Gamma Additivity at a Common Rate

Why a Shared Rate Matters

Gamma distributions are used to model positive quantities, including waiting times and accumulated amounts. A particularly useful fact is that independent gamma variables add to another gamma variable when they share a rate: their shape parameters add, while the common rate stays fixed. The transform proof makes the condition visible: independence multiplies Laplace transforms, and a shared rate gives those transforms the same denominator.

The words rate and scale describe reciprocal parameters. Confusing them changes the distribution, so the additivity rule must be stated with its parameterization explicit.

Gamma Distributions and Their Laplace Transforms

Definition: For shape $\alpha>0$ and rate $\lambda>0$, a random variable $X$ has a gamma distribution with shape $\alpha$ and rate $\lambda$, written $X\sim\operatorname{Gamma}(\alpha,\lambda)$, if it has density $$ f_X(x)=\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x},\qquad x>0, $$ and density zero for $x\leq 0$. Here $\Gamma(\alpha)=\int_0^\infty u^{\alpha-1}e^{-u}\,du$ is the gamma function.

The Laplace transform of a nonnegative random variable $X$ is $L_X(s)=E[e^{-sX}]$, for $s\geq 0$. For the gamma density, the transform follows directly from the defining integral for $\Gamma$.

Result: If $X\sim\operatorname{Gamma}(\alpha,\lambda)$, then for every $s\geq0$, $$ L_X(s)=\left(\frac{\lambda}{\lambda+s}\right)^\alpha. $$

Indeed, the integral converges because $\lambda+s>0$. Substitute $u=(\lambda+s)x$, so $x=u/(\lambda+s)$ and $dx=du/(\lambda+s)$:

\begin{aligned} L_X(s) &=\frac{\lambda^\alpha}{\Gamma(\alpha)} \int_0^\infty x^{\alpha-1}e^{-(\lambda+s)x}\,dx\\ &=\frac{\lambda^\alpha}{\Gamma(\alpha)} \frac{1}{(\lambda+s)^\alpha} \int_0^\infty u^{\alpha-1}e^{-u}\,du\\ &=\left(\frac{\lambda}{\lambda+s}\right)^\alpha. \end{aligned}

The rate $\lambda$ appears in the denominator as $\lambda+s$. It is not interchangeable with a scale parameter. If the scale is $\beta>0$, then $\lambda=1/\beta$, and the same density and transform can be written as

f_X(x)=\frac{x^{\alpha-1}e^{-x/\beta}}{\Gamma(\alpha)\beta^\alpha}, \qquad L_X(s)=(1+\beta s)^{-\alpha}.

The Additivity Theorem

Theorem: Let $X_1,\ldots,X_m$ be independent, with $X_i\sim\operatorname{Gamma}(\alpha_i,\lambda)$ for shapes $\alpha_i>0$ and a common rate $\lambda>0$. Then $$ \sum_{i=1}^m X_i\sim\operatorname{Gamma}\left(\sum_{i=1}^m\alpha_i,\lambda\right). $$

Two facts complete the argument: independence makes the Laplace transform of a sum a product, and Laplace transforms identify distributions of nonnegative random variables. The latter is a uniqueness result, not just a convenient pattern match.

Uniqueness fact: If $U$ and $V$ are nonnegative, finite random variables and $E[e^{-sU}]=E[e^{-sV}]$ for every $s\geq0$, then $U$ and $V$ have the same distribution.

To see why the uniqueness fact holds, set $A=e^{-U}$ and $B=e^{-V}$. Both take values in $(0,1]$, and equality of the transforms at every nonnegative integer $k$ gives $E[A^k]=E[B^k]$. Thus the two probability distributions on $[0,1]$ agree on integrals of all polynomials. By the Weierstrass approximation theorem, every continuous function on $[0,1]$ is a uniform limit of polynomials. Since both measures have total mass one, their integrals of every continuous function agree as well. Probability measures on $[0,1]$ with the same integrals for all continuous functions are equal. Therefore $A$ and $B$ have the same distribution. As $-\log(A)=U$ and $-\log(B)=V$, their distributions agree too. Neither $A$ nor $B$ has mass at zero, since $U$ and $V$ are finite.

Now let $S=\sum_{i=1}^m X_i$ and $a=\sum_{i=1}^m\alpha_i$. The variables are nonnegative, and independence gives, for $s\geq0$,

\begin{aligned} L_S(s) &=E\left[\prod_{i=1}^m e^{-sX_i}\right]\\ &=\prod_{i=1}^m E[e^{-sX_i}]\\ &=\prod_{i=1}^m\left(\frac{\lambda}{\lambda+s}\right)^{\alpha_i}\\ &=\left(\frac{\lambda}{\lambda+s}\right)^a. \end{aligned}

The last expression is the Laplace transform of $\operatorname{Gamma}(a,\lambda)$, by the calculation above. Both $S$ and a random variable with that gamma distribution are nonnegative and finite, so uniqueness of Laplace transforms gives the claimed distribution. The rate remains $\lambda$, while the exponents—and therefore the shapes—add. $\square$

Worked Applications

Worked Example: Adding Two Gamma Variables With a Common Rate

Suppose $X\sim\operatorname{Gamma}(2,4)$ and $Y\sim\operatorname{Gamma}(3,4)$ are independent, using the shape-rate parameterization. Find the distribution of $S=X+Y$, and write its density.

