Gamma Aggregation With A Common Scale

Why the Scale Parameter Determines Aggregation

Adding independent outcomes is straightforward for their means and variances: means add, and variances add when the outcomes are independent. But those two calculations alone do not determine the distribution of a sum. In particular, adding two Gamma outcomes does not always produce another Gamma outcome. The key condition for exact Gamma aggregation is a common scale parameter.

This distinction matters when a model describes several positive outcomes and an analyst wants a distribution for their total. The Gamma parameterization must be read carefully: a common scale is not the same condition as a common mean, a common shape, or merely a common dispersion parameter.

Definition: A random variable $X$ has a Gamma distribution with shape $\alpha>0$ and scale $\theta>0$, written $X\sim\operatorname{Gamma}(\alpha,\theta)$, if its density is $$ f_X(x)=\frac{x^{\alpha-1}e^{-x/\theta}}{\Gamma(\alpha)\theta^\alpha},\qquad x>0. $$ Its mean and variance are $E[X]=\alpha\theta$ and $\operatorname{Var}(X)=\alpha\theta^2$. In the rate parameterization, the rate is $\beta=1/\theta$; the same distribution is then written $\operatorname{Gamma}(\alpha,\beta)$, with density proportional to $e^{-\beta x}$.

Unless a rate parameterization is explicitly stated, this tutorial uses shape and scale. Independence is also essential: the exact aggregation result below assumes the Gamma variables are mutually independent. Dependence can change the distribution of the sum, even when all marginal distributions have the same scale.

The Common-Scale Aggregation Result

Theorem: If $X_1,\ldots,X_n$ are independent and $X_j\sim\operatorname{Gamma}(\alpha_j,\theta)$ for every $j$, with the same scale $\theta$, then $$ \sum_{j=1}^n X_j\sim\operatorname{Gamma}\left(\sum_{j=1}^n\alpha_j,\theta\right). $$ The shape parameters add; the shared scale remains unchanged.

One justification uses the moment-generating function. For $X\sim\operatorname{Gamma}(\alpha,\theta)$, the moment-generating function is $M_X(t)=(1-\theta t)^{-\alpha}$ for $t<1/\theta$. Independence makes the moment-generating function of a sum the product of the individual moment-generating functions. With a common scale, the product simplifies to the moment-generating function of a Gamma distribution.

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

The final expression is the moment-generating function of a Gamma variable with shape $\sum_j\alpha_j$ and scale $\theta$. Since the moment-generating function exists in a neighborhood of zero, it determines the distribution, giving the stated result.

The same conclusion follows from Laplace transforms, which are defined for nonnegative variables at $s\geq 0$. For a Gamma variable, the transform is $E[e^{-sX}]=(1+\theta s)^{-\alpha}$. The product of these transforms for independent variables with common $\theta$ is $(1+\theta s)^{-\sum_j\alpha_j}$, again the transform of the stated Gamma distribution.

Worked Example: Aggregating Three Outcomes With a Common Scale

In a hypothetical study, three independent positive measurements are modeled as $X_1\sim\operatorname{Gamma}(2,2)$, $X_2\sim\operatorname{Gamma}(3,2)$, and $X_3\sim\operatorname{Gamma}(4,2)$, using shape and scale. Find the distribution, mean, and variance of their total $T=X_1+X_2+X_3$. No discounting or mortality-contingent payments are involved.

Method. Check independence and compare the scales before adding any parameters. The variables are independent, and each has scale $\theta=2$, so the common-scale theorem applies.

Step 1: Add the shapes. The total shape is $2+3+4=9$. The scale remains $2$, so $T\sim\operatorname{Gamma}(9,2)$.

Step 2: Calculate the mean. The Gamma mean is shape times scale: $E[T]=9(2)=18$. As a check, the individual means are $2(2)=4$, $3(2)=6$, and $4(2)=8$, whose sum is $4+6+8=18$.

Step 3: Calculate the variance. The Gamma variance is shape times scale squared: $\operatorname{Var}(T)=9(2^2)=9(4)=36$. Independence gives the same result by adding component variances: $2(2^2)+3(2^2)+4(2^2)=8+12+16=36$.

