Poisson-Gamma Frequency Components

Why the Poisson-Gamma Model Is Useful

Claim counts vary for two related reasons. Even for a risk with a fixed underlying claim rate, the realized count fluctuates randomly from period to period. In addition, different risks may have different underlying rates. The Poisson-Gamma model represents both sources of variation: a Poisson distribution describes counts conditional on a risk’s rate, and a Gamma distribution describes how that rate varies across risks.

The model provides a direct way to calculate the Bühlmann components for frequency data. The key is to keep the quantity being analyzed clear. A claim count over a specified exposure period and a frequency per unit of exposure have different process variances, even though they share the same underlying risk-rate distribution.

Definition: Let $\Lambda$ be a risk’s claim rate per unit of exposure, with $\Lambda\sim\operatorname{Gamma}(\alpha,\beta)$ using the shape-rate parameterization, where $\alpha>0$ and $\beta>0$. Its density is $$ f_\Lambda(\lambda)=\frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta\lambda},\qquad \lambda>0. $$ For exposure $e>0$, the claim count $N_e$, conditional on $\Lambda$, satisfies $N_e\mid\Lambda\sim\operatorname{Poisson}(e\Lambda)$. The exposure-normalized frequency is $X_e=N_e/e$.

Under this parameterization, $\alpha$ is the shape and $\beta$ is the rate; $\beta$ is not the scale. The corresponding Gamma scale is $1/\beta$. We assume the same risk rate applies throughout the exposure period, and that the conditional count is Poisson with mean equal to exposure times that rate. Exposures must be measured in units consistent with $\Lambda$.

The Three Components for Frequency

For the exposure-normalized frequency $X_e$, the hypothetical mean is the expected frequency for a randomly selected risk. The expected process variance measures average fluctuation within a risk, conditional on its rate. The variance of hypothetical means measures the variation between risks in their underlying expected frequencies. These are the Bühlmann components often denoted by $m$, EPV, and VHM, respectively.

\begin{aligned} m &= E[X_e],\\ \mathrm{EPV} &= E[\operatorname{Var}(X_e\mid\Lambda)],\\ \mathrm{VHM} &= \operatorname{Var}(E[X_e\mid\Lambda]). \end{aligned}

The model gives especially simple conditional moments. Since $N_e\mid\Lambda$ is Poisson with mean and variance $e\Lambda$, dividing by $e$ gives $E[X_e\mid\Lambda]=\Lambda$ and $\operatorname{Var}(X_e\mid\Lambda)=\Lambda/e$. Thus the exposure-normalized hypothetical mean does not depend on $e$, while its expected process variance decreases as exposure increases.

Theorem: If $\Lambda\sim\operatorname{Gamma}(\alpha,\beta)$ in shape-rate form and $N_e\mid\Lambda\sim\operatorname{Poisson}(e\Lambda)$, then for $X_e=N_e/e$, $$ m=\frac{\alpha}{\beta},\qquad \mathrm{EPV}=\frac{\alpha}{\beta e},\qquad \mathrm{VHM}=\frac{\alpha}{\beta^2}. $$ For the unnormalized count $N_e$, the corresponding hypothetical mean, expected process variance, and variance of hypothetical means are $e\alpha/\beta$, $e\alpha/\beta$, and $e^2\alpha/\beta^2$, respectively.

Why the Component Formulas Hold

The Gamma moments under the stated parameterization are $E[\Lambda]=\alpha/\beta$ and $\operatorname{Var}(\Lambda)=\alpha/\beta^2$. For example, these follow by differentiating the Gamma moment-generating function $M_\Lambda(t)=(\beta/(\beta-t))^\alpha$, valid for $t<\beta$. Its first derivative at zero is $\alpha/\beta$, and its second derivative at zero, less the square of the first moment, is $\alpha/\beta^2$.

Now use the conditional moments of $X_e$. Taking the expectation of its conditional mean gives $E[X_e]=E[\Lambda]=\alpha/\beta$. Taking the expectation of its conditional variance gives $E[\Lambda/e]=\alpha/(\beta e)$. Finally, the conditional mean itself is $\Lambda$, so the variance of that conditional mean is $\operatorname{Var}(\Lambda)=\alpha/\beta^2$. This derives each component rather than treating the count’s unconditional variance as a single unexplained quantity.

\operatorname{Var}(X_e) =E[\operatorname{Var}(X_e\mid\Lambda)] +\operatorname{Var}(E[X_e\mid\Lambda]) =\frac{\alpha}{\beta e}+\frac{\alpha}{\beta^2}.

