Why Estimate a Severity Model?
A severity model describes the distribution of the amount of an individual loss. When complete claim amounts are available, an actuary can estimate the parameters of a selected continuous distribution and then use the fitted model to calculate quantities such as the expected loss, variance, or probability that a loss exceeds a threshold. Those calculations depend both on the chosen distribution and on the estimation method.
This tutorial considers complete, strictly positive observations \(x_1,\ldots,x_n\) modeled as independent and identically distributed draws from a continuous density \(f_X(x;\theta)\). “Complete” means each observation is the loss itself, rather than a censored, truncated, or policy-modified payment. Those other data structures require likelihoods that account for what was and was not observed. A fitted distribution is a model for the assumed population; it does not establish that the model is appropriate for the data.
For method-of-moments calculations, empirical moments use divisor \(n\), because they are averages of the observed powers. In particular, the empirical second central moment is \(m_2-m_1^2\). It is not the usual unbiased sample variance with divisor \(n-1\). A likelihood calculation instead follows from the assumed density and its parameter constraints.
Exponential Severity
Use the rate parameterization \(X\sim\operatorname{Exponential}(\lambda)\), with \(\lambda>0\) and density \(f_X(x)=\lambda e^{-\lambda x}\) for \(x\geq 0\). Under this model, \(E[X]=1/\lambda\), \(\operatorname{Var}(X)=1/\lambda^2\), and \(P(X>t)=e^{-\lambda t}\) for \(t\geq0\).
Equating the model mean to \(m_1=\bar{x}\) gives the method-of-moments estimate \(\widehat{\lambda}_{\mathrm{MOM}}=1/\bar{x}\). The log-likelihood is \(n\log\lambda-\lambda\sum_jx_j\); differentiating and setting the derivative to zero gives \(\widehat{\lambda}_{\mathrm{MLE}}=n/\sum_jx_j=1/\bar{x}\). The two estimates coincide for this one-parameter model. Because all observations are positive, \(\bar{x}>0\), so this estimate is a valid positive rate.
Worked Example: Exponential Fit and an Exceedance Probability
A hypothetical insurer observes four complete positive losses, in thousands of dollars: \(2,4,6,8\). Assume the losses are independent and identically distributed exponential random variables. Estimate the rate by moments and maximum likelihood, then calculate the fitted mean, variance, and probability a loss exceeds \(10\) thousand dollars.
Method. Use the exponential rate parameterization and the complete-data likelihood. The observed sum is \(2+4+6+8=20\), so the sample mean is \(\bar{x}=20/4=5\) thousand dollars.
Estimates. Moment matching gives \(\widehat{\lambda}_{\mathrm{MOM}}=1/5=0.2\) per thousand dollars. The likelihood estimate is \(n/\sum_jx_j=4/20=0.2\), the same value.
Fitted moments. The fitted mean is \(1/0.2=5\) thousand dollars. The fitted variance is \(1/(0.2)^2=25\) square thousand dollars.
Tail probability. At \(t=10\), the fitted probability is \(e^{-0.2(10)}=e^{-2}\approx0.1353\), to four decimal places. Thus the fitted probability of a loss exceeding \(10\) thousand dollars is \(0.1353\).
Check and trap avoided. The estimated mean \(5\) agrees with the observed mean. The estimate \(0.2\) is a rate, not a mean: using \(0.2\) as the fitted mean would invert the parameterization.
Gamma Severity: Moment Estimates and Likelihood
Use the gamma shape-rate parameterization \(X\sim\operatorname{Gamma}(\alpha,\beta)\), where \(\alpha>0\) is the shape and \(\beta>0\) is the rate. Its density is \(f_X(x)=\beta^\alpha x^{\alpha-1}e^{-\beta x}/\Gamma(\alpha)\) for \(x>0\), its mean is \(\alpha/\beta\), and its variance is \(\alpha/\beta^2\). Specifying “rate” matters: the scale parameter is \(1/\beta\), not \(\beta\).
Let \(\bar{x}=m_1\) and \(s_0^2=m_2-m_1^2\), the empirical central second moment with divisor \(n\). When \(s_0^2>0\), equating the gamma mean and variance to these empirical quantities gives \(\widehat{\alpha}_{\mathrm{MOM}}=\bar{x}^2/s_0^2\) and \(\widehat{\beta}_{\mathrm{MOM}}=\bar{x}/s_0^2\).
For maximum likelihood, the log-likelihood, apart from the data-independent terms, is \(n\alpha\log\beta-n\log\Gamma(\alpha)+(\alpha-1)\sum_j\log x_j-\beta\sum_jx_j\). For a fixed \(\alpha\), differentiating with respect to \(\beta\) gives \(\widehat{\beta}=\alpha/\bar{x}\). Substituting this into the shape score equation yields \(\log\alpha-\psi(\alpha)=\log\bar{x}-n^{-1}\sum_j\log x_j\), where \(\psi\) is the digamma function, the derivative of \(\log\Gamma(\alpha)\). For nonconstant positive observations, solve this equation numerically for a positive \(\widehat{\alpha}\), then set \(\widehat{\beta}=\widehat{\alpha}/\bar{x}\). The equation generally has no elementary closed-form solution.
