TVaR Estimation With Censored Or Truncated Loss Data

Why Incomplete Loss Data Matter

Loss records are not always complete. A policy limit may mean that a claim above a stated amount is recorded only at that amount, while a database may omit claims below a reporting threshold. These mechanisms change what the observed sample says about the loss distribution. In particular, a sample that conceals or excludes large losses can make the upper tail look lighter than it is.

TVaR depends on losses in that tail. When observations are censored or truncated, the observed losses alone may not identify the target TVaR. An estimate therefore needs either an appropriate adjustment supported by the observation process or a fitted model for the unobserved portion. Any model-based result should make clear which part is estimated from data and which part depends on assumptions.

Definition: A loss $X$ is right-censored at a threshold $c$ when a loss above $c$ is known to exceed $c$, but its exact value is not observed. The record may show $c$ together with an indicator that the loss was censored. A loss is left-truncated at $L$ when it is included in the data only if $X>L$. It is right-truncated at $R$ when it is included only if $X\le R$. Censoring records partial information about a loss; truncation changes which losses enter the sample.

For example, if a claim payment is capped at a policy limit $c$, the recorded amount may be $\min(X,c)$. A recorded amount equal to $c$ is not necessarily an actual loss of exactly $c$: it may represent any loss at least that large. By contrast, if a claims database contains only losses above $L$, the absence of smaller losses does not mean those losses were zero. They were not sampled.

TVaR and the Information the Tail Requires

Write $q_\alpha=\operatorname{VaR}_\alpha(X)$ for the loss quantile at probability level $\alpha$. For a continuous loss distribution with a finite mean above $q_\alpha$, TVaR is the expected loss conditional on exceeding that quantile. Its tail-integral form is

\operatorname{TVaR}_\alpha(X) =E[X\mid X>q_\alpha] =q_\alpha+\frac{1}{1-\alpha}\int_{q_\alpha}^{\infty} S_X(x)\,dx, \qquad S_X(q_\alpha)=1-\alpha.

The formula shows why censoring beyond the target quantile is especially consequential: TVaR requires the area under the survival curve above $q_\alpha$, not merely the number of observations that reach the quantile. The displayed conditional-expectation formula is for continuous distributions. If there is probability mass at VaR, use the fractional VaR-mass convention rather than conditioning only on $X>q_\alpha$.

Justification: For a continuous loss, write $X=q_\alpha+(X-q_\alpha)$ conditional on $X>q_\alpha$. The conditional mean of the excess is the integral of its conditional survival function: $$ E[X-q_\alpha\mid X>q_\alpha] =\int_0^\infty P(X-q_\alpha>y\mid X>q_\alpha)\,dy =\frac{1}{S_X(q_\alpha)}\int_{q_\alpha}^\infty S_X(x)\,dx. $$ Since $S_X(q_\alpha)=1-\alpha$, adding $q_\alpha$ gives the stated formula. A finite TVaR requires this tail integral to be finite.

Likelihoods for Censoring and Truncation

Suppose each loss $X_i$ is subject to an independent censoring threshold $C_i$, and the record includes $T_i=\min(X_i,C_i)$ and an indicator $\delta_i$, where $\delta_i=1$ means the loss was observed exactly and $\delta_i=0$ means it was censored. Under independent, non-informative censoring, the loss-model contribution to the likelihood is

L_i(\theta)=f_X(T_i;\theta)^{\delta_i}S_X(T_i;\theta)^{1-\delta_i}.

An exact observation contributes its density. A censored observation contributes the probability that the loss reached at least the censoring time. This likelihood does not require the censored value to be the true loss. It assumes the censoring mechanism does not depend on the unobserved loss after accounting for the stated data. If that assumption is unreasonable, the censoring mechanism must also be modeled.

For left truncation at a known threshold $L$, an observed loss has the conditional density

f_{X\mid X>L}(x;\theta) =\frac{f_X(x;\theta)}{S_X(L;\theta)}, \qquad x>L.

This conditional likelihood accounts for the fact that the sample contains only selected losses. It does not automatically reveal how many losses were excluded. If the threshold is unknown, selection depends on unrecorded factors, or the count of eligible but unobserved losses is needed, additional information or a model is required. The corresponding density under right truncation at $R$ is $f_X(x;\theta)/F_X(R;\theta)$ for $x\le R$.

Worked Examples

Worked Example: Fitting an Exponential Model to Right-Censored Losses

A hypothetical portfolio has loss times $T=2,4,4,5,5$, measured in years. The first two losses are observed exactly; the other three are censored at their recorded times. Assume independent, non-informative censoring and an exponential loss-time model with density $f_X(x)=\lambda e^{-\lambda x}$ and survival function $S_X(x)=e^{-\lambda x}$, for $x\ge0$. Estimate the 80th-percentile TVaR.

