Method-of-Moments Estimation for Gamma Losses

Why Estimate a Gamma Loss Model by Moments?

A gamma distribution can represent positive, right-skewed losses using two parameters. In the shape-and-scale parameterization, one parameter controls the distribution’s shape and the other sets the units and scale of losses. Method of moments (MoM) estimates these parameters by making the model’s first two population moments agree with corresponding moments calculated from observed losses.

This is a direct calculation when complete, uncensored losses are available. It is not a likelihood method: in particular, the raw moments of losses observed only above a truncation threshold generally do not equal the population moments of the full loss distribution. The method below applies to an independent, identically distributed sample from the full severity distribution.

Definition: Let $X_1,\ldots,X_n$ be independent observations from a gamma severity distribution with shape $\alpha>0$ and scale $\theta>0$. 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 first two raw sample moments are $m_1=\frac{1}{n}\sum_{j=1}^n X_j$ and $m_2=\frac{1}{n}\sum_{j=1}^n X_j^2$. In this tutorial, MoM estimates match these raw sample moments to the corresponding population raw moments.

Gamma Moments and the Estimator Formulas

For a gamma variable in this parameterization, the scale $\theta$ has the same units as $X$, while the shape $\alpha$ is dimensionless. Its mean is $\alpha\theta$, its variance is $\alpha\theta^2$, and its second raw moment is $\alpha(\alpha+1)\theta^2$. The distinction between variance and second raw moment matters: the latter includes the square of the mean.

Result: Equating $m_1$ and $m_2$ to the gamma population raw moments gives $$ \widehat{\alpha}_{\mathrm{MOM}}=\frac{m_1^2}{m_2-m_1^2}, \qquad \widehat{\theta}_{\mathrm{MOM}}=\frac{m_2-m_1^2}{m_1}. $$ Equivalently, define $v_n=m_2-m_1^2=\frac{1}{n}\sum_{j=1}^n(X_j-m_1)^2$. Then $\widehat{\alpha}_{\mathrm{MOM}}=m_1^2/v_n$ and $\widehat{\theta}_{\mathrm{MOM}}=v_n/m_1$. These formulas require $m_1>0$ and $v_n>0$.

The denominator $v_n$ is the sample central moment with divisor $n$, not the usual unbiased sample variance with divisor $n-1$. If data are summarized by the unbiased sample variance $s^2=\frac{1}{n-1}\sum_{j=1}^n(X_j-\bar X)^2$, convert it first:

m_1=\bar X,\qquad v_n=\frac{n-1}{n}s^2,\qquad m_2=\bar X^2+\frac{n-1}{n}s^2.

This follows from the definitions: $s^2$ uses the sum of squared deviations divided by $n-1$, whereas $v_n$ divides the same sum by $n$. Here $\bar X=m_1$.

Why the Formulas Hold

The moment equations can be derived directly from the gamma density. For $r>-\alpha$, integrate $x^r f_X(x)$ over positive $x$ and substitute $z=x/\theta$. The resulting gamma integral gives the raw moment:

E[X^r] =\frac{1}{\Gamma(\alpha)\theta^\alpha} \int_0^\infty x^{\alpha+r-1}e^{-x/\theta}\,dx =\theta^r\frac{\Gamma(\alpha+r)}{\Gamma(\alpha)}.

The gamma-function recurrence $\Gamma(q+1)=q\Gamma(q)$ therefore gives $E[X]=\alpha\theta$ and $E[X^2]=\alpha(\alpha+1)\theta^2$. Subtracting the square of the mean gives $\operatorname{Var}(X)=\alpha\theta^2$. To obtain the MoM estimates, match the first two raw moments and solve:

m_1=\alpha\theta,\qquad m_2=\alpha(\alpha+1)\theta^2.

Subtracting $m_1^2=\alpha^2\theta^2$ from the second equation yields $m_2-m_1^2=\alpha\theta^2$. Dividing this by $m_1=\alpha\theta$ gives $\theta=(m_2-m_1^2)/m_1$. Substituting that expression into $m_1=\alpha\theta$ gives $\alpha=m_1^2/(m_2-m_1^2)$. This also shows why the sample central moment must be positive for finite, positive MoM estimates.

