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Let the summands have gamma distribution with shape two and rate $\beta$, and let the independent positive-support geometric distribution count have parameter $p$. Put $a=\beta (1-\sqrt{1-p})$ and $b=\beta (1+\sqrt{1-p})$. The aggregate moment-generating function factors as $ab/((a-t)(b-t))$, so it is the sum of two independent exponential distributions of rates $a,b$. Its probability density function is $ab(e^{-as}-e^{-bs})/(b-a)$ for $s>0$, nonnegative and integrating to one. This contrasts with the geometric sum of exponential variables, which is itself exponential.

For the exponential distribution of mean $\mu$, $M_X(t)=(1-\mu t{)}^{-1}$. Substitution into the geometric-sum moment-generating function givesThis is the moment-generating function of an exponential distribution with rate $p/\mu$. By the uniqueness theorem for moment-generating functions, the geometric sum of exponential variables therefore has probability density functionIts expected value is $\mu /p$, agreeing with $\mathbb{E}[N]\mathbb{E}[X]=\mu /p$ in the random sum of independent claims formula.

The moment-generating function of a mixture distribution is the mixture of its component transforms, soFor the aggregate, positivity of every claim implies $S=0$ exactly when $N=0$. ChooseIf $q>0$, take $\tilde{N}$ to have the zero-truncated claim-count distribution, namely the conditional law of $N$ given $N>0$. ThenChoose this count independently of a fresh independent claim-size sequence and put $Z={\sum }_{i=1}^{\tilde{N}}{X}_i$. Its probability distribution is that of $S$ conditional on being positive. Thus the hurdle decomposition of a positive random sum givesand an independent Bernoulli random variable $B$ of success probability $q$ realizes $S\stackrel{d}{=}BZ$. This establishes the distributional representation, including that $Z$ is itself a positive random sum of independent claims. If $p=1$, the aggregate is identically zero; set $q=0$ and choose any positive $Z$, for example one claim. Conditioning the count on positivity is then unnecessary and would be undefined.

For the specified geometric distribution on the nonnegative integers,An exponential distribution of expected value $\mu$ has transform $M_X(r)=(1-\mu r{)}^{-1}$. Substitution yieldsHence $Z$ is exponential with rate $1/(2\mu )$ and expected value $2\mu$. This is the geometric sum of exponential variables with a rescaling of the claim mean. Identification can also use the uniqueness theorem for Laplace transforms of nonnegative random variables by taking $r\le 0$.

The resulting distribution function isIts jump of size $1/2$ at zero is important: the aggregate law is not a purely continuous exponential distribution.

Conditioning on $N$ and using the moment-generating function of a unit exponential variable,where the geometric series converges precisely when $\theta <p$. This is the moment-generating function of an exponential variable of rate $p$, so the geometric sum of exponential variables satisfies