Gamma Severity Estimation by Moments and Likelihood

Why Gamma Severity Estimation Matters

A gamma model is useful for positive, continuous severities when the data may be right-skewed and the shape of that skew matters. Estimating its parameters is not just a matter of calculating the sample mean: the method of moments uses both the mean and variance directly, while maximum likelihood also uses the sample’s logarithms. Those methods need not produce the same fit.

The tutorial assumes a sample of complete, uncensored, untruncated, independent observations from one gamma distribution. A complete observation is the full loss amount, not a deductible payment or a payment capped by a policy limit. Estimation from altered or incomplete observations requires a different likelihood.

Definition: A random severity $X$ has a gamma distribution with shape $\alpha>0$ and scale $\theta>0$, written $X\sim\operatorname{Gamma}(\alpha,\theta)$, if its density is $$ f_X(x)=\frac{x^{\alpha-1}e^{-x/\theta}}{\Gamma(\alpha)\theta^\alpha},\qquad x>0. $$ Here $\Gamma(\alpha)$ is the gamma function. The scale $\theta$ has the same units as $X$; it is not the rate $1/\theta$. The mean, variance, and skewness are $E[X]=\alpha\theta$, $\operatorname{Var}(X)=\alpha\theta^2$, and $2/\sqrt{\alpha}$, respectively.

Method of Moments

For positive observations $x_1,\ldots,x_n$, define the empirical first and second raw moments by $m'_1=\bar{x}$ and $m'_2=n^{-1}\sum_{j=1}^n x_j^2$. The empirical variance used in the method of moments is $m'_2-(m'_1)^2=n^{-1}\sum_{j=1}^n(x_j-\bar{x})^2$. Its divisor is $n$, not $n-1$: it is the empirical moment being matched, rather than an unbiased estimate of population variance.

The gamma mean and variance provide two equations for the two parameters. Solving those equations gives the moment estimates below.

\widehat{\alpha}_{\mathrm{MM}}= \frac{\bar{x}^{\,2}}{m'_2-\bar{x}^{\,2}}, \qquad \widehat{\theta}_{\mathrm{MM}}= \frac{m'_2-\bar{x}^{\,2}}{\bar{x}}.

These formulas require a positive sample mean and a positive empirical variance. If all observations are identical, the empirical variance is zero and the finite moment estimates do not exist.

Worked Example: Moment Estimates for Hypothetical Equipment Claims

A hypothetical insurer records four complete equipment-claim severities, in thousands of dollars: $2,4,6,8$. Assume the observations are independent and identically distributed gamma severities. Find the moment estimates and the fitted gamma skewness.

Step 1: Calculate the sample mean. The total is $2+4+6+8=20$, so $\bar{x}=20/4=5$ thousand dollars.

Step 2: Calculate the empirical variance. The squared deviations from 5 are $9,1,1,9$, with sum $20$. Using divisor $n=4$, $m'_2-\bar{x}^2=20/4=5$ in squared-thousand-dollar units.

Step 3: Estimate shape and scale. Thus $\widehat{\alpha}_{\mathrm{MM}}=5^2/5=5$, and $\widehat{\theta}_{\mathrm{MM}}=5/5=1$ thousand dollars.

Step 4: Interpret the fit. The fitted mean is $5(1)=5$ thousand dollars, and the fitted variance is $5(1)^2=5$ squared-thousand-dollar units. Its skewness is $2/\sqrt{5}\approx0.8944$, to four decimal places.

The fitted mean and variance reproduce the empirical moments by construction. The skewness is a model implication, not a third sample moment that was matched.

Maximum Likelihood for Complete Observations

Let $X_1,\ldots,X_n$ be independent and identically distributed gamma observations, and suppose every observed $x_j$ is a positive, complete severity. Independence makes the joint likelihood the product of the individual densities. Taking its logarithm gives

\ell(\alpha,\theta) =(\alpha-1)\sum_{j=1}^n\ln x_j -\frac{\sum_{j=1}^n x_j}{\theta} -n\ln\Gamma(\alpha)-n\alpha\ln\theta.

The score with respect to the scale is

\frac{\partial\ell}{\partial\theta} =\frac{\sum_{j=1}^n x_j}{\theta^2} -\frac{n\alpha}{\theta}.