Method. Use the censored likelihood. There are $d=2$ exact losses, and the total observed time at risk is $2+4+4+5+5=20$. Ignoring factors that do not depend on $\lambda$, the likelihood and log-likelihood are

L(\lambda)=\lambda^2 e^{-20\lambda}, \qquad \ell(\lambda)=2\ln(\lambda)-20\lambda.

Differentiate and set the derivative to zero: $2/\lambda-20=0$, giving $\hat{\lambda}=2/20=0.1$ per year. For an exponential distribution, $S_X(x)=e^{-\lambda x}$, so $q_{0.8}=-\ln(0.2)/\hat{\lambda}=\ln(5)/0.1=16.0944$ years.

The exponential distribution is memoryless: $E[X-q\mid X>q]=1/\lambda$. Thus the estimated TVaR is $16.0944+1/0.1=26.0944$ years, or 26.09 years to two decimal places. As a check, the fitted survival probability at the estimated quantile is $e^{-0.1(16.0944)}=0.2$, as required. The target quantile exceeds every censoring time, so this estimate relies on the exponential tail model beyond the observed times.

Mistake avoided: the three censored records contribute survival probabilities, not exact losses of 4, 5, and 5 years. Treating them as exact observations would estimate a different distribution and understate the fitted tail.

Worked Example: Estimating a Pareto Tail From Left-Truncated Losses

A hypothetical claims study records only losses above $200$ thousand dollars. Assume the population follows a Pareto model with known scale $100$ thousand dollars and unknown shape $k>0$: $S_X(x)=(100/x)^k$ for $x\ge100$. The three observed losses are $250$, $300$, and $400$ thousand dollars. Fit $k$ using the truncated sample, then estimate the population 90th-percentile TVaR.

Method. Given that a loss exceeds $200$, its conditional survival function is $S_X(x)/S_X(200)=(200/x)^k$, for $x\ge200$. The conditional density is $k(200)^k x^{-(k+1)}$. Its log-likelihood, apart from terms not involving $k$, is $3\ln(k)-k\sum_i\ln(x_i/200)$. Therefore the maximum-likelihood estimate is

\hat{k}=\frac{3}{\ln(250/200)+\ln(300/200)+\ln(400/200)} =\frac{3}{\ln(1.25)+\ln(1.5)+\ln(2)} \approx 2.2697.

The scale $100$ is known, so the fitted model specifies the full population tail, including the probability below the truncation threshold. For a Pareto model, $S_X(q)=\alpha_{\text{tail}}=(1-\alpha)$ gives $q_\alpha=100(1-\alpha)^{-1/\hat{k}}$. At $\alpha=0.90$, this is $q_{0.90}=100(10)^{1/2.2697}\approx275.79$ thousand dollars. This is greater than $200$, so the target quantile lies within the observed range of the truncated population.

For $k>1$, the conditional mean above $q_\alpha$ is $kq_\alpha/(k-1)$. Hence the estimated TVaR is $(2.2697/1.2697)(275.79)\approx493.00$ thousand dollars, rounded to two decimal places. The check $S_X(275.79)\approx0.1000$ confirms the quantile. The finite mean condition is met because $\hat{k}>1$.

Mistake avoided: fitting an ordinary, untruncated likelihood to these three losses ignores how they were selected. Also, without a known scale or other information about losses below $200$, a truncated sample alone need not identify the full population quantile.

Worked Example: TVaR Beyond a Policy Limit Using a Fitted Tail

A hypothetical insurer records claim payments only up to a policy limit of $200$ thousand dollars. Above a tail threshold of $100$ thousand dollars, a fitted Pareto model has shape $k=2.5$ and survival function $S_X(x)=(100/x)^{2.5}$. Estimate the 95th-percentile TVaR. Assume this fitted tail model is appropriate beyond the limit and that losses are continuous.

Method. Payments at the limit are right-censored; they do not reveal the amounts above $200$. Under the stated Pareto model, the 95th percentile solves $S_X(q)=0.05$:

q_{0.95}=100(0.05)^{-1/2.5} =100(20)^{0.4} \approx331.45 \text{ thousand dollars}.

The model predicts $S_X(200)=(100/200)^{2.5}\approx0.1768$, so the target quantile is indeed above the policy limit. Since $k=2.5>1$, the Pareto conditional mean above $q$ is $kq/(k-1)$. Thus $\operatorname{TVaR}_{0.95}\approx(2.5/1.5)(331.45)=552.42$ thousand dollars, rounded to two decimal places. The estimate is consistent with the model’s tail probability: 5% of losses exceed the estimated quantile.

