Gamma Severities With Poisson Frequency

Model and Notation

When claim frequency is Poisson and claim amounts are gamma distributed, the aggregate loss is a mixture of gamma distributions: its conditional distribution depends on the number of claims. This structure makes it possible to calculate conditional probabilities and moments directly, without treating the aggregate as gamma with one fixed set of parameters. The methods here are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Let $N$ be the number of claims in a fixed period, with $N\sim\operatorname{Poisson}(\lambda)$. Assume claim amounts $X_1,X_2,\ldots$ are independent and identically distributed, independent of $N$, and each has a gamma distribution with shape $\alpha>0$ and scale $\theta>0$. The scale convention matters: the mean is $\alpha\theta$, not $\alpha/\theta$. Define the aggregate loss by $Z=\sum_{j=1}^{N}X_j$, with an empty sum equal to zero.

Definition: Under the shape-scale convention, $X\sim\operatorname{Gamma}(\alpha,\theta)$ has density $f_X(x)=x^{\alpha-1}e^{-x/\theta}/(\Gamma(\alpha)\theta^\alpha)$ for $x>0$. Its mean is $E[X]=\alpha\theta$, its variance is $\operatorname{Var}(X)=\alpha\theta^2$, and its second moment is $E[X^2]=\alpha(\alpha+1)\theta^2$. Here $\Gamma(\alpha)$ is the gamma function.

The Poisson assumption gives $P(N=n)=e^{-\lambda}\lambda^n/n!$ for $n=0,1,2,\ldots$. A gamma severity is strictly positive with probability one, so $Z=0$ exactly when $N=0$. Consequently, the aggregate distribution has a probability mass $e^{-\lambda}$ at zero as well as a continuous part for positive losses.

Aggregate Distribution Given the Claim Count

For a fixed positive count, the aggregate is a sum of independent gamma variables. The key fact is that gamma variables with a common scale add by adding their shapes. Thus, conditional on $N=n$, the aggregate has shape $n\alpha$ and scale $\theta$. The scale does not get multiplied by $n$.

Theorem: If $X_1,\ldots,X_n$ are independent with $X_j\sim\operatorname{Gamma}(\alpha,\theta)$, then $\sum_{j=1}^{n}X_j\sim\operatorname{Gamma}(n\alpha,\theta)$. Therefore, for $n\geq 1$, $Z\mid N=n\sim\operatorname{Gamma}(n\alpha,\theta)$. For $n=0$, $Z=0$ with probability one; it is a point mass, not a gamma distribution.

One justification uses moment-generating functions. For $X\sim\operatorname{Gamma}(\alpha,\theta)$, $M_X(t)=(1-\theta t)^{-\alpha}$ for $t<1/\theta$. Independence means the moment-generating function of a sum is the product of the individual functions. The product is the moment-generating function of a gamma variable with the summed shape and the same scale:

M_{\sum_{j=1}^{n}X_j}(t) =\prod_{j=1}^{n}(1-\theta t)^{-\alpha} =(1-\theta t)^{-n\alpha}.

This argument requires independent gamma variables with a common scale. If scales differ, this product does not generally describe a gamma distribution with the summed shape and one unchanged scale.

For $n\geq1$, the conditional distribution function can be written using the lower incomplete gamma function $\gamma(r,y)$:

P(Z\leq z\mid N=n) =\frac{\gamma(n\alpha,z/\theta)}{\Gamma(n\alpha)},\qquad z\geq0.

When $n\alpha=m$ is a positive integer, this is an Erlang distribution and the calculation simplifies to $1-e^{-z/\theta}\sum_{j=0}^{m-1}(z/\theta)^j/j!$. This finite-sum form is useful for hand calculations. It does not apply in that form when the shape is not an integer; use the incomplete-gamma distribution function in that case.

Worked Example: A Conditional Probability for Three Claims

A hypothetical insurer models each claim amount, in thousands of dollars, as $\operatorname{Gamma}(2,1)$. Given that exactly three claims occur in the period, find the probability that their aggregate is at most 6 thousand dollars. Also find its conditional mean and variance.

Method. Condition on the stated count and use gamma additivity. The frequency parameter is not needed because the claim count is already specified. A common mistake is to multiply the scale by three; additivity changes the shape, not the common scale.