Check and trap avoided. Both the distribution parameters and the moments agree. A common error is to add the scales and report $\operatorname{Gamma}(9,6)$. Scales do not add: each component has the same scale, so the total retains scale $2$.

Why Unequal Scales Generally Do Not Aggregate

If the scales differ, independence still lets us multiply the component Laplace transforms, but the result is generally not a single Gamma transform. For two independent variables with shapes $\alpha_1,\alpha_2$ and scales $\theta_1,\theta_2$, the transform of their sum is

E[e^{-s(X_1+X_2)}] =(1+\theta_1s)^{-\alpha_1}(1+\theta_2s)^{-\alpha_2}.

A single Gamma transform has the form $(1+\theta s)^{-\alpha}$, with one scale factor. When the component scales are unequal and both shapes are positive, the product has distinct scale factors and generally cannot be reduced to that single-factor form. Thus the sum is generally not Gamma. Special or degenerate cases may simplify, but one must establish such a case rather than assume it.

Moments alone can help expose the problem, but they do not prove a sum is Gamma. For independent Gamma variables with shapes $\alpha_j$ and scales $\theta_j$,

E[T]=\sum_j\alpha_j\theta_j,\qquad \operatorname{Var}(T)=\sum_j\alpha_j\theta_j^2,\qquad \kappa_3(T)=\sum_j2\alpha_j\theta_j^3,

where $T=\sum_jX_j$ and $\kappa_3$ is the third cumulant. A Gamma variable with mean $\mu$ and variance $V$ must have third cumulant $2V^2/\mu$. If the sum's actual third cumulant differs from that value, the sum cannot be Gamma. Agreement of these three quantities, however, is not by itself a general proof that the full distribution is Gamma.

Worked Example: Unequal Scales and a Failed Gamma Fit

Consider independent hypothetical measurements $X_1\sim\operatorname{Gamma}(2,3)$ and $X_2\sim\operatorname{Gamma}(4,1)$, with parameters in shape-scale order. Determine whether their sum $T$ can be represented as a Gamma variable. No interest or mortality assumption is needed for these measurements.

Step 1: Compute the mean. The component means are $2(3)=6$ and $4(1)=4$, so $E[T]=6+4=10$.

Step 2: Compute the variance. The component variances are $2(3^2)=18$ and $4(1^2)=4$. By independence, $\operatorname{Var}(T)=18+4=22$.

Step 3: Find the only possible Gamma parameters matching these moments. If $T$ were Gamma with shape $\alpha$ and scale $\theta$, then $\alpha\theta=10$ and $\alpha\theta^2=22$. Dividing the variance by the mean gives $\theta=22/10=11/5$. Then $\alpha=10/(11/5)=50/11$.

Step 4: Check the third cumulant. The actual third cumulant of the sum is $2(2)(3^3)+2(4)(1^3)=108+8=116$. The Gamma distribution forced by the mean and variance would instead have third cumulant $2V^2/\mu=2(22^2)/10=968/10=484/5=96.8$.

The cumulants disagree: $116\ne96.8$. Therefore the sum is not Gamma. The moment-matched Gamma distribution is only an approximation, not an exact aggregation result.

Check and trap avoided. Adding shapes and choosing either component's scale would incorrectly claim an exact Gamma distribution. The scales are $3$ and $1$, so the common-scale theorem does not apply; the cumulant check confirms that no Gamma distribution can match even the first three cumulants.

Common Dispersion Does Not Always Mean Common Scale

A useful connection to Gamma regression is the mean-dispersion parameterization. In a common Gamma GLM convention, a response with mean $\mu_j$ and dispersion $\phi>0$ has variance $\phi\mu_j^2$. Written in shape-scale form, its parameters are

\alpha_j=\frac{1}{\phi},\qquad \theta_j=\phi\mu_j.

Thus, having the same dispersion $\phi$ gives the same shape, but the scale depends on the mean. If the means differ, the scales differ as well. Independence and a common dispersion alone therefore do not establish an exact Gamma distribution for the sum. If the means are also equal, then the scales match and the common-scale result does apply.

Worked Example: Equal Means Versus Unequal Means Under a Shared Dispersion

Suppose three independent hypothetical positive responses follow the mean-dispersion Gamma parameterization with $\phi=0.25$. First, let all three means be $8$. Then consider means $8,10,12$ instead. Determine whether the total is Gamma in each case and calculate the relevant moments. There is no interest rate or mortality-contingent benefit in this statistical model.

