=e^{-2x}\). The selection rule is strict truncation at \(a\), so the retention probability is \(S_X(a)=e^{-2(\ln 2)/2}=e^{-\ln 2}=1/2\).
Conditional density. For \(x>a\), \(f_{X\mid X>a}(x)=2e^{-2x}/(1/2)=4e^{-2x}\). Its integral over the allowable range is \(\int_a^\infty 4e^{-2x}\,dx=2e^{-2a}=2(1/2)=1\). Thus the conditional contributions are normalized.
Contribution at the stated record. Since \(\ln 2>(\ln 2)/2\), the value is in the retained range. The conditional density there is \(4e^{-2\ln 2}=4/4=1\) per loss unit. This is a density value, not a point probability.
Check and common mistake avoided. The conditional interval probability from \(a\) to \(\ln 2\) is \([S_X(a)-S_X(\ln 2)]/S_X(a)=[1/2-1/4]/(1/2)=1/2\). A common mistake is to use \(f_X(x)\) alone for the retained record; that leaves the density’s integral over \(x>a\) equal to \(1/2\), not one. If there is no truncation, \(a=0\) and \(S_X(0)=1\), so the contribution reduces to \(f_X(x)\).
Checking Contributions Under Right Censoring
Consider a continuous loss \(X\) observed under a fixed right-censoring threshold \(c\). Suppose an exact value is reported when \(X\leq c\), while a censored record is reported when \(X>c\). For an exact observed value \(x\leq c\), the contribution is \(f_X(x)\); for a censored record, the contribution is the probability of the event \(X>c\), namely \(S_X(c)\).
These two types of contributions have different forms because they describe different information. An exact continuous value contributes a density. A censored record represents a range of possible losses and contributes the probability of that range. The usual formulas assume the censoring threshold is fixed or independent of \(X\), and that the censoring status and threshold are recorded correctly. If the censoring mechanism depends on the loss in an unmodeled way, these contributions alone need not describe the observation process.
Worked Example: Checking a Continuous Right-Censoring Likelihood
A hypothetical portfolio study has independent exponential losses with rate \(\lambda=0.5\) per loss unit. One loss is observed exactly as \(x=1\); a second is reported as exceeding the censoring threshold \(c=2\). Calculate the two contributions and their joint likelihood, then verify the record categories normalize.
Method. For this exponential model, \(f_X(x)=0.5e^{-0.5x}\) and \(S_X(x)=e^{-0.5x}\). The exact observation contributes \(f_X(1)=0.5e^{-0.5}\approx 0.3033\). The censored observation contributes \(S_X(2)=e^{-1}\approx 0.3679\).
By independence, the likelihood for these two records is the product: \(L=(0.5e^{-0.5})(e^{-1})=0.5e^{-1.5}\approx 0.1116\). The displayed decimals are rounded to four places.
Normalization check. The total probability of an exact record at or below 2 is \(\int_0^2 0.5e^{-0.5x}\,dx=1-e^{-1}\approx 0.6321\). The probability of the other record category, a loss exceeding 2, is \(e^{-1}\approx 0.3679\). Their sum is \(1-e^{-1}+e^{-1}=1\).
Check and common mistake avoided. A censored record contributes \(S_X(2)\), not \(f_X(2)\). The density at 2 describes an exact value near 2 per unit of loss; it does not give the probability of exceeding 2. If the loss were fully observed as \(x=1\), its contribution would simply be \(f_X(1)\), with no censoring survival term.
Atoms and the Meaning of a Threshold
For a distribution with atoms, the records at a censoring threshold must be defined explicitly. If exact observations are reported for \(X\leq c\) and censoring is reported only for \(X>c\), then an atom at \(c\) belongs to the exact-observation category. Its contribution is its probability mass, not a density. The censored contribution is \(P(X>c)\). If instead the reporting rule censors \(X\geq c\), the atom at \(c\) belongs to the censored category and the event probability must include it.
