Left Truncation in Deductible-Selected Claims

Why a Deductible Changes the Observed Severity Sample

A severity data set may include only claims whose ground-up losses exceed a deductible. The observed claims are therefore not an ordinary sample from the ground-up severity distribution: losses below the selection threshold could not have entered the data. Fitting the unmodified ground-up density to the observed amounts treats the missing lower losses as if they had never been possible, which can distort the fitted severity model.

The appropriate likelihood depends on what is recorded. Here, each recorded amount is the full ground-up loss $X$, and a claim is included only when $X>d$. The objective is to model the distribution of those selected amounts while accounting for the fact that each one had to exceed its deductible.

Definition: If $X$ has continuous density $f_X(x;\theta)$ and survival function $S_X(d;\theta)=P(X>d)$, then the density of a selected loss, conditional on exceeding deductible $d$, is $$ f_{X\mid X>d}(x;\theta) = \frac{f_X(x;\theta)}{S_X(d;\theta)},\qquad x>d. $$ It is zero for $x\le d$. This is a left-truncated severity distribution: the values below the threshold are absent from the observed sample, rather than recorded as values at the threshold.

For independent selected claims with observed ground-up losses $x_i$ and known thresholds $d_i$, the conditional likelihood is

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

The denominator is claim-specific when deductibles differ. It is the probability that a claim from the ground-up model could have appeared in the data for that observation. A common deductible $d$ gives $S_X(d;\theta)^n$ as the denominator of the product.

Why the Conditional Density Has This Form

For a continuous loss, the probability that $X$ lies in a small interval from $x$ to $x+\Delta x$ is approximately $f_X(x;\theta)\Delta x$. Among claims selected by the event $X>d$, only the proportion $P(X>d)=S_X(d;\theta)$ of all claims is eligible to be observed. Dividing the interval probability by the selection probability gives the conditional interval probability:

P(x<X\le x+\Delta x\mid X>d) \approx \frac{f_X(x;\theta)\Delta x}{S_X(d;\theta)} \quad\text{for }x>d.

Dividing by $\Delta x$ and taking the limit as $\Delta x$ approaches zero yields $f_X(x;\theta)/S_X(d;\theta)$. The conditional density integrates to one over the only possible observed range:

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

This normalization is the reason the survival probability appears in the likelihood. Omitting it uses a density that assigns probability to losses that the data collection process could never have recorded.

Specify the ground-up model.
Write its density and survival function using the same parameters.

Identify the selection event.
For a full ground-up loss selected above deductible $d_i$, the event is $X_i>d_i$.

Condition each observation on selection.
Use $f_X(x_i;\theta)/S_X(d_i;\theta)$, then multiply across independent claims.

Check the observed target.
Use this form only when the recorded amount is the full loss and inclusion is determined by exceeding the stated threshold.

Worked Likelihoods for Common Severity Models

For an exponential ground-up severity with mean $\theta>0$, the density and survival function are $f_X(x;\theta)=\theta^{-1}e^{-x/\theta}$ for $x>0$ and $S_X(d;\theta)=e^{-d/\theta}$. Thus

f_{X\mid X>d}(x;\theta) = \frac{1}{\theta}e^{-(x-d)/\theta}, \qquad x>d.

This is a density for the full observed loss $x$, not for a loss shifted to zero. Its form shows that, under an exponential model, the excess $X-d$ has an exponential distribution with mean $\theta$. This special memoryless property does not hold for a general severity distribution.

Worked Example: Two Exponential Claims With Different Deductibles

Assume independent ground-up losses have an exponential distribution with mean $\theta=\$1{,}000$. A hypothetical data set records full claim losses only when they exceed their respective deductibles. It contains $x_1=\$900$ above $d_1=\$400$ and $x_2=\$1{,}200$ above $d_2=\$700$. Find the conditional likelihood at the stated parameter.

Method. Use the exponential density conditional on exceeding each claim's own deductible. The two excesses are $x_1-d_1=\$500$ and $x_2-d_2=\$500$.

