Likelihood Contributions Under Left Truncation

Why the Entry Threshold Changes the Likelihood

Suppose an insurer records a loss only when it exceeds a known reporting threshold. The recorded losses are then not an ordinary sample from the full loss distribution: every recorded value has already passed the threshold. A likelihood that uses the unconditional density alone ignores that selection and generally gives the wrong model for the observed data.

This situation is called left truncation. A loss below the threshold is absent from the data, rather than being recorded as a value equal to the threshold. For an exact observed loss, the appropriate likelihood contribution is its density conditional on exceeding the threshold. The key adjustment is the survival-probability normalizer.

Definition: Let $X$ be a continuous loss with density $f_X$, distribution function $F_X$, and survival function $S_X(t)=P(X>t)$. If a record is observed only when $X>t$, then its conditional density at an observed value $x>t$ is $$ f_{X\mid X>t}(x)=\frac{f_X(x)}{S_X(t)},\qquad x>t. $$ The conditional likelihood contribution is $f_X(x)/S_X(t)$, not $f_X(x)$.

The threshold $t$ is assumed known. For $n$ independent exact losses $x_1,\ldots,x_n$, each with its own known threshold $t_i$, the conditional likelihood is

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

Here $\theta$ denotes the unknown model parameter or parameters. Each observation is conditioned on passing its own threshold; one does not generally use a single common denominator when the thresholds differ. The corresponding log-likelihood is

\ell(\theta)=\sum_{i=1}^{n}\left[\log f_X(x_i;\theta)-\log S_X(t_i;\theta)\right].

This form assumes that the observed losses are independent conditional on their known thresholds and that the thresholds are noninformative for the loss beyond the fact that each loss had to exceed its threshold. If the threshold mechanism depends on the unobserved loss in some additional way, or if the threshold is unknown, this conditional likelihood alone may not describe the sampling process.

Why the Survival Normalizer Appears

For a continuous loss, the conditional probability that $X$ falls in a small interval around $x$, given that it exceeded $t$, is the unconditional probability of that interval divided by the probability of passing the threshold. Taking the interval width to zero gives the conditional density.

\begin{aligned} P(x<X\le x+h\mid X>t) &=\frac{P(x<X\le x+h,\ X>t)}{P(X>t)}\\ &=\frac{F_X(x+h)-F_X(x)}{S_X(t)},\qquad x>t,\\ f_{X\mid X>t}(x) &=\lim_{h\to 0^+}\frac{P(x<X\le x+h\mid X>t)}{h} =\frac{f_X(x)}{S_X(t)}. \end{aligned}

The denominator is not an optional correction: it makes the conditional density integrate to one over the values that can be observed:

\int_t^\infty \frac{f_X(x)}{S_X(t)}\,dx =\frac{S_X(t)}{S_X(t)}=1.

For a discrete loss, use conditional probability mass instead: $P(X=x\mid X>t)=P(X=x)/P(X>t)$ for $x>t$. If the model has probability mass exactly at the threshold, the inclusion rule must be stated carefully: conditioning on $X\ge t$ uses $P(X\ge t)$, whereas conditioning on $X>t$ uses $P(X>t)$. For a continuous loss, the distinction at a single point does not change the probabilities.

Worked Likelihoods Under Different Loss Models

Worked Example: Exponential Losses With Different Thresholds

A hypothetical portfolio records three independent exact losses only when they exceed their known reporting thresholds. The thresholds are $t=(100,200,100)$ dollars and the observed losses are $x=(250,350,180)$ dollars. Assume $X\sim\operatorname{Exponential}(\lambda)$, with density $f_X(x;\lambda)=\lambda e^{-\lambda x}$ and survival function $S_X(t;\lambda)=e^{-\lambda t}$. There is no interest or mortality assumption in this loss-only model. Find the conditional likelihood and its maximum-likelihood estimate of $\lambda$.

Method and contributions. For each record, divide the exponential density by its survival probability at that record's threshold:

\frac{f_X(x_i;\lambda)}{S_X(t_i;\lambda)} =\frac{\lambda e^{-\lambda x_i}}{e^{-\lambda t_i}} =\lambda e^{-\lambda(x_i-t_i)}.

The excesses above threshold are $250-100=150$, $350-200=150$, and $180-100=80$ dollars. Thus the joint conditional likelihood is

L(\lambda)=\lambda^3e^{-\lambda(150+150+80)} =\lambda^3e^{-380\lambda}.