This is the law of total variance applied to the frequency. For the raw count $N_e=eX_e$, multiplying by $e$ multiplies its mean by $e$ and its variance by $e^2$. In particular, its process component becomes $e^2\mathrm{EPV}=e\alpha/\beta$, and its between-risk component becomes $e^2\mathrm{VHM}=e^2\alpha/\beta^2$. This scaling is an important check on calculations.

Identify the quantity.
Decide whether the problem asks for the count $N_e$ or the frequency $N_e/e$, and record the exposure.

Put the Gamma parameters in the correct form.
For shape-rate parameters, use $E[\Lambda]=\alpha/\beta$ and $\operatorname{Var}(\Lambda)=\alpha/\beta^2$. If a scale is supplied, invert it to obtain the rate.

Calculate the components.
For frequency, use $m=E[\Lambda]$, $\mathrm{EPV}=E[\Lambda]/e$, and $\mathrm{VHM}=\operatorname{Var}(\Lambda)$. Scale to count components only if the requested variable is the count.

Check the total variance.
Add EPV and VHM for the frequency, or their count-scale counterparts, and confirm the result agrees with the total variance decomposition.

Worked Applications

Worked Example: Components for a Half-Year Claim Count

A hypothetical class of risks has rate $\Lambda\sim\operatorname{Gamma}(4,2)$, with shape 4 and rate 2. The rate is measured in claims per year. A policy is observed for $e=0.5$ years, and its count satisfies $N_e\mid\Lambda\sim\operatorname{Poisson}(e\Lambda)$. Calculate the components for both the count and the annualized frequency.

Method and parameter check. The Gamma parameters are shape-rate, not shape-scale. Therefore, $$ E[\Lambda]=\frac{4}{2}=2,\qquad \operatorname{Var}(\Lambda)=\frac{4}{2^2}=1. $$

Annualized frequency. Since $X_e=N_e/e$, its hypothetical mean is $m=2$. Its expected process variance is $$ \mathrm{EPV}=\frac{E[\Lambda]}{e}=\frac{2}{0.5}=4, $$ and its variance of hypothetical means is $\mathrm{VHM}=1$. Its total variance is $4+1=5$.

Count over half a year. The expected count is $eE[\Lambda]=0.5(2)=1$, and the count EPV is also 1. The count VHM is $$ e^2\operatorname{Var}(\Lambda)=(0.5)^2(1)=0.25. $$ Thus the count variance is $1+0.25=1.25$. This agrees with multiplying the annualized-frequency variance by $e^2$: $(0.5)^2(5)=1.25$.

Check and common mistake avoided. The count mean is 1 and its variance is 1.25, so the variance exceeds the mean because rates vary across risks. Using the frequency EPV of 4 as the count EPV would mix units and give the wrong result; the count EPV is 1.

Worked Example: Converting a Gamma Scale

For a hypothetical book of risks, $\Lambda$ has a Gamma distribution with shape 6 and scale $1/3$ claims per year. A risk is observed for $e=2$ years, with $N_e\mid\Lambda\sim\operatorname{Poisson}(2\Lambda)$. Find the components for the two-year count and the average annual frequency.

Method and parameter conversion. The supplied Gamma parameter is a scale, so the equivalent rate is its reciprocal: $\beta=1/(1/3)=3$. Hence $$ E[\Lambda]=6\left(\frac13\right)=2,\qquad \operatorname{Var}(\Lambda)=6\left(\frac13\right)^2=\frac{2}{3}. $$

Two-year count. Its hypothetical mean and EPV are each $eE[\Lambda]=2(2)=4$. Its VHM is $$ e^2\operatorname{Var}(\Lambda)=2^2\left(\frac23\right)=\frac83. $$ The total count variance is $4+8/3=20/3$, or approximately $6.67$.

Average annual frequency. For $X_e=N_e/2$, the hypothetical mean is 2, the EPV is $2/2=1$, and the VHM is $2/3$. Its total variance is $1+2/3=5/3$, or approximately $1.67$. Multiplying this by $e^2=4$ gives $20/3$, the count variance calculated above.

Check and common mistake avoided. Both calculations give the same mean rate of 2 after the count is divided by its exposure. Treating the scale $1/3$ as a rate would incorrectly produce $E[\Lambda]=18$; converting the scale to its reciprocal avoids that parameterization error.

Worked Example: Comparing Two Exposure Periods

Suppose hypothetical risk rates follow $\Lambda\sim\operatorname{Gamma}(3,1.5)$ in shape-rate form, measured per year. Compare the frequency components for exposures of $0.5$ and $1.5$ years. Also calculate the corresponding count variances.