With $n=3$, $\alpha=2$, and $\theta=1$, the conditional aggregate distribution is $\operatorname{Gamma}(3\cdot2,1)=\operatorname{Gamma}(6,1)$. Its mean and variance are

E[Z\mid N=3]=6(1)=6,\qquad \operatorname{Var}(Z\mid N=3)=6(1)^2=6.

For the probability, the shape is the integer $m=6$, and $z/\theta=6/1=6$. The Erlang formula gives

\begin{aligned} P(Z\leq6\mid N=3) &=1-e^{-6}\left(1+6+\frac{6^2}{2!}+\frac{6^3}{3!} +\frac{6^4}{4!}+\frac{6^5}{5!}\right)\\ &=1-e^{-6}(1+6+18+36+54+64.8)\\ &=1-e^{-6}(179.8)\\ &\approx 0.5543. \end{aligned}

The result is a probability between zero and one. In the stated units, the conditional mean is 6 thousand dollars and the conditional variance is 6 thousand-squared dollars; these agree with the parameters of $\operatorname{Gamma}(6,1)$.

Conditional and Unconditional Moments

The conditional moments follow directly from the gamma parameters. For $n\geq1$, $E[Z\mid N=n]=n\alpha\theta$ and $\operatorname{Var}(Z\mid N=n)=n\alpha\theta^2$. Both are zero when $n=0$, so the same formulas also hold for that case. To obtain unconditional moments, average over the Poisson count and account for the variation in the conditional mean.

Average the conditional mean.
Since $E[N]=\lambda$, $E[Z]=E[E[Z\mid N]]=\lambda\alpha\theta$.

Apply total variance.
$\operatorname{Var}(Z)=E[\operatorname{Var}(Z\mid N)]+\operatorname{Var}(E[Z\mid N])$.

Substitute the Poisson moments.
The first term is $\lambda\alpha\theta^2$; the second is $\operatorname{Var}(N\alpha\theta)=\lambda\alpha^2\theta^2$.

E[Z]=\lambda\alpha\theta,\qquad \operatorname{Var}(Z)=\lambda\alpha(1+\alpha)\theta^2 =\lambda E[X^2].

The two contributions to variance have distinct meanings. The term $\lambda\alpha\theta^2$ comes from variation in claim amounts while the count is fixed; $\lambda\alpha^2\theta^2$ comes from variation in the number of claims. Their sum is also $\lambda E[X^2]$, a useful check against the general compound Poisson moment formula.

Worked Example: Unconditional Moments and the Zero-Loss Probability

A hypothetical repair portfolio has $N\sim\operatorname{Poisson}(2)$. Independent claim amounts, also independent of $N$, follow $\operatorname{Gamma}(3,200)$ in dollars, using shape and scale. Find the aggregate mean, aggregate variance, probability of zero aggregate loss, and the aggregate distribution given two claims.

Method. Use the Poisson-gamma moment formulas for the unconditional moments, and gamma additivity for the conditional distribution. The frequency parameter belongs in the unconditional moments, but not in the distribution conditional on a specified count.

Here $\lambda=2$, $\alpha=3$, and $\theta=200$. The individual claim mean and second moment are

E[X]=3(200)=600,\qquad E[X^2]=3(3+1)(200)^2=12(40{,}000)=480{,}000.

Therefore,

E[Z]=2(600)=\$1{,}200,\qquad \operatorname{Var}(Z)=2(480{,}000)=960{,}000\text{ dollars}^2.

Zero aggregate loss occurs exactly when there are no claims. Hence $P(Z=0)=P(N=0)=e^{-2}\approx0.1353$. Given two claims, the conditional aggregate distribution is $\operatorname{Gamma}(2\cdot3,200)=\operatorname{Gamma}(6,200)$, with conditional mean $6(200)=\$1{,}200$ and variance $6(200)^2=240{,}000\text{ dollars}^2$.

As a check, the unconditional mean equals the expected count times the expected claim amount, $2(\$600)=\$1{,}200$. The unconditional variance exceeds the conditional variance given two claims because the unconditional calculation also includes uncertainty in the count.

Worked Example: A Noninteger Severity Shape

In a hypothetical equipment pool, each claim amount follows $\operatorname{Gamma}(1.5,400)$ dollars. Given exactly two independent claims, calculate the probability that their aggregate exceeds \$1,200, together with the conditional mean and variance.

