Why Compare Fits?
A severity model turns a set of observed losses into a distribution that can be used to describe future losses. Different estimation methods can produce different parameter values even when they use the same distribution family and data. In particular, the method of moments (MM) matches selected sample moments, while maximum likelihood estimation (MLE) chooses parameters that make the observed sample most likely under the model.
This tutorial compares both methods for gamma and lognormal severity models. The central lesson is that matching the sample mean and variance does not generally make the fitted distributions identical: their shapes and estimated quantiles can still differ. The calculations below assume complete observations of individual losses, not censored, truncated, or otherwise modified payments.
Parameterizations and Estimation Targets
Use a gamma distribution with shape \(\alpha>0\) and scale \(\theta>0\). Its density is \(f(x)=x^{\alpha-1}e^{-x/\theta}/(\Gamma(\alpha)\theta^\alpha)\) for \(x>0\). Its mean and variance are \(\alpha\theta\) and \(\alpha\theta^2\), respectively. The scale \(\theta\) has the same units as the loss.
For a lognormal distribution, write \(\ln X\sim N(\mu,\sigma^2)\), where \(\sigma^2>0\) is the variance on the log scale. The original-scale mean and variance are \(e^{\mu+\sigma^2/2}\) and \((e^{\sigma^2}-1)e^{2\mu+\sigma^2}\). These are not \(\mu\) and \(\sigma^2\): those parameters describe the distribution of log loss.
Define the sample mean and the empirical second central moment with divisor \(n\) by
The divisor \(n\) is used here because MM equates distribution moments to empirical moments, and because the complete-sample likelihood estimates below use the corresponding maximum-likelihood log-scale variance. Do not substitute the unbiased sample variance with divisor \(n-1\) into these particular formulas.
Worked Example: Moment Fits to One Loss Sample
Worked Example: Moment Fits to One Loss Sample
A hypothetical insurer records three complete claim severities of \(2\), \(4\), and \(8\) thousand dollars. Fit gamma and lognormal models by moments, and check each fitted model's original-scale mean and variance. Treat the observations as independent and identically distributed.
Step 1: Calculate the sample moments. The sample mean is
The deviations are \(-8/3\), \(-2/3\), and \(10/3\). Therefore
The data are positive and nonconstant, as required.
Step 2: Fit the gamma model. The shape estimate is \(\widehat{\alpha}_{\mathrm{MM}}=(14/3)^2/(56/9)=196/56=3.5\). The scale estimate is \(\widehat{\theta}_{\mathrm{MM}}=(56/9)/(14/3)=4/3=1.3333\) thousand dollars. The fitted mean is \(3.5(4/3)=14/3=4.6667\) thousand dollars. The fitted variance is \(3.5(4/3)^2=56/9=6.2222\) squared thousands of dollars, matching both sample moments as it must.
Step 3: Fit the lognormal model. The squared coefficient of variation is
Thus \(\widehat{\sigma}^{\,2}_{\mathrm{MM}}=\ln(1+2/7)=\ln(9/7)=0.2513\), and \(\widehat{\mu}_{\mathrm{MM}}=\ln(14/3)-\tfrac12\ln(9/7)=1.4148\). By construction, the fitted original-scale mean and variance are \(14/3=4.6667\) thousand dollars and \(56/9=6.2222\) squared thousands of dollars. These are moments of loss, not log loss.
Check. Both MM fits reproduce \(\bar{x}=4.6667\) and \(m_2=6.2222\). The fitted parameters differ because the gamma and lognormal families describe different distribution shapes.
Method and trap avoided. This is a direct match of theoretical and empirical moments. A common error is to misadd the squared deviations or to use a different empirical-variance divisor. The squared deviations sum to \(56/3\), so dividing by \(n=3\) gives \(56/9\), not \(19/3\) or \(6.3333\).
Maximum Likelihood Fits
For gamma observations, the log likelihood, apart from terms not involving the parameters, is
For each fixed \(\alpha\), differentiating with respect to \(\theta\) and setting the derivative to zero gives \(\widehat{\theta}=\bar{x}/\alpha\). Substituting this scale estimate into the shape score equation gives a single equation for \(\alpha\):
where \(\psi\) is the digamma function, the derivative of \(\ln\Gamma(\alpha)\). In general this equation is solved numerically. For a nonconstant positive sample its right-hand side is positive, and the equation has a positive solution; the scale estimate then follows from \(\widehat{\theta}=\bar{x}/\widehat{\alpha}\).
For the lognormal model, the likelihood is the normal likelihood for \(y_j=\ln x_j\), together with the density transformation factor \(1/x_j\). That factor does not depend on \(\mu\) or \(\sigma^2\), so maximizing over those parameters gives the normal maximum-likelihood estimates on the log scale.
