Conditioning on Sample Size in Truncated-Loss Studies

Why the Sample-Size Condition Matters

A loss study that records only claims above a reporting threshold has two potentially useful forms of information: the sizes of the recorded losses and the number of losses that passed the threshold. Whether both forms belong in the likelihood depends on how the study was sampled. If investigators selected a fixed number of reported losses, the count is set by design and contains no information about the severity parameters. If they observed a known exposure and retained every loss above the threshold, the observed count can be informative—but only relative to a specified model for the underlying claim frequency.

The distinction is particularly important because the probability that a loss is included generally depends on the severity parameters. A conditional likelihood accounts for this through a truncated density. A likelihood that also uses the count must include the probability of observing that count under the exposure and frequency assumptions. Adding a count term without specifying those assumptions is not justified.

Definition: Let $X$ have density $f_\theta$, distribution function $F_\theta$, and survival function $S_\theta(c)=P(X>c)$, where $\theta$ is a severity parameter and $c$ is a fixed reporting threshold. A loss is observed only if $X>c$. The density of an observed loss, conditional on inclusion, is $$ f_\theta(x\mid X>c)=\frac{f_\theta(x)}{S_\theta(c)},\qquad x>c, $$ provided $S_\theta(c)>0$. This is the left-truncated severity density.

Likelihood Conditional on a Fixed Number of Observations

Suppose $m$ losses $x_1,\ldots,x_m$ are sampled independently from the population of losses that exceed $c$. The number $m$ is fixed by the sampling design: for example, an analyst might be given a specified number of qualifying records. Conditional on the event that each sampled loss exceeds $c$, the likelihood for $\theta$ is

L_{\mathrm{cond}}(\theta) =\prod_{i=1}^{m}\frac{f_\theta(x_i)}{S_\theta(c)},\qquad \ell_{\mathrm{cond}}(\theta) =\sum_{i=1}^{m}\log f_\theta(x_i)-m\log S_\theta(c).

The denominator is repeated $m$ times because each observed value is drawn from the conditional distribution above the threshold. It corrects for the fact that the sample excludes losses at or below $c$. In this design, the likelihood does not contain a probability for the count $m$: there is no random count being modeled.

Worked Example: Estimating an Exponential Rate From a Fixed Truncated Sample

A hypothetical study supplies exactly three losses above the reporting threshold $c=1$: $x_1=2$, $x_2=3$, and $x_3=4$. Assume the underlying severity distribution is exponential with rate $\theta>0$, so $f_\theta(x)=\theta e^{-\theta x}$ and $S_\theta(c)=e^{-\theta c}$. Estimate $\theta$ by maximum likelihood, conditional on the fixed sample size.

Method. Use the truncated density. For $x>1$, $$ f_\theta(x\mid X>1) =\frac{\theta e^{-\theta x}}{e^{-\theta}} =\theta e^{-\theta(x-1)}. $$ Therefore the log-likelihood for the three observations is $$ \ell_{\mathrm{cond}}(\theta) =3\log\theta-\theta\big[(2-1)+(3-1)+(4-1)\big] =3\log\theta-6\theta. $$

Differentiate and set the score equal to zero: $$ \frac{d\ell_{\mathrm{cond}}}{d\theta} =\frac{3}{\theta}-6=0, \qquad \hat{\theta}_{\mathrm{cond}}=\frac{3}{6}=0.5000. $$ The second derivative is $-3/\theta^2<0$, so this stationary point is a maximum. The fitted probability that a loss exceeds the threshold is $e^{-0.5000}=0.6065$, to four decimal places. That probability is not itself used as a separate count contribution because the sample size was fixed by design.

Check and common mistake. The observed excesses over the threshold are $1,2,3$, whose sum is $6$, so the estimate is $m/\sum_i(x_i-c)=3/6$. Using the unadjusted sum $2+3+4=9$ in the conditional likelihood would incorrectly give $3/9$; it would omit the truncation correction.

Adding the Count Under a Known Frequency Model

Now suppose the study covers a known exposure $E$, and claims occur according to a Poisson process with rate $\lambda$ per unit exposure. Assume the claim count and claim severities follow this model, severities are independent and identically distributed with density $f_\theta$, and every claim with $X>c$ is observed. The probability of inclusion is $p_\theta=S_\theta(c)$. The observed count $K$ is then Poisson with mean $\lambda E p_\theta$.

If $K=m$ and the observed severities are $x_1,\ldots,x_m>c$, the joint likelihood, apart from factors not depending on the parameters, is

L_{\mathrm{joint}}(\lambda,\theta) \propto e^{-\lambda E S_\theta(c)}(\lambda E)^m \prod_{i=1}^{m} f_\theta(x_i).

The count term uses the inclusion probability, while the individual severity terms use the original, untruncated density. Equivalently, one can multiply the Poisson likelihood for $K$ by the conditional severity likelihood given $K=m$; the factors $S_\theta(c)^m$ then cancel against the $m$ truncated-density denominators. Do not include the count likelihood and also treat the severity observations as unconditionally sampled without checking that the resulting factors represent this same joint model.

