Why zeros and large losses deserve separate attention
A GLM can have residuals that average near zero and still describe important parts of the outcome distribution poorly. In ratemaking, two patterns warrant particular attention: more zero outcomes than the fitted model predicts, and a small number of unusually large positive losses. Either pattern can be obscured by a reasonable overall average. Both can also arise for reasons other than a wrong distribution, so the diagnostic task is to identify a persistent mismatch rather than label an observation as problematic from one residual alone.
The response residual \(y_i-\hat{\mu}_i\) compares an observed outcome with its fitted mean. A deviance residual preserves the sign of that difference while measuring its size using the model’s likelihood-based deviance contribution. A zero outcome therefore commonly produces a negative residual, whereas a large outcome relative to its fitted mean produces a positive one. The sizes and repetition of those residuals, interpreted against the assumed distribution and fitted means, are more informative than their signs alone.
Residual patterns under count models
For a Poisson GLM, assume observations are conditionally independent and \(Y_i\mid x_i\sim\operatorname{Poisson}(\mu_i)\), with fitted mean \(\hat{\mu}_i\). The response residual is \(y_i-\hat{\mu}_i\). The Poisson deviance contribution and signed deviance residual are
At \(y_i=0\), the term \(y_i\log(y_i/\hat{\mu}_i)\) is defined by its limit, zero. Thus \(D_i=2\hat{\mu}_i\) and \(r_{D,i}=-\sqrt{2\hat{\mu}_i}\). A zero for a risk with a very small predicted mean is not as surprising as a zero for one with a large predicted mean. Residual plots should therefore be read against fitted values or relevant rating variables, not as though every zero had the same expected behavior.
A Poisson model also makes a direct prediction about the number of zeros. For independent observations with fitted means \(\hat{\mu}_i\), the probability of a zero for observation \(i\) is \(e^{-\hat{\mu}_i}\). Consequently, the expected number of zeros is \(\sum_i e^{-\hat{\mu}_i}\), and the variance of that number, conditional on the fitted means, is \(\sum_i e^{-\hat{\mu}_i}(1-e^{-\hat{\mu}_i})\). These calculations provide a useful reference, although fitted means have been estimated from data, so the resulting standardized difference is a diagnostic approximation rather than a formal test that accounts for all model fitting uncertainty.
Worked Example: More Zeros and Larger Positive Counts Than a Poisson Model Predicts
A hypothetical insurer fits a Poisson frequency GLM to 100 independent, equal-exposure policies. The fitted mean is \(0.4\) claims per policy for every policy. The observed counts are 80 zeros and 20 policies with two claims each. Assume the Poisson model is the proposed conditional distribution and there are no exposure differences or offsets omitted from the fitted means. There is no discounting or mortality assumption in this frequency analysis. Assess the zero pattern and calculate the residuals for a zero and for a two-claim observation.
Step 1: Check the overall mean. The total observed claim count is \(80(0)+20(2)=40\), so the observed mean is \(40/100=0.4\), equal to the fitted mean. The mean alone therefore does not reveal the distributional pattern.
Step 2: Calculate the expected number of zeros. Each policy has fitted zero probability \(e^{-0.4}=0.6703\). The expected zero count is \(100(0.6703)=67.03\). The variance of the zero count is \(100(0.6703)(1-0.6703)=22.11\), so its standard deviation is \(\sqrt{22.11}=4.70\). The observed count exceeds expectation by \(80-67.03=12.97\), or approximately \(12.97/4.70=2.76\) standard deviations.
Step 3: Calculate the residuals. For a zero, the response residual is \(0-0.4=-0.4\). Its deviance contribution is \(2(0.4)=0.8\), so its deviance residual is \(-\sqrt{0.8}=-0.8944\). For a two-claim observation, the response residual is \(2-0.4=1.6\). Its deviance contribution is \(2[2\log(2/0.4)-(2-0.4)]=2[2\log(5)-1.6]=3.2378\), giving deviance residual \(+\sqrt{3.2378}=1.7994\).
Interpretation and check. The total count agrees with the fitted mean, but the data put that total into fewer, higher-count policies than the Poisson model’s pattern suggests. The zero-count comparison and the positive residuals together indicate a possible distributional mismatch, such as unmodeled heterogeneity or a process that generates extra zeros and a heavier positive-count tail. They do not establish which explanation is correct. As a check, the observed total remains \(40\), and the fitted total is \(100(0.4)=40\). Method and mistake avoided: compare both the mean and the distribution of counts; do not conclude that the Poisson model fits merely because the totals match.