Their transforms are $$ L_X(s)=\left(\frac{4}{4+s}\right)^2,\qquad L_Y(s)=\left(\frac{4}{4+s}\right)^3. $$ Independence gives $$ L_S(s)=L_X(s)L_Y(s)=\left(\frac{4}{4+s}\right)^5. $$ This is the transform of $\operatorname{Gamma}(5,4)$. Its density, for $x>0$, is $$ f_S(x)=\frac{4^5}{\Gamma(5)}x^4e^{-4x} =\frac{1024}{24}x^4e^{-4x} =\frac{128}{3}x^4e^{-4x}. $$ The result retains rate $4$ and adds the shapes $2+3=5$.

Worked Example: Adding Gamma Variables Given by Their Scale

Let $U$ and $V$ be independent gamma variables with shapes $1/2$ and $3/2$, respectively, and common scale $\beta=2$. Find the distribution of $T=U+V$.

Convert the common scale to a rate: $\lambda=1/\beta=1/2$. Equivalently, each transform in scale notation is $(1+2s)^{-\alpha}$. Thus $$ L_T(s)=(1+2s)^{-1/2}(1+2s)^{-3/2}=(1+2s)^{-2}. $$ This is the transform of a gamma distribution with shape $2$ and scale $2$, or rate $1/2$. Its density is $$ f_T(t)=\frac{t^{2-1}e^{-t/2}}{\Gamma(2)2^2} =\frac{t}{4}e^{-t/2},\qquad t>0. $$ The scale stays at $2$; the shapes add to $1/2+3/2=2$.

Worked Example: A Sum of Three Gamma Variables

Suppose $X_1,X_2,X_3$ are independent with $X_1\sim\operatorname{Gamma}(3/4,2)$, $X_2\sim\operatorname{Gamma}(1,2)$, and $X_3\sim\operatorname{Gamma}(1/2,2)$. Identify the sum $R=X_1+X_2+X_3$.

The common rate is $2$, and the total shape is $$ \frac{3}{4}+1+\frac{1}{2}=\frac{3}{4}+\frac{4}{4}+\frac{2}{4}=\frac{9}{4}. $$ Multiplying the transforms verifies the identification: $$ L_R(s)=\left(\frac{2}{2+s}\right)^{3/4} \left(\frac{2}{2+s}\right)^1 \left(\frac{2}{2+s}\right)^{1/2} =\left(\frac{2}{2+s}\right)^{9/4}. $$ Therefore $R\sim\operatorname{Gamma}(9/4,2)$. The rule applies even though the shapes are not all integers.

Why the Common-Rate Condition Cannot Be Dropped

Having gamma-shaped summands is not, by itself, enough for the theorem. The shared rate is what allows the transform factors to combine into a single power of one expression. For example, let $X\sim\operatorname{Gamma}(1,2)$ and $Y\sim\operatorname{Gamma}(1,3)$ be independent. Their sum has transform

L_{X+Y}(s)=\frac{2}{2+s}\frac{3}{3+s} =\frac{6}{(s+2)(s+3)}.

This is not the transform of a gamma distribution. If it were $\operatorname{Gamma}(\alpha,\lambda)$, its transform would be $(\lambda/(\lambda+s))^\alpha$. As $s$ tends to infinity, the displayed transform is asymptotic to $6s^{-2}$, while a gamma transform is asymptotic to $\lambda^\alpha s^{-\alpha}$. Equality would require $\alpha=2$ and $\lambda^2=6$. It would then require $$ \frac{6}{(s+\lambda)^2}=\frac{6}{(s+2)(s+3)}, $$ which is impossible: $(s+\lambda)^2$ has a repeated root, whereas $(s+2)(s+3)$ has two distinct roots. This example illustrates the failure of the common-rate additivity rule when rates differ; it does not assert that the sum has no distribution.

Common Mistake: Switching Rate and Scale

A frequent error is to add parameters called “rates” when the given parameters are actually scales, or to keep a scale fixed while reporting a rate. The correction is to write down the parameterization before applying the rule:

With a common rate $\lambda$, add shapes and retain $\lambda$.
With a common scale $\beta$, add shapes and retain $\beta$, equivalently retaining rate $1/\beta$.
If neither rate nor scale is common, this additivity theorem does not apply.

For instance, a common scale of $2$ means a common rate of $1/2$, not $2$. In the transform, that distinction is visible: scale $2$ gives $(1+2s)^{-\alpha}$, while rate $2$ gives $(2/(2+s))^\alpha$. These are different transforms and describe different distributions.

Key takeaway: For independent gamma variables with one shared rate (or, equivalently, one shared scale), the sum is gamma with shape equal to the sum of the shapes and with the common rate (or scale) unchanged. The Laplace-transform product proves the rule, and transform uniqueness identifies the distribution.

Check Your Understanding

Work through these questions using the shape-rate convention unless a scale is specified.

Independent $X\sim\operatorname{Gamma}(2,5)$ and $Y\sim\operatorname{Gamma}(1/2,5)$ are added. What is the sum’s shape and rate?
Write the Laplace transform of a gamma variable with shape $3$ and scale $4$, first using scale notation and then using rate notation.
Why does independence matter when multiplying the Laplace transforms of gamma summands?
For independent variables with shapes $1$ and $2$ and a common rate $3$, write the transform of their sum and identify its distribution.
What feature of the transform of $\operatorname{Gamma}(\alpha,\lambda)$ makes a shared rate necessary for the additivity argument?