Estimating Severity from Deductible and Limited Payments

From Ground-Up Loss to Insurer Payment

A deductible and a payment limit change what an insurer observes about the underlying loss. A zero payment does not identify the loss amount; it says the loss did not exceed the deductible. A payment at the limit also does not identify the loss amount; it says the ground-up loss reached or exceeded the amount needed to exhaust coverage. Payments strictly between those endpoints, by contrast, identify the ground-up loss exactly when the policy terms are known.

This distinction matters when estimating a severity model. A zero or capped payment contributes a probability to the likelihood, while an interior payment contributes a density. Treating all observed payments as exact ground-up losses discards the information encoded in the policy terms and generally gives the wrong likelihood.

Definition: Let $X$ be a nonnegative, continuously distributed ground-up loss with distribution function $F_X$, survival function $S_X(x)=P(X>x)$, and density $f_X$. Under an ordinary deductible $d\geq 0$ and a limit $u>0$ on the insurer’s payment per loss, the insurer payment is $$ Y^L=\min\bigl((X-d)_+,u\bigr),\qquad (z)_+=\max(z,0). $$ A payment of zero occurs when $X\leq d$; an interior payment $y$, where $0<y<u$, corresponds to $X=d+y$; and a payment of $u$ occurs when $X\geq d+u$.

The limit $u$ here is the maximum insurer payment, not a maximum ground-up loss. The amount of ground-up loss that exhausts this coverage is $d+u$. The continuous-loss assumption avoids boundary distinctions between strict and weak inequalities. If the ground-up distribution has atoms, the endpoint probabilities must instead be computed using the exact event definitions and the distribution’s jumps.

The Mixed Distribution of Payments

The payment distribution has two possible endpoint masses and a continuous interior component. The endpoint masses can be positive, but need not be: their sizes depend on the ground-up distribution and policy terms. For example, if the ground-up support lies entirely above $d$, the zero-payment probability is zero.

P(Y^L=0)=F_X(d),\qquad f_{Y^L}(y)=f_X(d+y)\quad(0<y<u),\qquad P(Y^L=u)=S_X(d+u).

For the interior density, the transformation is a shift: $X=d+y$, so the change of variable has derivative $dX/dy=1$. Thus a small interval of payments from $y$ to $y+dy$ corresponds to a ground-up loss interval from $d+y$ to $d+y+dy$, giving density $f_X(d+y)$. The endpoint events follow directly from the payment mapping: $Y^L=0$ exactly when $X\leq d$, and $Y^L=u$ exactly when $X\geq d+u$.

These components sum to one. The total probability of an interior payment is $F_X(d+u)-F_X(d)$, so $$ F_X(d)+\bigl(F_X(d+u)-F_X(d)\bigr)+S_X(d+u)=1. $$ This identity is also a useful check on a proposed payment distribution.

Worked Example: Checking the Endpoint Probabilities

A hypothetical portfolio has ground-up losses $X\sim\operatorname{Uniform}(0,10)$. Each loss has deductible $d=2$ and insurer payment limit $u=3$. Find the probabilities of a zero payment, an interior payment, and a payment at the limit. Check that they add to one.

For this uniform model, $F_X(x)=x/10$ for $0\leq x\leq 10$, and the density is $f_X(x)=0.1$ on that interval.

Zero payment. A zero payment corresponds to $X\leq 2$, so $P(Y^L=0)=F_X(2)=2/10=0.2$.

Payment at the limit. The insurer pays the full limit when $X\geq d+u=5$. Therefore, $P(Y^L=3)=S_X(5)=P(X>5)=5/10=0.5$. The equality at 5 has probability zero under this continuous model.

Interior payment. The remaining probability is $1-0.2-0.5=0.3$. Equivalently, it is $P(2<X<5)=(5-2)/10=0.3$. The payment density is $f_{Y^L}(y)=0.1$ for $0<y<3$, and its integral over this interval is $0.1(3)=0.3$.

Check. The three components sum to $0.2+0.3+0.5=1.0$. Here both endpoint probabilities are positive; that is a consequence of this particular model and policy, not a general requirement.

