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.
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One thought on “Compound Poisson distribution

  1. Thx for your article. Could you write me something more about this?

    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.

    I do not understand the indication.

    Thank you!

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