Likelihood Contributions Under Double Truncation

Why Double Truncation Changes the Likelihood

A loss database may include a record only when the loss lies between two thresholds. For example, a hypothetical data system might omit losses below a reporting threshold and also fail to retain losses above a separate upper threshold. The observed losses are therefore not an unconditional sample from the severity distribution: they are values selected by an observation window.

For likelihood calculations, the key is to condition on a loss being inside that window. The likelihood contribution is not just the underlying density evaluated at the observed loss. It is that density divided by the probability that a loss could have been observed in the first place.

Definition: Let $X$ be a continuous loss with parameterized density $f_\theta$ and distribution function $F_\theta$. A loss is double-truncated to the known window $(L,U)$, where $L<U$, if it is recorded only when $L<X<U$. For an observed value $x$ inside that window, its conditional density contribution is $$ g_\theta(x\mid L<X<U) = \frac{f_\theta(x)}{F_\theta(U)-F_\theta(L)}, \qquad L<x<U. $$

The denominator is the probability of inclusion. For a continuous distribution, whether the endpoints are included does not change this probability. The formula assumes that the thresholds are known, fixed with respect to $\theta$, and that records are analyzed conditional on having been observed within their stated windows. It does not describe a censored value: under truncation, losses outside the window are not recorded as observations.

Deriving the Conditional Density

The contribution follows directly from conditional probability. For a small interval of width $h$ around an observed value $x$, the probability that a loss lies in that interval and inside the observation window is approximately $f_\theta(x)h$. The probability of being inside the window is $F_\theta(U)-F_\theta(L)$. Dividing the first probability by the second gives the conditional probability of falling near $x$, given inclusion.

P(x<X<x+h\mid L<X<U) = \frac{P(x<X<x+h)}{P(L<X<U)} \approx \frac{f_\theta(x)h}{F_\theta(U)-F_\theta(L)}.

Divide by $h$ and take the limit as $h$ approaches zero. The resulting conditional density is the stated contribution. This also explains why the denominator cannot be omitted: the conditional density must describe the distribution among losses that passed the observation rule, not among all possible losses.

If $n$ independent observed losses $x_1,\ldots,x_n$ have respective known windows $(L_i,U_i)$, their conditional likelihood is the product of their conditional densities:

L(\theta) = \prod_{i=1}^{n} \frac{f_\theta(x_i)} {F_\theta(U_i)-F_\theta(L_i)}, \qquad L_i<x_i<U_i.

When all records share one window, set $L_i=L$ and $U_i=U$ for every $i$. If windows differ, each observation has its own inclusion probability. This conditional likelihood is appropriate for inference based on the observed records and their known windows; its form should not be replaced by an unconditional likelihood that treats the observed values as an unrestricted sample.

Worked Example: One Exponential Loss in a Fixed Window

A hypothetical portfolio has exponential loss severity with rate $\lambda=0.25$ per loss unit. A loss is recorded only if it is between $L=1$ and $U=5$. Find the likelihood contribution for an observed loss $x=2$. There is no interest or mortality component; the severity model and rate are the stated exponential assumptions.

Step 1: Check the observation window. The observed value satisfies $1<2<5$, so it is a possible record under the truncation rule.

Step 2: Evaluate the unconditional density at the observation. For an exponential distribution, $f_\lambda(x)=\lambda e^{-\lambda x}$. Thus $f_{0.25}(2)=0.25e^{-0.5}\approx0.1516327$.

Step 3: Find the probability of inclusion. The exponential distribution function is $F_\lambda(x)=1-e^{-\lambda x}$ for $x\geq0$. Therefore

F_{0.25}(5)-F_{0.25}(1) = e^{-0.25}-e^{-1.25} \approx 0.7788008-0.2865048 = 0.4922960.

Step 4: Divide by the inclusion probability. The conditional density contribution is $0.1516327/0.4922960\approx0.3080112$, or 0.3080 to four decimal places.

This is the conditional contribution, not the unconditional density $0.1516327$. The method avoids treating a loss that was required to fall inside $(1,5)$ as though it had been sampled without restriction.

Using the Conditional Likelihood to Estimate a Parameter

When the severity model has an unknown parameter, the inclusion probability is part of the likelihood and generally depends on that parameter. It must therefore remain inside the likelihood when maximizing. The next example shows this adjustment for an exponential model with a common window.

Worked Example: Estimating an Exponential Rate Under Truncation

Two hypothetical losses are observed only when they fall in $(0,\ln 4)$. Their values are $x_1=0.500000$ and $x_2=0.727411$. Assume independent exponential losses with unknown rate $\lambda>0$, and treat the window as fixed. There is no interest or mortality component. Estimate $\lambda$ by maximizing the conditional likelihood.

Step 1: Write the conditional likelihood. For one observation $x$ in $(0,b)$, the exponential inclusion probability is $1-e^{-\lambda b}$. With $b=\ln4$, the two-record likelihood is

L(\lambda) = \prod_{i=1}^{2} \frac{\lambda e^{-\lambda x_i}}{1-e^{-\lambda b}}, \qquad b=\ln4.

Step 2: Differentiate the log-likelihood. Let $\sum x_i=0.500000+0.727411=1.227411$. Then

\ell(\lambda) = 2\ln\lambda-\lambda(1.227411)-2\ln(1-e^{-\lambda b}),

and

\ell'(\lambda) = \frac{2}{\lambda}-1.227411 -\frac{2b}{e^{\lambda b}-1}.