Important condition: The count can inform $\theta$ through $S_\theta(c)$ when the frequency rate $\lambda$ is known, or when it is constrained by additional information or assumptions. If $\lambda$ is an unrestricted unknown parameter, the count alone does not generally identify the severity inclusion probability separately from the frequency rate.

Worked Example: A Known Claim Rate Changes the Severity Estimate

Use the same hypothetical losses $2,3,4$ above $c=1$, but suppose they came from known exposure $E=10$, with a known Poisson claim rate $\lambda=0.2$ per exposure unit. The severity is exponential with rate $\theta$. Estimate $\theta$ using both the observed count $m=3$ and the observed severities.

Method. The expected number of claims before applying the threshold is $\lambda E=0.2(10)=2$. The expected number above the threshold is $2e^{-\theta}$. The joint log-likelihood, omitting terms constant in $\theta$, is $$ \ell_{\mathrm{joint}}(\theta) =-2e^{-\theta}+3\log\theta-\theta(2+3+4) =-2e^{-\theta}+3\log\theta-9\theta. $$ Its score equation is $$ \frac{3}{\theta}-9+2e^{-\theta}=0. $$

Solving this equation numerically gives $\hat{\theta}\approx0.3922$ (the root is approximately $0.392214$). For example, the score at $0.3922$ is approximately $0.0003$; the small nonzero value is due to rounding the root. The score is decreasing for $\theta>0$, since its derivative is $-3/\theta^2-2e^{-\theta}<0$, so the root is the unique maximum. The fitted inclusion probability is $e^{-0.3922}\approx0.6756$, to four decimal places.

Comparison and check. Conditioning on the three observed losses alone gave $\hat{\theta}_{\mathrm{cond}}=0.5000$. With the known frequency and exposure, the count of three qualifying losses also contributes: a larger inclusion probability makes that count more plausible at the stated underlying claim rate. The joint estimate is therefore different. A common mistake is to use the conditional estimate $0.5000$ while claiming to use the count; that calculation discards the count information.

What If the Frequency Rate Is Unknown?

The distinction between a known exposure and a known frequency rate is essential. Known exposure specifies how much risk was observed; it does not, by itself, specify how many claims should have occurred before truncation. If $\lambda$ is unknown and unrestricted, maximizing the joint likelihood over it can remove the count's direct information about $\theta$.

For $m>0$, the terms involving $\lambda$ in the joint log-likelihood are $$ -\lambda E S_\theta(c)+m\log\lambda. $$ Setting their derivative to zero gives $$ -E S_\theta(c)+\frac{m}{\lambda}=0, \qquad \hat{\lambda}(\theta)=\frac{m}{E S_\theta(c)}. $$ Substituting this profiled estimate into the joint likelihood leaves, up to factors independent of $\theta$, the conditional likelihood: $$ \ell_{\mathrm{joint}}\big(\hat{\lambda}(\theta),\theta\big) =\sum_{i=1}^{m}\log f_\theta(x_i)-m\log S_\theta(c)+\text{constant}. $$ Thus, with a free Poisson frequency rate, the profiled severity estimate is the same as the estimate from the conditional truncated sample. This cancellation is not a reason to ignore the count in every model; it follows specifically from profiling an unrestricted $\lambda$ in this Poisson setup.

If $m=0$, the interior score argument does not apply. The likelihood as a function of $\lambda$ is proportional to $e^{-\lambda E S_\theta(c)}$, which is maximized at the boundary $\hat{\lambda}=0$ when $E S_\theta(c)>0$. With no observed severities, this boundary fit does not provide a severity estimate through the conditional severity likelihood.

Worked Example: Profiling an Unknown Frequency Rate

Suppose the three exponential losses $2,3,4$ above $c=1$ were observed over $E=10$ exposure units, but now $\lambda$ is unknown. Find the profiled frequency estimate at the conditional severity estimate, and determine whether the severity estimate changes.

Method. From the fixed-sample calculation, $\hat{\theta}_{\mathrm{cond}}=0.5000$ and $S_{\hat{\theta}}(1)=e^{-0.5}\approx0.6065$. For $m=3>0$, the profiled frequency estimate is $$ \hat{\lambda}(0.5000) =\frac{3}{10e^{-0.5}} \approx\frac{3}{6.0653} \approx0.4946 $$ claims per exposure unit.

The profiled likelihood for $\theta$ is proportional to the conditional likelihood, as shown above. Therefore $\hat{\theta}=0.5000$ remains the maximum likelihood estimate. The rate estimate adjusts to the observed count and the fitted inclusion probability. This is the expected result when a free frequency parameter can compensate for changes in the proportion of claims that exceed the threshold.

