Likelihoods With Individual Entry Times

Why Individual Entry Thresholds Matter

A severity study may include a loss only if it exceeds a threshold, and that threshold may vary by policy, contract, or subject. A further complication is that some included losses are not observed exactly: the record may only establish that the loss exceeds a censoring point. To estimate a severity distribution correctly, each record must be conditioned on the threshold that allowed that particular subject to enter the data.

The key distinction is that an entry threshold determines which losses could appear in the data, while a right-censoring point determines how much is known about an included loss. Treating several different entry thresholds as one common threshold generally changes the likelihood and can change the estimate.

Definition: For record \(i\), let \(a_i\) be its known left-entry threshold, and suppose the loss \(X_i\) is observed only when \(X_i>a_i\). An exact record has observed loss \(x_i>a_i\). A right-censored record has censoring point \(c_i\geq a_i\) and tells us that \(X_i>c_i\), but does not give the exact loss. Here \(S_X(t)=P(X>t)\) and \(f_X\) are the severity survival function and density.

Assume the losses are independent draws from a continuous severity model with parameter \(\theta\). The entry thresholds are known and fixed, or are generated by a process that is noninformative for the loss after conditioning on entry. For the usual severity likelihood, also assume censoring is noninformative: the censoring mechanism does not provide additional information about \(\theta\) once the observed threshold and censoring point are accounted for. If entry or censoring depends on the unobserved loss in a way not captured by these assumptions, the selection or censoring mechanism must be modeled rather than omitted.

Conditional Contributions Record by Record

An exact observation at \(x_i\) contributes a conditional density, not the unconditional density \(f_X(x_i)\). A right-censored record contributes the conditional probability that the loss exceeds its censoring point. Both contributions use that record’s own entry threshold.

L_i(\theta)= \begin{cases} \dfrac{f_X(x_i;\theta)}{S_X(a_i;\theta)}, & \text{exact observation at }x_i,\\[8pt] \dfrac{S_X(c_i;\theta)}{S_X(a_i;\theta)}, & \text{right-censored at }c_i. \end{cases}

For independent records, the likelihood is the product of these conditional contributions. Writing \(\delta_i=1\) for an exact record and \(\delta_i=0\) for a right-censored record gives a compact expression. For exact records, \(x_i\) is observed; for censored records, \(c_i\) is observed.

L(\theta)= \prod_{i=1}^n \left\{ \left[\frac{f_X(x_i;\theta)}{S_X(a_i;\theta)}\right]^{\delta_i} \left[\frac{S_X(c_i;\theta)}{S_X(a_i;\theta)}\right]^{1-\delta_i} \right\}.

The denominator is not a single cohort-wide adjustment: it is \(S_X(a_i;\theta)\) for record \(i\). The likelihood can be written as a sum of log contributions for computation, but the same record-specific conditioning remains essential.

\ell(\theta)= \sum_{i:\,\delta_i=1} \left[\log f_X(x_i;\theta)-\log S_X(a_i;\theta)\right] + \sum_{i:\,\delta_i=0} \left[\log S_X(c_i;\theta)-\log S_X(a_i;\theta)\right].

Why the Conditioning Formula Holds

The formulas follow directly from conditioning on having crossed the entry threshold. For an exact observation, the conditional probability of a loss in a small interval around \(x_i\) is the probability of being in that interval and exceeding \(a_i\), divided by the probability of exceeding \(a_i\). Since \(x_i>a_i\), the numerator has density \(f_X(x_i;\theta)\) locally.

P(x_i<X_i\leq x_i+dx\mid X_i>a_i) \approx \frac{f_X(x_i;\theta)\,dx}{S_X(a_i;\theta)}.

Dividing by \(dx\) gives the conditional density. For a censored record, the observed event is \(X_i>c_i\). Since \(c_i\geq a_i\), this event already implies entry, so its conditional probability is the ratio of the two survival probabilities. Independence across subjects then makes the sample likelihood the product of the per-record terms.

P(X_i>c_i\mid X_i>a_i) =\frac{P(X_i>c_i)}{P(X_i>a_i)} =\frac{S_X(c_i;\theta)}{S_X(a_i;\theta)}.

Worked Conditional Likelihoods

Worked Example: Weibull Contributions at Two Entry Thresholds

A hypothetical study uses a Weibull severity model with known shape \(2\) and scale \(5\), in loss units. Thus \(S_X(x)=\exp[-(x/5)^2]\) and \(f_X(x)=(2x/25)\exp[-(x/5)^2]\), for \(x\geq 0\). One subject entered above \(a_1=3\) and had an exact loss \(x_1=5\). Another entered above \(a_2=1\) and was right-censored at \(c_2=5\). Find their joint conditional likelihood at these parameter values.

