Gamma Additivity and Beta Integrals

Why Gamma Additivity Has a Condition

The gamma distribution is a flexible model for positive quantities, including waiting times and accumulated amounts. When independent gamma variables have the same rate, their sum is also gamma, with the shapes added. This result can simplify a distribution calculation substantially—but the shared rate is essential. If the rates differ, the same conclusion does not follow.

The key calculation is a convolution integral. The beta integral evaluates the part of that integral that depends on how a fixed total is split between the two variables. Keeping the shape and rate parameters distinct makes clear both why the result works and where it stops working.

Definition: A random variable \(X\) has a gamma distribution with shape \(\alpha>0\) and rate \(\lambda>0\), written \(X\sim\operatorname{Gamma}(\alpha,\lambda)\), if its density is

f_X(x)=\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x},\qquad x>0,

and \(f_X(x)=0\) otherwise. Here \(\Gamma(\alpha)=\int_0^\infty t^{\alpha-1}e^{-t}\,dt\). The rate is the parameter multiplying \(x\) in the exponential; it is not the scale \(1/\lambda\).

The Beta Integral

For positive \(a\) and \(b\), the beta function is defined by an integral over the unit interval. It describes the integral that arises when a positive total \(s\) is divided into parts \(u\) and \(s-u\).

Definition: For \(a>0\) and \(b>0\), the beta function is

B(a,b)=\int_0^1 t^{a-1}(1-t)^{b-1}\,dt.

It satisfies \(B(a,b)=\dfrac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\).

The identity connecting the beta and gamma functions follows by multiplying their defining integrals and changing variables. In the resulting double integral, let \(r=x+y\) and \(t=x/(x+y)\). Then \(x=rt\), \(y=r(1-t)\), and the absolute value of the Jacobian is \(r\). Thus

\begin{aligned} \Gamma(a)\Gamma(b) &=\int_0^\infty\int_0^\infty x^{a-1}y^{b-1}e^{-(x+y)}\,dx\,dy\\ &=\int_0^\infty r^{a+b-1}e^{-r}\,dr \int_0^1 t^{a-1}(1-t)^{b-1}\,dt\\ &=\Gamma(a+b)B(a,b). \end{aligned}

The change of variables covers the positive quadrant exactly once, with \(r>0\) and \(0<t<1\). The integrals converge for \(a,b>0\), so dividing by \(\Gamma(a+b)\) gives the stated identity.

Deriving the Common-Rate Sum

Suppose \(X\sim\operatorname{Gamma}(\alpha,\lambda)\) and \(Y\sim\operatorname{Gamma}(\beta,\lambda)\) are independent, with positive shapes and the same positive rate \(\lambda\). Independence allows the density of \(S=X+Y\) to be found by convolution. Since both variables are positive, the possible values of \(X\), for a fixed \(S=s>0\), run from \(0\) to \(s\).

Write the convolution.
Independence gives \(f_S(s)=\int_0^s f_X(u)f_Y(s-u)\,du\) for \(s>0\).

Collect the factors.
The two exponential terms combine as \(e^{-\lambda u}e^{-\lambda(s-u)}=e^{-\lambda s}\), because the rates are equal.

Rescale the integral.
Set \(t=u/s\), so \(u=st\), \(du=s\,dt\), and the convolution integral becomes a beta integral.

Recognize the gamma density.
Use \(B(\alpha,\beta)=\Gamma(\alpha)\Gamma(\beta)/\Gamma(\alpha+\beta)\) to identify the resulting density.

In full, the calculation is

\begin{aligned} f_S(s) &=\frac{\lambda^{\alpha+\beta}e^{-\lambda s}} {\Gamma(\alpha)\Gamma(\beta)} \int_0^s u^{\alpha-1}(s-u)^{\beta-1}\,du\\ &=\frac{\lambda^{\alpha+\beta}e^{-\lambda s}s^{\alpha+\beta-1}} {\Gamma(\alpha)\Gamma(\beta)} \int_0^1 t^{\alpha-1}(1-t)^{\beta-1}\,dt\\ &=\frac{\lambda^{\alpha+\beta}}{\Gamma(\alpha+\beta)} s^{\alpha+\beta-1}e^{-\lambda s},\qquad s>0. \end{aligned}

Theorem: If \(X\sim\operatorname{Gamma}(\alpha,\lambda)\) and \(Y\sim\operatorname{Gamma}(\beta,\lambda)\) are independent, where \(\alpha,\beta,\lambda>0\), then

X+Y\sim\operatorname{Gamma}(\alpha+\beta,\lambda).

