Compound Poisson-Gamma Representation

Why a Compound Representation?

For a Tweedie model with power \(1<p<2\), the mean and variance are linked by \(\operatorname{Var}(X)=\phi\mu^p\), where \(\mu>0\) and \(\phi>0\). This variance relationship can seem abstract. A compound representation gives it a concrete interpretation: draw a Poisson number of independent, positive gamma amounts and add them. The expected sum and its variance can then be checked directly against the Tweedie parameters.

The representation is useful for understanding how a continuous response can arise from random event counts and random event sizes. It also identifies exactly how the power index affects the event rate and the distribution of each amount. Here, the focus is the construction and its parameter mapping, not likelihood calculations or separate analyses of zero and positive responses.

Definition: Let \(N\sim\operatorname{Poisson}(\lambda)\), and let \(Z_1,Z_2,\ldots\) be independent and identically distributed random variables, independent of \(N\). The random sum \(S=\sum_{j=1}^{N}Z_j\), defined as \(S=0\) when \(N=0\), is a compound Poisson random variable. In a compound Poisson–Gamma model, each amount \(Z_j\) has a gamma distribution with shape \(\alpha>0\) and scale \(\beta>0\), with density \(z^{\alpha-1}e^{-z/\beta}/(\Gamma(\alpha)\beta^\alpha)\) for \(z>0\). Its mean is \(\alpha\beta\).

The scale convention matters: \(\beta\) is the gamma scale, not its rate. Thus the gamma rate is \(1/\beta\), and the moment generating function of one amount is \(E[e^{tZ_j}]=(1-\beta t)^{-\alpha}\) for \(t<1/\beta\). We will use this convention throughout.

Deriving the Component Parameters

Let \(X\) have a Tweedie exponential-dispersion distribution with mean \(\mu\), dispersion \(\phi\), and power \(p\), where \(1<p<2\). Its variance is \(\phi\mu^p\). We seek a compound Poisson–Gamma sum with three parameters: the Poisson rate \(\lambda\), gamma shape \(\alpha\), and gamma scale \(\beta\). These parameters must make the sum’s mean and variance equal to \(\mu\) and \(\phi\mu^p\), respectively.

For a compound Poisson sum with gamma amounts, the mean is \(\lambda\alpha\beta\). Its variance is \(\lambda E[Z_1^2]=\lambda\alpha(\alpha+1)\beta^2\). The first equality follows by conditioning on \(N\); the second uses the gamma second moment. Matching these to the Tweedie mean and variance, together with matching the full exponential-dispersion form, gives the following parameter map.

Theorem: For \(\mu>0\), \(\phi>0\), and \(1<p<2\), a Tweedie random variable with mean \(\mu\), dispersion \(\phi\), and power \(p\) has the compound Poisson–Gamma representation

\(X\overset{d}{=}\sum_{j=1}^{N}Z_j\), where \(N\sim\operatorname{Poisson}(\lambda)\), \(Z_j\sim\operatorname{Gamma}(\alpha,\beta)\), and \(N,Z_1,Z_2,\ldots\) are independent, with

\(\displaystyle \lambda=\frac{\mu^{2-p}}{\phi(2-p)},\qquad \alpha=\frac{2-p}{p-1},\qquad \beta=\phi(p-1)\mu^{p-1}.\)

Here the gamma distribution is parameterized by shape \(\alpha\) and scale \(\beta\).

First, the mapping gives the required mean:

\lambda\alpha\beta = \frac{\mu^{2-p}}{\phi(2-p)} \frac{2-p}{p-1} \phi(p-1)\mu^{p-1} =\mu.

For the variance, use \(\alpha+1=1/(p-1)\), so that \(\alpha(\alpha+1)=(2-p)/(p-1)^2\). Substitution gives

\begin{aligned} \lambda\alpha(\alpha+1)\beta^2 &=\frac{\mu^{2-p}}{\phi(2-p)} \frac{2-p}{(p-1)^2} \phi^2(p-1)^2\mu^{2p-2}\\ &=\phi\mu^p. \end{aligned}

These moment checks are necessary, but by themselves they do not identify a whole distribution. The cumulant generating function provides the stronger verification.