Likelihood Contributions

Suppose $n$ independent ground-up losses share the same policy terms and severity model, and the data record the insurer payment for every loss, including zero payments. Write $n_0$ for the number of zero payments, $n_u$ for the number equal to $u$, and $y_1,\ldots,y_m$ for the interior payments, each strictly between 0 and $u$. Then $n=n_0+n_u+m$. Under a parametric model with parameter $\theta$, the likelihood is

L(\theta) = \bigl[F_X(d;\theta)\bigr]^{n_0} \bigl[S_X(d+u;\theta)\bigr]^{n_u} \prod_{j=1}^{m} f_X(d+y_j;\theta).

This is a likelihood with respect to the appropriate mixture of point masses and continuous density. An interior payment is not an event with positive probability of occurring at that exact real number; its density supplies the likelihood contribution. A zero or capped payment, on the other hand, is an event with probability given by the corresponding endpoint mass. If an endpoint mass is zero and the data include an observation there, the likelihood is zero for that parameter value.

If policy terms vary by observation, use the deductible and limit belonging to each payment. For observation $j$, a zero contributes $F_X(d_j;\theta)$, an interior payment $y_j$ contributes $f_X(d_j+y_j;\theta)$, and a capped payment contributes $S_X(d_j+u_j;\theta)$. Independence gives the product of these individual contributions.

Worked Example: Estimating an Exponential Rate with All Payments Observed

A hypothetical dataset records all four insurer payments under a deductible $d=1$ and payment limit $u=2$: $0,\ 0.5,\ 1,\ 2$. Assume independent ground-up losses $X\sim\operatorname{Exponential}(\theta)$, with density $f_X(x;\theta)=\theta e^{-\theta x}$ for $x\geq0$, where $\theta>0$ is the rate per loss unit. Estimate $\theta$ by maximum likelihood.

Classify the observations. There is one zero payment, two interior payments $0.5$ and $1$, and one payment at the limit $2$. The interior payments correspond to ground-up losses $1+0.5=1.5$ and $1+1=2$. A capped payment corresponds to $X\geq d+u=3$.

Write the contributions. The zero-payment probability is $F_X(1;\theta)=1-e^{-\theta}$. The interior densities are $f_X(1.5;\theta)=\theta e^{-1.5\theta}$ and $f_X(2;\theta)=\theta e^{-2\theta}$. The capped-payment probability is $S_X(3;\theta)=e^{-3\theta}$. Multiplying,

L(\theta) =(1-e^{-\theta})(\theta e^{-1.5\theta})(\theta e^{-2\theta})e^{-3\theta} =\theta^2(1-e^{-\theta})e^{-6.5\theta}.

Find the likelihood equation. The log-likelihood is $\ell(\theta)=2\ln\theta+\ln(1-e^{-\theta})-6.5\theta$. Differentiating and setting the score to zero gives

\ell'(\theta)=\frac{2}{\theta}+\frac{1}{e^\theta-1}-6.5=0.

The score is strictly decreasing for $\theta>0$, since its derivative is $-2/\theta^2-e^\theta/(e^\theta-1)^2<0$. It tends to $+\infty$ as $\theta$ approaches zero and to $-6.5$ as $\theta$ grows without bound. Therefore it has one root, which is the unique maximum-likelihood estimate. Solving numerically gives $\hat{\theta}\approx0.4308$ per loss unit. Substitution gives $2/0.4308+1/(e^{0.4308}-1)-6.5\approx0$, as required. The corresponding fitted mean ground-up loss is $1/\hat{\theta}\approx2.321$ loss units.

Method and trap avoided. The method is to map each payment back to its event or ground-up loss, then multiply the correct probability or density contributions. Treating the capped payment of 2 as an exact ground-up loss of 3 would use a density there instead of the probability $S_X(3;\theta)$, changing the likelihood and the estimate.

When Only Positive Payments Are Recorded

The likelihood changes if the dataset contains only losses with positive insurer payments. This is a sampling condition: observations have been selected by $Y^L>0$, equivalently $X>d$ under the continuous model. The distribution of a recorded positive payment must then be conditioned on that event. Its interior density is $f_X(d+y;\theta)/S_X(d;\theta)$; its conditional probability of being at the limit is $S_X(d+u;\theta)/S_X(d;\theta)$.

Do not divide by $S_X(d;\theta)$ when the data include all losses, including zero payments. Conversely, omitting the conditioning factor for a positive-payment-only sample generally gives an incorrect likelihood. If the sample design or reporting rule differs from these two cases, its selection mechanism must be represented in the likelihood.

