Conditional Densities Under Left Truncation

Why Left Truncation Changes the Density

A loss database may include only claims above an entry threshold: for example, a hypothetical portfolio may record a claim only if it exceeds a specified amount. A loss not meeting that condition is absent from the data, rather than recorded as a small value. Consequently, the observed losses do not follow the original severity distribution without adjustment. They follow the distribution conditional on exceeding the threshold.

The central likelihood question is therefore: among losses that qualify for observation, how dense is the distribution at the exact amount recorded? The answer is the original density divided by the probability of qualifying for observation. This tutorial develops that result and applies it to several severity models. The treatment is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $X$ be a continuous loss with density $f_X(x;\theta)$, survival function $S_X(x;\theta)=P(X>x)$, and parameter or parameter vector $\theta$. Under left truncation at a known threshold $t$, a loss is observed only when $X>t$. The observed loss has the conditional density of $X$ given $X>t$, on the support $x>t$.

The threshold is an entry condition, not a deductible applied to the recorded loss. If the exact recorded value is $x$, the record says that the loss equals $x$, as well as that it passed the entry condition. In a continuous model, using $X\ge t$ instead of $X>t$ makes no probability difference at the boundary, since $P(X=t)=0$.

Deriving the Conditional Density

For any $x>t$, the conditional distribution function of an observed loss is

P(X\le x\mid X>t) =\frac{P(t<X\le x)}{P(X>t)} =\frac{F_X(x;\theta)-F_X(t;\theta)}{S_X(t;\theta)}.

Differentiating with respect to $x$ gives the conditional density. Since the derivative of $F_X(x;\theta)$ is $f_X(x;\theta)$, while $F_X(t;\theta)$ and $S_X(t;\theta)$ are constants with respect to $x$, the result is

f_{X\mid X>t}(x;\theta) = \begin{cases} \dfrac{f_X(x;\theta)}{S_X(t;\theta)}, & x>t,\\[6pt] 0, & x\le t. \end{cases}

The denominator is also required for the conditional density to integrate to one. Indeed,

\int_t^\infty \frac{f_X(x;\theta)}{S_X(t;\theta)}\,dx = \frac{P(X>t)}{S_X(t;\theta)} =1,

provided $S_X(t;\theta)>0$. This normalization explains why the density is divided by the probability of passing the threshold.

Theorem: Suppose $X_1,\ldots,X_n$ are independent losses from a continuous model with density $f_X(x;\theta)$, and the data consist of their exact values conditional on each loss exceeding the same known threshold $t$. The likelihood based on the observed exact values $x_1,\ldots,x_n$, all greater than $t$, is $$ L(\theta\mid x_1,\ldots,x_n,\ X_i>t\ \text{for all }i) = \prod_{i=1}^n \frac{f_X(x_i;\theta)}{S_X(t;\theta)}. $$

This likelihood is for a sample whose members are observed conditional on passing the threshold. It does not include a separate count of how many losses failed to pass. If the data also included information about the number of losses below the threshold or the total number exposed to selection, that information would describe a different sampling setup and would need to be represented accordingly.

When estimating $\theta$, the denominator generally depends on $\theta$, even if the threshold $t$ is fixed and known. It must not be dropped merely because it is the same threshold for every record. In contrast, for a fixed model with no parameters being varied, it is simply part of the numerical conditional density.

Worked Examples

Worked Example: Pareto Density Above an Entry Threshold

A hypothetical loss has a Pareto Type I distribution with scale $\theta=50$ dollars and shape $\alpha=3$. Its density and survival function are $f_X(x)=\alpha\theta^\alpha/x^{\alpha+1}$ and $S_X(x)=(\theta/x)^\alpha$ for $x\ge\theta$. Losses enter the data only if they exceed $t=100$ dollars. Find the conditional density for an exact observed loss of $x=200$ dollars. Give the answer per dollar, to six decimal places.

Method. Use the conditional density $f_X(x)/S_X(t)$, since the recorded value is exact and its observation requires exceeding the fixed threshold.

First calculate the original density at 200:

f_X(200) =\frac{3(50)^3}{200^4} =\frac{375{,}000}{1{,}600{,}000{,}000} =0.000234375\ \text{per dollar}.

Next calculate the probability that a loss passes the threshold:

S_X(100) =\left(\frac{50}{100}\right)^3 =(0.5)^3 =0.125.

Therefore the conditional density is

f_{X\mid X>100}(200) =\frac{0.000234375}{0.125} =0.001875\ \text{per dollar}.

Check. The same result follows by simplifying the conditional Pareto density to $\alpha t^\alpha/x^{\alpha+1}$: $3(100)^3/(200)^4=3{,}000{,}000/1{,}600{,}000{,}000=0.001875$. The answer is $0.001875$ per dollar to six decimal places. The common mistake this method avoids is using the original density at 200 as if every loss were observable.