Proof by Cumulant Generating Functions

The cumulant generating function of a random variable \(Y\) is \(K_Y(t)=\log E[e^{tY}]\) wherever this expectation is finite. For one gamma amount, \(M_Z(t)=(1-\beta t)^{-\alpha}\) when \(t<1/\beta\). Conditioning on the Poisson count and summing over its possible values gives

\begin{aligned} E[e^{tS}] &=\sum_{n=0}^{\infty}P(N=n)\,E[e^{tS}\mid N=n]\\ &=\sum_{n=0}^{\infty}e^{-\lambda}\frac{\lambda^n}{n!} \bigl(M_Z(t)\bigr)^n\\ &=\exp\{\lambda(M_Z(t)-1)\}. \end{aligned}

The series is the exponential series and converges for \(t<1/\beta\). Taking logarithms therefore yields

K_S(t)=\lambda\bigl((1-\beta t)^{-\alpha}-1\bigr),\qquad t<1/\beta.

To compare this with the Tweedie exponential-dispersion form, write its canonical parameter and cumulant function as

\theta=\frac{\mu^{1-p}}{1-p}, \qquad b(\theta)=\frac{\mu^{2-p}}{2-p}.

The cumulant generating function of an exponential-dispersion model at fixed \(\mu,\phi,p\) is \([b(\theta+\phi t)-b(\theta)]/\phi\), on the domain where \(\theta+\phi t\) is in the cumulant function’s domain. For the parameters above,

\theta+\phi t =\theta\bigl(1-\beta t\bigr).

Indeed, \(\phi t/\theta=-\phi(p-1)\mu^{p-1}t=-\beta t\). Because \(\theta<0\), the required domain corresponds to \(1-\beta t>0\). The power in \(b\) is \((2-p)/(1-p)=-\alpha\), so

\begin{aligned} \frac{b(\theta+\phi t)-b(\theta)}{\phi} &=\frac{b(\theta)}{\phi} \bigl((1-\beta t)^{-\alpha}-1\bigr)\\ &=\lambda\bigl((1-\beta t)^{-\alpha}-1\bigr). \end{aligned}

This is exactly the compound sum’s cumulant generating function. Both moment generating functions are finite on an open interval containing zero. The standard uniqueness theorem for moment generating functions therefore establishes equality in distribution. This proves the stated representation. \(\square\)

Worked Parameter Mappings

Worked Example: Power One and a Half

Suppose \(\mu=4\), \(\phi=0.2\), and \(p=1.5\). Compute the compound Poisson and gamma parameters, then check the mean and variance.

The shape is \(\alpha=(2-1.5)/(1.5-1)=0.5/0.5=1\). The scale is \(\beta=0.2(0.5)4^{0.5}=0.1(2)=0.2\). The Poisson rate is \(\lambda=4^{0.5}/[0.2(0.5)]=2/0.1=20\). Thus \(N\sim\operatorname{Poisson}(20)\), and each amount is gamma with shape \(1\) and scale \(0.2\).

The aggregate mean is \(20(1)(0.2)=4\). Since \(\alpha(\alpha+1)=1(2)=2\), its variance is \(20(2)(0.2)^2=20(2)(0.04)=1.6\). Directly from the Tweedie variance function, \(\phi\mu^p=0.2(4^{1.5})=0.2(8)=1.6\), as required.

Worked Example: A Higher Power Within the Interval

Take \(\mu=16\), \(\phi=0.125\), and \(p=1.75\). The power differences are \(2-p=0.25\) and \(p-1=0.75\). Hence \(\alpha=0.25/0.75=1/3\), \(\beta=0.125(0.75)16^{0.75}\), and \(\lambda=16^{0.25}/[0.125(0.25)]\).

