Compound geometric distribution ($M_S(t)=p/(1-(1-p)M_X(t))$)

Take an independent geometric distribution count with $\mathbb{P}(N=k)=p(1-p{)}^k$, $k\ge 0$, and sum its independent positive claims. Its moment-generating function is the displayed composition wherever finite. With $q=1-p$ and severity moments $m_j$, the third cumulant is $(q/p)m_3+3(q/p{)}^2{m}_1{m}_2+2(q/p{)}^3{m}_1^3$, which is strictly positive for nonzero positive claims and $0<p<1$. The law has an atom of mass $p$ at zero. Its positive-count convention would instead add a factor $M_X(t)$ to the numerator, as in the geometric-sum moment-generating function.

If independent summands have exponential distribution of mean $\mu$ and the independent count has a zero-based geometric distribution of parameter $p$, their random sum of independent claims has an atom $p$ at zero and an exponential positive component of rate $p/\mu$. Its Laplace transform of a nonnegative random variable is $p/[1-(1-p)/(1+\mu s)]=p+(1-p)p/(p+\mu s)$. Thus $\mathbb{P}(S>x)=(1-p)e^{-px/\mu }$ for $x>0$. Unlike the geometric sum of exponential variables with a positive count, this law is not purely exponential.