# 金融代写|风险理论代写Risk theory代考|MATH4128

## 金融代写|风险理论代写Risk theory代考|Poisson Compounding

In Chap. IV, we will meet geometric compounding, i.e. $p_n=(1-\rho) \rho^n$, as a technical tool in ruin theory. However, by far the most important case from the point of view of aggregate claims is Poisson compounding, $p_n=\mathrm{e}^{-\lambda} \lambda^n / n !$. As explained in Sect. I.5, the Poisson assumption can be motivated in a number of different, though often related, ways. For example:

1. Assume that the number of policy holders is $M$, and that the $m$ th produces claims according to a Poisson process with rate (intensity) $\lambda_m$. In view of the interpretation of the Poisson process as a model for events which occur in an ‘unpredictable’ and time-homogeneous way, this is a reasonable assumption for example in car insurance or other types of accident insurance. If further the policy holders behave independently, then the total claims process (the superposition of the individual claim processes) is again Poisson, now with rate $\lambda=\lambda_1+\cdots+\lambda_M$. In particular, if $N$ is the number of claims in a time period of unit length, then $N$ has a Poisson $(\lambda)$ distribution.
2. Assume again that the number of policy holders is $M$, and that the number of claims produced by each during the given time period is 0 or 1, w.p. $p$ for 1 . If $p$ is small and $M$ is large, the law of small numbers (Theorem 5.1) therefore implies that $N$ has an approximate Poisson( $\lambda$ ) distribution where $\lambda=M p$.

In the Poisson case, $\widehat{p}[z]=\mathrm{e}^{\lambda(z-1)}$, and therefore the c.g.f. of $A$ is (replace $z$ by $\widehat{F}[\theta]$, cf. Proposition I.5.1, and take logarithms)
$$\kappa_A(\theta)=\lambda(\widehat{F}[\theta]-1) .$$
Further, Proposition $5.1$ yields
$$\mu_A=\lambda \mu_V, \quad \sigma_A^2=\lambda \sigma_V^2+\lambda \mu_V^2=\lambda \mathbb{E} V^2$$

We consider the case of a Poisson $N$ where $\mathbb{E e}^{\alpha A}=\mathrm{e}^{\kappa(\alpha)}$ with $\kappa(\alpha)=\lambda(\widehat{F}[\alpha]-1)$. The exponential family generated by $A$ is given by
$$\mathbb{P}\theta(A \in \mathrm{d} x)=\mathbb{E}\left[\mathrm{e}^{\theta A-\kappa(\theta)} ; A \in \mathrm{d} x\right] .$$ In particular, $$\kappa\theta(\alpha)=\log \mathbb{E}\theta \mathrm{e}^{\alpha A}=\kappa(\alpha+\theta)-\kappa(\theta)=\lambda\theta\left(\widehat{F}\theta[\alpha]-1\right),$$ where $\lambda\theta=\lambda \widehat{F}[\theta]$ and $F_\theta$ is the distribution given by
$$F_\theta(\mathrm{d} x)-\frac{\mathrm{e}^{\theta x}}{\widehat{F}[\theta]} F(\mathrm{~d} x) .$$
This shows that the $\mathbb{P}\theta$-distribution of $A$ has a similar compound Poisson form as the $\mathbb{P}$-distribution, only with $\lambda$ replaced by $\lambda\theta$ and $F$ by $F_\theta$.

Following the lines of Sect. 3.4, we shall derive a tail approximation by exponential tilting. For a given $x$, we define the saddlepoint $\theta=\theta(x)$ by $\mathbb{E}\theta A=x$, i.e. $\kappa\theta^{\prime}(0)=\kappa^{\prime}(\theta)=x$.
Proposition 7.3 Assume that $\lim {r \uparrow r^} \widehat{F}^{\prime \prime}[r]=\infty$, $$\lim {r \uparrow r^} \frac{\widehat{F}^{\prime \prime \prime}[r]}{\left(\widehat{F}^{\prime \prime}[r]\right)^{3 / 2}}=0,$$
where $r^*=\sup {r: \widehat{F}[r]<\infty}$. Then as $x \rightarrow \infty$, $$\mathbb{P}(A>x) \sim \frac{\mathrm{e}^{-\theta x+\kappa(\theta)}}{\theta \sqrt{2 \pi \lambda \widehat{F}^{\prime \prime}[\theta]}} .$$

## 金融代写|风险理论代写Risk theory代考|Poisson Compounding

1. 假设保单持有人的数量为 $M$ ，并且那个 $m$ th 根据具有速率 (强度) 的 Poisson 过程产生声明 $\lambda_m$. 鉴于 将泊松过程解释为以“不可预测”和时间均匀的方式发生的事件的模型，这是一个合理的假设，例如在汽 车保险或其他类型的事故保险中。如果保单持有人进一步独立行事，那么总索赔过程 (个人索赔过程 的喗加）再次是泊松，现在有利率 $\lambda=\lambda_1+\cdots+\lambda_M$. 特别是，如果 $N$ 是单位长度的时间段内的索赔 数量，那么 $N$ 有泊松 $(\lambda)$ 分配。
2. 再次假设保单持有人的数量为 $M$ ，并且每个人在给定时间段内产生的索赔数量为 0 或 $1 ， w p p$ 为 1 。 如果 $p$ 很小而且 $M$ 很大，因此小数定律（定理 5.1) 意味着 $N$ 有一个近似泊松 ( $\lambda$ ) 分布在哪里 $\lambda=M p$.
在泊松情况下， $\hat{p}[z]=\mathrm{e}^{\lambda(z-1)}$ ，因此 $\operatorname{cgf}$ 的 $A$ 是（替换 $z$ 经过 $\widehat{F}[\theta]$ ，参见。命题 I.5.1，取对数)
$$\kappa_A(\theta)=\lambda(\widehat{F}[\theta]-1) .$$
此外，命题5.1产量
$$\mu_A=\lambda \mu_V, \quad \sigma_A^2=\lambda \sigma_V^2+\lambda \mu_V^2=\lambda \mathbb{E} V^2$$

$$\mathbb{P} \theta(A \in \mathrm{d} x)=\mathbb{E}\left[\mathrm{e}^{\theta A-\kappa(\theta)} ; A \in \mathrm{d} x\right] .$$

$$\kappa \theta(\alpha)=\log \mathbb{E} \theta \mathrm{e}^{\alpha A}=\kappa(\alpha+\theta)-\kappa(\theta)=\lambda \theta(\widehat{F} \theta[\alpha]-1),$$

$$F_\theta(\mathrm{d} x)-\frac{\mathrm{e}^{\theta x}}{\widehat{F}[\theta]} F(\mathrm{~d} x) .$$

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