Worked Example: Estimation from Positive Payments Only

A hypothetical report contains only positive insurer payments. The policy has deductible $d=2$ and payment limit $u=5$, and the three recorded payments are $1,\ 2,\ 3$; none is capped. Assume independent ground-up losses have an exponential distribution with rate $\theta>0$. Estimate $\theta$, accounting for the fact that zero payments were excluded.

Each recorded payment satisfies $0<y<5$, so it is interior. For an exponential ground-up loss, $f_X(x;\theta)=\theta e^{-\theta x}$ and $S_X(2;\theta)=e^{-2\theta}$. The conditional density of a positive interior payment is therefore

f_{Y^L\mid Y^L>0}(y;\theta) =\frac{f_X(2+y;\theta)}{S_X(2;\theta)} =\frac{\theta e^{-\theta(2+y)}}{e^{-2\theta}} =\theta e^{-\theta y},\qquad 0<y<5.

For these three observed payments, the conditional likelihood is $L(\theta)=(\theta e^{-\theta})(\theta e^{-2\theta})(\theta e^{-3\theta}) =\theta^3e^{-6\theta}$. Its log-likelihood is $3\ln\theta-6\theta$, so the score equation is $3/\theta-6=0$. Thus $\hat{\theta}=3/6=0.5$ per loss unit. The score changes from positive to negative at this value, and the log-likelihood has second derivative $-3/\theta^2<0$, confirming a maximum.

The result uses the positive-payment sampling rule. Although a payment limit exists, none of the observed payments is at that limit, so there is no capped-payment contribution in this particular sample. The interior observations still have support restricted to $0<y<5$.

Common Mistakes and Practical Checks

Assuming endpoint masses must be positive. They can be positive, with probabilities $F_X(d)$ and $S_X(d+u)$, but either may equal zero for some ground-up distributions and policy terms.
Using the wrong threshold for a capped payment. With a payment limit $u$, the cap is reached at ground-up loss $d+u$, not at $u$.
Using a density for an endpoint. A zero or capped payment is an event and contributes its probability. An interior payment contributes a density evaluated at the corresponding ground-up amount.
Ignoring the observation rule. A dataset containing all losses and one containing only positive payments have different likelihoods. Include the conditioning on positive payment only when the data were selected that way.
Confusing the payment limit with a ground-up loss limit. Here $u$ caps the insurer’s payment. A policy that instead caps the covered ground-up amount needs a different payment mapping.

A reliable calculation starts by classifying every observation, writing the ground-up event associated with it, and only then forming the likelihood. As a final check, verify that the zero mass, interior probability, and cap mass sum to one. This approach is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the observation mechanism before choosing probability or density contributions.

A hypothetical exponential ground-up loss has rate $\theta=0.25$ per unit. The deductible is $d=1$ and the insurer payment limit is $u=2$. Find the zero-payment probability and the probability of a payment at the limit. (Answer: $1-e^{-0.25}=0.2212$ and $e^{-0.75}=0.4724$, respectively.)
A hypothetical ground-up loss is uniform on $[0,10]$, with deductible $d=2$ and payment limit $u=3$. Give the zero-payment probability, the probability of an interior payment, and the capped-payment probability. (Answer: $0.2,\ 0.3,\ 0.5$, respectively.)
All payments are observed under a common deductible $d=1$ and payment limit $u=2$. The payments are $0,\ 0.5,\ 1.5,\ 2$. For independent exponential ground-up losses with unknown rate $\theta$, write the likelihood as a function of $\theta$. (Answer: $L(\theta)=(1-e^{-\theta})\theta^2e^{-7\theta}$.)
Only positive payments are recorded under deductible $d=4$ and payment limit $u=6$. The observed payments are $1,\ 2,\ 3$, all interior, and ground-up losses are exponential with unknown rate $\theta$. Find the conditional maximum-likelihood estimate. (Answer: $\hat{\theta}=3/(1+2+3)=0.5$ per loss unit.)
All payments are observed under deductible $d=2$ and payment limit $u=3$. A continuous ground-up severity model has distribution function $F_X$, density $f_X$, and survival function $S_X$. The payments are $0,\ 1,\ 3$. Write their joint likelihood contribution. (Answer: $F_X(2)f_X(3)S_X(5)$.)

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