Introduction
A likelihood is only as appropriate as the observation process it represents. An exact loss may contribute a density, a censored loss may contribute a probability, and a truncated sample requires conditioning on the event that allowed each loss to be observed. Confusing these mechanisms can produce a plausible-looking likelihood and a systematically incorrect estimate.
The practical task is to translate the sampling description into the information actually recorded. This tutorial develops a decision process for doing so, then uses exponential loss models to show how the contributions combine and how conditioning errors affect maximum likelihood estimates.
A Decision Process for Likelihood Contributions
Start with the sampling design, not with the formula. Ask whether every sampled risk is observed, or whether a loss must pass an inclusion threshold to appear in the data. Then ask whether each included loss is exact or only partially observed. In the examples below, \(X\) is a continuous loss with parameter \(\theta\), and the thresholds are known.
- Exact, untruncated observation: use \(f_X(x;\theta)\).
- Left truncation at \(a\): inclusion requires \(X>a\), so an exact included value \(x>a\) contributes \(f_X(x;\theta)/S_X(a;\theta)\).
- Right truncation at \(b\): inclusion requires \(X<b\), so an exact included value \(x<b\) contributes \(f_X(x;\theta)/F_X(b;\theta)\).
- Right censoring at \(c\): an observed exact loss \(x<c\) contributes \(f_X(x;\theta)\); a report that \(X\ge c\) contributes \(S_X(c;\theta)\), provided the risk was observed regardless of the loss.
If observations are independent and the observation mechanism is known or has a parameter-free contribution, multiply the individual contributions to form the likelihood. A threshold that varies by observation must be used in that observation’s own contribution. For a left-truncated and right-censored record, the censoring threshold must exceed its entry threshold; otherwise the record could not have entered under that design.
Why Truncation Requires a Conditional Density
Suppose the study includes only losses exceeding \(a\). The density of an observed loss is not the original density over the whole population: it is the density conditional on inclusion. For \(x>a\), the conditional density follows directly from the definition of conditional probability for a continuous variable.
The numerator describes how likely the observed value is under the population model. The denominator is the probability that an observation would have qualified for the sample at all. For right truncation, the inclusion event is \(X<b\), so the denominator is \(P(X<b;\theta)=F_X(b;\theta)\). These denominators are not optional adjustments: they are what make the likelihood describe the conditional sample.
Worked Likelihoods for an Exponential Loss
For the examples, use the exponential model with rate \(\lambda>0\):
The examples assume independent losses, known observation thresholds, and no additional parameter-dependent contribution from the censoring or selection mechanism. Each maximum likelihood estimate is obtained by maximizing the likelihood, or equivalently its log, over \(\lambda>0\).
Worked Example: Exact Losses in a Left-Truncated Sample
A hypothetical severity study records only losses above \(a=2\). The exact observed losses are \(4,6,\) and \(9\), in the same units. Assuming independent exponential losses with rate \(\lambda\), find the maximum likelihood estimate. Identify the conditioning error that would result from treating these as an untruncated sample.
Method. Each record is included only when \(X>2\). Its conditional density is \(f_X(x;\lambda)/S_X(2;\lambda)\), which simplifies to \(\lambda e^{-\lambda(x-2)}\). The three excesses over the threshold are \(4-2=2\), \(6-2=4\), and \(9-2=7\).
The likelihood and log likelihood are therefore
Differentiate and set the score equal to zero:
The second derivative is \(-3/\lambda^2<0\), so this stationary point is a maximum. The conditioning error is to use the untruncated likelihood \(\lambda^3e^{-19\lambda}\), based on the sum \(4+6+9=19\). That gives \(3/19=0.1579\), not the estimate from the stated sampling design. The correct estimate uses the excesses above the entry threshold. The estimates are rounded to four decimal places.
Worked Example: Exact Losses Under Right Truncation
A hypothetical data source retains a loss only if it is below \(b=6\). Three independent exponential losses are observed exactly: \(1,2,\) and \(3\). Write the likelihood and find the rate estimate numerically. Explain why dividing by the survival probability would be the wrong conditioning.
Method. Here inclusion requires \(X<6\). Each observed value therefore has conditional density \(f_X(x;\lambda)/F_X(6;\lambda)\), with \(F_X(6;\lambda)=1-e^{-6\lambda}\). The observed values sum to \(1+2+3=6\), so
The score equation is
This equation has no simple closed-form solution. Let \(t=6\lambda\). Dividing the score equation by \(3\) and substituting \(\lambda=t/6\) gives
Numerically, the positive solution is \(t\approx2.149\), giving \(\widehat{\lambda}=t/6\approx0.3582\). The score changes from positive near \(\lambda=0\) to negative for sufficiently large \(\lambda\), and the log likelihood has its interior maximum at this root. The wrong alternative is to divide by \(S_X(6;\lambda)\): that would condition on \(X>6\), the opposite of the inclusion rule. The rate is rounded to four decimal places.
