Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 41 / 1

Let $\Lambda ={\sum }_{i=1}^m{\lambda }_i$. For $\Lambda >0$, the resulting compound Poisson distribution hasThus the merged claim size probability distribution is a rate-weighted mixture distribution, not an equally weighted mixture distribution unless the rates agree.

To prove this without any moment-generating function assumption, let ${\chi }_i(t)$ be the characteristic function of a claim of type $i$. Conditioning on its Poisson distribution count givesIndependence of the aggregate risks therefore givesThe weighted characteristic function inside is precisely that of the displayed mixture distribution. This is the transform of a sum of independent claims with independent count $\mathrm{Pois}(\Lambda )$, proving the claim by uniqueness of characteristic functions. This is Poisson superposition of insurance portfolios. If $\Lambda =0$, all risks make no claims almost surely and $S=0$; the severity law is then irrelevant.

Use the preceding construction with type-$i$ claims of deterministic size $a_i$. For $\Lambda ={\sum }_i{\lambda }_i>0$, the answer isThe characteristic function verifies this directly: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.

Define $N_i={\sum }_{k=1}^N{1}_{\{X_k=a_i\}}$. Grouping equal claim sizes gives the pathwise identityConditional on $N=n$, the $n$ independent claims receive category labels with probabilities $p_i$. There are $n!/(n_1!\cdots n_m!)$ assignments producing prescribed category counts, and every assignment has probability ${\prod }_i{p}_i^{n_i}$. Thus the conditional probability distribution is a multinomial distribution:when the nonnegative counts sum to $n$, and is zero otherwise.

For $0\le z_i\le 1$, condition first on $N$ and use the probability generating function of a Poisson distribution:This factorization is the joint probability generating function of independent Poisson distribution variables. HenceA category with $p_i=0$ has zero count almost surely. The categorical thinning of a Poisson claim count produces independence after averaging over the random total, even though the counts are constrained to sum to $n$ conditional on that total.