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A compound mixed Poisson distribution is a random sum of independent claims whose count has a Poisson mixture law: conditional on a nonnegative random intensity $\Lambda$, $N$ has Poisson distribution with parameter $\Lambda$. The claim sizes are independent and identically distributed random variables, independent of the intensity and count mechanism. If $M_X$ is their moment-generating function, then $M_S(t)=\mathbb{E}\exp (\Lambda (M_X(t)-1))$ wherever finite. A gamma distribution intensity of shape $r$ and rate $\alpha$ gives a negative binomial distribution count with shape $r$ and success probability $\alpha /(\alpha +1)$. Independent intensities with common gamma rate add their shapes by additivity of independent gamma distributions with a common rate.

For the rate-parameterized Gamma distribution with shape $r>0$ and rate $\lambda >0$, integration of its probability density giveswhen $t<\lambda$, by the gamma function integral and a change of variables. The integral is infinite for $t\ge \lambda$. In particular the moment-generating function is finite on an open interval containing zero, so the uniqueness theorem for moment-generating functions applies. Multiplication of these moment-generating functions proves additivity of independent gamma distributions with a common rate.

The claim-size exponential distribution has expected value $\mu >0$ and moment-generating functionConditional on $N_i=k$, independence makes the aggregate moment-generating function equal to $M_X(t{)}^k$. Averaging over the Poisson distribution of the count givesPut $\Lambda ={\sum }_{i=1}^n{\lambda }_i$. For fixed intensities, independence between risks givesThis is precisely the transform of a compound Poisson distribution. Indeed the addition of independent Poisson random variables makes the total count Poisson distributed with parameter $\Lambda$, and every claim in the pooled portfolio has the same exponential distribution. Thus the portfolio aggregate is compound Poisson with intensity ${\sum }_i{\lambda }_i$ and exponential claim sizes of mean $\mu$.

Now regard the individual intensities as independent gamma distributed quantities, with shape $m\ge 1$ and rate $\alpha$. Their moment-generating functions are $(\alpha /(\alpha -t){)}^m$ for $t<\alpha$. Multiplying these transforms proves additivity of independent gamma distributions with a common rate:Conditional on the individual intensities, the total count $N$ has Poisson distribution with parameter $\Lambda$. Because this conditional law depends only on their sum, it is also the law of $N$ conditional on $\Lambda$. Integrating its probability mass function against the gamma distribution density yieldsThus the total count has a negative binomial distribution with shape $nm$ and success probability $\alpha /(\alpha +1)$, using the convention that counts failures before the specified number of successes. Its probability generating function is $(\alpha /(\alpha +1-z){)}^{nm}$.

Conditional on $\Lambda$, the claim sizes retain their common exponential distribution and the amount $S$ has a compound Poisson distribution of intensity $\Lambda$. Therefore the unconditional amount has a compound mixed Poisson distribution with gamma mixing intensity of shape $nm$ and rate $\alpha$. To make the mixture explicit, average the conditional aggregate moment-generating function:The displayed domain ensures both the severity transform and the intensity transform are finite. The mixing distribution describes the Poisson intensity, not the claim sizes. In particular, the aggregate has an atom at zero of mass $\mathbb{P}(N=0)=(\alpha /(\alpha +1){)}^{nm}$.

An exponential distribution of rate $\lambda >0$ has density $f_X(x)=\lambda e^{-\lambda x}$ for $x\ge 0$ and zero for $x<0$, with survival probability $\mathbb{P}(X>x)=e^{-\lambda x}$. For $s,t\ge 0$, its memoryless property follows fromFor the minimum, independence gives $\mathbb{P}(Z>z)=\mathbb{P}(X>z)\mathbb{P}(Y>z)=e^{-2\lambda z}$. HenceThe density is zero for negative $z$. The comparison probability follows from identical continuous laws and zero tie probability; directly it is ${\int }_0^{\infty }\lambda e^{-2\lambda y}dy=1/2$. These are competing exponential clocks.

For the last density, substitute $y=s^2$ into the normalizing integral to obtain $C=2{\int }_0^{\infty }{e}^{-s^2}ds=\sqrt{\pi }$, the Gaussian integral. Equivalently, each variable has a gamma distribution with shape $1/2$ and rate one. Independence permits convolution of the densities. For $z>0$,Putting $y=z{\sin }^2\theta$ turns the integral into $2{\int }_0^{\pi /2}d\theta =\pi$. ConsequentlyThis is also the additivity of independent gamma distributions with a common rate: their shapes add to one. Density values at the single endpoint zero are immaterial.

The displayed density is that of a gamma distribution with shape $k+1$ and rate one. PutBy the additivity of independent gamma distributions with a common rate,Equivalently, the independence of random variables makes the Laplace transform of a nonnegative random variable of the sum equalTherefore the required probability density function is

Independence makes the moment-generating function of the sum the product of the two moment-generating functions. For $t<\lambda$,The expectations are finite on, for example, $(-\lambda /2,\lambda /2)$, as is the moment-generating function of the candidate Gamma distribution. The uniqueness theorem for moment-generating functions therefore applies, provingThis is additivity of independent gamma distributions with a common rate; a common rate is essential to this specific formula.

An exponential distribution of rate $\lambda$ is a Gamma distribution with shape one. Repeated additivity of independent gamma distributions with a common rate therefore givesWith integer shape this is also called an Erlang distribution.