Check and common mistake. At the fitted values, the expected number of observed claims is $$ \hat{\lambda}E S_{\hat{\theta}}(1) \approx0.4946(10)(0.6065)\approx3.000. $$ Using $m/E=0.3$ as the estimate of the underlying claim rate would ignore that only claims above $c$ were observed. It estimates the observed qualifying-claim rate per exposure, not the underlying rate $\lambda$.

A Fixed Number of Opportunities Gives a Different Count Model

Not every known-exposure design has a Poisson claim count. For example, if there are $N$ independent policies and each can produce at most one loss during the study period, the number of included losses may have a binomial distribution. The appropriate count likelihood must match that sampling design. The next example isolates count information: the conditional severity distribution among included losses is uninformative about the parameter, but the number included is informative.

Worked Example: Using a Binomial Inclusion Count

A hypothetical portfolio has $N=10$ independent policies, each with exactly one possible loss. For each policy, $X=1$ with probability $\theta$ and $X=3$ with probability $1-\theta$, where $0<\theta<1$. Only losses strictly greater than $c=2$ are recorded. The study records six losses, all equal to $3$. Estimate $\theta$ using the known number of policies.

Method. The inclusion probability is $P(X>2)=1-\theta$. Therefore the observed count $K=6$ has a binomial likelihood, and the joint likelihood for the count and recorded severities is proportional to $$ \binom{10}{6}(1-\theta)^6\theta^4. $$ The observed severities add no further parameter information: conditional on inclusion, the loss must equal $3$, with probability one. Taking logarithms and differentiating gives $$ \ell(\theta)=\text{constant}+6\log(1-\theta)+4\log\theta, \qquad \ell'(\theta)=-\frac{6}{1-\theta}+\frac{4}{\theta}. $$ Setting the score to zero yields $4(1-\theta)=6\theta$, so $$ \hat{\theta}=\frac{4}{10}=0.4000. $$ The second derivative is $-6/(1-\theta)^2-4/\theta^2<0$, confirming a maximum. The fitted inclusion probability is $1-\hat{\theta}=0.6000$, and the expected included count is $10(0.6000)=6$, matching the observation.

Check and common mistake. If the six observed losses had instead been selected in advance as a fixed-size sample of qualifying losses, their values would all be $3$ for every $\theta$, and the conditional likelihood would be constant in $\theta$. Estimating $\theta=0.4$ in that design would incorrectly treat the sample count as random information.

Choosing the Likelihood Correctly

Before maximizing a likelihood in a truncated-loss study, identify what was random and what was fixed. The decision determines whether the inclusion probability appears only through the conditional density or also through a count model.

Identify the sampling design.
Was a fixed number of qualifying losses supplied, or were all qualifying losses observed over a stated exposure?

Write the inclusion probability.
For a threshold $c$, calculate $S_\theta(c)=P(X>c)$ under the stated severity model.

Specify the count model if needed.
Use the model implied by the design, such as Poisson with mean $\lambda E S_\theta(c)$ or binomial with the relevant number of opportunities and inclusion probability.

Check which parameters are known.
A known exposure is not the same as a known underlying claim rate. If frequency is freely estimated, profile it and check whether count information about severity cancels.

The practical lesson is that a count can shift a severity estimate only when the likelihood gives that count a meaningful probability that cannot be absorbed freely by another parameter. This is why a statement such as “the study observed losses over ten exposure units” is not sufficient by itself to determine the severity likelihood. The frequency assumptions and the way records entered the study must also be stated. Correctly distinguishing these designs is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify whether the count is fixed by design or modeled as random, then determine the requested likelihood result.

A fixed sample contains $4$ exponential losses above $c=2$, with observed excesses $1,1,2,4$. Write the conditional MLE of the exponential rate and give its value. (Answer: $\hat{\theta}=4/8=0.5000$.)
For an exponential severity with rate $\theta$, the threshold is $c=1$, the Poisson rate is known to be $\lambda=0.5$ per exposure unit, and exposure is $E=4$. If no losses exceed the threshold, write the likelihood factor involving $\theta$. (Answer: $e^{-2e^{-\theta}}$, up to factors constant in $\theta$.)
Three exponential losses $1.5,2.5,4.0$ are observed above $c=1$ over exposure $E=5$. The Poisson rate is known to be $\lambda=0.4$. Write the score equation for the joint MLE of $\theta$. (Answer: $3/\theta-8+2e^{-\theta}=0$.)
In a Poisson-thinning model, $m=5$, $E=8$, and the severity inclusion probability at a proposed parameter value is $0.25$. Find the profiled frequency estimate at that value. (Answer: $\hat{\lambda}=5/[8(0.25)]=2.5$ per exposure unit.)
There are $12$ independent policies, each with one possible loss. Losses equal $2$ with probability $\theta$ and $5$ with probability $1-\theta$; only losses greater than $3$ are recorded. If $9$ are recorded, find the joint MLE of $\theta$. (Answer: $\hat{\theta}=3/12=0.2500$.)

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