The shapes add; the common rate stays the same.

Worked Applications

Worked Example: Combining Two Waiting-Time Components

In a hypothetical service system, two independent waiting-time components have distributions \(X\sim\operatorname{Gamma}(2,1)\) and \(Y\sim\operatorname{Gamma}(3,1)\), with time measured in hours. Find the density of the total waiting time \(S=X+Y\), and calculate \(P(S\leq 2)\).

Method. Both variables are gamma with rate \(1\), and they are independent. Gamma additivity therefore gives \(S\sim\operatorname{Gamma}(2+3,1)=\operatorname{Gamma}(5,1)\).

The density is

f_S(s)=\frac{1^5}{\Gamma(5)}s^4e^{-s} =\frac{s^4e^{-s}}{24},\qquad s>0,

because \(\Gamma(5)=4!=24\). For an integer shape of \(5\), the cumulative probability at \(2\) is

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

The sum in parentheses is \(1+2+2+\frac{4}{3}+\frac{2}{3}=7\). Hence \(P(S\leq2)=1-7e^{-2}\approx 0.0527\), to four decimal places. The density at \(s=2\), if required, is \(2^4e^{-2}/24=(2/3)e^{-2}\approx0.0902\).

Check. The density has the required gamma form with shape \(5\) and rate \(1\); its support is positive, as a total of positive waiting times must be. The calculation avoids the mistake of adding rates instead of shapes.

Worked Example: Different Rates Change the Sum

Two independent positive waiting times have distributions \(X\sim\operatorname{Gamma}(1,1)\) and \(Y\sim\operatorname{Gamma}(1,2)\), with time measured in hours. Find the density of \(S=X+Y\), decide whether the common-rate result applies, and calculate \(P(S>1)\).

Method. These shape-one gamma variables are exponential: \(f_X(u)=e^{-u}\) and \(f_Y(v)=2e^{-2v}\) for positive arguments. Their rates are \(1\) and \(2\), not a common rate, so the gamma-additivity theorem cannot be used. We compute the convolution directly:

\begin{aligned} f_S(s) &=\int_0^s e^{-u}\,2e^{-2(s-u)}\,du\\ &=2e^{-2s}\int_0^s e^u\,du\\ &=2e^{-2s}(e^s-1) =2(e^{-s}-e^{-2s}),\qquad s>0. \end{aligned}

This is not a single gamma density. For instance, near zero, \(e^{-s}-e^{-2s}\) is approximately \(s\), so the density begins like \(2s\), as a gamma density with shape \(2\) would. But the displayed density decays like \(2e^{-s}\) for large \(s\), whereas a shape-two gamma density with rate \(1\) has a factor proportional to \(s e^{-s}\). The forms cannot agree.

To find the requested tail probability, integrate the density:

\begin{aligned} P(S>1) &=\int_1^\infty 2(e^{-s}-e^{-2s})\,ds\\ &=2e^{-1}-e^{-2}\\ &\approx 0.6004. \end{aligned}

The last value is rounded to four decimal places. The tail expression is positive, as required for a probability.

Check. The density is nonnegative for \(s>0\), and its integral over \((0,\infty)\) is \(2(1-1/2)=1\). The common mistake is to add the shapes and report a gamma distribution without checking that the rates agree.

Worked Example: Fractional Shapes and a Beta Integral

Two independent positive components of a hypothetical process have distributions \(X\sim\operatorname{Gamma}(\frac12,3)\) and \(Y\sim\operatorname{Gamma}(\frac32,3)\), with time measured in hours. Find the distribution of \(S=X+Y\), then calculate \(P(S>1)\).