Worked Example: Lognormal Loss With a Median Entry Threshold

A hypothetical loss satisfies $\ln X\sim N((\ln 100)^{} ,(\ln 2)^2)$, where the normal distribution’s second argument is its variance. Thus $\ln X$ has mean $\ln 100$ and standard deviation $\ln 2$. Only losses above $t=100$ dollars are observed. Find the conditional density at $x=200$ dollars, to seven decimal places.

Method. Standardize the logarithm of the exact value to obtain the original lognormal density, calculate the probability of passing the threshold, and divide.

For a lognormal loss with $\ln X\sim N(\mu,\sigma^2)$, its density at $x>0$ is $f_X(x)=\phi((\ln x-\mu)/\sigma)/(x\sigma)$, where $\phi$ is the standard normal density. Here $\mu=\ln100$ and $\sigma=\ln2$. At $x=200$,

\frac{\ln 200-\ln100}{\ln2} =\frac{\ln2}{\ln2} =1, \qquad f_X(200) =\frac{\phi(1)}{200\ln2} =\frac{0.2419707245}{200(0.6931471806)} \approx 0.001745450\ \text{per dollar}.

At the threshold, the standardized logarithm is $(\ln100-\ln100)/\ln2=0$. A standard normal variable exceeds zero with probability $0.5$, so

$$ S_X(100)=0.5. $$

The conditional density is therefore

f_{X\mid X>100}(200) =\frac{0.001745450}{0.5} \approx 0.0034909\ \text{per dollar}.

Check. Dividing by $0.5$ doubles the original density, as expected when conditioning on an event with probability one-half and considering a value above the threshold. Rounded only at the final step, the answer is $0.0034909$ per dollar to seven decimal places. The method avoids the error of using the lognormal density but forgetting that only half the model’s losses enter this sample.

Worked Example: Two Weibull Records and Their Likelihood

A hypothetical loss model has a Weibull distribution with shape $k=2$ and scale $\lambda=100$ dollars, so $S_X(x)=\exp[-(x/\lambda)^k]$ and $f_X(x)=(k/\lambda)(x/\lambda)^{k-1}\exp[-(x/\lambda)^k]$ for $x\ge0$. Two independent exact losses, 150 and 200 dollars, are recorded only when they exceed 100 dollars. Find the conditional likelihood for these two observations, in dollars$^{-2}$, to eight significant figures.

Method. Each record contributes its original density divided by the probability of exceeding 100. The joint likelihood is the product of the two conditional densities.

The common entry probability is

S_X(100)=\exp[-(100/100)^2]=e^{-1}.

At 150 dollars, the original density and conditional density are

f_X(150) =\frac{2}{100}\left(\frac{150}{100}\right)e^{-(1.5)^2} =0.03e^{-2.25}, \qquad f_{X\mid X>100}(150) =\frac{0.03e^{-2.25}}{e^{-1}} =0.03e^{-1.25} \approx 0.00859514\ \text{per dollar}.

At 200 dollars, they are

f_X(200) =\frac{2}{100}\left(\frac{200}{100}\right)e^{-2^2} =0.04e^{-4}, \qquad f_{X\mid X>100}(200) =\frac{0.04e^{-4}}{e^{-1}} =0.04e^{-3} \approx 0.00199148\ \text{per dollar}.

Independence gives the likelihood as the product:

L =(0.03e^{-1.25})(0.04e^{-3}) =0.0012e^{-4.25} \approx 0.000017117081\ \text{dollars}^{-2}.

Check. Each conditional density is the corresponding Weibull density multiplied by $e$, because the probability of entry is $e^{-1}$. The two-record likelihood is their product, not their sum. The answer is approximately $0.000017117081$ dollars$^{-2}$, to eight significant figures. This avoids the common error of applying the entry adjustment only once to the entire sample.

Likelihood Contributions and Common Traps

For $n$ independent exact records above a common threshold, the conditional likelihood can also be written as

L(\theta) = \frac{\prod_{i=1}^n f_X(x_i;\theta)} {\{S_X(t;\theta)\}^n}, \qquad x_i>t.

The exponent $n$ appears because there are $n$ separate conditioning events, one for each observed loss. This expression represents the conditional density of the observed sample. It is not the likelihood for a data set that includes exact observations below the threshold.

Two checks help prevent mistakes. First, verify that every recorded value satisfies $x_i>t$; otherwise the record is incompatible with the stated observation rule. Second, confirm that the denominator is the probability of entry, $S_X(t;\theta)=P(X>t)$, not the probability of falling below the threshold, $F_X(t;\theta)$.

