Fit Ground-Up Severity Models from Claim Data

Why Fit Ground-Up Severity?

A severity model describes the distribution of the amount of an individual loss before policy deductibles, limits, or reinsurance recoveries are applied. Fitting this distribution from claim data provides estimates of quantities such as the mean loss and high quantiles. Those quantities can inform a hypothetical reinsurance pricing analysis: a high fitted quantile, for example, signals the scale of losses a layer may need to accommodate. It does not, by itself, determine a reinsurance premium.

The fitting methods in this tutorial assume that each recorded observation is a complete ground-up claim amount. That condition matters. A payment capped at a policy limit is not the full loss, and claims observed only above a deductible do not form an ordinary sample from the ground-up distribution. Those data require a likelihood that accounts for the coverage or observation mechanism.

Definition: Let $X$ be a positive ground-up claim amount with distribution function $F_X$, density $f_X$, and survival function $S_X(x)=1-F_X(x)$. A complete ground-up sample $x_1,\ldots,x_n$ consists of independently observed values from the same distribution, with each full claim amount recorded and no truncation or censoring. A fitted model replaces unknown distribution parameters by estimates from this sample.

These assumptions should be checked before calculating an estimate. In particular, the data must be measured in a consistent monetary unit, and the observations must not be policy payments mistaken for full losses. The examples below use hypothetical claims and explicitly state their assumed distributions. No interest rate or mortality assumption is relevant to fitting these severity distributions.

Estimation Methods for Common Severity Models

For a specified model, maximum likelihood chooses parameter values that make the observed sample most likely. The method of moments instead matches sample moments to the corresponding model moments. An estimate is meaningful only when the model and the assumptions behind its estimator are appropriate.

Severity family	Parameterization	Useful estimation result
Exponential	$f_X(x)=\lambda e^{-\lambda x}$, $x\geq 0$	The maximum-likelihood estimate is $\widehat{\lambda}=1/\bar{x}$.
Pareto Type I	$S_X(x)=(x_m/x)^\alpha$, $x\geq x_m$, with known $x_m>0$	The maximum-likelihood estimate is $\widehat{\alpha}=n/\sum_j\ln(x_j/x_m)$.
Lognormal	$\ln X\sim N(\mu,\sigma^2)$	The maximum-likelihood estimates are the sample mean and the divisor-$n$ variance of the log observations.
Gamma	Shape $k$, scale $\theta$; mean $k\theta$, variance $k\theta^2$	Method of moments gives $\widehat{k}=\bar{x}^{\,2}/m_2$ and $\widehat{\theta}=m_2/\bar{x}$, where $m_2=n^{-1}\sum_j(x_j-\bar{x})^2$.

The gamma method-of-moments formulas follow by equating the model mean $k\theta$ to $\bar{x}$ and model variance $k\theta^2$ to $m_2$. Dividing the variance equation by the mean equation gives $\theta=m_2/\bar{x}$; substituting this into the mean equation gives $k=\bar{x}^{\,2}/m_2$. These are moment estimates, not the gamma maximum-likelihood estimates.

Worked Estimates and Their Interpretation

Worked Example: Exponential Estimate and a High Quantile

A hypothetical portfolio has complete ground-up claims of 20, 40, 60, and 80 thousand dollars. Assume independent claims from an exponential severity distribution with rate $\lambda$, measured per thousand dollars. Estimate the parameter, check the fitted mean, and calculate the fitted 95th percentile. There is no interest or mortality assumption.

Method and estimate. For a complete sample from this exponential model, the likelihood is $L(\lambda)=\lambda^n\exp(-\lambda\sum_j x_j)$. Its log likelihood is $n\ln\lambda-\lambda\sum_jx_j$; differentiating and setting the derivative to zero gives $n/\lambda-\sum_jx_j=0$, so $\widehat{\lambda}=n/\sum_jx_j=1/\bar{x}$.

The sample total is $20+40+60+80=200$ thousand dollars, and $n=4$. Thus $\bar{x}=200/4=50$ thousand dollars and $\widehat{\lambda}=4/200=0.02$ per thousand dollars.

Fitted moment check. An exponential distribution with rate $\widehat{\lambda}$ has mean $1/\widehat{\lambda}=1/0.02=50$ thousand dollars. This agrees with the sample mean, as it must for this estimate.

Fitted 95th percentile. The exponential quantile at probability $p$ is $-\ln(1-p)/\lambda$. Therefore,

\widehat{Q}_{0.95} =\frac{-\ln(0.05)}{0.02} =\frac{2.995732}{0.02} =149.7866\text{ thousand dollars}.

Rounded only at the final step, the fitted 95th percentile is $149,786.60. The calculation uses the rate per thousand dollars, so the result is also in thousands of dollars. The fitted quantile is a model-based tail estimate, not a claim that the sample itself contains a loss of that size.

