Likelihood Under Double Truncation

Why Double Truncation Changes the Likelihood

A loss record may enter a data set only when its value falls within a specified range. For example, a hypothetical audit might retain claims above a reporting threshold but below a separate review ceiling. Claims outside that range are absent from the data; their values are not recorded as censored observations. When the lower and upper bounds differ by subject, each observed loss has its own selection probability and therefore its own likelihood normalizer.

This tutorial treats exact losses observed between known, subject-specific bounds. It does not treat censored records, for which a value is known only to lie beyond a threshold or within an interval. A conditional likelihood is appropriate when the observed record is an exact loss and inclusion occurs precisely when that loss lies inside its stated interval.

Definition: Let $X_i$ be a continuous loss with density $f_X(x;\theta)$ and distribution function $F_X(x;\theta)$. For subject $i$, suppose the known bounds satisfy $L_i<U_i$, and the record is observed only if $L_i<X_i<U_i$. The conditional density of an observed exact value $x_i$ is $$ f_{X_i\mid L_i<X_i<U_i}(x_i;\theta) = \frac{f_X(x_i;\theta)} {F_X(U_i;\theta)-F_X(L_i;\theta)}, \qquad L_i<x_i<U_i, $$ provided $F_X(U_i;\theta)>F_X(L_i;\theta)$. It is zero outside the interval. The denominator is the probability that a loss from the model falls between the two bounds.

The bounds may differ across records. The numerator uses the density at the exact observed loss; the denominator uses the probability mass in that record’s own interval. The density and distribution function are evaluated under the same model and parameter value. When the observations are independent before selection and selection for each record depends only on its loss falling within its stated interval, the conditional likelihood is the product of these conditional densities:

L(\theta) = \prod_{i=1}^{n} \frac{f_X(x_i;\theta)} {F_X(U_i;\theta)-F_X(L_i;\theta)}.

This product is a likelihood for the observed exact values conditional on their having been selected. It is not the likelihood for the unobserved losses outside the intervals. The denominator must be positive for every record under the parameter values being considered.

Why the Interval Probability Is the Normalizer

The conditional-density formula follows directly from conditional probability. For $L_i<x<U_i$, the event $X_i\le x$ together with $L_i<X_i<U_i$ is the event $L_i<X_i\le x$. Its probability is $F_X(x;\theta)-F_X(L_i;\theta)$. Dividing by the probability of selection gives the conditional distribution function:

P(X_i\le x\mid L_i<X_i<U_i) = \frac{F_X(x;\theta)-F_X(L_i;\theta)} {F_X(U_i;\theta)-F_X(L_i;\theta)}.

Differentiating with respect to $x$ gives the conditional density $f_X(x;\theta)/[F_X(U_i;\theta)-F_X(L_i;\theta)]$. This also verifies that it is a valid density: its integral over the interval is

\int_{L_i}^{U_i} \frac{f_X(x;\theta)} {F_X(U_i;\theta)-F_X(L_i;\theta)}\,dx = \frac{F_X(U_i;\theta)-F_X(L_i;\theta)} {F_X(U_i;\theta)-F_X(L_i;\theta)} =1.

For a continuous model, whether an endpoint is included does not change the probability because a single point has probability zero. If a model has point masses, however, endpoint conventions can matter; the continuous-density formula must not be applied blindly to a discrete or mixed distribution.

Write the selection event.
For record $i$, the observed value must satisfy $L_i<x_i<U_i$. Check this before calculating a likelihood contribution.

Find the exact-value density.
Evaluate $f_X(x_i;\theta)$ at the observed loss, not at a bound or at an interval endpoint.

Find that record’s selection probability.
Calculate $F_X(U_i;\theta)-F_X(L_i;\theta)$, using the record’s own two bounds.

Normalize and combine.
Divide the density by the interval probability for each record, then multiply the contributions if the records are conditionally independent.

Worked Likelihoods Under Double Truncation

Worked Example: Two Pareto Losses With Different Observation Windows

A hypothetical portfolio records losses in thousands of dollars. Assume independent losses have a Pareto Type I distribution with minimum $\theta=1$ thousand dollars and shape $\alpha=2$, so $f_X(x)=2x^{-3}$ and $F_X(x)=1-x^{-2}$ for $x\ge1$. The first exact loss is $x_1=2$, observed only when $1<X_1<3$. The second is $x_2=3$, observed only when $2<X_2<4$. Find each conditional density and the joint conditional likelihood at these parameter values.

First record. Its density at the observed value is $f_X(2)=2(2)^{-3}=2/8=1/4$. Its interval probability is $F_X(3)-F_X(1)=(1-1/9)-(1-1)=8/9$. Therefore its conditional density is

\frac{1/4}{8/9} = \frac{1}{4}\frac{9}{8} = \frac{9}{32} = 0.28125 \quad\text{per thousand dollars}.

Second record. Its density is $f_X(3)=2(3)^{-3}=2/27$. The interval probability is $F_X(4)-F_X(2)=(1-1/16)-(1-1/4)=15/16-3/4=3/16$. Thus

\frac{2/27}{3/16} = \frac{2}{27}\frac{16}{3} = \frac{32}{81} \quad\text{per thousand dollars}.

The records are independent under the stated model, so their joint conditional likelihood is $(9/32)(32/81)=9/81=1/9$, in units of inverse squared thousands of dollars. Both observations lie strictly within their stated intervals, and each conditional density uses the corresponding interval probability.

Worked Example: Uniform Losses and Record-Specific Normalizers

A hypothetical severity model is uniform on $(0,\theta)$, with $\theta=10$ thousand dollars. Two independent exact losses are recorded: $x_1=2$ with bounds $1<X_1<4$, and $x_2=5$ with bounds $2<X_2<6$. Calculate the conditional densities and their product.

