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\nWe have seen that the natural exponential family includes continuous distributions, such as the normal and gamma, as well as discrete distributions, such as the binomial and Poisson. It also includes distributions that are mixtures of discrete and continuous components. In insurance claims modeling, the most widely used mixture is the Tweedie (1984) distribution. It has a positive mass at zero representing no claims and a continuous component for positive values representing the amount of a claim. <\/p>\n
The Tweedie distribution is defined as a Poisson sum of gamma random variables, known as an aggregate loss<\/em> in actuarial science. Specifically, suppose that \\(N\\) has a Poisson distribution with mean \\(\\lambda\\), representing the number of claims. Let {\\(y_j\\)} be an i.i.d. sequence, independent of \\(N\\), with each \\(y_j\\) having a gamma distribution with parameters \\(\\alpha\\) and \\(\\gamma\\), representing the amount of a claim. Then, \\(S_N = y_1 + \\ldots + y_N\\) is Poisson sum of gammas. Section 16.5 will discuss aggregate loss models in further detail. This section focuses on the important special case of the Tweedie distribution. <\/p>\n To understand the mixture aspect of the Tweedie distribution, first note that it is straightforward to compute the probability of zero claims as At first glance, this density does not appear to be a member of the linear exponential family. To see the relationship, we first calculate the moments using iterated expectations as Thus, the Tweedie distribution is a member of the linear exponential family. Easy calculations show that \\begin{equation}
\n\\begin{equation*}
\n\\Pr ( S_N=0) = \\Pr (N=0) = e^{-\\lambda}.
\n \\end{equation*} The distribution function can be computed using conditional expectations,
\n\\begin{equation*}
\n\\Pr ( S_N \\leq y) = e^{-\\lambda}+\\sum_{n=1}^{\\infty} \\Pr(N=n) \\Pr(S_n \\leq y), ~~~~~y \\geq 0.
\n\\end{equation*} Because the sum of i.i.d. gammas is a gamma, \\(S_n\\) (not \\(S_N\\)) has a gamma distribution with parameters \\(n\\alpha\\) and \\(\\gamma\\). Thus, for \\(y>0\\), the density of the Tweedie distribution is
\n\\begin{equation}
\n \\mathrm{f}_{S}(y) = \\sum_{n=1}^{\\infty} e^{-\\lambda} \\frac{\\lambda ^n}{n!} \\frac{\\gamma^{n \\alpha}}{\\Gamma(n \\alpha)} y^{n \\alpha -1} e^{-y \\gamma}.
\n\\end{equation} <\/p>\n
\n\\begin{equation}
\n \\mathrm{E~}S_N = \\lambda \\frac{\\alpha}{\\gamma}~~~~~~\\mathrm{and}~~~~~ \\mathrm{Var~}S_N = \\frac{\\lambda \\alpha}{\\gamma^2} (1+\\alpha).
\n \\end{equation} Now, define three parameters \\(\\mu, \\phi, p\\) through the relations
\n\\begin{equation*}
\n\\lambda = \\frac{\\mu^{2-p}}{\\phi (2-p)},~~~~~~\\alpha = \\frac{2-p}{p-1}~~~~~~\\mathrm{and}~~~~~ \\frac{1}{\\gamma} = \\phi(p-1)\\mu^{p-1}.
\n\\end{equation*} Inserting these new parameters yields
\n\\begin{equation}
\n \\mathrm{f}_{S}(y) = \\exp \\left[ \\frac{-1}{\\phi} \\left( \\frac{\\mu^{2-p}}{2-p} + \\frac{y}{(p-1)\\mu^{p-1}} \\right) +S(y,\\phi) \\right].
\n\\end{equation} We leave the calculation of \\(S(y,\\phi)\\) as an exercise. <\/p>\n
\n\\mathrm{E~}S_N = \\mu~~~~~~\\mathrm{and}~~~~~ \\mathrm{Var~}S_N = \\phi \\mu^p,
\n \\end{equation} where 1 < p < 2. Examining our variance function Table 13.1, the Tweedie distribution can also be viewed as a choice that is intermediate between the Poisson and the gamma distributions. <\/p>\n