Method. The common rate is \(3\), so the sum has shape \(\frac12+\frac32=2\) and rate \(3\). To see the beta integral explicitly, the convolution contains

\int_0^s u^{-1/2}(s-u)^{1/2}\,du =sB\left(\frac12,\frac32\right) =s\frac{\Gamma(\frac12)\Gamma(\frac32)}{\Gamma(2)} =s\frac{\pi}{2}.

Here \(\Gamma(\frac12)=\sqrt{\pi}\), \(\Gamma(\frac32)=\frac12\sqrt{\pi}\), and \(\Gamma(2)=1\). The density constants in the convolution are

\frac{3^{1/2}}{\Gamma(\frac12)} \frac{3^{3/2}}{\Gamma(\frac32)} =\frac{18}{\pi}.

Multiplying this constant by \(e^{-3s}\) and the integral \(s\pi/2\) gives

f_S(s)=9se^{-3s},\qquad s>0,

which is the density of \(\operatorname{Gamma}(2,3)\). Integrating the density above \(1\), or using the shape-two gamma tail, gives

P(S>1)=e^{-3}(1+3)=4e^{-3}\approx0.1991.

The answer is rounded to four decimal places. The density is normalized because \(\int_0^\infty 9se^{-3s}\,ds=1\), confirming the rate and leading constant.

Check. Fractional shapes are valid because the gamma definition requires only a positive shape, not an integer. The beta integral remains valid for these parameters.

When the Result Applies—and Common Traps

For two nondegenerate gamma variables, independence and a common rate are the conditions for the additivity result above. A mismatch in rates changes the exponential factors inside the convolution: \(e^{-\lambda_1u}e^{-\lambda_2(s-u)}\) cannot be simplified to a factor depending only on \(s\). The beta-integral step then no longer produces the density of a single gamma variable. The sum can still be analyzed by convolution, but it should not be labeled gamma merely because both components are gamma.

Confusing rate and scale. This tutorial uses rate \(\lambda\), so the exponential factor is \(e^{-\lambda x}\). If parameters are provided as scales, convert each scale to its reciprocal rate before checking for a match.
Adding rates. With a common rate, add the shapes and retain that rate. Do not add the rates.
Ignoring independence. The convolution formula used here requires independent variables. Gamma marginals alone do not establish the distribution of their sum.
Using an invalid convolution range. For two positive gamma variables and total \(s>0\), the integration range is \(0\leq u\leq s\), not the whole real line.

This result is useful whenever a total is built from independent positive components that share a common rate—for example, successive stages with a common exponential scale. It converts a sum-distribution problem into a parameter update. When rates differ, check the model rather than forcing the familiar gamma form. These methods are aligned with the SOA VEE Mathematical Statistics syllabus.

Check Your Understanding

For each problem, identify whether the common-rate result applies before calculating the requested quantity.

Independent \(U\sim\operatorname{Gamma}(3,2)\) and \(V\sim\operatorname{Gamma}(4,2)\) represent two positive components, measured in the same time units. Give the density of \(U+V\). (Answer: \(f(s)=\frac{8}{45}s^6e^{-2s}\) for \(s>0\).)
Independent \(A\sim\operatorname{Gamma}(1,3)\) and \(B\sim\operatorname{Gamma}(2,3)\) are measured in hours. Calculate \(P(A+B>1)\). (Answer: \(e^{-3}(1+3+\frac92)\approx0.4232\).)
Independent \(C\sim\operatorname{Gamma}(\frac12,4)\) and \(D\sim\operatorname{Gamma}(\frac12,4)\) have a common rate. Calculate \(P(C+D>0.5)\). (Answer: \(e^{-2}\approx0.1353\).)
Independent \(E\sim\operatorname{Gamma}(1,2)\) and \(F\sim\operatorname{Gamma}(1,5)\) have different rates. Use convolution to find the density of \(E+F\), then evaluate it at \(s=1\). (Answer: \(f(1)=\frac{10}{3}(e^{-2}-e^{-5})\approx0.4287\).)
Independent \(G\sim\operatorname{Gamma}(2,1)\) and \(H\sim\operatorname{Gamma}(3,1)\) are positive accumulated amounts. Calculate \(P(G+H<3)\). (Answer: \(1-\frac{131}{8}e^{-3}\approx0.1847\).)

What You'll Learn