Compound Poisson distribution-discrete example

We present a discrete example of a compound Poisson distribution. A random variable $Y$ has a compound distribution if $Y=X_1+ \cdots +X_N$ where the number of terms $N$ is a discrete random variable whose support is the set of all nonnegative integers (or some appropriate subset) and the random variables $X_i$ are identically distributed (let $X$ be the common distribution). We further assume that the random variables $X_i$ are independent and each $X_i$ is independent of $N$. When $N$ follows the Poisson distribution, $Y$ is said to have a compound Poisson distribution. When the common distribution for the $X_i$ is continuous, $Y$ is a mixed distribution if $P[N=0]$ is nonzero. When the common distribution for the $X_i$ is discrete, $Y$ is a discrete distribution. In this post we present an example of a compound Poisson distribution where the common distribution $X$ is discrete. The compound distribution has a natural insurance interpretation (see the following links).

General Discussion
In general, the distribution function of a compound Poisson random variable $Y$ is the weighted average of all the $n^{th}$ convolutions of the common distribution function of the individual claim amount $X$. The following shows the form of such a distribution function:

$\displaystyle F_Y(y)=\sum \limits_{n=0}^{\infty} F^{*n}(y) P[N=n]$

where $\displaystyle F$ is the common distribution of the $X_n$ and $F^{*n}$ is the $n^{th}$ convolution of $F$.

If the distribution of the individual claim $X$ is discrete, we can obtain the probability mass function of $Y$ by convolutions as follows:

$\displaystyle f_Y(y)=P[Y=y]=\sum \limits_{n=0}^{\infty} p^{*n}(y) P[N=n]$

where $\displaystyle p^{*1}(y)=P[X=y]$
and $\displaystyle p^{*n}=p^* \cdots p^{*}(x)=P[X_1+X_2+ \cdots +X_n=y]$
and $\displaystyle p^{*0}(y)=\left\{\begin{matrix}0&\thinspace y \ne 0\\{1}&\thinspace x=0\end{matrix}\right.$

Example
Suppose the number of claims generated by a portfolio of insurance policies over a fixed time period has a Poisson distribution with parameter $\lambda$. Individual claim amounts will be 1 or 2 with probabilities 0.6 and 0.4, respectively. For the compound Poisson aggregate claims $Y=X_1+ \cdots +X_N$, find $P[Y=k]$ for $k=0,1,2,3,4$.

The probability mass function of $N$ is: $\displaystyle f_N(n)=\frac{\lambda^n e^{-\lambda}}{n!}$ where $n=0,1,2, \cdots$. The individual claim amounnt $X$ has a Bernoulli distribution since it is a two-valued discrete random variable. For convenience, we let $p=0.4$ (i.e. we consider $X=2$ is a success). Then the sum $X_1+ \cdots + X_n$ has a Binomial distribution. Consequently, the $n^{th}$ convolution $p^{*n}$ is simply the distribution function of Binomial(n,p). The following shows $p^{*n}$ for $n=1,2,3,4$.

$\displaystyle p^{*1}(1)=0.6, \thinspace p^{*1}(2)=0.4$

$\displaystyle p^{*2}(2)=\binom{2}{0} (0.4)^0 (0.6)^2=0.36$
$\displaystyle p^{*2}(3)=\binom{2}{1} (0.4)^1 (0.6)^1=0.48$
$\displaystyle p^{*2}(4)=\binom{2}{2} (0.4)^2 (0.6)^0=0.16$

$\displaystyle p^{*3}(3)=\binom{3}{0} (0.4)^0 (0.6)^3=0.216$
$\displaystyle p^{*3}(4)=\binom{3}{1} (0.4)^1 (0.6)^2=0.432$
$\displaystyle p^{*3}(5)=\binom{3}{2} (0.4)^2 (0.6)^1=0.288$
$\displaystyle p^{*3}(6)=\binom{3}{3} (0.4)^3 (0.6)^0=0.064$

$\displaystyle p^{*4}(4)=\binom{4}{0} (0.4)^0 (0.6)^4=0.1296$
$\displaystyle p^{*4}(5)=\binom{4}{1} (0.4)^1 (0.6)^3=0.3456$
$\displaystyle p^{*4}(6)=\binom{4}{2} (0.4)^2 (0.6)^2=0.3456$
$\displaystyle p^{*4}(7)=\binom{4}{3} (0.4)^3 (0.6)^1=0.1536$
$\displaystyle p^{*4}(8)=\binom{4}{4} (0.4)^4 (0.6)^0=0.0256$

Since we are interested in finding $P[Y=y]$ for $y=0,1,2,3,4$, we only need to consider $N=0,1,2,3,4$. The following matrix shows the relevant values of $p^{*n}$. The rows are for $y=0,1,2,3,4$. The columns are $p^{*0}$, $p^{*1}$, $p^{*2}$, $p^{*3}$, $p^{*4}$.

$\displaystyle \begin{pmatrix} 1&0&0&0&0 \\{0}&0.6&0&0&0 \\{0}&0.4&0.36&0&0 \\{0}&0&0.48&0.216&0 \\{0}&0&0.16&0.432&0.1296\end{pmatrix}$

To obtain the probability mass function of $Y$, we simply multiply each row by $P[N=n]$ where $n=0,1,2,3,4$.

$\displaystyle P[Y=0]=e^{-\lambda}$
$\displaystyle P[Y=1]=0.6 \lambda e^{-\lambda}$
$\displaystyle P[Y=2]=0.4 \lambda e^{-\lambda}+0.36 \frac{\lambda^2 e^{-\lambda}}{2}$
$\displaystyle P[Y=3]=0.48 \frac{\lambda^2 e^{-\lambda}}{2}+0.216 \frac{\lambda^3 e^{-\lambda}}{6}$
$\displaystyle P[Y=4]=0.16 \frac{\lambda^2 e^{-\lambda}}{2}+0.432 \frac{\lambda^3 e^{-\lambda}}{6}+0.1296 \frac{\lambda^4 e^{-\lambda}}{24}$