Identifiability of Power and Dispersion

Why Identifiability Is a Practical Issue

For a Tweedie model with power \(p\) between 1 and 2, the power controls how the variance grows with the mean, while the dispersion scales the variance. Both parameters also affect the distribution’s pattern of zeros and positive values. This creates a practical challenge: different power-dispersion pairs can describe some prominent features of the same data quite similarly.

It is useful to separate two questions. Structural identifiability asks whether different parameter values imply different population distributions. Practical, or weak, identification asks whether the available data distinguish those distributions precisely. A parameter pair can be structurally identifiable and still have a nearly flat likelihood profile, broad uncertainty, or strong dependence between its estimates.

Definition: A parameter is structurally identifiable in a model if equality of the probability distributions it specifies implies equality of that parameter. It is weakly identified by a particular data set when the likelihood changes little across a substantial range of its values, often because another parameter can adjust to compensate.

The discussion below concerns independent Tweedie observations with means \(\mu_i>0\), a common power \(1<p<2\), and dispersion \(\phi>0\), where the means are treated as fixed when describing the likelihood. In a fitted model, estimating the mean structure adds further uncertainty and can make identification weaker.

What the Power and Dispersion Control

In this range of powers, a Tweedie response has a point mass at zero and a continuous distribution over positive values. Its mean and variance are

E[Y]=\mu,\qquad \operatorname{Var}(Y)=\phi\mu^p.

The compound Poisson-gamma representation makes the zero probability explicit. A response is the sum of a Poisson number of gamma-distributed positive increments. If \(N\) is the number of increments, then

N\sim\operatorname{Poisson}(\lambda),\qquad \lambda=\frac{\mu^{2-p}}{\phi(2-p)}.

The response is zero exactly when \(N=0\). Thus the probability of zero is \(P(Y=0)=e^{-\lambda}\), and the probability of a positive response is \(1-e^{-\lambda}\). Increasing \(\lambda\) makes a zero less likely. Both \(p\) and \(\phi\) affect \(\lambda\), so the zero frequency alone need not distinguish them.

The positive increments also depend on \(p\) and \(\phi\). In particular, their gamma shape is \((2-p)/(p-1)\), and their scale is \(\phi(p-1)\mu^{p-1}\). Consequently, the distribution of positive response sizes contains information beyond the probability of observing a zero.

Key distinction: Zero indicators principally reveal the Poisson intensity \(\lambda\). The magnitudes of positive outcomes contribute information about the positive-value distribution as well. Reliable separation of \(p\) and \(\phi\) generally depends on using both kinds of information.

Structural Identifiability from the Full Distribution

For a single Tweedie distribution with \(1<p<2\), the power and dispersion are structurally identifiable when the full distribution is known. The mean, variance, and zero probability together determine them.

Theorem: Suppose \(Y\) follows a Tweedie distribution with \(\mu>0\), \(\phi>0\), and \(1<p<2\). The distribution of \(Y\) uniquely determines \(\mu\), \(p\), and \(\phi\).

Proof. The distribution determines its mean \(\mu\) and variance \(V=\operatorname{Var}(Y)\). It also determines its zero probability \(q=P(Y=0)\), which satisfies \(0<q<1\). The Poisson intensity is therefore \(\lambda=-\log(q)>0\). Using \(V=\phi\mu^p\),

\lambda V =\frac{\mu^{2-p}}{\phi(2-p)}\phi\mu^p =\frac{\mu^2}{2-p}.

It follows that

p=2-\frac{\mu^2}{\lambda V}, \qquad \phi=\frac{V}{\mu^p}.

Both expressions are uniquely determined by the distribution’s mean, variance, and zero probability. Hence no two distinct admissible pairs \((p,\phi)\) give the same full distribution at the same mean. \(\square\)

This theorem concerns a population distribution, not the precision of estimates from a finite sample. A sample may contain too few zeros, too few positive outcomes, or too little variation among positive values to estimate both parameters precisely.

Worked Examples: What the Data Can and Cannot Separate

Worked Example: Equal Zero Probabilities, Different Positive-Value Variation

Let \(\mu=2\), and choose parameter pairs so that \(\lambda=1\). Since \(\phi=\mu^{2-p}/((2-p)\lambda)\), at \(p=1.5\) the dispersion is

\phi=\frac{2^{0.5}}{0.5} \approx 2.8284.

At \(p=1.8\), it is

\phi=\frac{2^{0.2}}{0.2}\approx 5.7435.

Both models have \(P(Y=0)=e^{-1}\approx0.3679\), because they have the same \(\lambda\). Their variances, however, differ. For the first pair,

\operatorname{Var}(Y)=2.8284(2^{1.5})=2.8284(2.8284)\approx8.

For the second pair, the variance is \(20\). This follows directly from \(\operatorname{Var}(Y)=\phi\mu^p\), or from \(\lambda\operatorname{Var}(Y)=\mu^2/(2-p)\): with \(\lambda=1\), \(\mu=2\), and \(p=1.8\), it is \(4/0.2=20\). Thus the same zero probability can accompany very different positive-response variation. The observed positive magnitudes can help distinguish the models; the zero pattern alone cannot.

Worked Example: Translating a Fitted Distribution into Positive-Response Moments

Take \(\mu=2\), \(p=1.8\), and

\phi=\frac{2^{0.2}}{0.2\log(2.5)}\approx6.26820.

Then \(\lambda=\log(2.5)\), so \(P(Y=0)=e^{-\lambda}=0.40\), and the positive-response probability is \(r=P(Y>0)=0.60\). The marginal variance is

\operatorname{Var}(Y)=\frac{2^{0.2}}{0.2\log(2.5)}2^{1.8} \approx21.8271.