Worked Estimation Examples

Worked Example: Estimating from Three Complete Losses

A hypothetical insurer observes three complete losses of $2$, $4$, and $6$ thousand dollars, modeled as independent observations from a gamma distribution. Estimate the shape and scale by MoM, then calculate the fitted mean and variance. Report monetary quantities in thousands of dollars.

Method. Calculate the first two raw sample moments, form their difference $m_2-m_1^2$, and substitute in the shape-and-scale formulas. The common mistake avoided here is treating $m_2$ itself as the variance.

First raw moment. $m_1=(2+4+6)/3=12/3=4$.

Second raw moment. $m_2=(2^2+4^2+6^2)/3=(4+16+36)/3=56/3$.

Moment variance. $m_2-m_1^2=56/3-16=56/3-48/3=8/3$.

Shape estimate. $\widehat{\alpha}_{\mathrm{MOM}}=4^2/(8/3)=16(3/8)=6$.

Scale estimate. $\widehat{\theta}_{\mathrm{MOM}}=(8/3)/4=2/3$ thousand dollars.

Fitted mean and variance. The fitted mean is $6(2/3)=4$ thousand dollars. The fitted variance is $6(2/3)^2=6(4/9)=8/3$ thousand-dollars squared.

Check. The fitted second raw moment is $8/3+4^2=8/3+48/3=56/3$, equal to $m_2$. Both matched raw moments agree with the data.

Worked Example: Using Sufficient Summaries of the Sample

A hypothetical portfolio study records five complete losses, in thousands of dollars. The reported summaries are $\sum X_j=50$ and $\sum X_j^2=625$. Fit a gamma model by MoM and state its fitted mean, variance, and second raw moment.

Method. Convert the supplied totals to raw sample moments before estimating. This avoids using the sum of squares as if it were already an average.

First raw moment. $m_1=50/5=10$ thousand dollars.

Second raw moment. $m_2=625/5=125$ thousand-dollars squared.

Moment variance. $m_2-m_1^2=125-10^2=25$ thousand-dollars squared.

Shape estimate. $\widehat{\alpha}_{\mathrm{MOM}}=10^2/25=4$.

Scale estimate. $\widehat{\theta}_{\mathrm{MOM}}=25/10=2.5$ thousand dollars.

Fitted moments. The fitted mean is $4(2.5)=10$ thousand dollars. The fitted variance is $4(2.5)^2=4(6.25)=25$ thousand-dollars squared. The fitted second raw moment is $25+10^2=125$ thousand-dollars squared.

Check. These fitted values reproduce $m_1=10$ and $m_2=125$, the two sample moments used in estimation.

Worked Example: Converting an Unbiased Sample Variance

Four hypothetical complete losses, measured in thousands of dollars, are $3$, $5$, $7$, and $9$. Estimate the gamma shape and scale using the reported unbiased sample variance, then calculate the fitted mean and variance.

Method. Convert the unbiased variance to the divisor-$n$ central moment before applying the MoM equations. The common mistake avoided is substituting $s^2$ directly for $m_2-m_1^2$.

Sample mean. $\bar X=(3+5+7+9)/4=24/4=6$ thousand dollars.

Unbiased sample variance. The deviations from 6 are $-3,-1,1,3$, whose squares sum to $9+1+1+9=20$. Thus $s^2=20/(4-1)=20/3$ thousand-dollars squared.

Convert to the moment variance. $v_n=((4-1)/4)(20/3)=(3/4)(20/3)=5$ thousand-dollars squared. Therefore $m_2=\bar X^2+v_n=36+5=41$ thousand-dollars squared.

Shape estimate. $\widehat{\alpha}_{\mathrm{MOM}}=6^2/5=36/5=7.2$.

Scale estimate. $\widehat{\theta}_{\mathrm{MOM}}=5/6$ thousand dollars.

