Why the Pattern of Residuals Matters
A Poisson frequency model can show more claim-count variation than the model predicts, but one overall measure of dispersion does not explain why. The deviations might be concentrated in particular rating classes, increase with exposure, or follow a pattern over time. Each pattern points to different questions about the specification or the data. Diagnosis begins by examining where the model misses, not only how much the residual variation adds up to.
This tutorial assumes a fitted Poisson frequency model with a log link and an exposure offset. Conditional on the modeled rating variables and exposure, the model treats observations as independent Poisson counts. The earlier material on Poisson frequency models and overdispersion establishes the count mean and variance used here. The goal here is to interpret residual patterns: they can suggest a source of excess variation, but do not by themselves prove a cause.
For a Poisson observation, the model specifies \(E[Y_i\mid\text{model predictors}]=\hat{\mu}_i\) and \(\operatorname{Var}(Y_i\mid\text{model predictors})=\hat{\mu}_i\). The residual therefore expresses the observed-minus-fitted difference in units of the model's predicted standard deviation. A residual near zero is a small discrepancy relative to that predicted variability; its sign and its pattern across observations are also important.
What Residual Patterns Can Reveal
For independent Poisson counts, a common overall descriptive measure is the Pearson sum of squares, \(\sum_i r_i^2=\sum_i (Y_i-\hat{\mu}_i)^2/\hat{\mu}_i\). It can be compared with residual degrees of freedom as a broad check on dispersion. But this sum discards the signs, ordering, and labels of the residuals. Different patterns can produce a similar total, so it cannot identify a source of excess variation on its own.
Useful views keep that information visible. Examine residuals by rating class, by fitted mean or exposure, and in time order. In these views, look for sustained positive or negative deviations, systematic changes across a variable, unusually many large residuals, and sequences of residuals with the same sign. Also check that the model's predicted mean has been calculated for the relevant exposure: with a log link and an exposure offset, the expected count grows with exposure, so raw counts are not directly comparable across unequal terms.
One justification for the Pearson scaling is direct. Under the fitted Poisson model, \(E[Y_i-\hat{\mu}_i]=0\) and \(\operatorname{Var}(Y_i-\hat{\mu}_i)=\hat{\mu}_i\), treating the fitted mean as specified for this calculation. Consequently, dividing the difference by \(\sqrt{\hat{\mu}_i}\) gives a residual with model-based variance one:
This standardization makes residuals from observations with different fitted means more comparable. It does not make them exactly normally distributed, especially when expected counts are small, and it does not remove the effects of estimating model parameters. Treat residual magnitudes as diagnostics rather than as automatic, exact significance tests.
Worked Examples: Read the Residuals in Context
Worked Example: Opposing Deviations Across Rating Classes
A hypothetical insurer groups policies into two rating classes. A fitted Poisson model predicts 40 claims in Class A and 60 in Class B; the observed counts are 50 and 50. Calculate the class residuals and their combined Pearson sum of squares. What does the total count conceal?
Method. Treat each class count as a Poisson count with its stated fitted mean. Calculate each observed-minus-fitted difference first, then divide by the square root of the fitted mean.
For Class A, the difference is \(50-40=10\), so \(r_A=10/\sqrt{40}=10/6.3246=1.5811\).
For Class B, the difference is \(50-60=-10\), so \(r_B=-10/\sqrt{60}=-10/7.7460=-1.2910\).
The Pearson sum of squares is \(r_A^2+r_B^2=10^2/40+(-10)^2/60=2.5000+1.6667=4.1667\). The observed total is \(50+50=100\); the fitted total is \(40+60=100\). Thus, the totals agree exactly even though each class has a nonzero residual.
Interpretation. The total alone hides opposing class deviations: Class A is above its fitted count and Class B is below. This pattern warrants checking the class-level fit and the specification of the class relativities; it does not establish why those differences occurred.
Common mistake avoided. The Class A residual numerator is not 50; it is the observed-minus-fitted difference \(50-40=10\). Using 50 would calculate the wrong residual and the wrong Pearson sum of squares.