Since \(16^{0.75}=8\) and \(16^{0.25}=2\), these reduce to \(\beta=0.09375(8)=0.75\) and \(\lambda=2/0.03125=64\). The aggregate mean is \(64(1/3)(0.75)=16\). For the variance, \(\alpha(\alpha+1)=(1/3)(4/3)=4/9\), so \(64(4/9)(0.75)^2=64(4/9)(9/16)=16\). The Tweedie variance is also \(0.125(16^{1.75})=0.125(128)=16\).

Worked Example: Power Four-Thirds

Let \(\mu=27\), \(\phi=1/9\), and \(p=4/3\). Here \(2-p=2/3\), \(p-1=1/3\), \(27^{1/3}=3\), and \(27^{2/3}=9\). The component parameters are \(\alpha=(2/3)/(1/3)=2\), \(\beta=(1/9)(1/3)(3)=1/9\), and \(\lambda=9/[(1/9)(2/3)]=9/(2/27)=243/2=121.5\).

The mean is \(121.5(2)(1/9)=27\). The variance is \(121.5[2(3)](1/9)^2=121.5(6/81)=9\). The target Tweedie variance agrees: \((1/9)27^{4/3}=(1/9)(27\cdot3)=9\). The non-integer Poisson rate is valid: \(\lambda\) is a positive rate parameter, not a count that must itself be an integer.

How the Power Changes the Components

The parameter map separates the effects of \(\mu,\phi,\) and \(p\). The gamma shape \(\alpha=(2-p)/(p-1)\) depends only on the power. The gamma scale \(\beta=\phi(p-1)\mu^{p-1}\) depends on the dispersion, mean, and power. The Poisson rate \(\lambda=\mu^{2-p}/[\phi(2-p)]\) also depends on all three. The aggregate mean and variance arise from their combination; neither the component mean nor the event rate alone is the Tweedie mean.

As \(p\) varies within the interval, the gamma shape varies as well as the rate and scale. Consequently, it is generally incorrect to interpret power as changing only how often events occur while leaving event sizes unchanged. For a fixed \(\mu\) and \(\phi\), all three component parameters must be recalculated when \(p\) changes.

Common Mistakes

Using the gamma rate as though it were the scale. The formula here gives \(\beta\), the scale. The corresponding rate is \(1/\beta\); substituting \(1/\beta\) into a scale-based gamma mean \(\alpha\beta\) would give the wrong component mean.
Checking only the aggregate mean. Many choices of rate and jump distribution can have the same mean. Check the variance as well, and use the cumulant generating function to verify the full Tweedie distribution.
Assuming the event rate is the mean response. The rate is \(\lambda=\mu^{2-p}/[\phi(2-p)]\), while the mean response is \(\lambda\alpha\beta=\mu\). A jump has a nonzero expected size, so the two quantities are not interchangeable.
Requiring \(\lambda\) to be an integer. A Poisson rate can be any positive real number. It is the realized count \(N\), not its parameter \(\lambda\), that takes nonnegative integer values.

Key takeaway: For \(1<p<2\), the Tweedie parameters map to a Poisson rate \(\lambda=\mu^{2-p}/[\phi(2-p)]\), gamma shape \(\alpha=(2-p)/(p-1)\), and gamma scale \(\beta=\phi(p-1)\mu^{p-1}\). The resulting random sum has mean \(\mu\), variance \(\phi\mu^p\), and the Tweedie cumulant generating function.

Check Your Understanding

Use the scale parameterization of the gamma distribution throughout.

For \(\mu=9\), \(\phi=0.1\), and \(p=1.5\), calculate \(\lambda,\alpha,\beta\), then verify the aggregate mean and variance.
Why does matching the first two moments alone not establish equality with the Tweedie distribution?
State the domain of the compound sum’s cumulant generating function in terms of \(\beta\).
For fixed \(\mu\) and \(\phi\), which compound Poisson–Gamma parameter depends only on \(p\)?
Explain why a Poisson rate of \(12.5\) is mathematically valid.