Common mistake avoided: treating $0.02$ as a rate per dollar would produce a quantile 1,000 times too small when the data are in thousands. Keep the data unit and rate unit paired throughout.

Worked Example: Pareto Type I Fit with a Known Lower Bound

A hypothetical set of three complete ground-up claims is 1, 1.5, and 2 thousand dollars. Assume a Pareto Type I distribution with known lower bound $x_m=0.5$ thousand dollars and unknown shape $\alpha$. Estimate $\alpha$, calculate the fitted 90th percentile, and check whether the fitted mean exists. There is no interest or mortality assumption.

Method and likelihood. For $x\geq x_m$, the density is $f_X(x)=\alpha x_m^\alpha x^{-(\alpha+1)}$. For independent complete observations, the log likelihood is

\ell(\alpha) =n\ln\alpha+n\alpha\ln x_m-(\alpha+1)\sum_{j=1}^n\ln x_j.

Differentiating gives $d\ell/d\alpha=n/\alpha+n\ln x_m-\sum_j\ln x_j$. Setting this to zero yields

\widehat{\alpha} =\frac{n}{\sum_{j=1}^n\ln(x_j/x_m)}.

The observed ratios $x_j/x_m$ are $1/0.5=2$, $1.5/0.5=3$, and $2/0.5=4$. The denominator is $\ln 2+\ln 3+\ln 4=\ln 24=3.1780538$, so $\widehat{\alpha}=3/\ln 24=0.94398$, or approximately 0.9440.

For a Pareto Type I model, $S_X(x)=(x_m/x)^\alpha$. Setting this equal to $1-p$ and solving for $x$ gives $Q_p=x_m(1-p)^{-1/\alpha}$. Hence

\widehat{Q}_{0.90} =0.5(0.1)^{-1/\widehat{\alpha}} =0.5(0.1)^{-\ln(24)/3} \approx 5.73\text{ thousand dollars}.

The fitted mean of a Pareto Type I distribution is finite only when $\alpha>1$, in which case it is $\alpha x_m/(\alpha-1)$. Here $\widehat{\alpha}=0.94398<1$, so the fitted model has an infinite mean. This is an important diagnostic for interpretation: the fitted 90th percentile is defined, but a finite fitted expected loss is not.

Common mistake avoided: the denominator is the sum of all three log ratios: $\ln 2+\ln 3+\ln 4=\ln 24$. Replacing it with $\ln 12$ omits a factor from the data and gives the wrong shape estimate and quantile.

Worked Example: Lognormal Fit on the Log Scale

A hypothetical portfolio has complete ground-up claims of 1, 2, and 4 thousand dollars. Assume independent claims from a lognormal model, so $\ln X\sim N(\mu,\sigma^2)$, with unknown $\mu$ and $\sigma^2$. Estimate both parameters and calculate the fitted mean and 90th percentile. There is no interest or mortality assumption.

Transform and estimate. In thousands of dollars the observations are $1,2,4$, so their natural logarithms are $0,\ln 2,\ln 4=0, L, 2L$, where $L=\ln 2=0.693147$. The lognormal maximum-likelihood estimates are the mean and divisor-$n$ variance of these log observations:

\widehat{\mu}=\frac{0+L+2L}{3}=L=0.693147, \qquad \widehat{\sigma}^2 =\frac{(0-L)^2+(L-L)^2+(2L-L)^2}{3} =\frac{2L^2}{3}=0.320302.

Thus $\widehat{\sigma}=\sqrt{0.320302}=0.56595$. The fitted median is $\exp(\widehat{\mu})=\exp(L)=2$ thousand dollars.

Fitted mean. If $\ln X\sim N(\mu,\sigma^2)$, then $E[X]=\exp(\mu+\sigma^2/2)$. Substituting the estimates gives

\widehat{E[X]} =\exp(0.693147+0.320302/2) =\exp(0.853298) \approx 2.35\text{ thousand dollars}.

Fitted 90th percentile. If $z_{0.90}=1.2816$ is the 90th percentile of the standard normal distribution, then $Q_{0.90}=\exp(\mu+\sigma z_{0.90})$. Therefore,

\widehat{Q}_{0.90} =\exp(0.693147+0.56595(1.2816)) \approx 4.13\text{ thousand dollars}.

Rounded at the final step to two decimal places, the fitted mean is $2,350.00 and the fitted 90th percentile is $4,130.00. These amounts are approximate because the calculations use rounded intermediate values.

Common mistake avoided: the lognormal parameters describe $\ln X$, not $X$. In particular, $\widehat{\mu}=0.693147$ is not the fitted mean claim amount; the fitted mean on the original scale is $\exp(\widehat{\mu}+\widehat{\sigma}^2/2)$.