Worked Example: Residuals Across Exposure Levels
A hypothetical book is modeled at an annual claim frequency of 0.12, with expected count equal to rate times exposure. For low-, medium-, and high-exposure groups, exposures are 100, 200, and 400 policy-years. The observed counts are 8, 25, and 70. Calculate each fitted count and Pearson residual, then assess the pattern.
Method. Apply the exposure offset through \(\hat{\mu}=0.12e\), where \(e\) is policy-years. Compare each observed count with its own exposure-adjusted fitted count.
For low exposure, \(\hat{\mu}=0.12(100)=12\). The residual is \((8-12)/\sqrt{12}=-4/3.4641=-1.1547\).
For medium exposure, \(\hat{\mu}=0.12(200)=24\). The residual is \((25-24)/\sqrt{24}=1/4.8990=0.2041\).
For high exposure, \(\hat{\mu}=0.12(400)=48\). The residual is \((70-48)/\sqrt{48}=22/6.9282=3.1754\).
The observed rates are \(8/100=0.0800\), \(25/200=0.1250\), and \(70/400=0.1750\) claims per policy-year, compared with the modeled rate of 0.12. The residual pattern moves from negative to strongly positive as exposure increases. The Pearson contributions are \(16/12=1.3333\), \(1/24=0.0417\), and \(484/48=10.0833\), for a total of \(11.4583\).
Interpretation. The high-exposure group contributes most of the discrepancy, and the increasing observed rates invite investigation of exposure-related mix, the exposure definition, and any rating differences not represented in the model. Three grouped observations do not identify which explanation is correct.
Common mistake avoided. Comparing the raw counts 8, 25, and 70 would make the high-exposure group look larger simply because it has more exposure. The diagnosis compares each count with its own offset-adjusted mean, or equivalently examines rates while accounting for their different statistical precision.
Worked Example: A Time-Ordered Pattern
A hypothetical portfolio has four quarterly claim counts. A fitted model predicts 20 claims in each quarter after accounting for exposure. The observed counts are 15, 18, 25, and 32. Calculate the residual sequence and Pearson sum of squares. What should be investigated?
Method. Since each fitted mean is 20, each residual is the observed-minus-fitted difference divided by \(\sqrt{20}=4.4721\).
Quarter 1: \((15-20)/\sqrt{20}=-5/4.4721=-1.1180\).
Quarter 2: \((18-20)/\sqrt{20}=-2/4.4721=-0.4472\).
Quarter 3: \((25-20)/\sqrt{20}=5/4.4721=1.1180\).
Quarter 4: \((32-20)/\sqrt{20}=12/4.4721=2.6833\).
The residuals rise in order from negative to positive. Their squared sum is \(25/20+4/20+25/20+144/20=1.25+0.20+1.25+7.20=9.90\).
Interpretation. The sequence is more informative than the sum alone: later quarters increasingly exceed the fitted counts. Investigate whether the model has omitted a time-related change, whether exposure or data processing changed, or whether the sequence is consistent with chance variation. The pattern may motivate further analysis, but four quarters do not prove a trend or establish temporal clustering.
Common mistake avoided. Rearranging these four counts would leave the Pearson sum of squares unchanged but destroy the time-order information. Preserve the order when diagnosing a possible temporal pattern.
Checking an Excess of Zero Counts
An excess of zero-claim risks is one possible residual pattern, but the zero indicator has a different variance from the claim count itself. Under the assumptions that each risk has an independent Poisson count with mean \(\mu\), the probability of zero claims is \(p=e^{-\mu}\). The indicator that a risk has zero claims is Bernoulli with variance \(p(1-p)\). Therefore, for \(n\) independent risks with the same fitted mean, the number of zero-count risks has expected value \(np\) and variance \(np(1-p)\).
Worked Example: Is the Zero Count Unusually High?
A hypothetical portfolio contains 100 independent risks, each with a fitted Poisson mean of 0.20 claims. There are 90 observed zero-claim risks. Calculate the model's expected number of zeros and the standardized deviation of the observed zero count.
