Left Truncation With Right Censoring

Why Both Mechanisms Matter

A loss study may include only claims above a reporting threshold, while also following each included claim for only a limited period. The first feature is left truncation: losses at or below the threshold are not in the observed sample. The second is right censoring: for some included claims, the exact loss is unknown, but it is known to exceed a specified censoring point.

These mechanisms affect different parts of the likelihood. An exact observation contributes a density, while a censored observation contributes a survival probability. Because every subject is observed only after exceeding the entry threshold, both contributions must be conditional on exceeding that threshold. This tutorial is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $X$ be a continuous loss with density $f_X(x)$ and survival function $S_X(x)=P(X>x)$. Suppose a record is included only if $X>L$, where $L$ is a fixed threshold. For an included record, let $y$ be the observed value and let $\delta=1$ indicate an exact observation; $\delta=0$ indicates right censoring at $y$. Conditional on inclusion, the record contributes $$ \left(\frac{f_X(y)}{S_X(L)}\right)^\delta \left(\frac{S_X(y)}{S_X(L)}\right)^{1-\delta}, \qquad y>L. $$

The exact value is known to equal $y$, so its contribution is the conditional density at $y$. A censored record tells us only that $X>y$, so its contribution is the conditional probability of surviving beyond $y$. In either case, the record is already known to have passed the entry threshold. That conditioning produces the denominator $S_X(L)$.

Building the Conditional Likelihood

Assume $n$ independent losses come from the same continuous model, and each is included only if it exceeds the common threshold $L$. For record $i$, write $y_i$ for its exact value or censoring point and $\delta_i$ for its exact-observation indicator. Conditional on the observed sample size $n$, the likelihood is

L(\theta)= \prod_{i=1}^{n} \left(\frac{f_X(y_i;\theta)}{S_X(L;\theta)}\right)^{\delta_i} \left(\frac{S_X(y_i;\theta)}{S_X(L;\theta)}\right)^{1-\delta_i}.

Here $\theta$ denotes the model parameter or parameters. If $m=\sum_{i=1}^n\delta_i$ records are exact, the likelihood can also be written

L(\theta)= \frac{ \displaystyle\prod_{i:\delta_i=1} f_X(y_i;\theta) \displaystyle\prod_{i:\delta_i=0} S_X(y_i;\theta) }{ S_X(L;\theta)^n }.

There is one normalizing factor for every included record, not just for exact records. In particular, a censored record is still known to have passed the entry threshold. The formula assumes that censoring is non-informative for the loss model: conditional on the stated observation process, censoring does not provide additional information about the unobserved loss. If the censoring mechanism is informative or its parameter is linked to $\theta$, this likelihood alone is not sufficient.

Condition on inclusion.
For every record, the information that $X>L$ contributes the normalizer $S_X(L;\theta)$ to its conditional density or probability.

Classify the record.
Use $f_X(y_i;\theta)$ for an exact loss and $S_X(y_i;\theta)$ for a loss known only to exceed its censoring point.

Multiply, then estimate.
Form the product of conditional contributions and maximize it over the allowed parameter values.

Why the Survival Normalizer Appears

The normalizer follows directly from conditional probability. For an exact observation in a small interval of width $dx$ near $x>L$,

P(x<X\leq x+dx\mid X>L) = \frac{P(x<X\leq x+dx)}{P(X>L)} \approx \frac{f_X(x)\,dx}{S_X(L)}.

Dividing by $dx$ gives the conditional density $f_X(x)/S_X(L)$. For a right-censored record at $c>L$, the observed event is $X>c$, which is a subset of $X>L$. Therefore,

P(X>c\mid X>L)=\frac{P(X>c)}{P(X>L)} =\frac{S_X(c)}{S_X(L)}.

This also shows why the normalizer cannot generally be discarded: it may depend on the unknown parameter. In an exponential model with a fixed threshold, it combines with the numerator in a particularly simple way. In other distributions, the conditional likelihood may retain a nontrivial parameter-dependent denominator.

Worked Examples

Worked Example: Exponential Losses With Mixed Records

A hypothetical study includes losses only when they exceed $L=2$ years. The loss $X$ is exponential with unknown rate $\lambda>0$ per year, so $f_X(x)=\lambda e^{-\lambda x}$ and $S_X(x)=e^{-\lambda x}$. Two losses are exact, at 6 and 8 years; two others are right-censored at 5 and 7 years. Assume independent losses and non-informative censoring. No interest or mortality assumption is involved. Find the conditional likelihood and the maximum likelihood estimate of $\lambda$.

