= Solution
The <moment-generating function> of a <mixture distribution> is the mixture of its component transforms, so
$$
\boxed{M_T(r)=1-q+qM_Z(r).}
$$
For the aggregate, positivity of every claim implies $S=0$ exactly when $N=0$. Choose
$$
p=\mathbb P(N=0)=G_N(0),\qquad q=1-p.
$$
If $q>0$, take $\widetilde N$ to have the <zero-truncated claim-count distribution>, namely the conditional law of $N$ given $N>0$. Then
$$
\mathbb P(\widetilde N=n)=\frac{\mathbb P(N=n)}{1-p}\quad(n\ge1),
\qquad G_{\widetilde N}(z)=\frac{G_N(z)-p}{1-p}.
$$
Choose this count independently of a fresh independent claim-size sequence and put $Z=\sum_{i=1}^{\widetilde N}X_i$. Its <probability distribution> is that of $S$ conditional on being positive. Thus the <hurdle decomposition of a positive random sum> gives
$$
M_S(r)=p+(1-p)G_{\widetilde N}(M_X(r))=1-q+qM_Z(r),
$$
and an independent <Bernoulli random variable> $B$ of success probability $q$ realizes $S\overset d=BZ$. 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,
$$
G_N(z)=\frac{1}{2-z},\qquad p=q=\frac12,\qquad
G_{\widetilde N}(z)=\frac{z}{2-z}.
$$
An <exponential distribution> of <expected value> $\mu$ has transform $M_X(r)=(1-\mu r)^{-1}$. Substitution yields
$$
M_Z(r)=\frac{(1-\mu r)^{-1}}{2-(1-\mu r)^{-1}}
=\frac{1}{1-2\mu r},\qquad r<\frac{1}{2\mu}.
$$
Hence \b[$Z$ 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 $r\le0$.
The resulting <distribution function> is
$$
\boxed{F_S(x)=\begin{cases}
0,&x<0,\\
1-\tfrac12e^{-x/(2\mu)},&x\ge0.
\end{cases}}
$$
Its jump of size $1/2$ at zero is important: the aggregate law is not a purely continuous <exponential distribution>.
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