Equal means. For each response, the shape is $1/\phi=1/0.25=4$, and the scale is $\phi\mu=0.25(8)=2$. The three independent responses therefore share scale $2$. Their sum is $\operatorname{Gamma}(4+4+4,2)=\operatorname{Gamma}(12,2)$. Its mean is $12(2)=24$, and its variance is $12(2^2)=48$. These agree with adding component means, $8+8+8=24$, and component variances, $0.25(8^2)$ three times, or $16+16+16=48$.

Unequal means. The shapes remain $4$, but the scales are $0.25(8)=2$, $0.25(10)=2.5$, and $0.25(12)=3$. Since the scales differ, the exact common-scale result does not apply. The total mean is $8+10+12=30$, and its variance is $0.25(8^2)+0.25(10^2)+0.25(12^2)=16+25+36=77$.

To check whether a Gamma distribution could nevertheless match the unequal-mean sum, calculate its third cumulant. For each response it is $2\alpha\theta^3$, so the total's third cumulant is $2(4)(2^3)+2(4)(2.5^3)+2(4)(3^3)=64+125+216=405$. A Gamma variable with mean $30$ and variance $77$ would have third cumulant $2(77^2)/30=11858/30=5929/15$, approximately $395.27$. Since $405\ne5929/15$, the sum is not Gamma.

Check and trap avoided. The dispersion is shared in both situations. It is the equal means in the first case that make the scales common. Treating shared dispersion alone as sufficient would wrongly classify the unequal-mean total as Gamma.

Exam Checks and Practical Interpretation

When asked to aggregate Gamma outcomes, identify the parameterization before combining anything. A symbol called a “rate” is the reciprocal of scale, so a common rate is equivalent to a common scale; adding rate parameters is not the aggregation rule. Then verify independence and compare the actual scale parameters. Only when those conditions hold does the shape-addition result give an exact Gamma distribution.

When scales are unequal, still calculate means and variances by their usual rules, but label a moment-matched Gamma distribution as an approximation unless exactness has been established. A cumulant mismatch is a decisive way to rule out a Gamma sum. This topic is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Key takeaway: For independent Gamma variables in shape-scale form, a shared scale makes the sum exactly Gamma, with shapes added and scale unchanged. Unequal scales generally prevent that exact simplification, even when the variables share a dispersion parameter.

Check Your Understanding

For each problem, assume independence and use shape-scale order unless rate is explicitly specified. Check whether the conditions for exact aggregation are satisfied before assigning a Gamma distribution.

Let $A\sim\operatorname{Gamma}(2,4)$ and $B\sim\operatorname{Gamma}(5,4)$. Find the distribution, mean, and variance of $A+B$. (Answer: $\operatorname{Gamma}(7,4)$, mean $28$, variance $112$.)
Three independent variables have distributions $\operatorname{Gamma}(3,2)$, $\operatorname{Gamma}(6,2)$, and $\operatorname{Gamma}(1,2)$. Find the distribution, mean, and variance of their sum. (Answer: $\operatorname{Gamma}(10,2)$, mean $20$, variance $40$.)
Let $C\sim\operatorname{Gamma}(3,2)$ and $D\sim\operatorname{Gamma}(2,1)$. Find the mean and variance of $C+D$, and decide whether the common-scale theorem gives an exact Gamma distribution. (Answer: mean $8$, variance $14$; no, the scales differ.)
Four independent responses have Gamma mean-dispersion parameterization with $\phi=0.2$ and equal means $10$. Find the distribution, mean, and variance of their sum. (Answer: each has shape $5$ and scale $2$; the sum is $\operatorname{Gamma}(20,2)$, with mean $40$ and variance $80$.)
Let $U\sim\operatorname{Gamma}(2,1)$ and $V\sim\operatorname{Gamma}(3,2)$ independently. Use the first three cumulants to rule out an exact Gamma distribution for $U+V$, if possible. (Answer: mean $8$, variance $14$, and actual third cumulant $52$; a Gamma distribution matching the mean and variance would have third cumulant $2(14^2)/8=49$, so the sum is not Gamma.)

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