Why the Observation Mechanism Matters
A severity sample may contain exact loss amounts, losses known only to fall below a reporting threshold, and losses known to exceed a follow-up threshold. If the study also includes only losses above a specified entry threshold, each likelihood contribution must account for that selection. A censoring probability calculated for the original population is not, by itself, the probability of the observed record in a left-truncated sample.
The task is to write the likelihood of the observations that were actually available. This tutorial develops that likelihood for a continuous severity model, first deriving the contribution for each record type and then maximizing a model parameter. The examples use hypothetical severity studies. They assume a fixed observed sample size and known truncation and censoring thresholds.
Deriving the Conditional Contributions
Assume each underlying loss follows the stated severity model, truncation thresholds are known, and censoring thresholds are fixed by the observation design or are noninformative about the unobserved loss after conditioning on the design. Treat records as independent. The likelihood below is conditional on the observed sample having entered the study; it does not model how many losses were excluded below the truncation thresholds.
For an exact observation \(x>a\), the conditional density is the original density divided by the probability of passing the entry threshold. For a left-censored observation, use the conditional probability of the interval between entry and the censoring threshold. For a right-censored observation, use the conditional probability of exceeding the censoring threshold.
These expressions follow directly from conditioning. For example, the left-censored contribution is \(P(a<X\le c\mid X>a)\). The conditional-probability formula gives \(P(a<X\le c)/P(X>a)=[F_\theta(c)-F_\theta(a)]/S_\theta(a)\). The exact-record formula is a conditional density: for a small interval of width \(h\) around \(x\), the conditional probability is approximately \(f_\theta(x)h/S_\theta(a)\), so the density is \(f_\theta(x)/S_\theta(a)\). The same conditioning gives the right-censored contribution.
For independent records with possibly different entry thresholds \(a_i\), multiply the appropriate contribution for each record using its own \(a_i\). In a continuous model, including or excluding a single interval endpoint does not change these probabilities. If the severity model has atoms, endpoint conventions can matter and the event probabilities must be written to match the observation rule.
Worked Likelihood Under Exponential Severity
For the first example, let losses follow an exponential distribution with rate \(\lambda>0\), so \(f_\lambda(x)=\lambda e^{-\lambda x}\) and \(S_\lambda(x)=e^{-\lambda x}\) for \(x\ge0\). For a record truncated at \(a\), an exact observation contributes \(\lambda e^{-\lambda(x-a)}\); a left-censored interval ending at \(c\) contributes \(1-e^{-\lambda(c-a)}\); and a right-censored record at \(r\) contributes \(e^{-\lambda(r-a)}\). These are conditional contributions, not unconditional density or tail probabilities.
Worked Example: One Record of Each Type
A hypothetical study includes losses only when \(X>2\). The observed records are: an exact loss of \(3\); a left-censored loss known to satisfy \(2<X\le2+\ln 2\); and a right-censored loss known to exceed \(2+\ln 2\). Assume independent exponential losses with unknown rate \(\lambda\). Find the maximum likelihood estimate and check that it is a maximum.
Method and contributions. Condition each record on passing the common entry threshold \(a=2\). The exact loss has excess \(3-2=1\), so its contribution is \(\lambda e^{-\lambda}\). The left-censored interval has width \((2+\ln 2)-2=\ln 2\), so its contribution is \(1-e^{-\lambda\ln 2}\). The right-censored record contributes \(e^{-\lambda((2+\ln 2)-2)}=e^{-\lambda\ln 2}\).
Multiplication gives the likelihood and log-likelihood:
Differentiate. The derivative of \(\ln(1-e^{-\lambda b})\) is \(b/(e^{\lambda b}-1)\), with \(b=\ln 2\). Thus the score equation is \(\ell'(\lambda)=1/\lambda-1-\ln 2+\ln 2/(e^{\lambda\ln 2}-1)=0\). At \(\lambda=1\), \(e^{\lambda\ln 2}=2\), so the score is \(1-1-\ln 2+\ln 2/(2-1)=0\).
To check the stationary point, differentiate again: \[ \ell''(\lambda)=-\frac{1}{\lambda^2} -\frac{(\ln 2)^2e^{\lambda\ln 2}}{(e^{\lambda\ln 2}-1)^2}<0 \quad\text{for every }\lambda>0. \] The log-likelihood is strictly concave, so its stationary point is the unique maximum. Therefore \(\hat\lambda=1.0000\), to four decimal places. As a likelihood check, \(L(1)=e^{-1}(1/2)(1/2)=e^{-1}/4\approx0.0920\).
