Worked Complete-Data Severity Estimation

Complete-Data Severity Likelihoods

When severity observations are exact, each recorded loss contributes its probability density to the likelihood. The model is fitted to the observed loss amounts, not to grouped counts or incomplete records. The main work is to choose the correct density, multiply the individual contributions, and maximize that likelihood over the model parameters.

This tutorial fits three standard positive-loss models: the exponential distribution, a Pareto Type I distribution with a fixed known threshold, and the lognormal distribution. Each example assumes independent, identically distributed exact observations from the stated severity model. There is no truncation or censoring. Interest discounting and mortality are not part of these severity likelihoods. These methods are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: For independent exact losses $x_1,\ldots,x_n$ with density $f_X(x;\theta)$, the likelihood is $L(\theta)=\prod_{i=1}^n f_X(x_i;\theta)$. The maximum likelihood estimate (MLE) is a parameter value that maximizes this likelihood over its allowed parameter space. Since the logarithm is strictly increasing, maximizing the log-likelihood $\ell(\theta)=\ln L(\theta)$ gives the same MLE.

The examples below use fixed model forms and estimate their unknown parameters. For a continuous severity model, an observed exact value contributes a density, not the probability of observing that exact value. The likelihood is a function of the parameters with the data held fixed.

Exponential Severity

Use the exponential distribution parameterized by its rate $\lambda>0$, so that $f_X(x)=\lambda e^{-\lambda x}$ for $x\ge 0$. Its mean is $1/\lambda$. With independent exact losses, the likelihood and log-likelihood are

L(\lambda)=\prod_{i=1}^n\lambda e^{-\lambda x_i} =\lambda^n e^{-\lambda\sum_{i=1}^n x_i}, \qquad \ell(\lambda)=n\ln\lambda-\lambda\sum_{i=1}^n x_i.

Differentiating gives the score $n/\lambda-\sum_i x_i$. Setting it equal to zero yields $\widehat{\lambda}=n/\sum_i x_i=1/\bar{x}$. The second derivative is $-n/\lambda^2<0$ for every $\lambda>0$, so the log-likelihood is strictly concave and this stationary point is the unique maximum.

Worked Example: Exponential Rate and Mean Severity

A hypothetical insurer records three exact claim amounts, in thousands of dollars: $2, 3,$ and $7$. Assume the losses are independent and identically distributed exponential observations with unknown rate $\lambda$ per thousand dollars. There is no interest or mortality assumption because neither is used in this severity fit.

Method and likelihood. The density is $f_X(x)=\lambda e^{-\lambda x}$, for $x\ge 0$. Here $n=3$ and $\sum_i x_i=2+3+7=12$. Thus

L(\lambda)=\lambda^3 e^{-12\lambda}, \qquad \ell(\lambda)=3\ln\lambda-12\lambda.

Estimate and check. The score is $3/\lambda-12$. Equating it to zero gives $\widehat{\lambda}=3/12=0.25$ per thousand dollars. The second derivative is $-3/\lambda^2<0$, confirming a maximum. The fitted mean severity is $1/\widehat{\lambda}=4$ thousand dollars, which agrees with the observed average $(2+3+7)/3=4$ thousand dollars.

Common mistake avoided. The observed average, $4$, estimates the exponential mean, not its rate. Reporting $4$ as the rate reverses the parameterization; the rate is $1/4=0.25$ per thousand dollars.

Pareto Severity With a Fixed Threshold

For this tutorial, the Pareto model is Type I with known threshold $\theta>0$ and unknown shape $\alpha>0$. Its density is $f_X(x)=\alpha\theta^\alpha x^{-(\alpha+1)}$ for $x\ge\theta$, and zero otherwise. The threshold is fixed in advance, not estimated from these observations. All recorded losses must be at least $\theta$; otherwise the likelihood under this model is zero.

For observations in the support, the log-likelihood can be written in terms of the threshold ratios:

\ell(\alpha) =n\ln\alpha-\alpha\sum_{i=1}^n\ln(x_i/\theta) -\sum_{i=1}^n\ln x_i.

