Admissible Parameters and Response Support

Power Ranges Restrict the Response

A power variance function specifies how variability changes with the mean, but it does not by itself guarantee that a probability distribution with that variance exists. In a Tweedie exponential-dispersion model, the variance function is \(V(\mu)=\mu^p\), so the response variance is \(\operatorname{Var}(Y)=\phi\mu^p\). The power \(p\) selects a family with a particular response support; the mean \(\mu\) and dispersion \(\phi\) must also be admissible for that family.

This connection is practical, not just mathematical. A model whose support excludes zero cannot assign positive probability to an exact zero response. A model whose support includes zero may still assign zero probability to that exact value if it is continuous there. The distinction matters whenever the observed response includes zeros.

Definition: The support of a response distribution is the set of values on which the distribution is concentrated. A point mass at zero means \(P(Y=0)>0\). Including zero in the support does not, by itself, imply a point mass: a continuous distribution can have zero in its support while assigning probability zero to the single value \(0\).

Standard Power Ranges and Their Supports

For the four standard cases considered here, the admissible power ranges and response characteristics are as follows. The parameter \(\mu\) denotes \(E[Y]\), and \(\phi\) is the dispersion parameter.

Family	Power range	Admissible parameters	Response support and zero behavior
Gaussian	\(p=0\)	\(\mu\in\mathbb{R}\), \(\phi>0\)	Real line; continuous, so \(P(Y=0)=0\)
Poisson	\(p=1\)	\(\mu>0\); ordinary Poisson has \(\phi=1\)	Nonnegative integers; \(P(Y=0)=e^{-\mu}>0\)
Compound Poisson-gamma	\(1<p<2\)	\(\mu>0\), \(\phi>0\)	Zero and positive values; a point mass at zero and a continuous positive part
Gamma	\(p=2\)	\(\mu>0\), \(\phi>0\)	Strictly positive real values; \(P(Y=0)=0\)

These are distinct parameter cases, not interchangeable labels for the same support. At \(p=0\), \(\mu\) may be any real number. In the other three cases, the response is nonnegative and its mean must be positive. For the ordinary Poisson distribution, the variance equals the mean, which fixes \(\phi=1\); allowing an arbitrary positive dispersion while retaining the standard Poisson support and mass function would change the model.

The open interval \(1<p<2\) is important: it gives the compound Poisson-gamma family, not the Poisson or gamma boundary case. In that family, a random number of positive gamma contributions are added. No contributions produces an exact zero; one or more contributions produce a positive response. The detailed distribution of its positive part is not needed to understand the support distinction.

Theorem: In the standard Gaussian, Poisson, compound Poisson-gamma, and gamma Tweedie cases, the power ranges are respectively \(p=0\), \(p=1\), \(1<p<2\), and \(p=2\). Their supports and zero probabilities are as shown in the table. In particular, a continuous Gaussian response has no point mass at zero, whereas the Poisson and compound Poisson-gamma responses do.

Why the Zero-Outcome Distinction Holds

The table’s zero statements follow directly from the standard distributions. For a Gaussian response, a density describes probabilities over intervals; a single point has probability zero. For a Poisson response, zero is one of the possible integer counts and its probability is obtained from the mass function. For a gamma response, the density is concentrated strictly above zero. In the compound Poisson-gamma case, the event of zero total contribution is exactly the event that the count of contributions is zero.

Result: For a standard compound Poisson-gamma response with \(1<p<2\), write \(Y=\sum_{j=1}^{N} Z_j\), where \(N\) is Poisson and each \(Z_j\) is a strictly positive gamma contribution. Then \(Y=0\) exactly when \(N=0\). Consequently, \(Y\) has a point mass at zero; its positive outcomes come from one or more contributions.

The compound representation also verifies that the mean and variance match the power variance function. For this calculation, let \(N\sim\operatorname{Poisson}(\lambda)\), and let the independent contributions \(Z_j\) have a gamma distribution with shape \(\alpha\) and scale \(\gamma\). For \(1<p<2\), set

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

A gamma contribution has \(E[Z_j]=\alpha\gamma\) and \(\operatorname{Var}(Z_j)=\alpha\gamma^2\), so \(E[Z_j^2]=\alpha(\alpha+1)\gamma^2\). Conditional on \(N\), the sum has mean \(N E[Z_j]\) and variance \(N\operatorname{Var}(Z_j)\). Taking expectations and using \(\operatorname{Var}(N)=E[N]=\lambda\) gives

E[Y]=\lambda E[Z_j],\qquad \operatorname{Var}(Y)=\lambda E[Z_j^2].

Substituting the stated parameters yields the required moments:

\begin{aligned} E[Y] &=\frac{\mu^{2-p}}{\phi(2-p)} \frac{2-p}{p-1}\phi(p-1)\mu^{p-1} =\mu,\\ \operatorname{Var}(Y) &=\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}

Since every \(Z_j\) is strictly positive, a sum with \(N\geq1\) is positive, while the empty sum for \(N=0\) is zero. Thus \(P(Y=0)=P(N=0)=e^{-\lambda}>0\). This proves both the support statement and the presence of the zero point mass. \(\square\)

Worked Examples

Worked Example: Choosing a Model for Count Data

Suppose a response is the number of clinic visits in a fixed period. A Poisson model has mean \(\mu=2.5\). Find its variance and probability of zero visits.

