Some examples of compound distributions

We present two examples of compound distributions to illustrate the general formulas presented in the previous post (An introduction to compound distributions).

For the examples below, let N be the number of claims generated by either an individual insured or a group of independent insureds. Let X be the individual claim amount. We consider the random sum Y=X_1+ \cdots + X_N. We discuss the following properties of the aggregate claims random variable Y:

  1. The distribution function F_Y
  2. The mean and higher moments: E[Y] and E[Y^n]
  3. The variance: Var[Y]
  4. The moment generating function and cumulant generating function:M_Y(t) and \Psi_Y(t).
  5. Skewness: \gamma_Y.

Example 1
The number of claims for an individual insurance policy in a policy period is modeled by the binomial distribution with parameter n=2 and p. The individual claim, when it occurs, is modeled by the exponential distribution with parameter \lambda (i.e. the mean individual claim amount is \frac{1}{\lambda}).

The distribution function F_Y is the weighted average of a point mass at y=0, the exponential distribution and the Erlang-2 distribution function. For x \ge 0, we have:

\displaystyle F_Y(x)=(1-p)^2+2p(1-p)(1-e^{-\lambda x})+p^2(1-\lambda x e^{-\lambda x}-e^{-\lambda x})

The mean and variance are are follows:

\displaystyle E[Y]=E[N] \thinspace E[X]=\frac{2p}{\lambda}

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

\displaystyle =\frac{2p}{\lambda^2}+\frac{2p(1-p)}{\lambda^2}=\frac{4p-2p^2}{\lambda^2}

The following calculates the higher moments:

\displaystyle E[Y^n]=(1-p)^2 0 + 2p(1-p) \frac{n!}{\lambda^n}+p^2 \frac{(n+1)!}{\lambda^n}

\displaystyle = \frac{2p(1-p)n!+p^2(n+1)!}{\lambda^n}

The moment generating function M_Y(t)=M_N[ln \thinspace M_X(t)]. So we have:

\displaystyle M_Y(t)=\biggl(1-p+p \frac{\lambda}{\lambda -t}\biggr)^2

\displaystyle =(1-p)^2+2p(1-p) \frac{\lambda}{\lambda -t}+p^2 \biggl(\frac{\lambda}{\lambda -t}\biggr)^2

Note that \displaystyle M_N(t)=(1-p+p e^{t})^2 and \displaystyle M_X(t)=\frac{\lambda}{\lambda -t}.

For the cumulant generating function, we have:

\displaystyle \Psi_Y(t)=ln M_Y(t)=2 ln\biggl(1-p+p \frac{\lambda}{\lambda -t}\biggr)

For the measure of skewness, we rely on the cumulant generating function. Finding the third derivative of \Psi_Y(t) and then evaluate at t=0, we have:

\displaystyle \Psi_Y^{(3)}(0)=\frac{12p-12p^2+4p^3}{\lambda^3}

\displaystyle \gamma_Y=\frac{\Psi_Y^{(3)}(0)}{Var(Y)^{\frac{3}{2}}}=\frac{12p-12p^2+4p^3}{(4p-2p^2)^{\frac{3}{2}}}

Example 2
In this example, the number of claims N follows a geometric distribution. The individual claim amount X follows an exponential distribution with parameter \lambda.

One of the most interesting facts about this example is the moment generating function. Note that \displaystyle M_N(t)=\frac{p}{1-(1-p)e^t}. The following shows the derivation of M_Y(t):

\displaystyle M_Y(t)=M_N[ln \thinspace M_X(t)]=\frac{p}{1-(1-p) e^{ln M_X(t)}}

\displaystyle =\frac{p}{1-(1-p) \frac{\lambda}{\lambda -t}}=\cdots=p+(1-p) \frac{\lambda p}{\lambda p-t}

The moment generating function is the weighted average of a point mass at y=0 and the mgf of an exponential distribution with parameter \lambda p. Thus this example of compound geometric distribution is equivalent to a mixture of a point mass and an exponential distribution. We make use of this fact and derive the following basic properties.

Distribution Function
\displaystyle F_Y(y)=p+(1-p) (1-e^{\lambda p y})=1-(1-p) e^{-\lambda p y} for y \ge 0

Density Function
\displaystyle f_Y(y)=\left\{\begin{matrix}p&\thinspace y=0\\{(1-p) \lambda p e^{-\lambda p y}}&\thinspace 0 < y\end{matrix}\right.

Mean and Higher Moments
\displaystyle E[Y]=(1-p) \frac{1}{\lambda p}=\frac{1-p}{p} \frac{1}{\lambda}=E[N] E[X]

\displaystyle E[Y^n]=p 0 + (1-p) \frac{n!}{(\lambda p)^n}=(1-p) \frac{n!}{(\lambda p)^n}

Variance
\displaystyle Var[Y]=\frac{2(1-p)}{\lambda^2 p^2}-\frac{(1-p)^2}{\lambda^2 p^2}=\frac{1-p^2}{\lambda^2 p^2}

Cumulant Generating Function
\displaystyle \Psi_Y(t)=ln \thinspace M_Y(t)=ln\biggl(p+(1-p) \frac{\lambda p}{\lambda p-t}\biggr)

Skewness
\displaystyle E\biggl[\biggl(Y-\mu_Y\biggr)^3\biggr]=\Psi_Y^{(3)}(0)=\frac{2-2p^3}{\lambda^3 p^3}

\displaystyle \gamma_Y=\frac{\Psi_Y^{(3)}(0)}{(Var[Y])^{\frac{3}{2}}}=\frac{2-2p^3}{(1-p^2)^{\frac{3}{2}}}

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One thought on “Some examples of compound distributions

  1. Dan, how did you arrive at the exponential distribution distribution function in this line… \displaystyle F_Y(x)=(1-p)^2+2p(1-p)(1-e^{-\lambda x})+p^2(1-\lambda x e^{-\lambda x}-e^{-\lambda x})

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