Estimate. The log-likelihood is $\ell(\lambda)=3\log\lambda-380\lambda$. Its derivative is $3/\lambda-380$, so setting it equal to zero gives $\hat{\lambda}=3/380=0.0078947$ per dollar. The second derivative is $-3/\lambda^2<0$, confirming a maximum. At this estimate, $L(\hat{\lambda})=(3/380)^3e^{-3}\approx 2.45\times10^{-8}$, in density units per dollar cubed.

Check and trap avoided. The estimate uses three observations and their total excess, $380$, not their total loss, $780$. Using the unconditional likelihood $\prod_i\lambda e^{-\lambda x_i}$ would give $3/780$, incorrectly treating the sample as if losses below the thresholds could have been observed.

Worked Example: Pareto Losses and Threshold Ratios

A hypothetical excess-loss study records two independent exact losses, $x_1=200$ and $x_2=400$ dollars, after entry thresholds $t_1=100$ and $t_2=200$ dollars. Assume a Pareto Type I model with known scale $\theta=50$ dollars and unknown shape $\alpha>0$: $f_X(x;\alpha)=\alpha\theta^\alpha x^{-(\alpha+1)}$ and $S_X(t;\alpha)=(\theta/t)^\alpha$ for $x,t\ge\theta$. There is no interest or mortality assumption. Find the conditional likelihood and the maximum-likelihood estimate of $\alpha$.

Condition each density. For a record with threshold $t$ and observed loss $x>t$,

\frac{f_X(x;\alpha)}{S_X(t;\alpha)} =\frac{\alpha\theta^\alpha x^{-(\alpha+1)}}{(\theta/t)^\alpha} =\frac{\alpha}{x}\left(\frac{t}{x}\right)^\alpha.

The scale parameter cancels because both the density and the probability of passing the threshold contain $\theta^\alpha$. Here both observed-to-threshold ratios are $2$, so

L(\alpha) =\left[\frac{\alpha}{200}\left(\frac12\right)^\alpha\right] \left[\frac{\alpha}{400}\left(\frac12\right)^\alpha\right] =\frac{\alpha^2}{80{,}000}\,2^{-2\alpha}.

Estimate. The log-likelihood is $2\log\alpha-\log(80{,}000)-2\alpha\log 2$. Differentiating gives $2/\alpha-2\log 2=0$, hence $\hat{\alpha}=1/\log 2\approx1.4427$. The second derivative is $-2/\alpha^2<0$, so this is the maximum. Substitution gives $L(\hat{\alpha})=(1/\log2)^2e^{-2}/80{,}000\approx3.52\times10^{-6}$.

Check and trap avoided. The result depends on the ratios $x_i/t_i$, as the conditional density shows. Keeping the unconditional Pareto density without dividing by $S_X(t_i;\alpha)$ would leave a spurious dependence on the known scale and ignore the fact that only losses above each threshold entered the data.

Worked Example: Uniform Losses With Unequal Entry Thresholds

A hypothetical inspection program records an exact loss only if it exceeds the applicable threshold. Two independent recorded losses are $x_1=12$ and $x_2=16$ dollars, with thresholds $t_1=4$ and $t_2=10$ dollars. Assume $X\sim\operatorname{Uniform}(0,20)$, with known upper endpoint $20$ dollars. There is no interest or mortality assumption. Calculate the conditional likelihood and compare it with the incorrect unconditional likelihood.

Conditional densities. For $0\le t<20$ and $t<x<20$, the uniform density is $1/20$ and $S_X(t)=(20-t)/20$. Therefore

f_{X\mid X>t}(x)=\frac{1/20}{(20-t)/20}=\frac{1}{20-t}.

For the first observation, the contribution is $1/(20-4)=1/16$. For the second, it is $1/(20-10)=1/10$. The joint conditional likelihood is $L=(1/16)(1/10)=1/160=0.00625$ per dollar squared.

Comparison. The unconditional density would assign $1/20$ to each observation and give $1/400=0.0025$ per dollar squared. That is not the likelihood for this selected sample: among losses known to exceed 10 dollars, for example, the conditional density is $1/10$, not $1/20$. The conditional densities also pass a normalization check: the density for the second record integrates to one over $(10,20)$, since $\int_{10}^{20}(1/10)\,dx=1$.

Check and trap avoided. A threshold changes the distribution of the recorded loss even though it does not change the underlying uniform model. Dividing by the survival probability restores a valid density on the range eligible for observation.