For the first claim, $$ \frac{f_X(900;1000)}{S_X(400;1000)} = \frac{(1/1000)e^{-900/1000}}{e^{-400/1000}} = 0.001e^{-0.5} = 0.0006065307. $$

For the second claim, $$ \frac{f_X(1200;1000)}{S_X(700;1000)} = \frac{(1/1000)e^{-1200/1000}}{e^{-700/1000}} = 0.001e^{-0.5} = 0.0006065307. $$

Independence gives the conditional likelihood $$ L_c=(0.0006065307)^2 =0.0000003678794 \approx 3.679\times 10^{-7}. $$ The likelihood is a product of two densities, so its units are dollars$^{-2}$. As a check, both selected claims have the same excess over their own deductible and therefore have equal conditional densities under this exponential model.

Common mistake avoided: Multiplying the two unconditioned densities would omit the selection probabilities $S_X(400)$ and $S_X(700)$. The likelihood must condition separately on each claim's threshold.

For a gamma ground-up severity with shape $k$ and scale $\theta$, the density for $x>0$ is

f_X(x;k,\theta) = \frac{x^{k-1}e^{-x/\theta}}{\Gamma(k)\theta^k}.

Its survival function is $S_X(d;k,\theta)=1-F_X(d;k,\theta)$. For the special case $k=2$, the survival function simplifies to $e^{-d/\theta}(1+d/\theta)$. In general, the conditional density is the gamma density divided by the gamma survival probability at the deductible; it is not obtained by simply subtracting the deductible from $x$.

Worked Example: Gamma Claims Above Unequal Thresholds

Suppose two independent hypothetical ground-up losses follow a gamma distribution with shape $k=2$ and scale $\theta=3$, in consistent loss units. The observed losses are $x_1=4$ above $d_1=1$, and $x_2=6$ above $d_2=3$. Calculate the conditional likelihood at these parameter values.

Method. For $k=2$, $\Gamma(2)=1$, so $f_X(x)=xe^{-x/3}/9$ and $S_X(d)=e^{-d/3}(1+d/3)$.

For the first observation, $$ f_X(4)=\frac{4e^{-4/3}}{9}, \qquad S_X(1)=e^{-1/3}\left(1+\frac{1}{3}\right) =\frac{4}{3}e^{-1/3}. $$ Therefore, $$ \frac{f_X(4)}{S_X(1)} = \frac{4e^{-4/3}/9}{(4/3)e^{-1/3}} = \frac{1}{3}e^{-1} \approx 0.1226265. $$

For the second observation, $$ f_X(6)=\frac{6e^{-2}}{9}, \qquad S_X(3)=e^{-1}\left(1+\frac{3}{3}\right) =2e^{-1}. $$ Therefore, $$ \frac{f_X(6)}{S_X(3)} = \frac{6e^{-2}/9}{2e^{-1}} = \frac{1}{3}e^{-1} \approx 0.1226265. $$

Multiplying the two conditional densities gives $$ L_c = \left(\frac{1}{3}e^{-1}\right)^2 = \frac{1}{9}e^{-2} \approx 0.01504. $$ This is the likelihood for the two independent observations. The result is checked by simplifying each ratio before multiplying: each ratio contributes $e^{-1}/3$.

Common mistake avoided: The first ratio is not $(4/9)e^{-1}$. Its coefficient is $(4/9)/(4/3)=1/3$. Likewise, the second coefficient is $(6/9)/2=1/3$. Keeping both denominator calculations explicit prevents a factor-of-four error in the product.

Estimating a Parameter From Selected Claims

The truncation adjustment affects parameter estimation, not just the numerical value of a likelihood. For an exponential model with unknown mean $\theta$, observations $x_i>d_i$ have conditional likelihood

L_c(\theta) = \prod_{i=1}^{n}\frac{1}{\theta} e^{-(x_i-d_i)/\theta} = \theta^{-n} \exp\left(-\frac{\sum_{i=1}^{n}(x_i-d_i)}{\theta}\right).

Taking logarithms and differentiating gives the maximum-likelihood estimate. The derivative is zero when $\theta$ equals the average observed excess over the deductible. This follows because conditioning on selection leaves each exponential excess with mean $\theta$.