State the selection rule.
Write the known threshold $t$ and confirm that an observation is included only when $X>t$.

Evaluate the original density.
For each exact value $x_i$, calculate $f_X(x_i;\theta)$ under the stated severity model.

Normalize for entry.
Calculate $S_X(t;\theta)$ and divide each exact-value density by it.

Combine records.
For independent observations, multiply the conditional densities. Retain any parameter-dependent denominator when forming a likelihood.

A frequent exam trap is to use $f_X(x)/F_X(t)$. That denominator has the wrong event: it is the probability of being at or below the threshold, whereas inclusion requires exceeding it. Another trap is to divide the product of $n$ densities by $S_X(t)$ only once. The correct denominator is $S_X(t)^n$, because every record is conditionally observed after passing the same selection rule.

Worked Example: Diagnosing a Wrong Pareto Normalizer

Suppose a hypothetical Pareto Type I loss has scale $\theta=40$ dollars and shape $\alpha=2$. Exact losses are recorded only above $t=80$ dollars. For a recorded loss of 160 dollars, a candidate writes $f_{X\mid X>80}(160)=f_X(160)/F_X(80)$. Identify the error and calculate the correct density to six decimal places.

Method. Check which event determines observation. The entry event is $X>80$, so its probability is $S_X(80)$, not $F_X(80)$.

The original density at 160 is

f_X(160) =\frac{2(40)^2}{160^3} =\frac{3{,}200}{4{,}096{,}000} =0.00078125\ \text{per dollar}.

The probability of entry is

S_X(80)=\left(\frac{40}{80}\right)^2=0.25.

Thus

f_{X\mid X>80}(160) =\frac{0.00078125}{0.25} =0.003125\ \text{per dollar}.

Wrong calculation and correction. Since $F_X(80)=1-S_X(80)=0.75$, the proposed calculation would give $0.00078125/0.75\approx0.001042$ per dollar. It is wrong because it conditions on being below 80, not above 80. The correct answer, checked by the simplified conditional Pareto density $2(80)^2/(160)^3$, is $0.003125$ per dollar to six decimal places.

What the Conditional Likelihood Represents

Left truncation changes the population represented by the observed sample. In the original loss population, values below $t$ may occur; in the observed sample, they are absent by design. The conditional density describes the relative concentration of possible values among the losses that did enter. Its units remain inverse dollars when $X$ is measured in dollars, and a multi-record likelihood has the corresponding product units.

This distinction is important when interpreting fitted models and likelihood comparisons. A model that assigns substantial probability to losses below the threshold must account for that probability through the survival denominator. Omitting the denominator makes the data appear to have been sampled directly from the unconditioned severity model. Conversely, inserting a survival denominator when observations were not selected by a threshold would impose a selection mechanism that did not occur.

The derivation here assumes a continuous loss distribution, a known fixed threshold, exact observed loss amounts, and independent observations from the same model. If the recorded value is only known to exceed a limit, it is not an exact observation and its likelihood contribution is not the density at a point. If records have different entry thresholds or an additional censoring mechanism, the likelihood must reflect those features separately.

Key takeaway: Under left truncation at a fixed threshold $t$, an exact observed loss $x>t$ contributes $$ \frac{f_X(x;\theta)}{S_X(t;\theta)} $$ to the conditional likelihood. For $n$ independent records with a common threshold, multiply the $n$ contributions, keeping the survival probability in every denominator.

Check Your Understanding

For each problem, identify the entry probability before calculating the conditional density or likelihood.

A Pareto Type I loss has scale 40 dollars and shape 2. Losses are observed only above 80 dollars. Find the conditional density at 160 dollars, per dollar, to six decimal places. (Answer: 0.003125 per dollar)
A loss is uniformly distributed on the interval from 0 to 300 dollars. It is recorded only above 100 dollars. Find the conditional density at 200 dollars. (Answer: 0.005 per dollar)
A lognormal loss satisfies $\ln X\sim N(\ln80,(\ln2)^2)$. Only losses above 80 dollars enter the data. Find the conditional density at 160 dollars, per dollar, to seven decimal places. (Answer: approximately 0.0043636 per dollar)
A Weibull loss has shape $k=2$, scale $\lambda=100$ dollars, and survival function $S_X(x)=\exp[-(x/\lambda)^k]$. If it is observed only above 100 dollars, find the conditional density at 200 dollars to six decimal places. (Answer: approximately 0.001991 per dollar)
Two independent exact losses from a Pareto Type I model with scale 30 dollars and shape 2 are observed above 60 dollars. The observed values are 120 and 180 dollars. Find their joint conditional likelihood in dollars$^{-2}$, to eight significant figures. (Answer: approximately 0.0000051440329 dollars$^{-2}$)

What You'll Learn