For any nonnegative response with mean \(\mu\) and positive probability \(r\), \(E[Y\mid Y>0]=\mu/r\). Here that conditional mean is \(2/0.60\approx3.3333\). Also, because \(Y=0\) off the positive event,

\operatorname{Var}(Y\mid Y>0) =\frac{\operatorname{Var}(Y)+\mu^2}{r} -\left(\frac{\mu}{r}\right)^2.

Substituting the values gives

\operatorname{Var}(Y\mid Y>0) =\frac{20/\log(2.5)+4}{0.60} -\left(\frac{2}{0.60}\right)^2 \approx31.9341.

The zero probability summarizes how often outcomes are positive, while the conditional mean and variance describe the positive amounts. In data, these are useful fitted summaries to inspect together. The calculation is a moment identity; in a sample, substituting empirical moments gives descriptive estimates, not an independent proof that the fitted Tweedie parameters are correct.

Worked Example: A Zero-Indicator Likelihood Has a Power-Dispersion Ridge

Suppose \(n=20\) observations share the known mean \(\mu=2\), and 8 are zero while 12 are positive. Consider only the zero-versus-positive indicators. If \(r=P(Y>0)=1-e^{-\lambda}\), their likelihood, apart from a binomial coefficient that does not depend on the parameters, is

L_{\mathrm{ind}}(r)\propto(1-r)^8r^{12}.

The log-likelihood is \(8\log(1-r)+12\log r\). Its derivative is \(-8/(1-r)+12/r\), which is zero at \(r=12/20=0.60\); its second derivative is negative for \(0<r<1\). Thus the indicator likelihood is maximized at \(r=0.60\), or \(\lambda=-\log(0.40)=\log(2.5)\).

For every \(p\) in \((1,2)\), choosing

\phi(p)=\frac{2^{2-p}}{(2-p)\log(2.5)}

gives this same intensity and hence the same zero-indicator likelihood. For example, the corresponding dispersions at \(p=1.5\) and \(p=1.8\) are approximately \(3.087\) and \(6.268\), respectively. The positive values themselves may favor one pair over the other, but the indicator component cannot do so. This is a concrete likelihood ridge: the indicators identify \(\lambda\), not \(p\) and \(\phi\) separately.

Power Profiles and Fitted Behavior

A profile likelihood for \(p\) fixes the power in turn and maximizes the likelihood over the dispersion and any other parameters being treated as nuisance parameters. In notation, with fixed means, one may write

\ell_{\mathrm{prof}}(p)=\sup_{\phi>0}\ell(p,\phi).

The profile records how well each power can fit after dispersion has had a chance to adjust. A sharply peaked profile indicates that nearby powers cannot be rescued by changing dispersion without a substantial loss of fit. A broad, flat, or irregular profile indicates that the observed data distinguish the candidate powers weakly, though it does not by itself establish exact non-identifiability.

The profile should be considered alongside the fitted behavior that produced it. Useful checks include:

Zero frequency: compare observed and fitted proportions of zeros, overall and across groups with different fitted means.
Positive amounts: inspect whether fitted positive-response means and spread reflect the observed magnitudes, rather than only matching the number of zeros.
Parameter trade-off: record the fitted \(\phi\) at each profiled \(p\). A large change in dispersion accompanying small profile-likelihood changes is evidence of weak separation.
Uncertainty and stability: examine the profile’s curvature and whether reasonable changes in the data or starting values produce materially different estimates.

Zero counts can be informative when there are enough observations and the fitted means are well determined. Positive responses are especially important for distinguishing models that give similar zero probabilities but different positive-value distributions. If there are few zeros, or if positive values occupy a narrow range, the information that separates the parameters may be limited.

A Useful Moment Identity and a Common Pitfall

The conditional-moment calculations used above follow from a general identity. Let \(Y\geq0\), let \(r=P(Y>0)>0\), and suppose \(E[Y^2]\) is finite. Since \(Y=0\) whenever \(Y\) is not positive,

E[Y\mid Y>0]=\frac{E[Y]}{r}, \qquad E[Y^2\mid Y>0]=\frac{E[Y^2]}{r}.

Subtracting the square of the conditional mean from the conditional second moment gives

\operatorname{Var}(Y\mid Y>0) =\frac{\operatorname{Var}(Y)+(E[Y])^2}{r} -\left(\frac{E[Y]}{r}\right)^2.

This identity explains why the positive-response spread depends on both the marginal variance and the positive fraction; using the marginal variance as if it were the variance among positive values is generally wrong.

Common mistake: “The zero fraction determines the Tweedie power because the zero probability depends on \(p\).” This reasoning overlooks that the zero probability depends on \(\lambda=\mu^{2-p}/\{\phi(2-p)\}\), which also depends on \(\phi\). With fixed \(\mu\), a change in \(p\) can be offset by a change in \(\phi\) to preserve \(\lambda\). The correction is to assess zero behavior and positive-response magnitudes together, and to use the profile likelihood to see how much the full data distinguish the alternatives.

Key takeaway: The full Tweedie distribution identifies \(p\) and \(\phi\) in principle, but finite data may identify them weakly. Zero frequencies constrain the Poisson intensity; positive-response sizes and their variation help separate power from dispersion. A profile likelihood makes the remaining trade-off visible.

Check Your Understanding

Use the definitions and relationships in this tutorial to answer the following.

For fixed \(\mu\), why can two different power-dispersion pairs have the same probability of zero?
Which three features of the full distribution determine \(\mu\), \(p\), and \(\phi\) in the structural-identifiability argument?
What does a broad profile likelihood for \(p\) suggest, and what does it not prove?
Why can a model match the observed zero proportion yet describe positive amounts poorly?
State the relationship between marginal variance and variance conditional on a positive response, including the role of \(r=P(Y>0)\).