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).

Compound Poisson distribution
Some examples of compound distributions
An introduction to compound distributions

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}

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One thought on “Compound Poisson distribution-discrete example

  1. Reblogged this on Vidhya writes… and commented:
    Very helpful and saved my time. I am yet to read this concept in another paper, however, my current exam tests me on this. I would have died scratching my brain, as how to resolve such problems without this.

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