Zero-Loss Policies in Tweedie Fitting

Why Zero-Loss Policies Belong in the Data

A Tweedie pure-premium model describes aggregate loss across policies, including policies with no positive loss in the observation period. Those zero-loss policies are not empty records: under the model, they are evidence about how often positive loss occurs. Removing them changes both the observed population and the target being estimated.

The focus here is the compound Poisson-Gamma form of the Tweedie model, with variance power \(1<p<2\). The response is aggregate covered loss, and the mean is expressed per exposure unit. Exposure is incorporated explicitly, so a policy with more exposure has a different probability of recording zero loss.

In this model, a zero is a genuine outcome of the aggregate-loss distribution. It can occur because no positive covered payment arises during the period. Depending on the product and the definition of the response, a policy might have had an event but still have zero covered payment—for example, if the event produced no payment under the policy terms. The model concerns the recorded aggregate payment, not necessarily the count of all events.

For a policy with exposure \(e\), the compound Poisson construction gives $$ P(Y=0)=P(N=0)=\exp(-e\lambda). $$ This probability is implied by the model's \(\mu\), \(\phi\), and \(p\); it is not a separate zero probability freely chosen independently of those parameters. A change in any of these quantities can change the predicted zero proportion.

Deriving the Zero Probability and Aggregate Moments

The zero-probability formula follows directly from the Poisson distribution. For a Poisson count with mean \(e\lambda\), the probability of zero payments is $$ P(N=0)=\frac{(e\lambda)^0\exp(-e\lambda)}{0!}=\exp(-e\lambda). $$ Because every Gamma payment in the construction is strictly positive, the aggregate \(Y\) is zero exactly when \(N=0\). This equivalence is why the Poisson probability is the Tweedie zero probability.

The parameterization also links the compound model to the usual Tweedie mean-variance relationship. If a positive payment \(Z\) has Gamma shape and scale parameters $$ \alpha=\frac{2-p}{p-1}, \qquad \theta=\phi(p-1)\mu^{p-1}, $$ then \(E[Z]=\alpha\theta\). The compound Poisson mean per exposure unit is \(\lambda E[Z]\), which equals \(\mu\):

For a compound Poisson sum, its variance per exposure unit is \(\lambda E[Z^2]\). Substituting the Gamma second moment gives \(\phi\mu^p\). Thus the aggregate loss and the loss rate per exposure satisfy

These relationships assume the stated compound Poisson-Gamma model and a common \(\mu,\phi,p\) for the policy being considered. In a rating model, \(\mu\) may vary with policy characteristics. The model's implied zero probability then varies accordingly, and the fitted mean structure and zero pattern should be considered together.

How Zero Policies Contribute to a Fit

In an ordinary Tweedie fit to policy-level aggregate losses, a zero observation contributes to the likelihood just as a positive observation does. For a zero response, the model assigns probability \(\exp(-e\lambda)\). A collection of zero observations therefore informs whether the model predicts a plausible number of policies with no positive payment.

The zero observations also affect estimation of the pure-premium mean. To see this in a simple case, suppose all policies share one mean \(\mu\), the variance power and dispersion are fixed, and policy \(i\) has exposure \(e_i\) and loss rate \(y_i=Y_i/e_i\). The Tweedie score for the common mean is proportional to $$ \sum_i \frac{e_i(y_i-\mu)}{\phi\mu^p}. $$ Setting this score to zero yields $$ \widehat{\mu} = \frac{\sum_i e_i y_i}{\sum_i e_i} = \frac{\sum_i Y_i}{\sum_i e_i}. $$

A zero policy has \(Y_i=0\), so it adds exposure to the denominator without adding loss to the numerator. It therefore contributes to the estimated loss per exposure unit. If the zero policies are removed, the numerator and denominator are both calculated from a selected, positive-only group, and the resulting estimate targets that group rather than all policies.

This simple intercept-only result does not mean that every coefficient in a model with rating variables must move in one predictable direction when zeros are deleted. The direction and size of coefficient changes can depend on the covariate mix. The general point is that fitting only positive observations changes the data population and does not estimate the same unconditional pure-premium structure.

Exposure and Zero-Frequency Diagnostics

Exposure affects the probability of zero through the product \(e\lambda\). Two policies with the same modeled per-exposure parameters but different exposure amounts do not have the same zero probability: the policy with greater exposure has more opportunity for positive payments. A zero probability must therefore be calculated using the policy's exposure, not just \(\lambda\).

For a group of independent policies with the same exposure and model parameters, the number of zero-loss policies has a binomial distribution. Its expected value is the number of policies multiplied by the common zero probability. This gives a straightforward model diagnostic, although an observed count differing from its expectation does not by itself establish that the model is wrong.

What Changes When Zero Policies Are Removed?

Deleting zeros and fitting an ordinary Tweedie model to the remaining observations treats a selected positive-only sample as if it were the full population. That discards information about zero frequency and generally shifts the estimated response level upward. In the intercept-only example, the change is explicit: the all-policy mean is total loss divided by total exposure, whereas the positive-only mean divides the same positive losses by a smaller amount of exposure.

If the data were deliberately sampled only from policies with positive loss, the appropriate probability model is conditional on \(Y>0\), not the ordinary Tweedie distribution. For a fixed exposure and parameter set, $$ E[Y\mid Y>0] = \frac{E[Y]}{P(Y>0)} = \frac{e\mu}{1-\exp(-e\lambda)}. $$ The first equality holds because \(Y=0\) whenever it is not positive, so \(E[Y]=P(Y>0)E[Y\mid Y>0]\). This conditional expectation is not the unconditional aggregate mean \(e\mu\). A positive-only dataset therefore needs a sampling-aware or truncated likelihood if the goal is to model that selected population correctly.

A related trap is replacing zero losses with a small positive amount so a logarithm can be taken. That changes the observed response and does not preserve the Tweedie model's point mass at zero. Tweedie models with \(1<p<2\) are specifically useful in part because they accommodate both an exact zero probability and a continuous positive-loss component.

Common Exam Traps

• Dropping zero responses. The zero is a valid Tweedie outcome, not a missing observation. Removing it generally changes the population and the estimated unconditional pure premium.
• Forgetting exposure in the zero probability. Use \(\exp(-e\lambda)\), not \(\exp(-\lambda)\), when \(e\) is not one exposure unit.
• Miscomputing \(\lambda\). Evaluate the full denominator \(\phi(2-p)\) and the exponent \(\mu^{2-p}\) before calculating the exponential.
• Calling a positive-only average the pure premium for all policies. The former is conditional on a positive observation; the latter includes zeros.
• Assuming zero payment means no event of any kind. The modeled response is the aggregate covered payment. Its zero value need not imply that no event occurred.

These distinctions are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus: the central task is to keep the response, exposure, and modeled population consistent. In practical fitting, inspect whether the data represent all eligible policies or a selected subset before interpreting either the zero pattern or the pure-premium estimate.