# Compound Poisson distribution

The compound distribution is a model for describing the aggregate claims arised in a group of independent insureds. Let $N$ be the number of claims generated by a portfolio of insurance policies in a fixed time period. Suppose $X_1$ is the amount of the first claim, $X_2$ is the amount of the second claim and so on. Then $Y=X_1+X_2+ \cdots + X_N$ represents the total aggregate claims generated by this portfolio of policies in the given fixed time period. In order to make this model more tractable, we make the following assumptions:

• $X_1,X_2, \cdots$ are independent and identically distributed.
• Each $X_i$ is independent of the number of claims $N$.

The number of claims $N$ is associated with the claim frequency in the given portfolio of policies. The common distribution of $X_1,X_2, \cdots$ is denoted by $X$. Note that $X$ models the amount of a random claim generated in this portfolio of insurance policies. See these two posts for an introduction to compound distributions (An introduction to compound distributions, Some examples of compound distributions).

When the claim frequency $N$ follows a Poisson distribution with a constant parameter $\lambda$, the aggreagte claims $Y$ is said to have a compound Poisson distribution. After a general discussion of the compound Poisson distribution, we discuss the property that an independent sum of compound Poisson distributions is also a compound Poisson distribution. We also present an example to illustrate basic calculations.

Compound Poisson – General Properties

Distribution Function
$\displaystyle F_Y(y)=\sum \limits_{n=0}^{\infty} F^{*n}(y) \frac{\lambda^n e^{-\lambda}}{n!}$

where $\lambda=E[N]$, $F$ is the common distribution function of $X_i$ and $F^{*n}$ is the n-fold convolution of $F$.

Mean and Variance
$\displaystyle E[Y]=E[N] E[X]= \lambda E[X]$

$\displaystyle Var[Y]=\lambda E[X^2]$

Moment Generating Function and Cumulant Generating Function
$\displaystyle M_Y(t)=e^{\lambda (M_X(t)-1)}$

$\displaystyle \Psi_Y(t)=ln M_Y(t)=\lambda (M_X(t)-1)$

Note that the moment generating function of the Poisson $N$ is $M_N(t)=e^{\lambda (e^t - 1)}$. For a compound distribution $Y$ in general, $M_Y(t)=M_N[ln M_X(t)]$.

Skewness
$\displaystyle E[(Y-\mu_Y)^3]=\Psi_Y^{(3)}(0)=\lambda E[X^3]$

$\displaystyle \gamma_Y=\frac{E[(Y-\mu_Y)^3]}{Var[Y]^{\frac{3}{2}}}=\frac{1}{\sqrt{\lambda}} \frac{E[X^3]}{E[X^2]^{\frac{3}{2}}}$

Independent Sum of Compound Poisson Distributions
First, we state the results. Suppose that $Y_1,Y_2, \cdots, Y_k$ are independent random variables such that each $Y_i$ has a compound Poisson distribution with $\lambda_i$ being the Poisson parameter for the number of claim variable and $F_i$ being the distribution function for the individual claim amount. Then $Y=Y_1+Y_2+ \cdots +Y_k$ has a compound Poisson distribution with:

• the Poisson parameter: $\displaystyle \lambda=\sum \limits_{i=1}^{k} \lambda_i$
• the distribution function: $\displaystyle F_Y(y)=\sum \limits_{i=1}^{k} \frac{\lambda_i}{\lambda} \thinspace F_i(y)$

The above result has an insurance interpretation. Suppose we have $k$ independent blocks of insurance policies such that the aggregate claims $Y_i$ for the $i^{th}$ block has a compound Poisson distribution. Then $Y=Y_1+Y_2+ \cdots +Y_k$ is the aggregate claims for the combined block during the fixed policy period and also has a compound Poisson distribution with the parameters stated in the above two bullet points.

To get a further intuitive understanding about the parameters of the combined block, consider $N_i$ as the Poisson number of claims in the $i^{th}$ block of insurance policies. It is a well known fact in probability theory (see [1]) that the indpendent sum of Poisson variables is also a Poisson random variable. Thus the total number of claims in the combined block is $N=N_1+N_2+ \cdots +N_k$ and has a Poisson distribution with parameter $\lambda=\lambda_1 + \cdots + \lambda_k$.

How do we describe the distribution of an individual claim amount in the combined insurance block? Given a claim from the combined block, since we do not know which of the constituent blocks it is from, this suggests that an individual claim amount is a mixture of the individual claim amount distributions from the $k$ blocks with mixing weights $\displaystyle \frac{\lambda_1}{\lambda},\frac{\lambda_2}{\lambda}, \cdots, \frac{\lambda_k}{\lambda}$. These mixing weights make intuitive sense. If insurance bock $i$ has a higher claim frequency $\lambda_i$, then it is more likely that a randomly selected claim from the combined block comes from block $i$. Of course, this discussion is not a proof. But looking at the insurance model is a helpful way of understanding the independent sum of compound Poisson distributions.

