Translated Gamma Approximation by Moment Matching

Why Translate a Gamma Distribution?

Aggregate losses are often right-skewed: many outcomes are near the center, while a smaller number are much larger. A normal approximation uses the aggregate mean and variance but cannot represent skewness. A translated gamma approximation incorporates a third moment as well, allowing the fitted distribution to reflect asymmetry.

The method fits a shifted gamma distribution to the aggregate loss’s mean, variance, and skewness. It is an approximation, not a statement that the aggregate loss is exactly gamma distributed. In particular, a compound loss may be discrete or may have a probability atom at zero, while the fitted gamma component is continuous.

Definition: Let $S$ be an aggregate loss with mean $m$, variance $\sigma^2>0$, and positive skewness $\gamma_1$. The translated gamma approximation represents $S$ by $c+G$, where $G$ has a gamma distribution with shape parameter $\alpha>0$ and scale parameter $\theta>0$. The parameters are chosen to match the mean, variance, and skewness of $S$.

Here, $c$ is a location shift, $\alpha$ is the gamma shape, and $\theta$ is the scale. With this scale convention, a gamma random variable $G$ has mean $\alpha\theta$, variance $\alpha\theta^2$, and skewness $2/\sqrt{\alpha}$. The translated variable $c+G$ has the same variance and skewness as $G$, but its mean is increased by $c$.

Matching the Three Moments

Suppose the aggregate moments are $E[S]=m$, $\operatorname{Var}(S)=\sigma^2$, and skewness $\gamma_1$. Equating these to the corresponding moments of $c+G$ gives three equations:

m=c+\alpha\theta,\qquad \sigma^2=\alpha\theta^2,\qquad \gamma_1=\frac{2}{\sqrt{\alpha}}.

The skewness equation determines the shape. The variance equation then determines the scale, and the mean equation determines the translation.

Match skewness.
From $\gamma_1=2/\sqrt{\alpha}$, obtain $\alpha=4/\gamma_1^2$.

Match variance.
Use $\sigma^2=\alpha\theta^2$, so $\theta=\sigma/\sqrt{\alpha}=\gamma_1\sigma/2$.

Match the mean.
Set $c=m-\alpha\theta=m-2\sigma/\gamma_1$.

Theorem: For positive aggregate variance $\sigma^2$ and positive skewness $\gamma_1$, the moment-matching translated gamma parameters are

\alpha=\frac{4}{\gamma_1^2},\qquad \theta=\frac{\gamma_1\sigma}{2},\qquad c=m-\frac{2\sigma}{\gamma_1}.

The fitted approximation is $S\approx c+G$, where $G\sim\operatorname{Gamma}(\alpha,\theta)$ uses the shape-scale parameterization.

Why the Matching Formulas Hold

A gamma variable with shape $\alpha$ and scale $\theta$ has cumulants $\kappa_r=\alpha(r-1)!\theta^r$ for positive integers $r$. In particular, its first three cumulants are $\alpha\theta$, $\alpha\theta^2$, and $2\alpha\theta^3$. Translation by $c$ adds $c$ to the first cumulant and leaves the second and higher cumulants unchanged. Therefore the translated gamma has mean $c+\alpha\theta$, variance $\alpha\theta^2$, and third central moment $2\alpha\theta^3$. Its skewness is consequently

\frac{2\alpha\theta^3}{(\alpha\theta^2)^{3/2}} =\frac{2}{\sqrt{\alpha}}.

Solving the three matching equations in order gives the stated parameters. This also explains why the location shift affects the mean but cannot repair a mismatch in variance or skewness.

Using the Fitted Distribution

Let $F_\Gamma(y;\alpha,\theta)$ denote the gamma cumulative distribution function with shape $\alpha$ and scale $\theta$. Since $G\geq 0$, the translated approximation has support beginning at $c$. For a threshold $x$,

P(S\leq x)\ \approx\ \begin{cases} 0, & x\leq c,\\ F_\Gamma(x-c;\alpha,\theta), & x>c. \end{cases}

For an upper-tail probability, use $P(S>x)\approx 1-P(S\leq x)$. When $\alpha$ is a positive integer, the gamma distribution is an Erlang distribution and its CDF has the finite-sum form

F_\Gamma(y;\alpha,\theta) =1-e^{-y/\theta}\sum_{j=0}^{\alpha-1}\frac{(y/\theta)^j}{j!}, \qquad y\geq 0.

For noninteger shape, the gamma CDF is evaluated using a gamma distribution function or table. The threshold must be shifted by $c$ before evaluating the gamma CDF; evaluating at $x$ itself would omit the translation.

Worked Example: A Poisson Aggregate With Fixed Claim Amounts

A hypothetical portfolio has a Poisson number $N$ of claims, with $N\sim\operatorname{Poisson}(1)$. Each claim is exactly $100, independently of the others, so $S=100N$. Fit a translated gamma approximation by moment matching and estimate $P(S\leq 100)$. Give the probability to four decimal places.

Method. First calculate the aggregate mean, variance, and skewness. For a compound Poisson sum with claim amount $X$, the first three cumulants are $\lambda E[X]$, $\lambda E[X^2]$, and $\lambda E[X^3]$. Here $\lambda=1$ and $X=100$ with probability 1, so

E[S]=1(100)=100,\quad \operatorname{Var}(S)=1(100^2)=10{,}000,\quad \kappa_3=1(100^3)=1{,}000{,}000.