Exact records. For an exact loss $x$, the conditional density is

\frac{f_X(x)}{S_X(2)} =\frac{\lambda e^{-\lambda x}}{e^{-2\lambda}} =\lambda e^{-\lambda(x-2)}.

Thus the exact contributions are $\lambda e^{-4\lambda}$ at 6 and $\lambda e^{-6\lambda}$ at 8. The censoring contributions are

\frac{S_X(5)}{S_X(2)}=e^{-3\lambda}, \qquad \frac{S_X(7)}{S_X(2)}=e^{-5\lambda}.

Multiplying all four contributions gives

L(\lambda) = \lambda^2 e^{-\lambda(4+6+3+5)} = \lambda^2e^{-18\lambda}.

The log likelihood is $\ell(\lambda)=2\ln\lambda-18\lambda$. Its derivative is $2/\lambda-18$, so the stationary point is $\widehat{\lambda}=2/18=1/9$ per year. The second derivative is $-2/\lambda^2<0$, confirming a maximum. The estimate is approximately $0.1111$ per year to four significant figures.

Check. The four times measured beyond entry are $6-2=4$, $8-2=6$, $5-2=3$, and $7-2=5$ years. Their sum is 18 years, matching the exponent in the likelihood. A common mistake is to total these as 14 years; that drops four years of observed exposure and would give the incorrect estimate $2/14$.

Worked Example: Rayleigh Losses and a Parameter-Dependent Normalizer

A hypothetical portfolio study includes losses above $L=1$ unit. Assume $X$ has a Rayleigh distribution with unknown scale $\theta>0$, density $f_X(x)=(2x/\theta^2)e^{-(x/\theta)^2}$, and survival function $S_X(x)=e^{-(x/\theta)^2}$. The exact observations are 2 and 3; two other losses are censored at 2 and 4. Assume independent observations and non-informative censoring. No interest or mortality assumption is involved. Estimate $\theta$.

For an exact value $x$, conditioning on $X>1$ gives

\frac{f_X(x)}{S_X(1)} = \frac{2x}{\theta^2} e^{-(x^2-1)/\theta^2}.

The two exact contributions have constants $2(2)/\theta^2=4/\theta^2$ and $2(3)/\theta^2=6/\theta^2$. Their exponential terms have squared-loss differences $2^2-1^2=3$ and $3^2-1^2=8$. The censoring contributions are $e^{-(2^2-1^2)/\theta^2}=e^{-3/\theta^2}$ and $e^{-(4^2-1^2)/\theta^2}=e^{-15/\theta^2}$. Hence

L(\theta) = \frac{24}{\theta^4} \exp\left(-\frac{3+8+3+15}{\theta^2}\right) = 24\theta^{-4}e^{-29/\theta^2}.

The log likelihood is $\ell(\theta)=\ln 24-4\ln\theta-29/\theta^2$. Setting its derivative $-4/\theta+58/\theta^3$ equal to zero gives $\theta^2=58/4=14.5$, so $\widehat{\theta}=\sqrt{14.5}\approx3.808$ units, to four significant figures. The second derivative at this point is $4/\theta^2-174/\theta^4<0$, confirming a maximum.

Check. The exponent total is $3+8+3+15=29$, and the likelihood has one $\theta^{-2}$ factor for each of the two exact records. Censored records contribute survival probabilities, not density factors.

Worked Example: Pareto Losses Above a Higher Entry Threshold

A hypothetical study observes only losses above $L=2$ units. In the original population, $X$ follows a Pareto Type I distribution with known minimum $x_m=1$ and unknown shape $\alpha>0$: $f_X(x)=\alpha x^{-{\alpha}-1}$ and $S_X(x)=x^{-\alpha}$ for $x\geq1$. Exact losses are 4 and 8, and a third loss is censored at 6. Assume independent losses and non-informative censoring. No interest or mortality assumption is involved. Find the conditional likelihood and the maximum likelihood estimate of $\alpha$.

