Support and Boundary Pitfalls in Tweedie Fits

Why Support Checks Matter

A fitted mean can look plausible while the model assigns probability zero to an observed response. This is not a minor numerical imperfection: it means the proposed distribution cannot generate the data. Tweedie models require particular care because changing the power can change the support from a lattice to a continuous positive range, or to a distribution that allows an atom at zero.

The response support is determined by the distribution, including its dispersion, not merely by the fitted mean or the link function. A log link makes a fitted mean positive; it does not make negative observations valid, create a point mass at zero, or turn a continuous density into a point probability.

Power Determines Which Outcomes Are Possible

For a Tweedie exponential-dispersion model, write the mean as \(\mu>0\), dispersion as \(\phi>0\), and variance function as \(\operatorname{Var}(Y)=\phi\mu^p\). The power \(p\) is not just a way to describe how variance grows: it also determines the response distribution’s form and support. The following cases are especially important when checking data.

• At \(p=1\), \(Y=\phi N\), where \(N\) is Poisson with mean \(\mu/\phi\). Thus the support is \(\{k\phi:k=0,1,2,\ldots\}\), not necessarily the nonnegative integers.
• For \(1<p<2\), the distribution is compound Poisson–Gamma. It has an atom at zero and a density on positive responses.
• At \(p=2\), the distribution is Gamma, with support \(y>0\); an exact zero has probability zero.
• For \(p>2\), the usual Tweedie distributions have continuous positive support. They do not include an atom at zero.

These are distinct support statements, not interchangeable descriptions of a generic “nonnegative” response. In particular, a value can be nonnegative and still be impossible under the selected model. Also, the model’s dispersion matters at \(p=1\): treating that case as an ordinary Poisson distribution with integer-valued outcomes is only correct when \(\phi=1\).

The Zero Atom and Positive Density

When \(1<p<2\), the compound Poisson–Gamma representation makes the zero outcome precise. Let \(\lambda=\mu^{2-p}/\{\phi(2-p)\}\), let \(N\sim\operatorname{Poisson}(\lambda)\), and, conditional on \(N=j\geq1\), let \(Y\) be the sum of \(j\) independent Gamma severities. Each severity has shape \(\alpha=(2-p)/(p-1)\) and scale \(\theta=\phi(p-1)\mu^{p-1}\). If \(N=0\), the sum is exactly zero; if \(N\geq1\), it is positive with a continuous distribution.

To verify the result, \(P(N=0)=e^{-\lambda}\), which gives the atom. For each \(j\geq1\), the sum of \(j\) independent Gamma variables with common scale \(\theta\) has a Gamma distribution with shape \(j\alpha\) and scale \(\theta\). Its density at \(y>0\) is \(y^{j\alpha-1}e^{-y/\theta}/\{\Gamma(j\alpha)\theta^{j\alpha}\}\). The event \(N=j\) has probability \(e^{-\lambda}\lambda^j/j!\). Multiplying that probability by the conditional density and summing over all \(j\geq1\) gives the displayed density. The \(j=0\) term is not part of this density: it is the separate probability mass at zero. \(\square\)

Densities Are Not Point Probabilities

For a continuous response, \(f(y)\) is a density, not \(P(Y=y)\). In fact \(P(Y=y)=0\) for every single point \(y\) under an absolutely continuous distribution, even when \(f(y)>0\). A density has units inverse to the response units and can be greater than one. To obtain a probability, integrate it over an interval. In a mixed distribution, use the probability mass for an exact zero and the density for a positive observation.

A useful formal way to understand the mixed case is to use a dominating measure that combines a point mass at zero with Lebesgue measure on the positive line. The likelihood contribution is then the mass \(P(Y=0)\) when the observation is zero, and the positive-part density when the observation is positive. Calling both contributions “densities” can conceal a support error; treating the continuous density as a point probability is also incorrect.

Boundary Values in Deviance and Likelihood Calculations

The Tweedie unit deviance is often used in fitting and diagnostics. For \(p\ne1,2\) and \(y\geq0\), its usual expression is \(d_p(y,\mu)=2\{y^{2-p}/[(1-p)(2-p)]-y\mu^{1-p}/(1-p)+\mu^{2-p}/(2-p)\}\), with the endpoint powers defined by their appropriate limits. At \(p=1\), \(d_1(y,\mu)=2\{y\log(y/\mu)-(y-\mu)\}\), using \(0\log(0/\mu)=0\). At \(p=2\), \(d_2(y,\mu)=2\{\log(\mu/y)+(y-\mu)/\mu\}\), for \(y>0\).

For \(1<p<2\), inserting \(y=0\) into the first expression is well-defined: \(y^{2-p}\) and \(y\mu^{1-p}\) both vanish because \(2-p>0\). Thus \(d_p(0,\mu)=2\mu^{2-p}/(2-p)\), a finite deviance value consistent with zero being allowed. As \(p\) approaches 2 from below, this expression diverges. At \(p=2\), zero is outside the Gamma support, so the divergence is not a numerical value to repair by substitution.

At \(p=1\), the limit \(y\log y=0\) as \(y\downarrow0\) gives \(d_1(0,\mu)=2\mu\). This finite deviance does not remove the \(p=1\) lattice restriction: a non-lattice observation still has probability zero. More generally, a finite deviance at a boundary does not by itself establish that an observation is in the distribution’s support.

The exact likelihood and the deviance are not interchangeable when comparing powers. A deviance expression may omit or suppress terms that vary with the power, and a density-series normalization cannot be discarded merely because it is inconvenient. At \(p=1\), use the appropriate Poisson probability on the scaled lattice; for \(1<p<2\), use the zero mass or the positive density as appropriate; at \(p=2\), use the Gamma density only for positive observations. Do not evaluate a formula derived for one support at an endpoint where the distribution has changed.

Common Mistakes and a Support-First Check

• Assuming \(p=1\) always means integer outcomes. With arbitrary \(\phi>0\), outcomes are multiples of \(\phi\). Check \(y/\phi\), not whether \(y\) itself is an integer.
• Using a positive density for an exact zero. In the compound Poisson–Gamma case, zero contributes \(e^{-\lambda}\); it is not evaluated by inserting zero into the positive-density series.
• Allowing zeros under a Gamma model because the mean is positive. A positive mean does not imply a point mass at zero. Gamma support is strictly positive.
• Reading a density value as a probability. The density at a point can exceed one; probabilities require an interval integral.
• Assuming finite deviance guarantees a possible observation. Support is a property of the distribution, and a finite deviance formula does not override it.
• Stabilizing a boundary by silently changing the data. Adding a small constant to zeros changes the response and the likelihood. It is not a correction to an incompatible model.

A reliable check proceeds in order: identify the candidate power and dispersion; write the corresponding support; check every observed value against that support; then choose the correct likelihood contribution for a zero, a lattice observation, or a positive continuous observation. Only after these checks should numerical optimization or model comparison be trusted.