Worked Example: Left Truncation Combined With Right Censoring
A hypothetical study includes a risk only after its loss has exceeded \(a=2\). Among three included risks, exact losses are \(3\) and \(4\), while the third is right-censored at \(c=5\), meaning its loss is known to be at least \(5\). Find the exponential rate estimate and compare it with the estimate from omitting the truncation conditioning.
Method. The exact records contribute \(f_X(3;\lambda)/S_X(2;\lambda)\) and \(f_X(4;\lambda)/S_X(2;\lambda)\). For the censored record, the relevant information is \(X\ge5\), conditional on having entered at \(X>2\), so its contribution is \(S_X(5;\lambda)/S_X(2;\lambda)\). Together, the likelihood is
The censored record contributes the conditional probability \(P(X\ge5\mid X>2)=S_X(5;\lambda)/S_X(2;\lambda)\). It contributes no density factor \(\lambda\), because its exact loss is not observed. The log likelihood is \(2\log\lambda-7\lambda\), so
The second derivative is \(-2/\lambda^2<0\), confirming a maximum. If the truncation condition were ignored, the likelihood would instead be proportional to \(\lambda^2e^{-\lambda(3+4+5)}\), yielding \(2/12=1/6=0.1667\). That likelihood uses the original-scale values and an unconditional survival probability; it does not describe risks selected by exceeding \(2\). Both estimates are rounded to four decimal places.
Checks Before Maximizing
A short contribution check often catches an error before any differentiation is needed. For each record, state the observation mechanism, the event or value it reports, and the conditioning event that determined inclusion. Then verify that the likelihood has one factor for every record and that its factors agree with those statements.
Write the inclusion event, such as \(X>a\) or \(X<b\), and determine whether each record was sampled regardless of its loss.
An exact value uses a density. A right-censored value uses an upper-tail probability. A truncated record uses a conditional contribution given inclusion.
Multiply the record-level contributions, take logs if useful, and check that the conditioning denominators and thresholds match the design.
A frequent exam trap is treating “not observed exactly” as though it always meant the same thing. A right-censored record says the loss reached at least a specified threshold, so its contribution is an upper-tail probability. A left-truncated record says the loss passed an entry condition before it could appear in the sample, so the distribution of its observed value is conditional on that condition. Replacing one mechanism with the other changes the likelihood.
Another useful check is to inspect the exponential log likelihood. With exact left-truncated observations, the exponential part involves excesses above the entry thresholds. With right-censored observations, the likelihood includes the time or loss up to the censoring threshold in the survival contribution, but no density factor for an unobserved exact value. These checks do not replace the likelihood derivation; they test whether the result is consistent with it.
Check Your Understanding
For each problem, identify the observation mechanism before forming or maximizing the likelihood.
- A hypothetical exponential-loss study includes only losses above \(3\). The exact observations are \(5,7,\) and \(10\). Find the maximum likelihood estimate of \(\lambda\). (Answer: \(\widehat{\lambda}=3/13\approx0.2308\).)
- In a hypothetical untruncated study, two exact exponential losses are \(1\) and \(3\), and a third record is right-censored at \(4\). Find the maximum likelihood estimate of \(\lambda\). (Answer: \(\widehat{\lambda}=2/(1+3+4)=0.2500\).)
- A data source retains only losses below \(5\), and its exact exponential observations are \(1,2,\) and \(3\). Write the likelihood and score equation for \(\lambda\). (Answer: \(L(\lambda)=\lambda^3e^{-6\lambda}/(1-e^{-5\lambda})^3\); score equation \(3/\lambda-6-15/(e^{5\lambda}-1)=0\).)
- A hypothetical study includes losses above \(1\); it records exact values \(2.5\) and \(4\), and a third loss is right-censored at \(5\). Assuming an exponential model, find \(\widehat{\lambda}\). (Answer: \(2/[(2.5-1)+(4-1)+(5-1)]=4/17\approx0.2353\).)
- A study includes risks only after their losses exceed their individual entry thresholds. It records an exact loss of \(4\) after entry at \(1\), an exact loss of \(6\) after entry at \(2\), and a loss right-censored at \(5\) after entry at \(1\). Find the exponential rate estimate. (Answer: \(2/[(4-1)+(6-2)+(5-1)]=2/11\approx0.1818\).)