Setting this score equal to zero gives the positive solution $\widehat{\theta}=\bar{x}/\alpha$. Substitute that expression into the likelihood and differentiate the resulting profile likelihood with respect to $\alpha$. Define the digamma function by $\psi(\alpha)=\frac{d}{d\alpha}\ln\Gamma(\alpha)$. The shape score equation becomes

\ln\widehat{\alpha}-\psi(\widehat{\alpha}) =\ln\bar{x}-\frac{1}{n}\sum_{j=1}^n\ln x_j, \qquad \widehat{\theta}_{\mathrm{ML}}=\frac{\bar{x}}{\widehat{\alpha}}.

The equation for shape generally has no closed-form solution. A numerical root finder can solve it; the scale estimate then follows directly. The right-hand side is nonnegative by the arithmetic-geometric mean inequality and is strictly positive unless all observations are identical.

Derivation: Substituting $\theta=\bar{x}/\alpha$ into the log likelihood and differentiating with respect to $\alpha$ yields the profile score $$ \sum_{j=1}^n\ln x_j-n\ln\bar{x} +n\ln\alpha-n\psi(\alpha)=0. $$ Rearranging and dividing by $n$ gives $\ln\alpha-\psi(\alpha)=\ln\bar{x}-n^{-1}\sum_j\ln x_j$. The scale equation follows because its score is $\sum_jx_j/\theta^2-n\alpha/\theta$; setting it to zero and using $\sum_jx_j=n\bar{x}$ gives $\theta=\bar{x}/\alpha$. This also shows why both the arithmetic mean and the mean of the log observations enter the likelihood fit.

For a positive right-hand side, the shape equation has one positive solution. The function $g(\alpha)=\ln\alpha-\psi(\alpha)$ decreases from infinity toward zero as $\alpha$ increases; its derivative is $1/\alpha-\psi_1(\alpha)<0$, where $\psi_1$ is the trigamma function. A numerical method such as Newton’s method can therefore target the unique root:

\alpha_{\mathrm{new}} =\alpha_{\mathrm{old}}- \frac{\ln\alpha_{\mathrm{old}}-\psi(\alpha_{\mathrm{old}})-D} {1/\alpha_{\mathrm{old}}-\psi_1(\alpha_{\mathrm{old}})}, \qquad D=\ln\bar{x}-\frac{1}{n}\sum_{j=1}^n\ln x_j.

A numerical routine may evaluate the digamma and trigamma functions directly. Check that the final shape is positive, that the score equation is satisfied to the solver’s tolerance, and that the scale is calculated using the same shape estimate.

Worked Example: Numerical Maximum-Likelihood Fit to Complete Losses

A hypothetical portfolio has four complete severities, in thousands of dollars: $1,2,3,4$. Assume independent observations from a common gamma distribution. Estimate shape and scale by maximum likelihood, then calculate the fitted skewness.

Step 1: Calculate the two summaries in the shape equation. The mean is $\bar{x}=(1+2+3+4)/4=2.5$. The mean log observation is $(\ln1+\ln2+\ln3+\ln4)/4=\ln24/4\approx0.79451346$. Also, $\ln(2.5)\approx0.91629073$. Therefore $D\approx0.91629073-0.79451346=0.12177727$.

Step 2: Solve for shape. Solve $\ln\alpha-\psi(\alpha)=0.12177727$. For example, Newton iteration starting at 4 gives a next value of about 4.2483 and then converges to $\widehat{\alpha}_{\mathrm{ML}}\approx4.2654$. Substitution gives $\ln(4.2654)-\psi(4.2654)\approx0.121777$, consistent with the target to the displayed precision.

Step 3: Calculate scale. $\widehat{\theta}_{\mathrm{ML}}=2.5/4.2654\approx0.5861$ thousand dollars.

Step 4: Calculate skewness. The fitted skewness is $2/\sqrt{4.2654}\approx0.9684$, to four decimal places. The scale changes the units and magnitude of losses but not the gamma skewness; skewness is determined by shape alone.

Reading and Comparing the Estimates

Method-of-moments and maximum-likelihood estimates use the same sample but answer different fitting criteria. The moment fit matches the empirical mean and variance. The likelihood fit chooses parameters that maximize the joint density of the observed complete sample. In particular, the maximum-likelihood shape equation depends on the gap between the log of the arithmetic mean and the mean of the log observations.

Feature	Method of moments	Maximum likelihood
Sample information used	Empirical mean and variance	Arithmetic mean and mean log observation
Shape calculation	Closed-form ratio	Numerical solution of a digamma equation
Scale calculation	Empirical variance divided by mean	Sample mean divided by fitted shape
Assumptions here	Positive complete sample from a gamma model	Independent, identically distributed positive complete observations from a gamma model

A useful interpretation follows from the skewness formula $2/\sqrt{\alpha}$. Smaller shape means greater gamma skewness; larger shape means less skewness. It does not mean that every individual observation must lie near the mean, and skewness alone does not determine the scale of the losses.