Common mistake avoided: treating the left-censored record as though it were known only to satisfy \(X\le2+\ln 2\). Its actual information is \(2<X\le2+\ln 2\), conditional on \(X>2\); the contribution is \(1-e^{-\lambda\ln 2}\), not \(F_\lambda(2+\ln 2)\).
Combining Records With Different Entry Thresholds
When entry thresholds differ by record, the denominators in the conditional contributions differ too. For exponential severity, those denominators simplify the calculation: the exact density divided by \(S_\lambda(a_i)\) depends on \(x_i-a_i\), and the censored contributions depend on the distance from \(a_i\) to the relevant threshold. The absolute threshold itself does not disappear from the data; it is the excess above that record’s threshold that enters these expressions.
The log-likelihood is the logarithm of this expression. Its score is \(m/\lambda-T+\sum_{i=1}^k b_i/(e^{\lambda b_i}-1)\), where \(T\) is the sum of the exact and right-censored excesses shown in braces, and \(b_i=c_i-a_i\) is the width of left-censored interval \(i\). When at least one left-censored interval is present, the score equation will generally require numerical solution. Its derivative is \(-m/\lambda^2-\sum_i b_i^2e^{\lambda b_i}/(e^{\lambda b_i}-1)^2\), which is negative when \(m>0\). Thus any positive score root is a unique maximum in that case.
Worked Example: Mixed Records With Varying Entry Thresholds
A hypothetical exponential-loss study records four independent losses. Two are exact: \(x_1=1.5\) after entry at \(a_1=1\), and \(x_2=3.5\) after entry at \(a_2=3\). One is left-censored: \(4<X\le4+\ln 2\), after entry at \(a_3=4\). One is right-censored: \(X>3+\ln 2\), after entry at \(a_4=2\). Estimate the rate \(\lambda\).
Method and contributions. The exact excesses are \(1.5-1=0.5\) and \(3.5-3=0.5\). Their contributions are \(\lambda e^{-0.5\lambda}\) each. The left interval has width \(\ln 2\), giving \(1-e^{-\lambda\ln 2}\). The right-censored excess is \((3+\ln 2)-2=1+\ln 2\), giving \(e^{-\lambda(1+\ln 2)}\). Therefore
The log-likelihood score is \[ \ell'(\lambda)=\frac{2}{\lambda}-(2+\ln 2) +\frac{\ln 2}{e^{\lambda\ln 2}-1}. \] At \(\lambda=1\), this equals \(2-(2+\ln 2)+\ln 2=0\). The second derivative is \(-2/\lambda^2-(\ln 2)^2e^{\lambda\ln 2}/(e^{\lambda\ln 2}-1)^2<0\); hence this root is the unique maximum. The estimate is \(\hat\lambda=1.0000\), to four decimal places. Substitution checks the likelihood: \(L(1)=e^{-(2+\ln 2)}(1/2)=e^{-2}/4\approx0.0338\).
Common mistake avoided: using the absolute loss \(3.5\) in the second exact contribution would ignore its entry threshold \(a_2=3\). The correct contribution is \(\lambda e^{-\lambda(3.5-3)}\), not \(\lambda e^{-3.5\lambda}\).
A Weibull Example Using a Reparameterization
The same conditioning logic applies beyond the exponential model. Consider a Weibull loss with known shape \(2\) and unknown scale \(\theta>0\), with survival function \(S(x)=e^{-(x/\theta)^2}\) and density \(f(x)=2x\theta^{-2}e^{-(x/\theta)^2}\). Set \(\eta=1/\theta^2\), so \(S(x)=e^{-\eta x^2}\). The transformation makes the exponents linear in \(\eta\); it does not change which event each record represents.
Worked Example: Known-Shape Weibull With Three Observation Types
A hypothetical sample is truncated at \(a=1\). It contains an exact loss \(X=\sqrt{2}\), a left-censored loss satisfying \(1<X\le\sqrt{1+\ln 2}\), and a right-censored loss satisfying \(X>\sqrt{1+\ln 2}\). Assume independent Weibull losses with known shape \(2\) and unknown scale \(\theta\). Estimate \(\theta\).
