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Power Variance Functions · Tutorial 632 of 981

Comparing Power Indices for Prediction

Compare candidate Tweedie powers by out-of-sample performance, then check whether their zero and positive-outcome predictions are calibrated.

Advanced 12 min read

What You'll Learn

  • Define held-out predictive scoring for candidate Tweedie powers
  • Explain when log predictive densities are comparable
  • Derive the Tweedie zero probability and conditional positive mean
  • Check calibration of predicted zero probabilities and positive amounts
  • Distinguish predictive power selection from likelihood-based inference

Prediction Is Not the Same Question as Inference

A Tweedie power affects the shape of the response distribution as well as how its variance changes with the mean. When several powers seem plausible, a useful question is which fitted model predicts new responses more accurately. That is a predictive question: its answer depends on performance for observations that were not used to fit the candidate model.

For semicontinuous outcomes with both exact zeros and positive amounts, a single overall score can conceal an important failure. A model may predict the positive amounts well but misstate how often zeros occur, or get the zero frequency right while predicting positive amounts poorly. Comparing powers therefore calls for both an overall held-out score and targeted calibration checks.

Definition: A held-out predictive score evaluates predictions for observations excluded from the fitting of the model being scored. A proper scoring rule is a score whose expected value is optimized, among the available predictive distributions, by reporting the true distribution. For negative log predictive density, smaller scores are better.

Write \(p\) for the Tweedie power and consider candidates \(1<p<2\), the range with a point mass at zero and a continuous distribution on positive values. For each candidate, estimate the regression and dispersion parameters using training data only. For a held-out response \(y_i\) and its covariates \(x_i\), let \(f_{p,i}(y_i)\) denote the candidate's predictive density or mass, evaluated at the response using the fitted parameters. The held-out mean negative log score is

$$ S_p=-\frac{1}{m}\sum_{i=1}^m \log f_{p,i}(y_i), $$

where \(m\) is the number of held-out observations. A smaller \(S_p\) indicates better average predictive performance on that held-out set. If using cross-validation, each observation is scored using a model fitted without that observation's fold; the fold scores can then be averaged.

What the Predictive Score Measures

For \(1<p<2\), the Tweedie response distribution has a mixed form: an atom at zero and a density for positive outcomes. A common dominating measure is the sum of a point mass at zero and Lebesgue measure on the positive half-line. With this convention, \(f(0)\) means the probability \(P(Y=0)\), while \(f(y)\) for \(y>0\) is a density. Both are evaluated as contributions to a log score. The positive density has units determined by the response scale, so scores are comparable only when candidates use the same response scale and scoring convention.

The score should assess the full predictive distribution, not just a fitted mean. A log score uses the Tweedie mass at zero for a zero response and the Tweedie density for a positive response. In practice, the density may be evaluated by an exact series or an approximation; whichever method is used should be applied consistently to all candidates.

Worked Example: Computing a Held-Out Log Score

Suppose four held-out responses are \(0,2,0,1\). Two candidate models have the following predictive probabilities or densities, already evaluated at the respective observations:

ResponseCandidate ACandidate B
00.300.20
20.120.16
00.300.20
10.200.15

For A, the sum of log predictive contributions is \(2\log(0.30)+\log(0.12)+\log(0.20)\), which is approximately \(-2.40795-2.12026-1.60944=-6.13765\). Thus \(S_A=6.13765/4\approx1.5344\).

For B, the sum is \(2\log(0.20)+\log(0.16)+\log(0.15)\), approximately \(-3.21888-1.83258-1.89712=-6.94858\). Thus \(S_B=6.94858/4\approx1.7371\). On these held-out observations, A has the smaller mean negative log score. This comparison alone does not establish that A will predict better in every new sample.

Why the Log Score Rewards Honest Predictive Distributions

The reason to use a proper score is that it does not reward a model merely for making confident predictions that happen to fit the training data. The following result states the population property of the negative log score. It applies to discrete, continuous, and mixed outcomes when their distributions have densities with respect to a common dominating measure.