Method. First calculate the risk-rate moments: $$ E[\Lambda]=\frac{3}{1.5}=2,\qquad \operatorname{Var}(\Lambda)=\frac{3}{1.5^2}=\frac43. $$ For either exposure, the frequency hypothetical mean is 2 and the frequency VHM is $4/3$.

Exposure $e=0.5$. The frequency EPV is $2/0.5=4$, giving total frequency variance $4+4/3=16/3$. On the count scale, the mean and EPV are both $0.5(2)=1$, and the VHM is $(0.5)^2(4/3)=1/3$. The count variance is $1+1/3=4/3$.

Exposure $e=1.5$. The frequency EPV is $2/1.5=4/3$, giving total frequency variance $4/3+4/3=8/3$. The count mean and EPV are both $1.5(2)=3$, and the count VHM is $(1.5)^2(4/3)=3$. The count variance is $3+3=6$.

Check and common mistake avoided. The longer observation period has a smaller frequency EPV because the same risk rate is averaged over more exposure. Its frequency VHM remains $4/3$, since changing exposure does not change the distribution of the underlying annual rates. The count VHM, however, grows with exposure squared. Applying that count-scale factor to the frequency VHM would confuse the two quantities.

Interpretation and Exam Traps

The components separate random variation in observed counts from variation in underlying rates. A larger $\alpha/\beta$ increases the hypothetical mean and, at fixed exposure, the expected process variance. A larger $\alpha/\beta^2$ means more heterogeneity in risk rates and therefore a larger VHM. At fixed mean rate, greater between-risk variation can increase the variability of observed counts without changing the expected frequency.

Exposure affects process variation because more exposure provides more opportunities for claims and also changes the scale of a frequency average. For an annualized frequency, the EPV is proportional to $1/e$; the VHM is unchanged. For a raw count, the EPV is proportional to $e$, while the VHM is proportional to $e^2$. Always check which variable the question defines before applying these relationships.

Using the wrong Gamma variance. For shape-rate parameters, the variance is $\alpha/\beta^2$, not $\alpha/\beta$. In the first example, $\alpha/\beta=2$ is the mean, while the variance is $4/4=1$.
Dropping the VHM. The unconditional count variance is not just its expected conditional variance. The total variance identity requires both EPV and VHM.
Mixing exposure units. A rate expressed per year must be paired with exposure in years. A half-year exposure is $0.5$, not 6 when entered in months, unless the rate is also converted to a monthly rate.
Using count formulas for frequency. Dividing a count by exposure changes the variance by the square of that divisor, not by the divisor itself.

Key takeaway: Under the Poisson-Gamma model, the frequency components are $m=\alpha/\beta$, $\mathrm{EPV}=\alpha/(\beta e)$, and $\mathrm{VHM}=\alpha/\beta^2$. Keep the shape-rate convention and the count-versus-frequency scale explicit. These calculations are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

For each problem, identify whether the requested components are on a count or frequency scale and use the stated Gamma parameterization.

A hypothetical rate has $\Lambda\sim\operatorname{Gamma}(5,2)$ in shape-rate form, and a risk has exposure $e=1.25$ years. Find the frequency hypothetical mean, EPV, VHM, and total variance. (Answer: $m=2.5$, EPV $=2.0$, VHM $=1.25$, total variance $=3.25$.)
A risk rate has a Gamma distribution with shape 3 and scale $0.4$ claims per year. For a two-year count, find the count mean, EPV, VHM, and total variance. (Answer: mean $=2.4$, EPV $=2.4$, VHM $=1.92$, total variance $=4.32$.)
Let $\Lambda\sim\operatorname{Gamma}(8,4)$ in shape-rate form, and observe a risk for $0.5$ years. Find the annualized frequency components and the count variance. (Answer: frequency $m=2$, EPV $=4$, VHM $=0.5$, total variance $=4.5$; count variance $=1.125$.)
For $\Lambda\sim\operatorname{Gamma}(2,1)$ in shape-rate form, a count is observed over three years. Find the frequency EPV and VHM, then the count EPV and VHM. (Answer: frequency EPV $=2/3$, frequency VHM $=2$; count EPV $=6$, count VHM $=18$.)
Two hypothetical rate distributions have the same mean: Model A is $\operatorname{Gamma}(4,2)$, and Model B is $\operatorname{Gamma}(10,5)$, both in shape-rate form. At one year of exposure, compare their frequency EPV, VHM, and total variance. (Answer: both have mean $2$ and EPV $=2$; Model A has VHM $=1$ and total variance $=3$, while Model B has VHM $=0.4$ and total variance $=2.4$.)

What You'll Learn