Method and contributions. Write \(\eta=1/\theta^2\). The exact record contributes \(f(\sqrt{2})/S(1)=2\sqrt{2}\eta e^{-\eta(2-1)} =2\sqrt{2}\eta e^{-\eta}\). For the left-censored interval, the difference in squared thresholds is \((1+\ln 2)-1=\ln 2\), so its conditional probability is \(1-e^{-\eta\ln 2}\). The right-censored record has the same squared-threshold excess, so its contribution is \(e^{-\eta\ln 2}\). Thus
The log-likelihood score is \(1/\eta-1-\ln 2+\ln 2/(e^{\eta\ln 2}-1)\). At \(\eta=1\), it is \(1-1-\ln 2+\ln 2=0\). Its derivative is \(-1/\eta^2-(\ln 2)^2e^{\eta\ln 2}/(e^{\eta\ln 2}-1)^2<0\), so this root is the unique maximum. Consequently \(\hat\eta=1\), and transforming back gives \(\hat\theta=1/\sqrt{\hat\eta}=1.00\), to two decimal places. As a check, \(L(1)=2\sqrt{2}e^{-1}(1/2)(1/2)=\sqrt{2}/(2e)\approx0.2601\).
Common mistake avoided: the Weibull exponent uses squared thresholds. The left-censoring interval width on the transformed scale is \(c^2-a^2=\ln 2\), not \(c-a\).
Conditioning Errors to Check Before Maximizing
A likelihood can be differentiated correctly and still give an invalid estimate if its contributions describe the wrong observation process. For the first worked example, the incorrect unconditional factors would include \(\lambda e^{-3\lambda}\) for the exact loss, \(F_\lambda(2+\ln 2)\) for the left-censored loss, and \(S_\lambda(2+\ln 2)\) for the right-censored loss. These expressions refer to the original population. They do not describe records conditional on having passed \(X>2\).
The correct exact factor is \(f_\lambda(3)/S_\lambda(2)=\lambda e^{-\lambda}\). The correct left-censored factor is \([F_\lambda(2+\ln 2)-F_\lambda(2)]/S_\lambda(2)=1-e^{-\lambda\ln 2}\). The correct right-censored factor is \(S_\lambda(2+\ln 2)/S_\lambda(2)=e^{-\lambda\ln 2}\). In each case, conditioning removes the portion of the original distribution below the entry threshold. This division is required even when the sample contains no exact value for a particular record.
Before maximizing a mixed likelihood, check that each exact value exceeds its entry threshold, each left-censoring interval begins at the entry threshold, and each right-censoring threshold is at least the entry threshold. Then verify that every contribution is a conditional density or probability, and that the product includes the correct excess or interval width for that record. If truncation thresholds are unknown, or the number selected into the sample is itself part of the model, the conditional likelihood used here may not be sufficient.
Check Your Understanding
For each problem, condition on the stated entry threshold. Treat losses as independent and the thresholds as fixed. Rates are in inverse loss units.
- Exponential losses have unknown rate \(\lambda\). One exact loss is \(3\) after truncation at \(1\), and one right-censored loss exceeds \(4\) after truncation at \(1\). Write the likelihood and find the maximum likelihood estimate. (Answer: \(\hat\lambda=0.2000\).)
- Exponential losses have unknown rate \(\lambda\). One record is left-censored at \(3\) after truncation at \(2\), and another is right-censored above \(4\) after truncation at \(2\). Write the conditional likelihood and solve its score equation. (Answer: \(\hat\lambda=\ln(1.5)\approx0.4055\).)
- Exponential losses have unknown rate \(\lambda\). The records are: one exact loss with excess above its own entry threshold equal to \(1\); one left-censored interval of width \(\ln 2\) above its entry threshold; and one right-censored record with excess above its entry threshold equal to \(\ln 2\). Find the estimate and verify it is a maximum. (Answer: \(\hat\lambda=1.0000\).)
- Exponential losses have unknown rate \(\lambda\). There is one exact record with excess above entry equal to \(1\), two left-censored records each with interval width \(\ln 2\), and one right-censored record with excess above entry equal to \(2\ln 2\). Find the estimate. (Answer: \(\hat\lambda=1.0000\).)
- Exponential losses have unknown rate \(\lambda\). Two exact records have excesses above their respective entry thresholds of \(0.5\) each, and one right-censored record has excess above its entry threshold of \(2\). There are no left-censored records. Form the conditional likelihood and find the estimate. (Answer: \(\hat\lambda=2/3\approx0.6667\).)