Method. Calculate each contribution conditional on that subject’s own entry threshold, then multiply. This avoids applying either entry normalizer to both subjects.

For the exact record, \(f_X(5)=(2\cdot5/25)e^{-1}=0.4e^{-1}\), while \(S_X(3)=e^{-9/25}\). Its conditional density contribution is

\frac{f_X(5)}{S_X(3)} =0.4e^{-1+9/25} =0.4e^{-16/25} =0.4e^{-0.64} \approx 0.2109.

For the censored record, \(S_X(5)=e^{-1}\) and \(S_X(1)=e^{-1/25}\). Its contribution is

\frac{S_X(5)}{S_X(1)} =e^{-1+1/25} =e^{-24/25} =e^{-0.96} \approx 0.3829.

Multiplying gives \(0.2109\times0.3829\), or, using the unrounded expressions, \(0.4e^{-0.64}e^{-0.96}=0.4e^{-1.60}\approx 0.0808\). The joint conditional likelihood is 0.0808, rounded to four decimal places. The calculation checks because the exact record uses \(S_X(3)\), whereas the censored record uses \(S_X(1)\).

Exponential Model With Individual Entry Times

For an exponential loss with rate \(\lambda>0\), \(S_X(x)=e^{-\lambda x}\) and \(f_X(x)=\lambda e^{-\lambda x}\), for \(x\geq0\). The conditional exact density simplifies to \(\lambda e^{-\lambda(x_i-a_i)}\); the conditional censored contribution is \(e^{-\lambda(c_i-a_i)}\). Thus each record contributes according to the distance between its observed endpoint and its own entry threshold.

L(\lambda)=\lambda^r \exp\left[-\lambda \left\{ \sum_{i:\,\delta_i=1}(x_i-a_i) + \sum_{i:\,\delta_i=0}(c_i-a_i) \right\}\right],

where \(r\) is the number of exact records. If the quantity in braces is positive and \(r>0\), differentiating the log likelihood gives the maximum likelihood estimate \(\widehat{\lambda}=r/\text{(quantity in braces)}\). The result is a special case for the exponential model; it is not a general formula for other severity distributions.

Worked Example: Exponential MLE With Four Different Entry Thresholds

A hypothetical portfolio records losses in thousands of dollars under an exponential severity model with unknown rate \(\lambda\) per thousand dollars. The records are: exact loss \(5\) after entry at \(2\); censored at \(6\) after entry at \(1\); exact loss \(4\) after entry at \(3\); and censored at \(8\) after entry at \(2\). Find the MLE.

Method. Use the conditional exponential likelihood, keeping all four entry amounts separate. This avoids replacing the actual entry thresholds with a pooled threshold.

The exact-record distances are \(5-2=3\) and \(4-3=1\). The censored-record distances are \(6-1=5\) and \(8-2=6\). Their total is \(3+1+5+6=15\), and there are \(r=2\) exact records. Therefore

L(\lambda) =(\lambda e^{-3\lambda})(e^{-5\lambda}) (\lambda e^{-\lambda})(e^{-6\lambda}) =\lambda^2e^{-15\lambda}.

The log likelihood is \(\ell(\lambda)=2\log\lambda-15\lambda\). Its score is \(2/\lambda-15\), so the stationary point is \(\widehat{\lambda}=2/15=0.1333\) per thousand dollars, rounded to four decimal places. The second derivative is \(-2/\lambda^2<0\) for \(\lambda>0\), confirming a maximum. As a check, substituting \(2/15\) into the score gives \(2/(2/15)-15=0\).

A pooled-threshold error would condition every record at the smallest entry, \(1\). The total endpoint amount is \(5+6+4+8=23\), so that incorrect construction gives \(\lambda^2e^{-19\lambda}\) and the incorrect estimate \(2/19\). It does not represent the actual subject-specific selection process; the correct estimate uses the total residual distance \(15\).

Pareto Example: Exact and Censored Records

The same record-by-record construction applies outside the exponential family. Consider a Pareto Type I loss with minimum value \(1\) and shape parameter \(\alpha>0\). For \(x\geq1\), its survival function is \(S_X(x)=x^{-\alpha}\) and its density is \(f_X(x)=\alpha x^{-\alpha-1}\). Entry thresholds in this model must also be at least \(1\).

Worked Example: Pareto Shape MLE With Individual Thresholds

A hypothetical sample has two exact records, \((a,x)=(2,4)\) and \((1,2)\), and two right-censored records, \((a,c)=(3,6)\) and \((4,8)\). Assume the Pareto model just defined. Estimate \(\alpha\) by maximum likelihood.