Checks Before Using a Fit in Pricing

A parameter estimate is not the end of the analysis. Check that the estimated parameters satisfy the model's restrictions, that the fitted moments being reported actually exist, and that the units are consistent. A sample mean can be finite even when a fitted heavy-tailed model has an infinite theoretical mean, as in the Pareto example.

For a hypothetical reinsurance pricing case, severity estimates may help describe the scale and tail of ground-up losses. The exponential example supplies a finite mean and a fitted high quantile; the Pareto example shows why the existence of the mean must be checked; and the lognormal example distinguishes the median from the mean. These are inputs for further analysis, not direct prices. A reinsurance layer's expected cost also depends on how its contract transforms each loss and on any frequency or expense assumptions. Do not apply a ground-up mean or quantile as though it were automatically a layer payment.

Key takeaway: Fit only the distribution represented by the observed data. For complete ground-up claims, derive estimates under the stated severity model, preserve the monetary unit, and check the implications of the fitted parameters before interpreting moments or quantiles.

Common Exam Traps

Using the wrong observation model. A capped payment is not a complete ground-up claim. An ordinary complete-data likelihood does not account for censoring, truncation, deductibles, or policy limits.
Dropping a log term. For a known-bound Pareto fit, every observation contributes $\ln(x_j/x_m)$ to the denominator.
Confusing scale and rate. For the exponential model, the mean is $1/\lambda$, not $\lambda$.
Using $n-1$ in a lognormal maximum-likelihood variance. The likelihood estimate uses the divisor $n$. The divisor-$(n-1)$ sample variance is a different estimator.
Reporting a moment without checking existence. A Pareto Type I fitted mean requires $\widehat{\alpha}>1$; the fitted variance requires $\widehat{\alpha}>2$.

Check Your Understanding

For each problem, assume complete, independent ground-up claims from the stated model. Amounts are hypothetical and are expressed in the units shown. No interest or mortality assumption is required.

A sample from an exponential severity model is 15, 25, and 35 thousand dollars. Find the maximum-likelihood rate per thousand dollars and fitted 80th percentile. (Answer: $\widehat{\lambda}=0.0400$ per thousand dollars; $\widehat{Q}_{0.80}=40.24$ thousand dollars.)
Three claims under a Pareto Type I model with known $x_m=2$ thousand dollars are 3, 4, and 8 thousand dollars. Find the maximum-likelihood estimate of $\alpha$, and state whether the fitted mean exists. (Answer: $\widehat{\alpha}=3/\ln(12)\approx1.2078$; the fitted mean exists because $\widehat{\alpha}>1$.)
Lognormal claims, in thousands of dollars, are 1, 2, and 8. Find the maximum-likelihood estimates of $\mu$ and $\sigma^2$, using natural logarithms. (Answer: $\widehat{\mu}=4\ln(2)/3\approx0.9242$, $\widehat{\sigma}^2=14(\ln 2)^2/9\approx0.7474$.)
A Pareto Type I fit has known $x_m=1$ thousand dollars and estimated shape $\widehat{\alpha}=2.5$. Calculate its fitted mean and state whether its variance is finite. (Answer: fitted mean $=2.5/(2.5-1)=1.6667$ thousand dollars; variance is finite because $2.5>2$.)
A lognormal model has fitted parameters $\widehat{\mu}=\ln(3)$ and $\widehat{\sigma}=0.5$, with losses measured in thousands of dollars. Calculate the fitted median and mean. (Answer: median $=3.00$ thousand dollars; mean $=\exp(\ln 3+0.5^2/2)\approx3.40$ thousand dollars.)

Severity family	Parameterization	Useful estimation result
Exponential	\(f_X(x)=\lambda e^{-\lambda x}\), \(x\geq 0\)	The maximum-likelihood estimate is \(\widehat{\lambda}=1/\bar{x}\).
Pareto Type I	\(S_X(x)=(x_m/x)^\alpha\), \(x\geq x_m\), with known \(x_m>0\)	The maximum-likelihood estimate is \(\widehat{\alpha}=n/\sum_j\ln(x_j/x_m)\).
Lognormal	\(\ln X\sim N(\mu,\sigma^2)\)	The maximum-likelihood estimates are the sample mean and the divisor-\(n\) variance of the log observations.
Gamma	Shape \(k\), scale \(\theta\); mean \(k\theta\), variance \(k\theta^2\)	Method of moments gives \(\widehat{k}=\bar{x}^{\,2}/m_2\) and \(\widehat{\theta}=m_2/\bar{x}\), where \(m_2=n^{-1}\sum_j(x_j-\bar{x})^2\).

What You'll Learn