Fitted mean and variance. The fitted mean is $(36/5)(5/6)=6$ thousand dollars. The fitted variance is $(36/5)(25/36)=5$ thousand-dollars squared.

Check. The fitted second raw moment is $5+6^2=41$, matching the converted sample second raw moment. For contrast, substituting $s^2=20/3$ directly would give shape $36/(20/3)=5.4$ and scale $(20/3)/6=10/9$. Those estimates match the unbiased variance rather than the raw-moment convention defined here.

Interpretation and Exam Traps

For a fitted gamma model, the formulas can also be read in terms of its mean $\mu=\alpha\theta$ and variance $\sigma^2=\alpha\theta^2$. Solving these equations gives $\alpha=\mu^2/\sigma^2$ and $\theta=\sigma^2/\mu$. Thus a larger variance relative to the square of the mean corresponds to a smaller shape estimate; the scale estimate has the units of the losses.

A useful final check is to recompute the fitted moments from the estimated parameters. The mean $\widehat{\alpha}_{\mathrm{MOM}}\widehat{\theta}_{\mathrm{MOM}}$ must equal $m_1$, and the variance $\widehat{\alpha}_{\mathrm{MOM}}\widehat{\theta}_{\mathrm{MOM}}^2$ must equal $m_2-m_1^2$. If a calculation does not satisfy both checks, revisit the moments, units, or parameterization.

Scale versus rate: This tutorial uses scale $\theta$, so the density contains $e^{-x/\theta}$. If another source uses a rate parameter, its parameter is the reciprocal of the scale; do not insert a rate into the scale formulas.
Raw second moment versus variance: $m_2$ is the average of squared observations. The central moment needed in the variance form of the estimator is $m_2-m_1^2$, not $m_2$.
Divisor in the sample variance: The MoM raw-moment convention uses divisor $n$ for the empirical central moment. Convert an unbiased $s^2$, which uses divisor $n-1$, before using the formulas.
Observation mechanism: The direct equations assume a sample from the full severity distribution. If observations are truncated or censored, applying these formulas to recorded values without accounting for the observation mechanism need not estimate the full-distribution moments.
Units: If losses are recorded in thousands of dollars, the estimated scale is also in thousands of dollars and the variance is in squared thousands of dollars. Converting units changes the scale but not the shape.

These moment estimates provide a compact fit that can be checked using the same two moments used to construct it. Correctly identifying the parameterization and sample-moment convention is essential for a reliable result, aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

Use the shape-and-scale gamma parameterization and the raw-moment convention described above. Treat all stated losses as complete observations from the full distribution.

A loss sample has $m_1=12$ and $m_2-m_1^2=36$, with amounts in thousands of dollars. Find the MoM shape and scale. (Answer: $\widehat{\alpha}_{\mathrm{MOM}}=4,\ \widehat{\theta}_{\mathrm{MOM}}=3$ thousand dollars.)
For four losses, $\sum X_j=32$ and $\sum X_j^2=320$. Find the MoM shape and scale. (Answer: $\widehat{\alpha}_{\mathrm{MOM}}=4,\ \widehat{\theta}_{\mathrm{MOM}}=2$, in the units of the losses.)
A sample of six losses has mean $15$ and unbiased sample variance $75$. Convert the variance, then estimate the gamma parameters. (Answer: $v_n=62.5,\ \widehat{\alpha}_{\mathrm{MOM}}=3.6,\ \widehat{\theta}_{\mathrm{MOM}}=25/6$, in the units of the losses.)
The first two raw sample moments are $m_1=20$ and $m_2=500$. Find the estimated parameters and verify the fitted variance. (Answer: $\widehat{\alpha}_{\mathrm{MOM}}=4,\ \widehat{\theta}_{\mathrm{MOM}}=5$, fitted variance $100$.)
A sample of five losses has mean $9$ and unbiased sample variance $18$. Find the MoM shape, scale, and fitted variance. (Answer: $v_n=14.4,\ \widehat{\alpha}_{\mathrm{MOM}}=5.625,\ \widehat{\theta}_{\mathrm{MOM}}=1.6$, fitted variance $14.4$.)

What You'll Learn