Worked Example: Likelihood Fits to the Same Sample
Worked Example: Likelihood Fits to the Same Sample
Use the hypothetical losses \(2,4,8\) thousand dollars from the previous calculation. Find the gamma and lognormal MLEs, then compare the fitted means and variances with the moment fits.
Gamma shape. The average log loss is \((\ln2+\ln4+\ln8)/3=\ln4=1.386294\). Hence the right-hand side of the gamma shape equation is \(\ln(14/3)-\ln4=\ln(7/6)=0.154151\). Solving \(\ln\alpha-\psi(\alpha)=0.154151\) numerically gives \(\widehat{\alpha}_{\mathrm{MLE}}\approx3.401\). Consequently,
The gamma MLE mean is \(\widehat{\alpha}\widehat{\theta}=\bar{x}=4.6667\) thousand dollars. Its variance is \(\widehat{\alpha}\widehat{\theta}^{\,2}=\bar{x}^{\,2}/\widehat{\alpha}\approx21.7778/3.401=6.4033\) squared thousands of dollars. Thus the gamma MLE matches the sample mean, but not the empirical variance.
Lognormal likelihood. Since the log observations are \(\ln2,\ln4,\ln8\), their mean is \(\widehat{\mu}_{\mathrm{MLE}}=\ln4=1.386294\). Their deviations from that mean are \(-\ln2,0,\ln2\), so
The fitted original-scale mean is \(e^{\widehat{\mu}+\widehat{\sigma}^2/2}\approx4.6947\) thousand dollars. The fitted original-scale variance is \((e^{0.320302}-1)e^{2(1.386294)+0.320302}\approx8.319\) squared thousands of dollars.
Check and interpretation. The gamma MLE mean equals the sample mean because its fitted scale is \(\bar{x}/\widehat{\alpha}\). The lognormal MLE instead matches the mean of log loss, not the arithmetic mean of loss. Neither MLE is required to match \(m_2\). The fitting method and the distribution family both affect the fitted variance.
Method and trap avoided. The gamma shape is obtained from the score equation, then the scale is recovered from the mean. For the lognormal MLE, calculate the variance of the logged observations using divisor \(n\), then transform back to the loss scale. Using the arithmetic sample variance as \(\widehat{\sigma}^2\) would mix the original and log scales.
Why the Estimates Differ
The distinction can be summarized by identifying exactly what each procedure matches. Gamma MM matches the arithmetic sample mean and variance. Lognormal MM does the same using formulas that convert those two original-scale moments into log-scale parameters. Gamma MLE matches the arithmetic sample mean through its scale estimate, while its shape is determined by the log-data score equation. Lognormal MLE matches the mean and variance of the logged data, so its original-scale moments need not equal the sample moments.
| Fit | What is matched or determined | Mean on loss scale |
|---|---|---|
| Gamma MM | Sample arithmetic mean and variance | Equals \(\bar{x}\) |
| Gamma MLE | Arithmetic mean through the scale; shape through the score equation | Equals \(\bar{x}\) |
| Lognormal MM | Sample arithmetic mean and variance | Equals \(\bar{x}\) |
| Lognormal MLE | Mean and variance of logged observations | May differ from \(\bar{x}\) |
A useful derivation explains the gamma mean match under MLE. The scale score is \(\partial\ell/\partial\theta=\sum_jx_j/\theta^2-n\alpha/\theta\). Setting it to zero gives \(\widehat{\theta}=\bar{x}/\alpha\). Multiplication by the fitted shape then yields \(\alpha\widehat{\theta}=\bar{x}\). This result depends on fitting the ordinary gamma model to complete observations; it should not be transferred without adjustment to censored, truncated, or modified-payment data.
Worked Example: Comparing Fitted Upper Quantiles
Worked Example: Comparing Fitted Upper Quantiles
For the same \(2,4,8\) thousand dollar sample, compare the fitted 90th-percentile losses. Use the exact lognormal quantile formula. For the gamma fits, use the Wilson–Hilferty approximation, and label those results as approximations rather than exact gamma quantiles.
For a lognormal fit, the \(p\)-quantile is \(e^{\mu+\sigma z_p}\), where \(z_p\) is the \(p\)-quantile of the standard normal distribution. At \(p=0.90\), use \(z_{0.90}=1.2816\). The MM fit gives
The lognormal MLE gives \[ \exp\!\left(1.386294+\sqrt{0.320302}(1.2816)\right) \approx8.262 \quad\text{thousand dollars}. \] These are direct evaluations of the fitted lognormal quantile formula.
