= Solution
Condition on $N$. Since the amounts are <independent and identically distributed random variables> and are independent of $N$, their <moment-generating functions> multiply:
$$
\mathbb E[e^{tS}\mid N=n]=M_X(t)^n.
$$
Summing against the positive-support <geometric distribution> gives the <geometric-sum moment-generating function>
$$
\boxed{M_S(t)=\sum_{n\geq1}p(1-p)^{n-1}M_X(t)^n
=\frac{pM_X(t)}{1-(1-p)M_X(t)}.}
$$
The finite-transform domain is $M_X(t)<\infty$ and $(1-p)M_X(t)<1$. The terms are nonnegative for real $t$, so if the latter condition fails the sum diverges. For $t\leq0$ both conditions always hold, since the claims are positive; thus the same calculation always provides the <Laplace transform of a nonnegative random variable>. Positivity alone does not guarantee a finite <moment-generating function> on a positive neighbourhood of zero.
There is at least one strictly positive amount in the portfolio. Hence \b[$\mathbb P(S=0)=0$]; zero carries no <atom of a measure> to add to the continuous laws in the next two parts.
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