Worked Example: Gamma Estimates by Moments and Likelihood
Suppose three hypothetical complete losses, measured in thousands of dollars, are \(1,2,3\). Assume an independent and identically distributed gamma model in the shape-rate parameterization. Find the moment estimates and the likelihood estimates, then calculate the fitted mean and variance for each fit.
Empirical moments. The mean is \(\bar{x}=(1+2+3)/3=2\). The empirical raw second moment is \(m_2=(1^2+2^2+3^2)/3=14/3\), so \(s_0^2=14/3-2^2=2/3\).
Method-of-moments fit. The estimates are \(\widehat{\alpha}_{\mathrm{MOM}}=2^2/(2/3)=6\) and \(\widehat{\beta}_{\mathrm{MOM}}=2/(2/3)=3\). The fitted mean is \(6/3=2\) thousand dollars, and the fitted variance is \(6/3^2=2/3\) square thousand dollars.
Likelihood fit. The right-hand side of the shape equation is \(\log 2-(\log1+\log2+\log3)/3=\log2-\log6/3\approx0.095894\). Numerically solving \(\log\alpha-\psi(\alpha)=0.095894\) gives \(\widehat{\alpha}_{\mathrm{MLE}}\approx5.375\). Therefore \(\widehat{\beta}_{\mathrm{MLE}}=5.375/2=2.6875\).
Fitted likelihood moments. The fitted mean is \(5.375/2.6875=2\) thousand dollars. The fitted variance is \(5.375/(2.6875)^2=4/5.375\approx0.7442\) square thousand dollars.
Check and trap avoided. Both fits reproduce the observed mean, but their variances differ. The moment estimate matches the empirical central second moment; the likelihood estimate does not generally have to do so. The variance \(2/3\) used for moment matching has divisor \(n=3\), not \(n-1=2\).
Lognormal Severity
A positive random variable \(X\) is lognormal when \(Y=\log X\) is normal. Write \(\log X\sim N(\mu,\sigma^2)\), with \(\sigma^2>0\). Then \(E[X]=e^{\mu+\sigma^2/2}\), \(\operatorname{Var}(X)=(e^{\sigma^2}-1)e^{2\mu+\sigma^2}\), and for \(t>0\), \(P(X>t)=1-\Phi((\log t-\mu)/\sigma)\), where \(\Phi\) is the standard normal cumulative distribution function.
For method of moments, let \(m_1\) and \(m_2\) be the observed raw moments of \(X\). Since \(E[X^2]/E[X]^2=e^{\sigma^2}\), matching the first two raw moments gives \(\widehat{\sigma}^2_{\mathrm{MOM}}=\log(m_2/m_1^2)\) and \(\widehat{\mu}_{\mathrm{MOM}}=\log m_1-\widehat{\sigma}^2_{\mathrm{MOM}}/2\), when the resulting variance parameter is positive.
For likelihood estimation, transform the observations to \(y_j=\log x_j\). The lognormal density includes the Jacobian factor \(1/x_j\), so its log-likelihood is the normal log-likelihood for the \(y_j\), minus \(\sum_j\log x_j\). That last term does not involve \(\mu\) or \(\sigma^2\). Consequently, when the log-observations are nonconstant, the estimates are \(\widehat{\mu}_{\mathrm{MLE}}=n^{-1}\sum_j\log x_j\) and \(\widehat{\sigma}^2_{\mathrm{MLE}}=n^{-1}\sum_j(\log x_j-\widehat{\mu}_{\mathrm{MLE}})^2\). The divisor is \(n\), not \(n-1\).
Worked Example: Comparing Lognormal Moment and Likelihood Fits
A hypothetical portfolio has four complete positive losses, in thousands of dollars: \(2,2,8,8\). Assume an independent and identically distributed lognormal model. Calculate both sets of estimates and the fitted probability that a loss exceeds \(8\) thousand dollars.
Method-of-moments fit. The observed mean is \(m_1=(2+2+8+8)/4=5\). The raw second moment is \(m_2=(4+4+64+64)/4=34\). Thus \(\widehat{\sigma}^2_{\mathrm{MOM}}=\log(34/25)=\log(1.36)\approx0.3075\), and \(\widehat{\mu}_{\mathrm{MOM}}=\log5-0.3075/2\approx1.4557\).
Likelihood fit. The log-observations are \(\log2,\log2,\log8,\log8\). Their mean is \((2\log2+2\log8)/4=\log4\approx1.3863\). Their deviations from this mean are \(-\log2,-\log2,\log2,\log2\). Therefore their mean squared deviation is \((4(\log2)^2)/4=(\log2)^2\approx0.4805\). The likelihood estimates are \(\widehat{\mu}_{\mathrm{MLE}}=\log4\) and \(\widehat{\sigma}^2_{\mathrm{MLE}}=(\log2)^2\).