Theorem: Let \(f\) be the true density with respect to a measure \(\nu\), and let \(q\) be a candidate predictive density. Assume the relevant expected log scores are defined, allowing an infinite score if \(q=0\) on a set of positive probability under \(f\). Then \(E_f[-\log q(Y)]\ge E_f[-\log f(Y)]\). Equality holds only when \(q=f\) almost everywhere under \(\nu\).
1
Express the score difference.
When \(q>0\) wherever \(f>0\), the difference is \(\int f\log(f/q)\,d\nu\). If \(q=0\) on a set with positive \(f\)-probability, the candidate's expected score is infinite and the inequality holds.
2
Apply a logarithm inequality.
For \(t>0\), \(\log t\le t-1\). On the set where \(f>0\), take \(t=q/f\). Then \(f\log(q/f)\le q-f\).
3
Integrate and conclude.
Integrating gives \(\int f\log(q/f)\,d\nu\le\int_{\{f>0\}}q\,d\nu-1\le0\), since \(q\) integrates to one. Hence \(\int f\log(f/q)\,d\nu\ge0\). Equality requires \(q=f\) almost everywhere; otherwise the logarithm inequality is strict on a set of positive \(f\)-probability, or \(q\) puts mass outside the support of \(f\). \(\square\)

This result concerns expected performance under the true distribution. A finite held-out sample is noisy, so a lower observed score is evidence for better prediction, not a guarantee. Repeated folds or an independent test set can help assess whether a difference is stable.

Separate Calibration Checks for Zeros and Positive Amounts

A log score summarizes the entire predictive distribution, but it does not by itself explain where a candidate succeeds or fails. In the compound Poisson-Gamma representation for \(1<p<2\), the number of positive jumps is Poisson with mean \(\lambda=\mu^{2-p}/\{\phi(2-p)\}\), where \(\mu=E[Y]\) and \(\phi\) is the dispersion parameter. The response is zero exactly when there are no jumps. This gives a direct zero-probability prediction.

Result: For a Tweedie response with \(1<p<2\), mean \(\mu>0\), and dispersion \(\phi>0\), \(P(Y=0)=\exp\{-\mu^{2-p}/[\phi(2-p)]\}\). Its conditional mean given a positive response is \(E[Y\mid Y>0]=\mu/\{1-P(Y=0)\}\).

To verify the first formula, if \(N\) is the Poisson jump count, then \(P(Y=0)=P(N=0)=e^{-\lambda}\). For the second, \(Y=0\) on \(N=0\), so \(E[Y]=E[Y\,1_{\{Y>0\}}]=P(Y>0)E[Y\mid Y>0]\). Since \(E[Y]=\mu\) and \(P(Y>0)=1-P(Y=0)\), rearrangement gives the stated conditional mean.

Worked Example: Deriving Zero and Positive Predictions

Consider \(\mu=4\), \(\phi=0.5\), and \(p=1.8\). The Poisson mean is

$$ \lambda=\frac{4^{0.2}}{0.5(0.2)} =\frac{1.3195079\ldots}{0.1} =13.195079\ldots . $$

Thus \(P(Y=0)=e^{-13.195079\ldots}\approx0.00000186\), and \(P(Y>0)\approx0.99999814\). The predicted conditional mean among positive outcomes is \(4/0.99999814\approx4.0000074\). The conditional mean is slightly above the unconditional mean because the unconditional mean also includes the zero outcomes.

For a set \(G\) of observations with similar predicted zero probabilities \(q_i=P(Y_i=0\mid x_i)\), compare the observed zero fraction with the average predicted probability, \(|G|^{-1}\sum_{i\in G}q_i\). Calibration means that these quantities agree across groups, not merely in the full sample. For positive amounts, compare observed positive outcomes with their predicted conditional distributions or conditional means. When predictions vary within a group, compare observed positive totals and counts with their corresponding expectations rather than treating a single unweighted average as exact.