For a uniform $(0,10)$ loss, the density is $1/10$ throughout the support and the probability of an interval is its width divided by 10. For the first record, the selection probability is $(4-1)/10=3/10$, so its conditional density is $(1/10)/(3/10)=1/3$ per thousand dollars. The observed value $2$ lies in $(1,4)$.

For the second record, the selection probability is $(6-2)/10=4/10$, giving conditional density $(1/10)/(4/10)=1/4$ per thousand dollars. The observed value $5$ lies in $(2,6)$. The joint conditional likelihood is therefore

\left(\frac{1}{3}\right)\left(\frac{1}{4}\right) = \frac{1}{12} \quad\text{per squared thousands of dollars}.

The unequal widths produce unequal conditional densities even though the original density is the same for both losses. As a check, a uniform distribution restricted to an interval has constant density equal to the reciprocal of that interval’s width: $1/3$ and $1/4$ here.

Worked Example: Exponential Losses With Different Bounds

Suppose two hypothetical independent losses, measured in thousands of dollars, have exponential density $f_X(x)=\lambda e^{-\lambda x}$ for $x\ge0$, with $\lambda=\ln 2$ per thousand dollars. The first exact loss is $x_1=2$ with bounds $1<X_1<3$; the second is $x_2=1$ with bounds $0<X_2<2$. Calculate both conditional densities and the joint likelihood.

At this rate, $e^{-\lambda}=1/2$, $e^{-2\lambda}=1/4$, and $e^{-3\lambda}=1/8$. For an exponential loss, the interval probability is $F_X(U)-F_X(L)=e^{-\lambda L}-e^{-\lambda U}$.

For the first record, the exact-value density is $f_X(2)=\lambda e^{-2\lambda}=\lambda/4$. The selection probability is $e^{-\lambda}-e^{-3\lambda}=1/2-1/8=3/8$. Its conditional density is $(\lambda/4)/(3/8)=2\lambda/3$, or approximately $0.4621$ per thousand dollars.

For the second record, $f_X(1)=\lambda/2$, and its interval probability is $1-e^{-2\lambda}=1-1/4=3/4$. Its conditional density is $(\lambda/2)/(3/4)=2\lambda/3$, also approximately $0.4621$ per thousand dollars. Hence the joint conditional likelihood is

\left(\frac{2\lambda}{3}\right)^2 = \frac{4(\ln 2)^2}{9} \approx 0.2135 \quad\text{per squared thousands of dollars}.

The equality of the two contributions is a consequence of these particular intervals and observed values under the specified rate; it is not a general property of exponential losses. The interval probabilities were calculated separately even though the resulting conditional densities happen to match.

Common Errors and When the Formula Applies

The most frequent errors are using the wrong density value, normalizing by only one bound, or applying the same interval probability to records with different bounds. For a continuous model, the correct denominator can also be written $S_X(L_i)-S_X(U_i)$, since $S_X(x)=1-F_X(x)$. It is the probability between the bounds, not the probability above the lower bound alone.

Do not confuse truncation with censoring. Under truncation, values outside the interval are not included as records. Under censoring, a record is present but its value is only partially known. These observations have different likelihood contributions.
Do not replace a record’s bounds with pooled bounds. If $L_i$ and $U_i$ vary, calculate a separate $F_X(U_i;\theta)-F_X(L_i;\theta)$ for each $i$.
Check support and inclusion. The observed value must be inside its interval and in the model’s support. The interval probability must be positive.
Keep density units in view. A conditional density has inverse-loss units; a product of $n$ densities has units equal to inverse loss units raised to the $n$th power.

The conditional likelihood relies on the stated observation rule: each exact loss is included precisely when it lies between its known bounds. If selection also depends on another variable or an unmodeled feature, conditioning only on the interval may not describe the sampling process. Likewise, if the bounds themselves are unknown or their generation carries information about $\theta$, this conditional likelihood alone may not capture all relevant information.

Key takeaway: For an exact loss observed only inside $(L_i,U_i)$, use its density at the observed value divided by the model probability of that same interval. With subject-specific bounds, form a separate normalizer for every record before multiplying the conditional contributions.

Check Your Understanding

For each problem, use the stated conditional observation rule and show the density and interval probability before forming the conditional contribution.

A Pareto Type I loss in thousands of dollars has minimum $\theta=1$ and shape $\alpha=3$. An exact loss $x=1.5$ is observed only when $1<X<2$. Find its conditional density. (Answer: $128/189\approx0.6772$ per thousand dollars.)
A loss is uniform on $(0,12)$ thousand dollars. An exact value $x=6$ is observed only when $3<X<9$. Find its conditional density. (Answer: $1/6$ per thousand dollars.)
An exponential loss has rate $\lambda=0.25$ per unit. An exact value $x=4$ is observed only when $2<X<6$. Calculate its conditional density to four decimal places. (Answer: $\frac{0.25e^{-1}}{e^{-0.5}-e^{-1.5}}\approx0.2399$ per unit.)
A uniform loss on $(0,15)$ thousand dollars gives independent exact observations $x_1=4$ with bounds $(0,5)$ and $x_2=8$ with bounds $(5,11)$. Find their joint conditional likelihood. (Answer: $(1/5)(1/6)=1/30$ per squared thousands of dollars.)
A Pareto Type I loss in thousands of dollars has minimum $\theta=1$ and shape $\alpha=2$. Independent exact losses $x_1=2$ and $x_2=3$ are observed within respective intervals $(1,4)$ and $(2,5)$. Find the joint conditional likelihood. (Answer: $(4/15)(200/567)=160/1701\approx0.0941$ per squared thousands of dollars.)

What You'll Learn