The last sum does not depend on $\alpha$. Differentiating the log-likelihood therefore gives $n/\alpha-\sum_i\ln(x_i/\theta)$. Equating this score to zero gives the estimate below. Its second derivative is $-n/\alpha^2<0$, so the log-likelihood is strictly concave for positive $\alpha$, provided at least one observation exceeds the threshold.

Result: For exact independent Pareto Type I losses with a fixed known threshold $\theta$, and with $\sum_i\ln(x_i/\theta)>0$, the MLE of the shape is $\widehat{\alpha}=n\big/\sum_{i=1}^n\ln(x_i/\theta)$. This result does not estimate an unknown threshold and does not apply unchanged to a different Pareto parameterization.

Worked Example: Pareto Shape With a Known Threshold

A hypothetical portfolio has three exact observed losses of $100$, $200$, and $400$ dollars. Assume independent observations from a Pareto Type I severity distribution with known threshold $\theta=\$100$ and unknown shape $\alpha$. Interest and mortality are not involved.

Method and likelihood. Each observation is at least the fixed threshold. The threshold ratios are $1,2,$ and $4$, so

\sum_{i=1}^3\ln(x_i/\theta) =\ln 1+\ln 2+\ln 4 =0+\ln 2+2\ln 2 =3\ln 2.

The likelihood is $L(\alpha)=\prod_i\alpha\theta^\alpha x_i^{-(\alpha+1)}$. Its log-likelihood, omitting no parameter-dependent terms, is $3\ln\alpha-\alpha(3\ln2)-\sum_i\ln x_i$. The score equation is $3/\alpha-3\ln2=0$, giving

\widehat{\alpha}=\frac{1}{\ln 2}\approx 1.4427.

Verification. The second derivative is $-3/\alpha^2<0$, and the estimate is positive, as required. As a further fitted-model check, the estimated survival probability above $\$400$ is $(100/400)^{\widehat{\alpha}}=(1/4)^{1/\ln2}=e^{-2}\approx 0.1353$.

Common mistake avoided. The Pareto estimate uses the sum of logarithms of ratios to the known threshold. Using $\sum_i\ln x_i$ alone in the denominator drops the threshold adjustment and produces the wrong shape estimate.

A fixed threshold matters because it determines the support and the ratio inside the score equation. If the threshold were unknown, or if observations were selected using a truncation rule, the likelihood would need a different treatment. The formula above is specifically for exact observations from the stated fixed-threshold model.

Lognormal Severity

A positive loss $X$ is lognormal when $Z=\ln X$ is normal. Write $\ln X\sim N(\mu,\sigma^2)$, where $\sigma^2>0$. The density of a loss $x>0$ is

f_X(x;\mu,\sigma^2) =\frac{1}{x\sigma\sqrt{2\pi}} \exp\left(-\frac{(\ln x-\mu)^2}{2\sigma^2}\right).

The factor $1/x$ is part of the lognormal density: it arises from changing variables from $Z=\ln X$ to $X$. For exact positive losses, the log-likelihood is

\ell(\mu,\sigma^2) =-\frac{n}{2}\ln(2\pi\sigma^2) -\frac{1}{2\sigma^2}\sum_{i=1}^n(\ln x_i-\mu)^2 -\sum_{i=1}^n\ln x_i.

For fixed $\sigma^2$, maximizing over $\mu$ gives the mean of the log observations. Substituting that value into the likelihood and maximizing over $\sigma^2$ gives the average squared deviation of the log observations. The divisor is $n$, because this is the maximum likelihood estimate, not the unbiased sample-variance estimate.

Result: For independent exact observations from a lognormal distribution with unknown $\mu$ and $\sigma^2$, let $y_i=\ln x_i$. If the observed log losses are not all identical, then $\widehat{\mu}=\bar{y}$ and $\widehat{\sigma}^2=n^{-1}\sum_{i=1}^n(y_i-\bar{y})^2$.

Worked Example: Lognormal Parameter Estimates and Median

A hypothetical insurer observes four exact losses of $\$10,\$20,\$40,$ and $\$80$. Assume these are independent observations from a lognormal distribution with unknown $\mu$ and $\sigma^2$, so $\ln X\sim N(\mu,\sigma^2)$. There is no interest or mortality assumption in this severity estimation.