Applying the Conditional Likelihood Correctly

A reliable calculation starts by identifying exactly what is observed and what determines inclusion. When the observation is an exact loss $x_i$ that was recorded because $x_i>t_i$, its contribution is a conditional density. The following sequence keeps the threshold and the model parameter in the right places.

State the observation rule.
Write the known entry threshold for each record and specify whether inclusion is $X>t_i$ or $X\ge t_i$. Confirm that the observed loss is exact, rather than merely known to exceed a value.

Write the unconditional model.
Identify $f_X(x;\theta)$ and $S_X(t;\theta)$ using the same parameterization and units. Check that each observed loss lies in the model support and exceeds its threshold.

Condition record by record.
For each $i$, form $f_X(x_i;\theta)/S_X(t_i;\theta)$. Retain a separate survival denominator whenever entry thresholds differ.

Combine and check.
Multiply the contributions under conditional independence, or sum their logarithms. Check units and confirm that each conditional density integrates to one over the eligible range.

Left truncation is different from censoring. Under truncation, a loss that fails the entry rule is not present in the observed sample at all. Under right censoring, a record is present but its exact loss is not known beyond a censoring point. For exact observations under left truncation, use the conditional density above; do not substitute a survival probability as though the loss were censored.

The likelihood here is conditional on the observed thresholds and inclusion. It estimates the loss distribution from the losses that entered the sample. If the objective also requires modeling how many losses were excluded, the number eligible, or the mechanism generating thresholds, that information requires a model beyond these conditional contributions.

Common Exam Traps

Omitting the normalizer. The incorrect contribution is $f_X(x_i;\theta)$. The correction is $f_X(x_i;\theta)/S_X(t_i;\theta)$, because the record is known to have passed the threshold.
Using one denominator for unequal thresholds. The correct likelihood has $S_X(t_i;\theta)$ in the denominator for observation $i$. A common denominator is appropriate only when all records share the same threshold.
Subtracting thresholds from every loss without checking the model. The exponential conditional density depends on $x_i-t_i$ because of its memoryless form. That simplification does not hold for every loss distribution; for a Pareto model, for example, the contribution depends on $t_i/x_i$.
Confusing truncation with censoring. A threshold is not an observed loss value. If only exact losses above the threshold are recorded, the contribution is conditional density, not the probability of exceeding the threshold alone.
Ignoring support and units. A contribution must be zero if the observation violates the model support or does not exceed its threshold. A density has units inverse to the loss unit; it is not itself the probability of observing an exact dollar amount.

Key takeaway: For an exact continuous loss observed only because it exceeded a known threshold, divide its unconditional density by the probability of exceeding that threshold. With varying thresholds, form and multiply a separate conditional contribution for every record.

Check Your Understanding

For each problem, write the contribution or likelihood conditional on the stated entry rule. Assume independent observations where multiple records are given.

A loss has an exponential distribution with known rate $\lambda=0.02$ per dollar. It is recorded only if it exceeds $50$ dollars, and the exact observed loss is $90$ dollars. Find its conditional density contribution. (Answer: $0.02e^{-0.8}\approx0.0089866$ per dollar.)
A Pareto Type I loss has known scale $\theta=40$ dollars and shape $\alpha=3$. A loss of $200$ dollars is observed after a threshold of $100$ dollars. Find its conditional density contribution. (Answer: $\frac{3}{200}(1/2)^3=0.001875$ per dollar.)
A loss is uniform on $(0,30)$ dollars. Two exact observations, $20$ and $25$, have thresholds $5$ and $15$ dollars, respectively. Find their joint conditional likelihood. (Answer: $(1/25)(1/15)=1/375$ per dollar squared.)
Three exact losses are $70,90,100$ dollars, with respective thresholds $20,50,80$ dollars. Assume an exponential model with unknown rate $\lambda$. Find the maximum-likelihood estimate using the conditional likelihood. (Answer: $\hat{\lambda}=3/(50+40+20)=3/110\approx0.02727$ per dollar.)
Under a Pareto Type I model with unknown shape $\alpha$ and known scale $40$ dollars, exact losses $200,400,500$ dollars are observed above thresholds $100,200,250$ dollars. Find the conditional maximum-likelihood estimate of $\alpha$. (Answer: $\hat{\alpha}=3/(3\log 2)=1/\log 2\approx1.4427$.)

What You'll Learn