The standard deviation is $\sigma=\sqrt{10{,}000}=100$. The skewness is $\gamma_1=\kappa_3/\sigma^3=1{,}000{,}000/1{,}000{,}000=1$. Thus

\alpha=\frac{4}{1^2}=4,\qquad \theta=\frac{1(100)}{2}=50,\qquad c=100-\frac{2(100)}{1}=-100.

The event $S\leq100$ corresponds under the approximation to $G\leq100-(-100)=200$. Hence $y/\theta=200/50=4$, and

P(S\leq100)\approx 1-e^{-4}\left(1+4+\frac{4^2}{2!}+\frac{4^3}{3!}\right) =1-e^{-4}(23.6667) \approx 0.5665.

The fitted distribution has mean $-100+4(50)=100$, variance $4(50^2)=10{,}000$, and skewness $2/\sqrt{4}=1$, as required. The exact aggregate has discrete outcomes, including a probability at zero; the answer is a continuous-distribution approximation, not its exact probability. The common mistake is to use $100/50=2$ as the gamma CDF argument. The translation requires $100-c=200$, not 100.

Worked Example: Estimating an Upper-Tail Probability

A hypothetical aggregate loss has mean $1,000, standard deviation $100\sqrt{2}$ dollars, and skewness $\sqrt{2}$. Use a translated gamma approximation to estimate the probability the aggregate exceeds $900. Give the probability to four decimal places.

Method. Match the three supplied moments:

\alpha=\frac{4}{(\sqrt{2})^2}=2,\qquad \theta=\frac{(\sqrt{2})(100\sqrt{2})}{2}=100,\qquad c=1{,}000-\frac{2(100\sqrt{2})}{\sqrt{2}}=800.

The threshold $900 corresponds to $G\leq900-800=100$. For shape 2 and scale 100, the standardized argument is $100/100=1$. The lower-tail probability is

P(S\leq900)\approx1-e^{-1}(1+1)=1-\frac{2}{e}=0.2642.

Therefore

P(S>900)\approx1-0.2642=\frac{2}{e}=0.7358.

As a check, the fitted mean is $800+2(100)=1{,}000$, its variance is $2(100^2)=20{,}000=(100\sqrt{2})^2$, and its skewness is $2/\sqrt{2}=\sqrt{2}$. The common mistake is to report the lower-tail probability as the requested upper tail; the exceedance probability is its complement.

Worked Example: A Threshold at the Fitted Mean

A hypothetical aggregate has mean $900, standard deviation $100\sqrt{3}$ dollars, and skewness $2/\sqrt{3}$. Estimate the probability the aggregate is at most $900. Give the answer to four decimal places.

Method. The shape is

\alpha=\frac{4}{(2/\sqrt{3})^2}=3.

The scale and shift are

\theta=\frac{(2/\sqrt{3})(100\sqrt{3})}{2}=100,\qquad c=900-\frac{2(100\sqrt{3})}{2/\sqrt{3}}=900-300=600.

At the threshold $x=900$, the gamma component is evaluated at $x-c=300$, so $y/\theta=300/100=3$. Therefore

P(S\leq900)\approx 1-e^{-3}\left(1+3+\frac{3^2}{2!}\right) =1-8.5e^{-3} \approx0.5768.

The fitted mean is $600+3(100)=900$, the variance is $3(100^2)=30{,}000=(100\sqrt{3})^2$, and the skewness is $2/\sqrt{3}$. A useful check is that the probability at the mean is not generally 0.5000: a right-skewed distribution is not symmetric about its mean. The common mistake is to assume that the mean divides the distribution into equal halves.

Interpretation and Exam Traps

The translated gamma approximation uses skewness to determine the gamma shape, variance to determine its scale, and the mean to locate it. For this reason, a small error in the skewness can change the fitted shape and the tail estimate even if the mean and variance are correct. Always keep the parameters distinct: $m$ is the mean of the aggregate approximation, while $\alpha\theta$ is the mean of the gamma component before translation.

Forgetting the shift. Evaluate the gamma CDF at $x-c$, not at $x$.
Mixing scale and rate. This tutorial uses scale $\theta$, so the gamma mean is $\alpha\theta$. A rate parameterization uses a different formula.
Reversing the tail. If the question asks for an exceedance probability, take one minus the lower-tail CDF.
Assuming an exact model. Moment matching does not make a discrete or zero-inflated aggregate exactly gamma distributed.

This method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. It is most useful when the aggregate moments are available and a quick probability estimate is needed; it should be treated as an approximation whose quality can depend on the aggregate distribution’s shape and support.

Check Your Understanding

For each problem, determine the translated gamma parameters before estimating the requested probability.

A hypothetical aggregate has mean $700, standard deviation $100, and skewness 1. Estimate $P(S\leq700)$. (Answer: $\alpha=4,\ \theta=50,\ c=500;\ P(S\leq700)\approx0.5665$.)
A hypothetical aggregate has mean $1,000, standard deviation $100\sqrt{2}$, and skewness $\sqrt{2}$. Estimate $P(S>900)$. (Answer: $\alpha=2,\ \theta=100,\ c=800;\ P(S>900)\approx0.7358$.)
A hypothetical aggregate has mean $1,200, standard deviation $100\sqrt{3}$, and skewness $2/\sqrt{3}$. Estimate $P(S\leq1{,}200)$. (Answer: $\alpha=3,\ \theta=100,\ c=900;\ P(S\leq1{,}200)\approx0.5768$.)
A hypothetical aggregate has mean $500, standard deviation $200, and skewness 2. Estimate $P(S\leq500)$. (Answer: $\alpha=1,\ \theta=200,\ c=300;\ P(S\leq500)\approx1-e^{-1}=0.6321$.)

What You'll Learn