Likelihood fitted moments. The fitted mean is \(e^{\log4+(\log2)^2/2}\approx5.088\) thousand dollars. The fitted variance is \((e^{(\log2)^2}-1)e^{2\log4+(\log2)^2}\approx16.51\) square thousand dollars.
Likelihood fitted tail. At \(t=8\), the standardized log threshold is \((\log8-\log4)/\log2=1\). Hence the fitted exceedance probability is \(1-\Phi(1)\approx0.1587\), to four decimal places.
Check and trap avoided. The moment fit is constructed to reproduce the observed raw first and second moments, whereas the likelihood fit is based on the mean and central second moment of the logarithms. These are different estimation targets. Using the divisor \(n-1\) for the likelihood variance, or substituting \(\sigma^2\) for \(\sigma\) in the standardized threshold, would give an incorrect result.
How to Choose and Apply an Estimator
The distribution is a modeling assumption; moments and likelihood are alternative ways to estimate its parameters. A reliable calculation keeps the assumptions and parameterization visible at each step. For complete observations, the likelihood includes one density contribution per observation. If an observation is instead known only to exceed a threshold, its contribution is a survival probability, not the density at an invented exact value.
Confirm that each recorded value is a complete positive loss and that the assumed observations are independent draws from the same severity distribution.
For example, state whether a gamma parameter is a rate or scale, and state any strict positivity conditions on the parameters.
For moments, calculate raw moments with divisor \(n\). For likelihood, form the log-likelihood from the specified density and check whether an interior maximum exists.
Substitute the fitted parameters into the model formulas for the mean, variance, or tail probability. Keep units consistent: a rate has reciprocal loss units, while a variance has squared loss units.
Common Exam Traps
- Confusing a rate and a scale. In the gamma shape-rate convention, the mean is \(\alpha/\beta\). Replacing \(\beta\) with a scale without changing the formulas reverses the parameter.
- Using the wrong variance divisor. The empirical central moment for method of moments and the normal MLE for transformed log-data use divisor \(n\). An unbiased sample variance with divisor \(n-1\) is a different statistic.
- Treating an estimate as evidence that the distribution fits. Parameter estimation alone does not assess model adequacy. Comparing fitted models and inspecting diagnostics are separate tasks.
- Ignoring the data structure. The complete-data likelihood is not appropriate for censored or truncated observations. In particular, a policy payment modified by a deductible or limit is not automatically the underlying loss observation.
- Accepting a boundary value as an interior MLE. If identical observations force a lognormal variance estimate of zero, the stated model requires positive variance, and there is no finite MLE. State the boundary issue rather than reporting an inadmissible parameter.
Check Your Understanding
For each problem, assume the stated complete observations are independent and identically distributed. Give rates in inverse thousands of dollars when losses are stated in thousands.
- Complete hypothetical losses are \(3,6,9\), in thousands of dollars. Under an exponential model, find the moment and likelihood rate estimates, the fitted mean and variance, and \(P(X>12)\). (Answer: both rate estimates are \(1/6\); fitted mean \(6\), variance \(36\), and exceedance probability \(e^{-2}\approx0.1353\).)
- Complete losses are \(2,3,7\), in thousands of dollars, and a gamma shape-rate model is assumed. Find the method-of-moments estimates using divisor \(n\), and give the fitted mean and variance. (Answer: \(\widehat{\alpha}=24/7\), \(\widehat{\beta}=6/7\), fitted mean \(4\), and fitted variance \(14/3\), all loss quantities in thousands or their corresponding squared units.)
- Complete losses are \(1,4,4,16\), in thousands of dollars. Under a lognormal model, find the likelihood estimates and the fitted probability that a loss exceeds \(16\). (Answer: \(\widehat{\mu}=\log4\), \(\widehat{\sigma}^2=(\log4)^2/2\), and \(P(X>16)=1-\Phi(\sqrt{2})\approx0.0787\).)
- Four complete positive losses are all \(5\), in thousands of dollars. Under a lognormal model with \(\sigma^2>0\), should the likelihood estimate be reported as \(\widehat{\sigma}^2=0\)? Explain the likelihood behavior. (Answer: No. Zero is outside the parameter space; the likelihood is unbounded as \(\sigma^2\) tends to zero, so there is no finite MLE.)
- For complete gamma observations \(1,2,3\), in thousands of dollars, the mean is \(2\) and the mean log-observation is \(\log6/3\). State the equation for the shape MLE, then find the corresponding rate and fitted variance. (Answer: solve \(\log\alpha-\psi(\alpha)=\log2-\log6/3\), giving \(\widehat{\alpha}\approx5.375\), \(\widehat{\beta}\approx2.6875\), and fitted variance \(\approx0.7442\) square thousand dollars.)