Mistake avoided: the capped payment is not the full claim amount. The calculation above estimates losses beyond the limit by extrapolating the stated model; the capped records alone cannot supply those severities.

Choosing and Reporting an Adjustment

The appropriate method depends on what the observation process leaves known. Censoring indicators retain information about how often losses reach a threshold, which a censored likelihood can use. Truncation removes records altogether; a conditional likelihood can adjust the distribution of observed values, but the missing selection fraction may remain unidentified. For TVaR, explicitly check whether the target quantile and the losses above it are represented in the observed data.

Data feature	What is observed	What is needed for TVaR
Right censoring at $c$	Exact losses below $c$, and that censored losses reached $c$	A supported adjustment or tail model for severities above $c$, if relevant to the target tail
Left truncation at $L$	Only losses exceeding $L$	A selection-aware fit; estimating a population quantile may also require the excluded fraction or a full-distribution model
Right truncation at $R$	Only losses at or below $R$	External information or a model for the unobserved upper tail

A transparent estimate should state the target population, censoring or truncation rule, fitted distribution and parameters, and whether the requested TVaR extends beyond the observed support. If estimates depend strongly on a tail assumption, report that dependence rather than presenting the result as directly observed.

Common Exam Trap

In the exponential example, replacing every censored time with an exact loss gives the naive estimate $\hat{\lambda}=5/(2+4+4+5+5)=0.25$. That would imply a 80th-percentile TVaR of $(\ln 5+1)/0.25\approx10.44$ years, rather than the censored-likelihood estimate of $26.09$ years. The correction is to count exact events in the density contribution and retain all observed time at risk in the survival contribution. A censored record says “at least this large,” not “exactly this large.”

Key takeaway: Censoring and truncation are different data mechanisms, and neither can be repaired by treating the observed values as a complete sample. Estimate TVaR with a likelihood or adjustment that reflects the mechanism, state assumptions for any unobserved tail, and check that the target quantile and tail mean are identifiable and finite under the fitted model.

Check Your Understanding

For each problem, identify the observation mechanism before calculating. Amounts are in the units stated.

A hypothetical exponential loss-time study has exact events at 3 and 7 years and right-censored observations at 4 and 6 years. Assume independent censoring and an exponential model. Find the maximum-likelihood rate and the 90th-percentile TVaR. (Answer: $\hat{\lambda}=2/20=0.1$ per year; $\operatorname{TVaR}_{0.90}=(\ln 10+1)/0.1\approx33.03$ years.)
Only losses exceeding $L=10$ are observed from an exponential distribution. Four observed excesses above 10 are 1, 2, 4, and 5. Fit the rate from the conditional truncated sample and estimate the population 95th-percentile TVaR. (Answer: $\hat{\lambda}=4/(1+2+4+5)=1/3$; $\operatorname{TVaR}_{0.95}=3(\ln 20+1)\approx11.99$.)
A continuous Pareto model has known scale $50$ and shape $k=3$. A reporting threshold censors all values at $100$. Estimate the 96th-percentile VaR and TVaR under the model, and state whether the VaR exceeds the censoring threshold. (Answer: $q_{0.96}=50(25)^{1/3}\approx146.20$; $\operatorname{TVaR}_{0.96}=1.5q_{0.96}\approx219.30$; the VaR exceeds 100.)
A database includes only claims above $L$, and the threshold $L$ is known. The claim distribution below $L$ and the fraction of claims exceeding $L$ are unknown. Can the selected sample by itself generally identify the full-population 90th-percentile TVaR? (Answer: No; it describes losses conditional on exceeding $L$, but without further assumptions or information it need not identify the population selection fraction or full-population tail quantile.)
A fitted Pareto tail has shape $k=0.8$. Can its unrestricted TVaR be finite? Explain using the tail-mean condition. (Answer: No; a Pareto tail has finite mean only when $k>1$, so the tail integral and unrestricted TVaR are infinite.)

Data feature	What is observed	What is needed for TVaR
Right censoring at \(c\)	Exact losses below \(c\), and that censored losses reached \(c\)	A supported adjustment or tail model for severities above \(c\), if relevant to the target tail
Left truncation at \(L\)	Only losses exceeding \(L\)	A selection-aware fit; estimating a population quantile may also require the excluded fraction or a full-distribution model
Right truncation at \(R\)	Only losses at or below \(R\)	External information or a model for the unobserved upper tail

What You'll Learn