The record categories must be mutually exclusive and collectively exhaustive. Once that partition is specified, adding their probabilities is a direct way to detect whether a threshold atom has been counted twice or omitted.
Worked Example: Normalizing Discrete Records at a Censoring Threshold
A hypothetical count-valued loss \(X\) has a geometric distribution on \(0,1,2,\ldots\), with \(P(X=k)=0.5^{k+1}\). The recording rule reports exact values \(X\leq 2\) and records “exceeds 2” when \(X>2\). Verify that the record probabilities sum to one and find the likelihood for one exact record \(X=2\) and one censored record.
Method. The exact-record probabilities are \(P(X=0)=1/2\), \(P(X=1)=1/4\), and \(P(X=2)=1/8\). The censored event is the tail \(P(X>2)=\sum_{k=3}^{\infty}0.5^{k+1}=1/8\).
Their sum is \(1/2+1/4+1/8+1/8=1\), as required. The exact record at 2 contributes \(1/8\), and the censored record contributes \(1/8\). Assuming independent losses, their joint likelihood is \(L=(1/8)(1/8)=1/64=0.015625\).
Check and common mistake avoided. The probability of \(X\geq 2\) is \(1/4\), but it is not the censored contribution under this rule: the value 2 is observed exactly. Using \(P(X\geq2)\) for the censored event would misclassify the threshold atom. With no censoring, a fully observed value \(k\) contributes its ordinary mass \(P(X=k)\).
A Reusable Validation Checklist
A likelihood contribution can be checked without solving the score equation or maximizing the likelihood. First state the record categories, then verify that each category receives the right type of contribution. Finally, compare the expression with the familiar complete-data likelihood when the observation restriction disappears.
Distinguish “observed only if selected” from “observed up to a censoring limit.” For threshold atoms, say which category includes equality.
Use a density for an exact continuous value, a mass for an exact discrete value, and an event probability for a censored range.
Integrate over exact continuous records and add the probabilities of discrete or censored categories. The result must be one under the model.
With no truncation, the selection probability is one. With no censoring, the record contribution is the ordinary density or mass of the fully observed value.
These checks are especially useful when a normalizer depends on an unknown parameter: dropping it may still leave an expression that looks plausible, but it will generally fail the conditional normalization check and alter the likelihood. They do not replace checking the sampling design. For example, a likelihood conditional on being retained is appropriate when the analysis conditions on the retained sample; a design that also uses information about the number of rejected losses may require additional likelihood terms.
Check Your Understanding
For each problem, identify the relevant record contribution and use normalization or the complete-data reduction as requested.
- A hypothetical \(X\sim\operatorname{Uniform}(0,10)\) loss is recorded only if \(X>4\). What is \(P(6<X\leq9\mid X>4)\)? (Answer: \(0.5000\))
- A hypothetical exponential loss has rate \(\lambda=0.25\) per loss unit and is censored at \(c=4\). What is the contribution for a record that only says the loss exceeds 4? (Answer: \(S_X(4)=e^{-1}\approx0.3679\))
- A hypothetical geometric loss has \(P(X=k)=0.4(0.6)^k\) for \(k=0,1,\ldots\). Exact values through 2 are reported, and \(X>2\) is censored. Give the contribution for an exact value of 2 and for a censored record, and verify the four record-category probabilities sum to one. (Answer: exact 2: \(0.1440\); censored: \(0.2160\); category probabilities \(0.4000+0.2400+0.1440+0.2160=1.0000\))
- A hypothetical exponential loss has rate \(0.5\) per loss unit and is observed only when \(X>2\). What is the conditional density contribution at an observed \(x=4\)? (Answer: \(f_{X\mid X>2}(4)=0.5e^{-1}\approx0.1839\) per loss unit)
- Two independent, fully observed hypothetical exponential losses are \(1\) and \(3\), with rate \(0.5\) per loss unit. What is their joint likelihood, with no truncation or censoring? (Answer: \(f_X(1)f_X(3)=0.25e^{-2}\approx0.0338\))