Method and log data. The log observations are $\ln 10,\ln 10+\ln2,\ln10+2\ln2,$ and $\ln10+3\ln2$. Their mean is $\bar{y}=\ln10+1.5\ln2$, approximately $3.3423$.

To calculate the MLE of the variance, subtract the mean from each log observation. The four deviations are $-1.5\ln2,-0.5\ln2,0.5\ln2,$ and $1.5\ln2$. Thus the sum of squared deviations is

\left(1.5^2+0.5^2+0.5^2+1.5^2\right)(\ln2)^2 =5(\ln2)^2.

Dividing by $n=4$ gives $\widehat{\sigma}^2=1.25(\ln2)^2\approx0.6006$. Therefore the fitted parameters are $\widehat{\mu}\approx3.3423$ and $\widehat{\sigma}^2\approx0.6006$, with $\widehat{\sigma}\approx0.7749$. The fitted median is $e^{\widehat{\mu}}=10\cdot2^{1.5}\approx\$28.28$.

Verification and common mistake avoided. The log deviations sum to zero, as required when their mean has been fitted. Their squared deviations sum to $5(\ln2)^2$, so the stated variance estimate follows directly. Dividing by $n-1=3$ would give the unbiased sample variance, not the MLE requested here.

What to Check Before Reporting an Estimate

A numerical answer is not enough: check that it follows from the specified model and the observed data. In these examples, the score equations provide the candidate estimates, while the second derivatives or likelihood structure verify that the candidates maximize the likelihood. Also check the parameter space and support: exponential losses must be nonnegative, Pareto observations must meet the fixed threshold, and lognormal observations must be positive.

Identify the model and parameterization.
For example, distinguish the exponential rate from its mean, and state whether a Pareto threshold is fixed and known.

Write the exact-data likelihood.
Multiply the densities of the independent observations, retaining any factors that depend on the unknown parameters.

Maximize and verify.
Differentiate the log-likelihood, solve the score equation, and check curvature and parameter constraints.

Interpret the fitted parameters.
Translate rates, shapes, or log-scale parameters into the quantity requested, preserving the original loss units.

A recurring exam trap is to reuse a familiar formula without checking its assumptions. The exponential estimate $\widehat{\lambda}=1/\bar{x}$ uses the rate parameterization. The Pareto estimate here assumes a fixed known threshold. The lognormal variance MLE uses divisor $n$. A changed parameterization, support, or data-collection mechanism can change the likelihood and the answer.

Key takeaway: For complete exact severity data, derive the likelihood from the stated density before using a closed-form estimate. Verify the estimate against the score equation, the model's support, and the parameterization.

Check Your Understanding

For each question, assume independent observations and the stated model. Interest and mortality are not relevant.

Exact hypothetical losses are $1,3,$ and $8$ thousand dollars from an exponential model with unknown rate. Find the rate MLE and fitted mean severity. (Answer: $\widehat{\lambda}=0.25$ per thousand dollars; fitted mean $=4$ thousand dollars.)
A Pareto Type I model has known threshold $\theta=\$50$. Exact losses are $\$50,\$100,\$200$. Find the shape MLE. (Answer: $\widehat{\alpha}=1/\ln2\approx1.4427$.)
Exact lognormal losses are $\$5,\$10,\$20$. Find the MLEs of $\mu$ and $\sigma^2$, using natural logarithms. (Answer: $\widehat{\mu}=\ln10\approx2.3026$; $\widehat{\sigma}^2=\frac{2}{3}(\ln2)^2\approx0.3203$.)
For exact exponential losses $2,2,6$ thousand dollars, calculate the rate MLE and verify whether the fitted mean equals the observed average. (Answer: $\widehat{\lambda}=0.30$ per thousand dollars; both fitted mean and observed average are $3.3333$ thousand dollars.)
A fixed-threshold Pareto Type I model has $\theta=\$25$, and the exact losses are $\$25,\$100,\$400$. Find the shape MLE, showing the log-ratio sum. (Answer: $\sum_i\ln(x_i/\theta)=\ln1+\ln4+\ln16=6\ln2$, so $\widehat{\alpha}=3/(6\ln2)=1/(2\ln2)\approx0.7213$.)

What You'll Learn