Compound mixed Poisson distribution

Let the random sum Y=X_1+X_2+ \cdots +Y_N be the aggregate claims generated in a fixed period by an independent group of insureds. When the number of claims N follows a Poisson distribution, the sum Y is said to have a compound Poisson distribution. When the number of claims N has a mixed Poisson distribution, the sum Y is said to have a compound mixed Poisson distribution. A mixed Poisson distribution is a Poisson random variable N such that the Poisson parameter \Lambda is uncertain. In other words, N is a mixture of a family of Poisson distributions N(\Lambda) and the random variable \Lambda specifies the mixing weights. In this post, we present several basic properties of compound mixed Poisson distributions. In a previous post (Compound negative binomial distribution), we showed that the compound negative binomial distribution is an example of a compound mixed Poisson distribution (with gamma mixing weights).

In terms of notation, we have:

  • Y=X_1+X_2+ \cdots +Y_N,
  • N \sim Poisson(\Lambda),
  • \Lambda \sim some unspecified distribution.

The following presents basic proeprties of the compound mixed Poisson Y in terms of the mixing weights \Lambda and the claim amount random variable X.

Mean and Variance

\displaystyle E[Y]=E[\Lambda] E[X]

\displaystyle Var[Y]=E[\Lambda] E[X^2]+Var[\Lambda] E[X]^2

Moment Generating Function

\displaystyle M_Y(t)=M_{\Lambda}[M_X(t)-1]

Cumulant Generating Function

\displaystyle \Psi_Y(t)=ln M_{\Lambda}[M_X(t)-1]=\Psi_{\Lambda}[M_X(t)-1]

Measure of Skewness
\displaystyle E[(Y-\mu_Y)^3]=\Psi_Y^{(3)}(0)

\displaystyle =\Psi_{\Lambda}^{(3)}(0) E[X]^3 + 3 \Psi_{\Lambda}^{(2)}(0) E[X] E[X^2]+\Psi_{\Lambda}^{(1)}(0) E[X^3]

\displaystyle =\gamma_{\Lambda} Var[\Lambda]^{\frac{3}{2}} E[X]^3 + 3 Var[\Lambda] E[X] E[X^2]+E[\Lambda] E[X^3]

Measure of skewness: \displaystyle \gamma_Y=\frac{E[(Y-\mu_Y)^3]}{(Var[Y])^{\frac{3}{2}}}

Previous Posts on Compound Distributions

An introduction to compound distributions
Some examples of compound distributions
Compound Poisson distribution
Compound Poisson distribution-discrete example
Compound negative binomial distribution

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Compound negative binomial distribution

In this post, we discuss the compound negative binomial distribution and its relationship with the compound Poisson distribution.

A compound distribution is a model for a random sum Y=X_1+X_2+ \cdots +X_N where the number of terms N is uncertain. To make the compund distribution more tractable, we assume that the variables X_i are independent and identically distributed and that each X_i is independent of N. The random sum Y can be interpreted the sum of all the measurements that are associated with certain events that occur during a fixed period of time. For example, we may be interested in the total amount of rainfall in a 24-hour period, during which the occurences of a number of events are observed and each of the events provides a measurement of an amount of rainfall. Another interpretation of compound distribution is the random variable of the aggregate claims generated by an insurance policy or a group of insurance policies during a fixed policy period. In this setting, N is the number of claims generated by the portfolio of insurance policies and X_1 is the amount of the first claim and X_2 is the amount of the second claim and so on. When N follows the Poisson distribution, the random sum Y is said to have a compound Poisson distribution. Even though the compound Poisson distribution has many attractive properties, it is not a good model when the variance of the number of claims is greater than the mean of the number of claims. In such situations, the compound negative binomial distribution may be a better fit. See this post (Compound Poisson distribution) for a basic discussion. See the links at the end of this post for more articles on compound distributons that I posted on this blog.

Compound Negative Binomial Distribution
The random variable N is said to have a negative binomial distribution if its probability function is given by the following:

\displaystyle P[N=n]=\binom{\alpha + n-1}{\alpha-1} \thinspace \biggl(\frac{\beta}{\beta+1}\biggr)^{\alpha}\biggl(\frac{1}{\beta+1}\biggr)^{n} \ \ \ \ \ \ \ \ \ \ \ \ (1)

where n=0,1,2,3, \cdots, \beta >0 and \alpha is a positive integer.

Our formulation of negative binomial distribution is the number of failures that occur before the \alpha^{th} success in a sequence of independent Bernoulli trials. But this interpretation is not important to our task at hand. Let Y=X_1+X_2+ \cdots +X_N be the random sum as described in the above introductory paragraph such that N follows a negative binomial distribution. We present the basic properties discussed in the post An introduction to compound distributions by plugging the negative binomial distribution into N.

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

where F is the common distribution function for X_i and F^{*n} is the n^{th} convolution of F. Of course, P[N=n] is the negative binomial probability function indicated above.

Mean and Variance
\displaystyle E[Y]=E[N] \thinspace E[X]=\frac{\alpha}{\beta} E[X]

\displaystyle Var[Y]=E[N] \thinspace Var[X]+Var[N] \thinspace E[X]^2

\displaystyle =\frac{\alpha}{\beta} Var[X]+\frac{\alpha (\beta+1)}{\beta^2} E[X]^2

Moment Generating Function
\displaystyle M_Y(t)=M_N[ln M_X(t)]=\biggl(\frac{p}{1-(1-p) M_X(t)}\biggr)^{\alpha}

\displaystyle M_Y(t)=\biggl(\frac{\beta}{\beta+1- M_X(t)}\biggr)^{\alpha}

where \displaystyle p=\frac{\beta}{\beta+1}, \displaystyle M_N(t)=\biggl(\frac{p}{1-(1-p) e^{t}}\biggr)^{\alpha}

Cumulant Generating Function
\displaystyle \Psi_Y(t)=\alpha \thinspace ln \biggl(\frac{\beta}{\beta+1- M_X(t)}\biggr)

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

\displaystyle =\frac{2}{\alpha^2} E[N]^3 E[X]^3 +\frac{3}{\alpha} E[N]^2 E[X] E[X^2]+E[N] E[X^3]

Measure of skewness: \displaystyle \gamma_Y=\frac{E[(Y-\mu_Y)^3]}{(Var[Y])^{\frac{3}{2}}}

Compound Mixed Poisson Distribution
In a previous post (Basic properties of mixtures), we showed that the negative binomial distribution is a mixture of a family of Poisson distributions with gamma mixing weights. Specifically, if N \sim \text{Poisson}(\Lambda) and \Lambda \sim \text{Gamma}(\alpha,\beta), then the unconditional distribution of N is a negative binomial distribution and the probability function is of the form (1) given above.

Thus the negative binomial distribution is a special example of a compound mixed Poisson distribution. When an aggregate claims variable Y=X_1+X_2+ \cdots +Y_N has a compound mixed Poisson distribution, the number of claims N follows a Poisson distribution, but the Poisson parameter \Lambda is uncertain. The uncertainty could be due to an heterogeneity of risks across the insureds in the insurance portfolio (or across various rating classes). If the information of the risk parameter \Lambda can be captured in a gamma distribution, then the unconditional number of claims in a given fixed period has a negative binomial distribution.

Previous Posts on Compound Distributions
An introduction to compound distributions
Some examples of compound distributions
Compound Poisson distribution
Compound Poisson distribution-discrete example

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}