Since $S_X(2)=2^{-\alpha}$, an exact observation at $x$ contributes

\frac{f_X(x)}{S_X(2)} = \frac{\alpha x^{-\alpha-1}}{2^{-\alpha}} = \frac{\alpha}{x}\left(\frac{2}{x}\right)^\alpha.

The exact contributions at 4 and 8 are $(\alpha/4)(1/2)^\alpha$ and $(\alpha/8)(1/4)^\alpha$. The censored contribution is

\frac{S_X(6)}{S_X(2)} = \left(\frac{2}{6}\right)^\alpha = \left(\frac{1}{3}\right)^\alpha.

Their product is

L(\alpha) = \frac{\alpha^2}{32} \left(\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{3}\right)^\alpha = \frac{\alpha^2}{32}\left(\frac{1}{24}\right)^\alpha.

Thus $\ell(\alpha)=2\ln\alpha-\ln32-\alpha\ln24$. The score equation $2/\alpha-\ln24=0$ gives $\widehat{\alpha}=2/\ln24\approx0.6293$, to four significant figures. Since $\ell''(\alpha)=-2/\alpha^2<0$, this is a maximum.

Check. The coefficient $1/32$ is $1/(4\cdot8)$, from the two exact densities. The three conditional survival ratios combine to $1/24$ raised to $\alpha$. The threshold normalizer is included for all three records.

Common Errors and Practical Checks

The conditional likelihood is easiest to audit record by record. Before simplifying, write one factor for each included subject and identify whether it is a density or a survival probability. Then check that every factor has been divided by $S_X(L)$.

Omitting the normalizer. Using $f_X(y)$ for an exact record or $S_X(y)$ for a censored record as though the sample were untruncated generally gives a different likelihood. Retain $S_X(L;\theta)$ unless it is constant in the parameter or cancels legitimately.
Normalizing only exact records. Censored records are also selected by $X>L$. There are $n$ denominator factors, not $m$.
Using a density for censoring. A censoring point $c$ records the event $X>c$, not $X=c$. Its loss-model contribution is $S_X(c)/S_X(L)$.
Mixing scales. In the exponential example, the conditional likelihood uses time beyond entry. For a general distribution, however, the correct factors are the ratios $f_X(x)/S_X(L)$ and $S_X(c)/S_X(L)$; one cannot simply subtract $L$ from every observation and assume the same model still applies.
Ignoring the sampling condition. This likelihood conditions on inclusion and on the observed sample size. If information about how many losses were screened out is also part of the data, that information requires a different likelihood formulation.

Key takeaway: With a common left-truncation threshold $L$, an exact record contributes $f_X(x)/S_X(L)$, and a right-censored record contributes $S_X(c)/S_X(L)$. The denominator appears once per included record because every observed subject is known to have exceeded $L$.

Check Your Understanding

For each problem, condition on inclusion above the stated common threshold and assume non-informative censoring.

Losses are exponential with unknown rate $\lambda$. The threshold is $L=1$; one loss is exact at 3 and one is censored at 2. Write the likelihood and find the MLE of $\lambda$. (Answer: $L(\lambda)=\lambda e^{-3\lambda}$; $\widehat{\lambda}=1/3$.)
A Rayleigh loss has unknown scale $\theta$. The threshold is $L=2$; one loss is exact at 3 and another is censored at 4. Find the MLE of $\theta^2$. (Answer: $\widehat{\theta^2}=17$.)
Losses are exponential with unknown rate $\lambda$, and $L=4$. One exact loss is 6, one exact loss is 10, and one loss is censored at 7. Find the likelihood and MLE. (Answer: $L(\lambda)=\lambda^2e^{-11\lambda}$; $\widehat{\lambda}=2/11$ per unit.)
A continuous loss model has density $f_X$ and survival function $S_X$. At threshold $L=2$, a record is exact at 5 and a second is censored at 4. Give their joint conditional likelihood contribution. (Answer: $f_X(5)S_X(4)/S_X(2)^2$.)
Two losses are exponential with unknown rate $\lambda>0$, and $L=2$. Both records are right-censored, at 3 and 5, with no exact losses. Does the likelihood have a finite positive interior maximum? (Answer: $L(\lambda)=e^{-4\lambda}$; no, it increases toward its supremum as $\lambda$ decreases to 0.)

What You'll Learn