To see why the stated result is true, let $M_i(t)$ be the moment generating function of the individual claim amount in the $i^{th}$ block of policies. Then the mgf of the aggregate claims $Y_i$ is $\displaystyle M_{Y_i}(t)=e^{\lambda_i (M_i(t)-1)}$. Consequently, the mgf of the independent sum $Y=Y_1+ \cdots + Y_k$ is:

$\displaystyle M_Y(t)=\prod \limits_{i=0}^{k} e^{\lambda_i (M_i(t)-1)}= e^{\sum \limits_{i=0}^{k} \lambda_i(M_i(t)-1)} \displaystyle = e^{\lambda \biggl[\sum \limits_{i=0}^{k} \frac{\lambda_i}{\lambda} M_i(t) - 1 \biggr]}$

The mgf of $Y$ has the form of a compound Poisson distribution where the Poisson parameter is $\lambda=\lambda_1 + \cdots + \lambda_k$. Note that the component $\displaystyle \sum \limits_{i=0}^{k} \frac{\lambda_i}{\lambda}M_i(t)$ in the exponent is the mgf of the claim amount distribution. Since it is the weighted average of the individual claim amount mgf’s, this indicates that the distribution function of $Y$ is the mixture of the distribution functions $F_i$.

Example
Suppose that an insurance company acquired two portfolios of insurance policies and combined them into a single block. For each portfolio the aggregate claims variable has a compound Poisson distribution. For one of the portfolios, the Poisson parameter is $\lambda_1$ and the individual claim amount has an exponential distribution with parameter $\delta_1$. The corresponding Poisson and exponential parameters for the other portfolio are $\lambda_2$ and $\delta_2$, respectively. Discuss the distribution for the aggregate claims $Y=Y_1+Y_2$ of the combined portfolio.

The aggregate claims $Y$ of the combined portfolio has a compound Poisson distribution with Poisson parameter $\lambda=\lambda_1+\lambda_2$. The amount of a random claim $X$ in the combined portfolio has the following distribution function and density function:

$\displaystyle F_X(x)=\frac{\lambda_1}{\lambda} (1-e^{-\delta_1 x})+\frac{\lambda_2}{\lambda} (1-e^{-\delta_2 x})$

$\displaystyle f_X(x)=\frac{\lambda_1}{\lambda} (\delta_1 \thinspace e^{-\delta_1 x})+\frac{\lambda_2}{\lambda} (\delta_2 \thinspace e^{-\delta_2 x})$

The rest of the discussion mirrors the general discussion earlier in this post.

Distribution Function
As in the general case, $\displaystyle F_Y(y)=\sum \limits_{n=0}^{\infty} F^{*n}(y) \frac{\lambda^n e^{-\lambda}}{n!}$

where $\lambda=\lambda_1 +\lambda_2$, $F=F_X$ and $F^{*n}$ is the n-fold convolution of $F_X$.

Mean and Variance
$\displaystyle E[Y]=\frac{\lambda_1}{\delta_1}+\frac{\lambda_2}{\delta_2}$

$\displaystyle Var[Y]=\frac{2 \lambda_1}{\delta_1^2}+\frac{2 \lambda_2}{\delta_2^2}$

Moment Generating Function and Cumulant Generating Function
To obtain the mgf and cgf of the aggregate claims $Y$, consider $\lambda [M_X(t)-1]$. Note that $M_X(t)$ is the weighted average of the two exponential mgfs of the two portfolios of insurance policies. Thus we have:

$\displaystyle M_X(t)=\frac{\lambda_1}{\lambda} \frac{\delta_1}{\delta_1 - t}+\frac{\lambda_2}{\lambda} \frac{\delta_2}{\delta_2 - t}$

$\displaystyle \lambda [M_X(t)-1]=\frac{\lambda_1 t}{\delta_1 - t}+\frac{\lambda_2 t}{\delta_2 - t}$

$\displaystyle M_Y(t)=e^{\lambda (M_X(t)-1)}=e^{\frac{\lambda_1 t}{\delta_1 - t}+\frac{\lambda_2 t}{\delta_2 - t}}$

$\displaystyle \Psi_Y(t)=\frac{\lambda_1 t}{\delta_1 -t}+\frac{\lambda_2 t}{\delta_2 -t}$

Skewness
Note that $\displaystyle E[(Y-\mu_Y)^3]=\Psi_Y^{(3)}(0)=\frac{6 \lambda_1}{\delta_1^3}+\frac{6 \lambda_2}{\delta_2^3}$

$\displaystyle \gamma_Y=\displaystyle \frac{\frac{6 \lambda_1}{\delta_1^3}+\frac{6 \lambda_2}{\delta_2^3}}{(\frac{2 \lambda_1}{\delta_1^2}+\frac{2 \lambda_2}{\delta_2^2})^{\frac{3}{2}}}$

Reference

1. Hogg R. V. and Tanis E. A., Probability and Statistical Inference, Second Edition, Macmillan Publishing Co., New York, 1983.