For the Poisson power \(p=1\), the ordinary model has \(\phi=1\), so \(\operatorname{Var}(Y)=\phi\mu^p=1(2.5)=2.5\). Its mass function gives

P(Y=0)=\frac{e^{-2.5}(2.5)^0}{0!}=e^{-2.5}\approx 0.0821.

The response takes nonnegative integer values, and zero is an ordinary possible count with positive probability. This is different from a continuous response that can be arbitrarily close to zero but has no point mass there.

Worked Example: A Compound Poisson-Gamma Response

Take \(p=1.5\), \(\mu=4\), and \(\phi=0.5\). Calculate the compound Poisson and gamma parameters, and check the mean, variance, and zero probability.

The formulas give

\lambda=\frac{4^{0.5}}{0.5(0.5)}=8,\qquad \alpha=\frac{0.5}{0.5}=1,\qquad \gamma=0.5(0.5)4^{0.5}=0.5.

Thus \(N\sim\operatorname{Poisson}(8)\), and each contribution is gamma with shape \(1\) and scale \(0.5\). The mean of one contribution is \(\alpha\gamma=0.5\), and its second moment is \(\alpha(\alpha+1)\gamma^2=1(2)(0.25)=0.5\). Therefore,

E[Y]=8(0.5)=4,\qquad \operatorname{Var}(Y)=8(0.5)=4.

The power variance formula independently gives \(\phi\mu^p=0.5(4^{1.5})=0.5(8)=4\), agreeing with the compound calculation. The total is zero exactly when \(N=0\), so

P(Y=0)=e^{-8}\approx 0.000335.

This probability is small but positive. The model can produce exact zeros as well as positive continuous outcomes.

Worked Example: A Gamma Response Cannot Equal Zero

Let \(Y\) have a gamma distribution with \(\mu=10\) and \(\phi=0.2\), corresponding to power \(p=2\). Express its shape and scale, and determine whether an exact zero is possible.

For a gamma distribution with shape \(k\) and scale \(\theta\), the mean is \(k\theta\) and the variance is \(k\theta^2\). Matching \(E[Y]=\mu\) and \(\operatorname{Var}(Y)=\phi\mu^2\) gives \(k=1/\phi=5\) and \(\theta=\phi\mu=2\). Indeed, \(k\theta=5(2)=10\) and \(k\theta^2=5(4)=20=\phi\mu^2\).

A gamma variable with these positive parameters takes values in \((0,\infty)\), so \(P(Y=0)=0\). The distribution has positive density at positive response values, but its support does not include zero as an attainable outcome.

Worked Example: Continuous Support Is Not a Zero Point Mass

Let \(Y\sim N(3,4)\), which is the Gaussian case \(p=0\), \(\mu=3\), and \(\phi=4\). The response can be negative, zero, or positive because its support is the real line. Nevertheless, a Gaussian variable has a density and no point mass, so \(P(Y=0)=0\).

For example, the probability of lying in a small interval around zero is positive:

P(-0.1<Y<0.1)= \int_{-0.1}^{0.1}\frac{1}{2\sqrt{2\pi}} \exp\left(-\frac{(y-3)^2}{8}\right)\,dy>0.

That interval probability is not \(P(Y=0)\). The distinction between an interval and a single value is essential when interpreting observed zeros.

Common Mistake: “Zero Is in the Support, So It Has Positive Probability”

This reasoning is incorrect for a continuous distribution. For a Gaussian response, zero lies in the support because every open interval around zero has positive probability, but the probability of the single value zero is still zero. Conversely, a gamma response has strictly positive support, so an exact zero is not an outcome of the model. Poisson and compound Poisson-gamma models differ from both: they assign a positive probability to zero.

A useful modeling check is therefore to ask two separate questions: Can the response take the value zero? and Does the model assign positive probability to exactly zero? Do not infer the second answer merely from the variance power, or from the fact that zero is a boundary or limit point.

Key takeaway: The standard power cases connect the variance function to both parameter constraints and response support: Gaussian at \(p=0\), Poisson at \(p=1\), compound Poisson-gamma for \(1<p<2\), and gamma at \(p=2\). Only the Poisson and compound Poisson-gamma cases among these assign positive probability to an exact zero.

Check Your Understanding

Use the power range, admissible parameters, and support together to answer each question.

Which of the four families has a real-valued response that can be negative, and can it assign positive probability to exactly zero?
For a Poisson response with mean \(\mu=3\), state the dispersion parameter and calculate \(P(Y=0)\).
Why does the compound Poisson-gamma range exclude the endpoints \(p=1\) and \(p=2\)? Name the standard family at each endpoint.
A continuous response is observed to equal zero exactly. Which of the four standard families can assign that observation positive probability?
For the compound Poisson-gamma model, explain why a zero total response occurs precisely when there are no positive contributions.