Step 3: Solve the likelihood equation. At $\lambda=0.5$, $e^{\lambda b}=e^{0.5\ln4}=2$, so

\ell'(0.5) = 4-1.227411-\frac{2(1.386294)}{2-1} = 4-1.227411-2.772588 \approx0.

The small nonzero remainder is due to rounding the second observation. Its unrounded value is $4-2\ln4-0.500000=0.727411278$, which makes the score exactly zero at $0.5$. Thus the maximum-likelihood estimate is $\hat{\lambda}=0.5$ per loss unit.

Step 4: Check that this is a maximum. The score per observation can be written as the conditional mean loss under the truncated exponential model minus the sample mean. As $\lambda$ increases, that conditional mean decreases at rate equal to minus its conditional variance, which is strictly negative. Therefore the score is decreasing in $\lambda$; its zero is unique and gives the maximum. The sample mean is $1.227411278/2=0.613705639$, matching the conditional mean at $\lambda=0.5$.

The common error is to maximize the unconditional likelihood $\lambda^2e^{-\lambda\sum x_i}$. That would omit the parameter-dependent inclusion probability and would not estimate the rate for the observed, truncated sample.

Different Windows Require Different Denominators

An observation’s window may vary by record—for example, because hypothetical contracts have different reporting thresholds. In that case, the likelihood contribution for each observation uses its own interval probability. A common denominator is justified only when the observation windows are in fact the same.

Worked Example: Combining Contributions From Two Windows

Two independent hypothetical losses follow an exponential distribution with known rate $\lambda=0.4$ per loss unit. The first is observed in $(1,4)$ and has value $x_1=2$; the second is observed in $(0.5,2)$ and has value $x_2=1.5$. Calculate the conditional likelihood. There is no interest or mortality component; the exponential severity model and its rate are specified.

First record. The value $2$ lies in $(1,4)$. Its density numerator is $0.4e^{-0.4(2)}=0.4e^{-0.8}\approx0.1797316$. Its inclusion probability is

F(4)-F(1) = e^{-0.4}-e^{-1.6} \approx 0.6703200-0.2018965 = 0.4684235.

The first conditional density contribution is $0.1797316/0.4684235\approx0.3837$.

Second record. The value $1.5$ lies in $(0.5,2)$. Its density numerator is $0.4e^{-0.4(1.5)}=0.4e^{-0.6}\approx0.2195247$. Its inclusion probability is

F(2)-F(0.5) = e^{-0.2}-e^{-0.8} \approx 0.8187308-0.4493290 = 0.3694018.

The second conditional density contribution is $0.2195247/0.3694018\approx0.5943$. Multiplying the contributions gives the conditional likelihood $0.3836947\times0.5942708\approx\mathbf{0.2280}$, rounded to four decimal places.

The calculation checks both observations against their own windows. Using the first record’s denominator for the second would condition on the wrong observation event.

Common Exam Traps

A reliable calculation follows the same sequence for every observed loss: identify its window, evaluate its density, calculate its probability of inclusion, and divide. When estimating parameters, form the full product before maximizing so that any parameter dependence in the denominator is retained.

Using the unconditional density as the contribution. In the first worked example, $0.1516327$ is $f(2)$, not the conditional contribution. The correct calculation is $0.1516327/0.4922960=0.3080$.
Using the wrong interval probability. For a window $(L,U)$, the denominator is $F(U)-F(L)$, not $F(U)$ alone or $1-F(L)$. Those expressions describe different events.
Reusing one denominator across different windows. If $L_i$ or $U_i$ changes with $i$, calculate $F_\theta(U_i)-F_\theta(L_i)$ for that record.
Dropping a parameter-dependent denominator. Even when the observed values are fixed, the inclusion probability can change with $\theta$. Omitting it changes the likelihood equation and generally the estimate.

Key takeaway: Under double truncation, an observed continuous loss contributes its underlying density divided by the probability of falling inside its known observation window. Multiply these conditional contributions across records, using each record’s own window when thresholds vary.

Check Your Understanding

For each problem, use the conditional likelihood contribution and retain the specified observation window.

A hypothetical exponential loss has rate $\lambda=0.2$, and only losses in $(2,7)$ are observed. Find the contribution for $x=4$. (Answer: 0.2121)
An exponential loss has rate $\lambda=0.5$ and is observed only in $(0,\ln4)$. Find the conditional density contribution at $x=\ln2$. (Answer: 0.7071)
Two independent exponential losses have rate $\lambda=0.3$. The first is observed as $2$ in $(1,4)$; the second is observed as $3$ in $(2,6)$. Find each conditional contribution and their product. (Answer: 0.3745 and 0.3180; likelihood 0.1191)
Two exponential observations are recorded only in $(0,\ln4)$, and their values are $0.400000$ and $1.6-2\ln4/3$. Find the conditional maximum-likelihood estimate of the rate. (Answer: $\hat{\lambda}=1$)
An exponential loss has rate $\lambda=0.5$, is observed only in $(1,3)$, and has observed value $2$. Calculate the inclusion probability and the conditional density contribution. (Answer: inclusion probability $0.3834$; contribution $0.4798$)

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