Method. Convert the fitted claim-count mean into a zero probability, then treat each risk's zero/nonzero status as a Bernoulli outcome.
For one risk, \(p=e^{-0.20}=0.8187308\). The expected number of zero-count risks is \(100p=81.8731\). The variance of that number is \(100p(1-p)=100(0.8187308)(0.1812692)=14.8411\). Its standard deviation is \(\sqrt{14.8411}=3.8524\). Thus the standardized deviation is \((90-81.8731)/3.8524=2.1096\).
Interpretation. The observed number of zeros is about 2.11 standard deviations above the model expectation under the stated assumptions. This is a reason to inspect the model and the risk mix, not proof of a separate zero-generating mechanism or of model failure.
Common mistake avoided. The denominator is not \(\sqrt{81.8731}\). That would treat the zero count as Poisson. It is a sum of Bernoulli indicators, so its variance is \(100p(1-p)\), not \(100p\).
A Practical Diagnostic Sequence
Check the model's rating variables, link, and exposure offset before interpreting residuals. An incorrect exposure basis can create apparent residual patterns.
Use the Pearson residual to retain both direction and scale. Keep observed-minus-fitted differences visible so sign errors are less likely.
Compare residuals across rating classes and exposure ranges, and preserve time order for temporal checks. Use fitted-count denominators for grouped independent Poisson counts.
Describe the evidence and plausible checks. Do not claim that a residual pattern uniquely identifies heterogeneity, clustering, or a particular model defect.
Unexplained differences concentrated in rating classes may indicate that the model does not distinguish risks adequately. Differences that grow with exposure may point to an exposure or composition issue, although large-exposure groups can also have more visible absolute count differences. A time sequence with runs or a trend can prompt checks for changes in the book or possible dependence. A Poisson model's assumption of independent counts matters to these interpretations: if events are related, the residual pattern may not behave as the model expects. Diagnosing dependence requires more than labeling a portfolio overdispersed.
A common exam and practice error is to treat an overall Pearson statistic as a diagnosis. It summarizes squared discrepancies, but loses whether residuals are positive or negative, which class contributes, and whether they occur in sequence. Another is to apply a Poisson variance to a different diagnostic quantity, such as the number of zero-count risks. First identify the random variable being summarized; then use its model-based variance.
Check Your Understanding
For each problem, calculate the requested quantity and state the main diagnostic implication.
- A fitted model predicts 36 claims for a rating class, and 42 are observed. Calculate the Pearson residual. (Answer: \(6/\sqrt{36}=1.0000\); the observed count is one fitted standard deviation above expectation.)
- Two independent Poisson class counts have fitted means 25 and 75 and observed counts 20 and 80. Calculate both residuals and their Pearson sum of squares. (Answer: \(-1.0000\) and \(5/\sqrt{75}=0.5774\); sum of squares \(1+25/75=1.3333\), with opposing class deviations.)
- A model has a rate of 0.08 claims per policy-year. Three exposure groups have 50, 150, and 250 policy-years, with observed counts 3, 16, and 17. Calculate fitted counts and residuals. (Answer: fitted counts \(4,12,20\); residuals \(-0.5000,\ 1.1547,\ -0.6708\). The middle group is above expectation, but the pattern is not steadily increasing with exposure.)
- Four successive periods have fitted counts of 12 each and observed counts of 8, 11, 14, and 19. Calculate the residual sequence and Pearson sum of squares. (Answer: residuals \(-1.1547,-0.2887,0.5774,2.0207\); sum of squares \(70/12=5.8333\). The sequence rises, warranting a check for temporal structure without proving its cause.)
- For 20 independent risks, each has fitted Poisson mean \(\ln 2\); 14 risks have zero claims. Calculate the expected number of zeros and the standardized deviation. (Answer: \(p=e^{-\ln 2}=0.5\), expected zeros \(20(0.5)=10\), variance \(20(0.5)(0.5)=5\), and standardized deviation \((14-10)/\sqrt{5}=1.7889\).)