Worked Example: Checking Zero-Probability Calibration

In a group of ten held-out cases, four responses are zero. Candidate A predicts \(q_i=0.4\) for each case, while candidate B predicts \(q_i=0.2\) for each. The observed zero fraction is \(4/10=0.4\), so A's average predicted zero probability matches this group exactly. B's is \(0.2\), which underpredicts the zero fraction by \(0.2\).

The Brier score for the zero indicator is the average squared error. With zero coded as 1, A's score is \([4(1-0.4)^2+6(0-0.4)^2]/10=[4(0.36)+6(0.16)]/10=2.4/10=0.24\). B's score is \([4(1-0.2)^2+6(0-0.2)^2]/10=[4(0.64)+6(0.04)]/10=2.8/10=0.28\). A is better on this zero-indicator score in this group. This does not compare the candidates' predictions for positive amounts.

Worked Example: Checking the Positive-Amount Prediction

Suppose five cases share \(\mu=2\), \(\phi=0.5\), and \(p=1.5\). Then \(\lambda=2^{0.5}/[0.5(0.5)]=1.4142136/0.25=5.6568542\), giving \(q=P(Y=0)=e^{-5.6568542}\approx0.00349\). The predicted positive conditional mean is \(2/(1-0.00349)\approx2.0070\). Across the five cases, the expected number of positive responses is approximately \(5(0.99651)=4.98255\), and the expected positive total is \(5(2)=10\).

Suppose the observed responses are \(1,2,2,3,2\). All five are positive, and their total is \(10\), with observed positive mean \(10/5=2\). These values are close to the model's expected positive total and conditional mean. The group is small, however, and the observed zero count of zero is not itself evidence of a calibration failure: the predicted expected zero count is only \(5(0.00349)\approx0.01745\).

Predictive Power Selection Versus Likelihood Inference

A power chosen to minimize a held-out score is a tuning choice for prediction. It need not equal the power that maximizes the likelihood for a particular observed sample, and neither choice automatically quantifies uncertainty about the power. Profile likelihood, likelihood-based estimation, and uncertainty assessment answer inferential questions; held-out scoring asks which fitted predictive procedure performs better on excluded observations.

For a fair predictive comparison, the full fitting procedure must be repeated using only the training portion of each split. This includes estimating regression coefficients and dispersion, and choosing any preprocessing or tuning settings. If the power itself is selected using the held-out responses and those same responses are then used to report performance, the reported score is optimistic: the holdout has influenced the choice it is supposed to evaluate.

A common mistake is to compare training log-likelihoods across powers and call the largest one the best predictor. A training likelihood rewards fit to the observations used for estimation and does not measure out-of-sample performance. Another mistake is to compare only predicted means. Different powers can yield similar means while assigning very different probabilities to zero or different distributions to positive values. Use a held-out score for overall prediction, then inspect zero and positive calibration to understand the result.

Key takeaway: Select a Tweedie power for prediction by comparing consistently computed held-out predictive scores, with model fitting confined to each training set. Separately check zero-probability and positive-outcome calibration. Treat the selected predictive power as a choice of prediction procedure, not as a likelihood-based estimate with inferential uncertainty.

Check Your Understanding

Use these questions to check that you can distinguish predictive comparison from calibration and inference.

  1. Why must the model being scored be fitted without using the held-out responses?
  2. For a mixed Tweedie response, what predictive contribution is used when the held-out response is zero, and what contribution is used when it is positive?
  3. How would you compare observed and predicted zero frequencies within a group of cases?
  4. Explain why a well-calibrated positive conditional mean does not guarantee that the zero probabilities are calibrated.
  5. What is the difference between selecting a power by held-out predictive score and estimating it by profile likelihood?
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