Residual patterns under severity models
For a Gamma GLM, assume positive observations are conditionally independent with mean \(\mu_i\) and variance \(\phi\mu_i^2\), where \(\phi>0\) is the dispersion parameter. Its unit deviance contribution is \(2[(y_i-\mu_i)/\mu_i-\log(y_i/\mu_i)]\). The conventional unscaled deviance residual is the signed square root of that contribution. A Pearson residual scaled by the model standard deviation is \((y_i-\mu_i)/(\sqrt{\phi}\mu_i)\). These formulas apply to positive Gamma responses; a Gamma model does not assign positive probability to an exact zero.
A large positive severity response residual identifies a loss above its fitted mean, but does not by itself show that the model has a heavy tail. The fitted mean may be wrong for the risk, a predictor effect may be missing, or the outcome may be recorded incorrectly. Evidence for a tail mismatch is stronger when unusually large positive residuals recur more often than the assumed distribution predicts, particularly across relevant rating groups. A small number of isolated extremes can also make a residual plot look striking without establishing a systematic pattern.
Worked Example: Interpreting a Large Severity Residual
A hypothetical property-severity model uses a Gamma distribution with conditional mean \(\mu=20\) thousand dollars and dispersion \(\phi=0.25\). For one policy, the observed loss is \(y=70\) thousand dollars. Assume the Gamma mean and variance specification is the proposed model, and the loss is positive and measured in thousands of dollars throughout. No interest-rate or mortality assumption is relevant. Calculate the response residual, scaled Pearson residual, and deviance residual, and interpret them.
Step 1: Response residual. The response residual is \(70-20=50\) thousand dollars. The loss is 3.5 times its fitted mean, since \(70/20=3.5\).
Step 2: Pearson residual. The assumed variance is \(\phi\mu^2=0.25(20^2)=100\) in squared thousands of dollars, so the standard deviation is \(\sqrt{100}=10\) thousand dollars. The scaled Pearson residual is \(50/10=5.00\).
Step 3: Deviance residual. The Gamma unit deviance is \(2[(70-20)/20-\log(70/20)]=2[2.5-\log(3.5)]=2[2.5-1.2528]=2.4945\). The unscaled deviance residual is \(+\sqrt{2.4945}=+1.5794\). If a dispersion-scaled deviance residual is reported, it is \(1.5794/\sqrt{0.25}=3.1589\). The positive sign follows from \(70>20\).
Interpretation and check. This observation is large relative to the model’s standard deviation: its scaled Pearson residual is 5.00. It merits investigation, but one large loss cannot establish that the severity distribution has a systematically heavy tail. Check whether comparable large positive residuals recur and whether the assumed distribution is plausible for the positive losses being modeled. The response residual, \(50\) thousand dollars, also agrees with the direct subtraction \(70-20\). Method and mistake avoided: state whether a deviance residual is scaled; do not compare an unscaled deviance residual directly with a dispersion-scaled Pearson residual as if the scales were identical.
Check whether the pattern persists across rating groups
When fitted means vary, an overall zero count can conceal where the mismatch occurs. Under independent Poisson outcomes, calculate the expected zeros from each fitted mean, then compare observed and expected zero counts by meaningful rating groups. A surplus concentrated in one group may point to a missing predictor effect, an exposure or data issue, or a conditional distribution problem. A surplus that persists across groups after the mean structure is credible gives stronger reason to question the assumed count distribution.
Worked Example: Separating a Group-Level Mean Check From a Zero Check
A hypothetical frequency model assigns 50 equal-exposure policies a fitted Poisson mean of \(0.2\) and another 50 a fitted mean of \(0.8\). Observed counts are 45 zeros in the first group and 29 in the second. The observed total counts are 10 and 40, respectively. Assume independent Poisson outcomes conditional on these fitted means, with no omitted exposure differences. There is no discounting or mortality assumption. Compare the zero pattern with the model’s prediction and check the overall mean.
Step 1: Calculate group-level expected zeros. In the first group, the expected number is \(50e^{-0.2}=50(0.8187)=40.94\). In the second group it is \(50e^{-0.8}=50(0.4493)=22.47\). The total expected count is \(40.94+22.47=63.40\), compared with \(45+29=74\) observed zeros.
