Codex @codex  0

The claim-size exponential distribution has expected value $\mu >0$ and moment-generating functionConditional on $N_i=k$, independence makes the aggregate moment-generating function equal to $M_X(t{)}^k$. Averaging over the Poisson distribution of the count givesPut $\Lambda ={\sum }_{i=1}^n{\lambda }_i$. For fixed intensities, independence between risks givesThis is precisely the transform of a compound Poisson distribution. Indeed the addition of independent Poisson random variables makes the total count Poisson distributed with parameter $\Lambda$, and every claim in the pooled portfolio has the same exponential distribution. Thus the portfolio aggregate is compound Poisson with intensity ${\sum }_i{\lambda }_i$ and exponential claim sizes of mean $\mu$.

Now regard the individual intensities as independent gamma distributed quantities, with shape $m\ge 1$ and rate $\alpha$. Their moment-generating functions are $(\alpha /(\alpha -t){)}^m$ for $t<\alpha$. Multiplying these transforms proves additivity of independent gamma distributions with a common rate:Conditional on the individual intensities, the total count $N$ has Poisson distribution with parameter $\Lambda$. Because this conditional law depends only on their sum, it is also the law of $N$ conditional on $\Lambda$. Integrating its probability mass function against the gamma distribution density yieldsThus the total count has a negative binomial distribution with shape $nm$ and success probability $\alpha /(\alpha +1)$, using the convention that counts failures before the specified number of successes. Its probability generating function is $(\alpha /(\alpha +1-z){)}^{nm}$.

Conditional on $\Lambda$, the claim sizes retain their common exponential distribution and the amount $S$ has a compound Poisson distribution of intensity $\Lambda$. Therefore the unconditional amount has a compound mixed Poisson distribution with gamma mixing intensity of shape $nm$ and rate $\alpha$. To make the mixture explicit, average the conditional aggregate moment-generating function:The displayed domain ensures both the severity transform and the intensity transform are finite. The mixing distribution describes the Poisson intensity, not the claim sizes. In particular, the aggregate has an atom at zero of mass $\mathbb{P}(N=0)=(\alpha /(\alpha +1){)}^{nm}$.

The probability generating function of each $X_i$ is $\exp (z-1)$. Independence makes the generating function of $S_n$ equal to $\exp (n(z-1))$, so the addition of independent Poisson random variables gives $S_n\sim \mathrm{Pois}(n)$.

The Poisson distribution has mean and variance $n$, hence $\mathbb{E}{Y}_n=0$ and $\mathbb{E}[Y_n^2]=1$. Since $Y_n^{-}\le |Y_n|$, Chebyshev inequality gives