Worked Example: Interpreting a Fitted Gamma Shape

For a hypothetical set of complete windscreen-claim severities measured in thousands of dollars, suppose a gamma maximum-likelihood fit gives $\widehat{\alpha}=2.5$ and $\widehat{\theta}=1.2$. Calculate the fitted mean, standard deviation, and skewness, and explain the role of each parameter.

Step 1: Mean. The fitted mean is $\widehat{\alpha}\widehat{\theta}=2.5(1.2)=3.0$ thousand dollars.

Step 2: Variance and standard deviation. The fitted variance is $\widehat{\alpha}\widehat{\theta}^{\,2}=2.5(1.2)^2=2.5(1.44)=3.6$ squared-thousand-dollar units. The standard deviation is $\sqrt{3.6}\approx1.8974$ thousand dollars.

Step 3: Skewness. The fitted skewness is $2/\sqrt{2.5}\approx1.2649$, to four decimal places. The shape controls this skewness, while the scale multiplies the severity level: changing scale from 1.2 to another positive value would change the mean and standard deviation but leave the skewness unchanged.

The units are consistent: the mean and standard deviation are in thousands of dollars, whereas variance is in squared-thousand-dollar units and skewness is dimensionless.

Common Mistakes and Applicability

Using the wrong scale convention. In this parameterization, the density contains $e^{-x/\theta}$ and the mean is $\alpha\theta$. If using a rate parameter instead, it is $1/\theta$, and the formulas must be changed consistently.
Using the unbiased sample variance in the moment formula without intending to. The method of moments matches the empirical variance with divisor $n$. Replacing it by the divisor-$n-1$ variance changes the estimates.
Mis-signing the scale score. Differentiating $-\sum_jx_j/\theta$ gives $+\sum_jx_j/\theta^2$, while differentiating $-n\alpha\ln\theta$ gives $-n\alpha/\theta$. Their sum is the scale score used above.
Using the moment shape in the likelihood scale formula. The likelihood scale estimate is $\bar{x}/\widehat{\alpha}_{\mathrm{ML}}$; the shape must come from the likelihood equation.
Applying the complete-data likelihood to altered observations. A recorded payment after a deductible, a capped payment, or a sample selected only above a threshold is not an unmodified complete severity. The likelihood must reflect the observation mechanism.

For a positive sample with nonzero variation, these methods produce positive parameter estimates. If the observations are not independent draws from a common gamma distribution, or are censored, truncated, or transformed payments, the estimates above do not address that data structure. This distinction is important in applied work and is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

Assume a gamma severity model and use the estimation convention stated in each question. Round numerical estimates to four decimal places where needed.

Complete hypothetical losses, in thousands of dollars, are $2,3,7$. Find the method-of-moments shape and scale, using divisor $n$ for the empirical variance. (Answer: $\widehat{\alpha}_{\mathrm{MM}}=24/7\approx3.4286$, $\widehat{\theta}_{\mathrm{MM}}=7/6\approx1.1667$ thousand dollars.)
A complete sample has mean $10$ dollars and empirical second raw moment $125$ dollars squared. Find the method-of-moments parameters and the fitted skewness. (Answer: empirical variance $25$, $\widehat{\alpha}_{\mathrm{MM}}=4$, $\widehat{\theta}_{\mathrm{MM}}=2.5$ dollars, skewness $1.0000$.)
For complete severities $1,2,6$, find the maximum-likelihood shape and scale numerically. (Answer: $\widehat{\alpha}_{\mathrm{ML}}\approx2.0004$, $\widehat{\theta}_{\mathrm{ML}}\approx1.4997$, in the same units as the observations.)
A gamma fit has shape $1.44$ and scale $500$ dollars. Find its mean, standard deviation, and skewness. (Answer: mean $720$ dollars, standard deviation $600$ dollars, skewness $1.6667$.)
A complete sample has $\bar{x}=10$ and $n^{-1}\sum_j\ln x_j=2.20$, in consistent positive units. Set up the maximum-likelihood shape equation and give the approximate shape and scale estimates. (Answer: $\ln\widehat{\alpha}-\psi(\widehat{\alpha})=\ln(10)-2.20\approx0.1026$; $\widehat{\alpha}\approx5.035$, $\widehat{\theta}\approx1.986$ in the observation units.)

What You'll Learn