Method. Divide each exact density by survival at its own entry, and each censored survival by survival at its own entry. This avoids treating the unconditional Pareto densities as though no truncation occurred.

For the exact record \((2,4)\), the conditional density is \(\alpha 2^\alpha/4^{\alpha+1}=(\alpha/4)2^{-\alpha}\). For \((1,2)\), it is \(\alpha/2^{\alpha+1}=(\alpha/2)2^{-\alpha}\). For the censored records, the contributions are \((3/6)^\alpha=2^{-\alpha}\) and \((4/8)^\alpha=2^{-\alpha}\). Hence

L(\alpha) =\left(\frac{\alpha}{4}2^{-\alpha}\right) \left(\frac{\alpha}{2}2^{-\alpha}\right) 2^{-\alpha}2^{-\alpha} =\frac{\alpha^2}{8}\,2^{-4\alpha}.

The log likelihood is \(2\log\alpha-\log 8-4\alpha\log2\). Its score is \(2/\alpha-4\log2\), which is zero at

\widehat{\alpha}=\frac{1}{2\log2}\approx 0.7213.

The second derivative is \(-2/\alpha^2<0\), so this stationary point is a maximum. Substitution into the score gives \(2(2\log2)-4\log2=0\), checking the result. The estimate is 0.7213, rounded to four decimal places.

Assumptions, Interpretation, and Exam Traps

The likelihood above is a conditional severity likelihood: it describes the observed losses given that each subject crossed the specified entry threshold. It does not, by itself, estimate how many subjects failed to enter or the distribution of their entry thresholds. If the thresholds are random but independent of loss, conditioning on their observed values can still be appropriate. If entry depends on the unobserved loss beyond the stated threshold, or censoring depends on loss in an unmodeled way, the conditional contributions may not give a valid likelihood for the intended model.

For continuous losses, a right-censored record at \(c_i\) represents \(X_i>c_i\); writing \(X_i\geq c_i\) gives the same probability because a continuous distribution assigns zero probability to a single point. A censoring point below the entry threshold is inconsistent with the stated observation scheme and should be checked rather than inserted mechanically. Exact records must satisfy \(x_i>a_i\).

Using \(f_X(x_i)\) for an exact record. That omits the fact that the subject was observed only after exceeding \(a_i\). Use \(f_X(x_i)/S_X(a_i)\).
Using \(S_X(c_i)\) alone for a censored record. The record is already conditioned on entry, so use \(S_X(c_i)/S_X(a_i)\).
Using one entry threshold for all subjects. The denominator, and thus the likelihood contribution, depends on the threshold attached to each record.
Confusing censoring with truncation. Censoring leaves a subject in the data with incomplete loss information; truncation excludes subjects who do not meet the entry condition.

For the exponential model, the compact MLE follows because each record contributes its observed distance above its own entry point. For more general severity models, retain the full conditional likelihood and optimize it with respect to the model parameters. This subject-specific construction is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the stated model and condition each record on its own entry threshold.

An exponential loss has rate \(0.4\) per loss unit. Find the contribution for an exact loss \(x=5\) after entry at \(a=2\), and for a separate record censored at \(c=7\) after entry at \(a=3\). Give both contributions to four decimal places. (Answer: exact \(0.4e^{-1.2}\approx0.1205\); censored \(e^{-1.6}\approx0.2019\).)
A Pareto Type I loss has minimum \(1\) and shape \(\alpha=2\). Find the conditional contribution for an exact loss \(x=6\) after entry at \(a=3\), and for a censored record at \(c=8\) after entry at \(a=4\). (Answer: exact \(1/12\approx0.0833\); censored \(1/4=0.2500\).)
Under an exponential model, records are exact \((a,x)=(1,4)\), right-censored \((a,c)=(2,5)\), and exact \((a,x)=(3,6)\). Find the MLE of \(\lambda\). (Answer: \(2/9\approx0.2222\) per loss unit.)
Under a Pareto Type I model with minimum \(1\), there are two exact records with \((a,x)=(3,6)\) and \((2,4)\), and one censored record with \((a,c)=(5,10)\). Find the MLE of \(\alpha\). (Answer: \(2/(3\log2)\approx0.9618\).)
For an exponential model with known \(\lambda=1\), one exact record has \((a,x)=(0,2)\) and another has \((a,x)=(1,3)\). Find their joint conditional likelihood, and the joint likelihood obtained if entry conditioning is incorrectly omitted. (Answer: correct \(e^{-4}\approx0.0183\); without entry conditioning \(e^{-5}\approx0.0067\).)

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