Method. Add the shapes while retaining the common scale, then use the integer-shape Erlang formula. The individual shape is not an integer, but the aggregate shape is $2(1.5)=3$, so the finite sum applies. A common mistake is to use the individual shape in the aggregate probability calculation.

The aggregate distribution conditional on $N=2$ is $\operatorname{Gamma}(3,400)$. Its conditional moments are

E[Z\mid N=2]=3(400)=\$1{,}200,\qquad \operatorname{Var}(Z\mid N=2)=3(400)^2=480{,}000\text{ dollars}^2.

At the threshold $z=\$1{,}200$, $z/\theta=1200/400=3$. Thus,

\begin{aligned} P(Z>\$1{,}200\mid N=2) &=e^{-3}\left(1+3+\frac{3^2}{2!}\right)\\ &=e^{-3}(1+3+4.5)\\ &=8.5e^{-3}\\ &\approx0.4232. \end{aligned}

The complementary conditional probability is $1-0.4232=0.5768$, consistent with the calculated upper tail. The threshold equals the conditional mean, but the gamma distribution is right-skewed, so the probability above the mean need not be one-half.

Interpreting the Mixture and Avoiding Errors

Unconditionally, the distribution is a Poisson-weighted mixture of the conditional distributions. For $z\geq0$, its distribution function can be expressed as

P(Z\leq z) =e^{-\lambda} +\sum_{n=1}^{\infty}e^{-\lambda}\frac{\lambda^n}{n!} \frac{\gamma(n\alpha,z/\theta)}{\Gamma(n\alpha)}.

The first term is the zero-claim probability. The other terms weight each positive-count conditional gamma distribution by the probability of that count. This formula also clarifies why the full aggregate is generally not a single gamma distribution: it includes a discrete mass at zero and a mixture over multiple gamma shapes.

Check the following points when solving a problem:

State the parameter convention. With shape and scale, the mean is $\alpha\theta$. A rate parameterization uses different formulas.
Keep the scale fixed when adding. For $n$ independent, common-scale gamma severities, the aggregate shape is $n\alpha$ and its scale is $\theta$.
Separate conditional from unconditional calculations. Given $N=n$, use the gamma distribution for that count. For unconditional moments, include the randomness in $N$.
Include the atom at zero. A positive gamma claim cannot produce zero loss, so the aggregate is zero precisely when the Poisson count is zero.
Check the scope of the Erlang formula. Its finite sum requires a positive integer shape. For other shapes, use the gamma distribution function rather than rounding the shape.

Key takeaway: For Poisson frequency with independent $\operatorname{Gamma}(\alpha,\theta)$ severities, the aggregate given $n$ claims is $\operatorname{Gamma}(n\alpha,\theta)$ when $n\geq1$, and is zero when $n=0$. The unconditional moments are $E[Z]=\lambda\alpha\theta$ and $\operatorname{Var}(Z)=\lambda\alpha(1+\alpha)\theta^2$.

Check Your Understanding

For each problem, use the shape-scale convention and treat claim amounts as independent of the Poisson count. Give probabilities to four decimal places.

Given four claims, each with severity $\operatorname{Gamma}(1,250)$ dollars, find the conditional aggregate distribution, mean, and variance. (Answer: $\operatorname{Gamma}(4,250)$; mean \$1,000; variance 250,000 dollars-squared.)
Given three claims with severity $\operatorname{Gamma}(2,100)$ dollars, find $P(Z\leq \$600\mid N=3)$. (Answer: 0.5543.)
Suppose $N\sim\operatorname{Poisson}(1.5)$ and claim severity is $\operatorname{Gamma}(2,300)$ dollars. Find the unconditional mean and variance of aggregate loss. (Answer: mean \$900; variance 810,000 dollars-squared.)
Given two claims with severity $\operatorname{Gamma}(1.5,500)$ dollars, find the probability their aggregate exceeds \$1,500. (Answer: 0.4232.)
Suppose $N\sim\operatorname{Poisson}(0.8)$ and severity is $\operatorname{Gamma}(1,1{,}000)$ dollars. Find the probability of zero aggregate loss, mean, and variance. (Answer: $P(Z=0)=0.4493$; mean \$800; variance 1,600,000 dollars-squared.)

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