= Zero-based geometric sum of exponential variables
{title2=$\mathcal L(S)=p\delta_0+(1-p)\operatorname{Exp}(p/\mu)$}
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.
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