Step 2: Quantify the overall discrepancy. The variance of the zero count is \(50(0.8187)(1-0.8187)+50(0.4493)(1-0.4493)=7.42+12.37=19.79\). Its standard deviation is \(\sqrt{19.79}=4.45\). The observed excess is \(74-63.40=10.60\), or \(10.60/4.45=2.38\) standard deviations.
Step 3: Check the mean and inspect the groups. The observed total is \(10+40=50\), and the fitted total is \(50(0.2)+50(0.8)=50\), so the overall mean matches. The zero excess occurs in both groups: \(45-40.94=4.06\) in the first and \(29-22.47=6.53\) in the second. The overall excess is a useful signal, not a definitive significance test, because model parameters were estimated and the fitted mean structure may still be wrong.
Interpretation and check. Matching totals do not remove the zero-pattern concern: the count is distributed differently between zero and positive outcomes than the Poisson predictions suggest. Because the surplus appears in both groups, a single group-specific mean issue is less immediately evident, but the model and data still require investigation. The total observed count and fitted expected count both equal 50. Method and mistake avoided: use each group’s own fitted zero probability; do not apply one average fitted mean to policies with different predicted means.
How to use these patterns responsibly
A residual plot can reveal where a proposed model struggles, but it cannot uniquely identify the replacement distribution. Before attributing a pattern to excess zeros or heavy tails, check the response definition, exposure and offset, and whether the fitted means are systematically wrong across rating variables. Then compare the observed pattern with what the assumed conditional distribution predicts. For counts, that includes the frequency of zeros and the distribution among positive counts. For severity, examine positive-tail residuals and the assumed variance and tail behavior.
- Many negative residuals at zero outcomes: compare the number and location of zeros with their predicted probabilities; a negative residual for one zero is not itself evidence of excess zeros.
- A few large positive severity residuals: investigate the observations and examine whether large residuals recur; do not let a single extreme value decide the distribution.
- Mean calibration with a residual pattern: recognize that matching the average does not guarantee a correct outcome distribution.
- A pattern in one segment: check the mean structure and relevant rating variables before treating the pattern as a global distributional failure.
Check Your Understanding
For each problem, use the stated model assumptions and distinguish a residual calculation from a conclusion about overall fit.
- A Poisson model predicts \(\hat{\mu}=0.3\) for each of 60 independent policies. There are 50 observed zeros. Calculate the expected number of zeros, its standard deviation under the fitted model, and the observed excess in standard-deviation units. (Answer: Expected zeros \(=60e^{-0.3}=44.45\); standard deviation \(=\sqrt{60e^{-0.3}(1-e^{-0.3})}=3.39\); excess \(=(50-44.45)/3.39=1.64\) standard deviations.)
- For a Poisson GLM with fitted mean \(0.6\), calculate the response and deviance residuals for an observation of zero claims. (Answer: Response residual \(=-0.6\); deviance residual \(=-\sqrt{2(0.6)}=-1.0954\).)
- A Gamma severity model has \(\mu=15\) thousand dollars and \(\phi=0.16\). An observed loss is \(60\) thousand dollars. Calculate the scaled Pearson residual and the dispersion-scaled deviance residual. (Answer: Scaled Pearson residual \(=(60-15)/(\sqrt{0.16}\,15)=7.50\); unit deviance \(=2[3-\log(4)]=3.2274\), so the dispersion-scaled deviance residual is \(\sqrt{3.2274}/\sqrt{0.16}=4.49\), positive.)
- A Poisson model with fitted mean \(0.5\) observes a count of 4. Calculate its response, Pearson, and deviance residuals. What does their sign say, and what can it not establish by itself? (Answer: Response residual \(=3.5\); Pearson residual \(=3.5/\sqrt{0.5}=4.9497\); deviance residual \(=\sqrt{2[4\log(8)-3.5]}=3.104\), positive. The observation exceeds its fitted mean; one observation alone cannot establish a systematic heavy-tail mismatch.)
- A frequency model has equal exposure and fitted Poisson mean \(0.2\) for one group and \(0.8\) for another. Each group contains 50 policies. The observed zero counts are 45 and 29. State the expected total number of zeros and one appropriate diagnostic interpretation. (Answer: \(50e^{-0.2}+50e^{-0.8}=63.40\) expected zeros versus 74 observed. The surplus occurs in both groups and merits follow-up, but does not alone prove a distributional mismatch.)