= Solution
Use the preceding construction with type-$i$ claims of deterministic size $a_i$. For $\Lambda=\sum_i\lambda_i>0$, the answer is
$$
\boxed{N\sim\operatorname{Pois}(\Lambda),\qquad
\mathbb P(X=a_i)=\frac{\lambda_i}{\Lambda},\qquad
F_X(x)=\sum_{i:a_i\le x}\frac{\lambda_i}{\Lambda}.}
$$
The <characteristic function> verifies this directly:
$$
\mathbb Ee^{itT}=\prod_i\exp\{\lambda_i(e^{ita_i}-1)\}
=\exp\{\Lambda(\mathbb Ee^{itX}-1)\}.
$$
Hence $T$ has the stated <compound Poisson distribution>. Distinctness of the $a_i$ makes the atomic <probabilities> unambiguous.
A zero coefficient is allowed. It corresponds to zero-sized claims in this representation and contributes nothing to $T$. If a representation using only positive claims is preferred, remove that category: its count parameter becomes $\Lambda_+=\sum_{i:a_i>0}\lambda_i$, with <probabilities> $\lambda_i/\Lambda_+$ on the positive values. The <zero claim sizes in a compound Poisson representation> explain why the count parameter need not be unique. If $\Lambda=0$, or if all positive-size categories have zero rate, the aggregate is identically zero.
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