Worked Example: Exponential Mean From Selected Claim Excesses

Assume three independent hypothetical ground-up losses come from an exponential distribution with unknown mean $\theta$. Full losses are recorded only above the listed thresholds: $x_1=700$ above $d_1=200$, $x_2=900$ above $d_2=500$, and $x_3=1{,}800$ above $d_3=1{,}000$. Estimate $\theta$ by conditional maximum likelihood.

Method. Compute the observed excesses: $700-200=500$, $900-500=400$, and $1{,}800-1{,}000=800$. Their sum is $1{,}700$, and $n=3$.

The conditional log-likelihood is $$ \ell_c(\theta) = -3\log(\theta)-\frac{1700}{\theta}. $$ Its derivative is $$ \frac{d\ell_c}{d\theta} = -\frac{3}{\theta}+\frac{1700}{\theta^2}. $$ Setting this equal to zero gives $1700=3\theta$, so $$ \widehat{\theta} = \frac{1700}{3} = 566.6667. $$

The second derivative at this value is $-3/\widehat{\theta}^{\,2}<0$, confirming a local maximum. The log-likelihood tends to negative infinity as $\theta$ approaches zero or infinity, so this stationary point is the maximum. As an additional check, the fitted mean excess is $\widehat{\theta}=566.6667$, which equals the sample mean excess $1700/3$.

If the untruncated exponential likelihood were incorrectly applied to the full observed losses, its estimate would be their mean: $(700+900+1800)/3=1133.3333$. That calculation ignores the fact that the sample contains only losses above positive thresholds and is not the conditional maximum-likelihood estimate.

Scope, Interpretation, and Exam Traps

A conditional likelihood is appropriate when the data consist of independent full ground-up severities selected because they exceed known deductibles, and inference is conditional on the observed selected sample. If the threshold varies by claim, use the corresponding survival probability for each threshold. In a regression model, the same construction applies claim by claim: both the density and the survival probability use that claim's modeled parameters and deductible.

Conditioning on the observed number of selected claims means this likelihood alone does not model how many claims were selected from the underlying exposure. If the number of selected claims or the unselected exposure is also part of the information being modeled, the selection probability may need to enter a joint frequency-and-severity analysis. Do not add such a component without specifying that sampling design.

The formula also depends on the response being the full loss $X$. If the recorded response is instead a payment amount defined relative to a deductible, its support and density must be specified for that target; substituting that response into the full-loss formula without changing the model is incorrect. Similarly, this tutorial concerns truncation by selection: only above-threshold observations are present. It does not treat observations recorded at a policy limit or other forms of censoring.

Key takeaway: For a full ground-up loss observed only when it exceeds $d_i$, use $f_X(x_i;\theta)/S_X(d_i;\theta)$, not $f_X(x_i;\theta)$. The survival denominator accounts for the selection event and must use the same severity parameters as the density.

Check Your Understanding

For each problem, use the full ground-up loss as the observed response and condition on the stated selection threshold.

A ground-up loss is exponential with mean $\$400$, and a full loss of $\$700$ is observed only because it exceeds a $\$300$ deductible. Calculate its conditional density. (Answer: $e^{-1}/400\approx0.0009197$ per dollar.)
Ground-up loss is exponential with mean $\$800$. Given that a claim exceeds a $\$200$ deductible, find the conditional probability that it exceeds $\$1{,}000$. (Answer: $e^{-1}\approx0.3679$.)
A ground-up loss has gamma shape $k=2$ and scale $\theta=4$. A loss of $8$ is observed above deductible $4$. Calculate its conditional density. (Answer: $e^{-1}/4\approx0.0920$ per loss unit.)
Assume independent exponential ground-up losses with unknown mean, selected above thresholds $100,300,500$. The observed full losses are $500,900,1{,}100$, respectively. Find the conditional maximum-likelihood estimate of the mean. (Answer: $1600/3\approx533.3333$.)
Two independent exponential ground-up losses have mean $\theta=500$. A loss of $600$ is observed above $100$, and a loss of $900$ is observed above $400$. Write and evaluate the conditional likelihood. (Answer: $500^{-2}e^{-2}\approx5.4134\times10^{-7}$ per squared loss unit.)

What You'll Learn