For a gamma distribution with shape \(\alpha\), scale \(\theta\), and mean \(\alpha\theta\), the Wilson–Hilferty approximation to its \(p\)-quantile is
For gamma MM, \(\alpha=3.5\) and \(\alpha\theta=4.6667\). The bracket is \(1-1/31.5+1.2816/(3\sqrt{3.5})\approx1.1967\), giving an approximate 90th percentile of \(4.6667(1.1967)^3\approx8.00\) thousand dollars. For gamma MLE, \(\alpha=3.401\) and the fitted mean remains \(4.6667\). The bracket is approximately \(1-1/(9(3.401))+1.2816/(3\sqrt{3.401})=1.1990\), giving an approximate 90th percentile of \(4.6667(1.1990)^3\approx8.04\) thousand dollars.
The four estimates are therefore about \(7.825\), \(8.262\), \(8.00\), and \(8.04\) thousand dollars for lognormal MM, lognormal MLE, gamma MM, and gamma MLE, respectively. The gamma values use an approximation; an exact gamma quantile is obtained by numerically inverting the fitted gamma distribution function. The comparison illustrates why similar fitted means do not imply identical upper-tail estimates.
Method and trap avoided. Use the quantile formula belonging to the fitted family and parameter estimates. Do not call a gamma approximation an exact quantile, and do not use \(\mu\) itself as a lognormal loss quantile: \(\mu\) is on the log scale.
Common Exam Traps and Practical Interpretation
- Using the wrong variance divisor. The formulas here use \(m_2\) with divisor \(n\). Replacing it with an \(n-1\) estimate changes the MM parameters and the lognormal MLE variance.
- Confusing the log and loss scales. For a lognormal model, \(\mu\) and \(\sigma^2\) describe \(\ln X\). Original-scale moments require the exponential transformations.
- Assuming every MLE matches the sample mean and variance. The gamma MLE matches the arithmetic mean, but its variance is not generally the empirical variance. The lognormal MLE matches moments on the log scale.
- Ignoring the sampling mechanism. These likelihoods describe complete independent loss observations. A sample altered by a deductible, policy limit, censoring, or truncation requires a likelihood that accounts for that mechanism.
- Reporting unjustified precision. Gamma shape estimates and gamma quantiles often require numerical methods. Report numerical solutions to a sensible number of significant figures and identify approximations.
The fitted distribution matters because actuarial calculations often depend on more than the mean. A model used to estimate a high loss quantile may give a different answer from another model that matches the same first two sample moments. Comparison should therefore include the purpose of the fit and the part of the loss distribution relevant to that purpose, not just the parameter estimates.
Check Your Understanding
For each problem, use complete independent observations and the parameterizations defined above. Answers are given for checking.
- A hypothetical sample of losses, in thousands of dollars, is \(3,6,9\). Calculate \(\bar{x}\), \(m_2\) with divisor \(n\), and the gamma MM shape and scale. (Answer: \(\bar{x}=6\), \(m_2=6\), \(\widehat{\alpha}_{\mathrm{MM}}=6\), \(\widehat{\theta}_{\mathrm{MM}}=1\).)
- A positive, nonconstant sample has arithmetic mean \(10\) and empirical variance \(25\). Find the lognormal MM estimates and state the fitted original-scale mean and variance. (Answer: \(\widehat{\sigma}^{\,2}_{\mathrm{MM}}=\ln(1.25)\approx0.2231\), \(\widehat{\mu}_{\mathrm{MM}}=\ln(10)-\tfrac12\ln(1.25)\approx2.1902\); fitted mean \(10\), variance \(25\).)
- For complete losses \(1,2,4\), compute the lognormal MLEs. (Answer: \(\widehat{\mu}_{\mathrm{MLE}}=\ln2\), \(\widehat{\sigma}^{\,2}_{\mathrm{MLE}}=2(\ln2)^2/3\approx0.3203\); the fitted original-scale mean is \(2e^{(\ln2)^2/3}\approx2.347\).)
- For the losses \(2,4,8\), write the equation for the gamma MLE shape and give the fitted gamma mean. Explain why the mean does not require the numerical shape solution. (Answer: \(\ln\widehat{\alpha}-\psi(\widehat{\alpha})=\ln(7/6)\); the fitted mean is \(14/3\) thousand dollars because \(\widehat{\theta}=\bar{x}/\widehat{\alpha}\).)
- A proposed complete-sample fit uses observations \(5,5,5,5\). Explain why the nondegenerate formulas in this tutorial do not provide finite, positive-variance MM fits for either family. (Answer: \(m_2=0\), so the gamma MM shape formula divides by zero and the lognormal MM formula gives \(\widehat{\sigma}^{\,2}=0\); both correspond to a degenerate boundary case, not a regular nondegenerate fit.)