Influence From Claims, Exposure, and Weights

Why Influence Has More Than One Source

An observation can matter to a ratemaking GLM for different reasons. A large claim count may be surprising relative to its expected count. A high-exposure policy may contribute substantial information because many risk-years were observed. A severity observation may receive a weight representing several claims, increasing its contribution to the fit and diagnostics. These are distinct mechanisms, even when they occur together.

The central distinction is between how unusual an observation is under the fitted model and how much it can affect the fitted model. Residuals primarily describe lack of fit at an observation; exposure and weights also affect the fitting information. A large raw count is not automatically a large residual, and a small residual does not guarantee negligible influence.

Definition: In a frequency GLM, an observation's exposure is the amount of time or risk-unit basis over which its claim count is observed. It is commonly included as an offset: for count $N_i$, exposure $e_i$, and rate $\lambda_i$, the fitted mean is $\mu_i=e_i\lambda_i$. A weight is a separate multiplier in the fitting or diagnostic contribution, with its meaning determined by the model and data construction. Exposure is not automatically an observation weight.

Frequency Models: Counts, Exposure, and Residuals

Consider independent Poisson claim counts $N_i\sim\operatorname{Poisson}(\mu_i)$, with $\log(\mu_i)=\log(e_i)+\boldsymbol{x}_i^\mathsf{T}\boldsymbol{\beta}$. The term $\log(e_i)$ is an offset: its coefficient is fixed at one. The remaining linear predictor models the claim rate per unit exposure. This model assumes that, conditional on the rating variables, the count mean is proportional to exposure and the Poisson variance equals the mean. If overdispersion or other dependence is material, the Poisson variance assumption may be inadequate.

For this model, the response residual is $N_i-\mu_i$, the Pearson residual is $(N_i-\mu_i)/\sqrt{\mu_i}$, and the Poisson deviance contribution is $2[N_i\log(N_i/\mu_i)-(N_i-\mu_i)]$ when $N_i>0$. When $N_i=0$, the term $N_i\log(N_i/\mu_i)$ is defined as zero. The signed deviance residual is the square root of the contribution, with the sign of $N_i-\mu_i$.

A high-exposure observation can have a high expected count simply because it represents more risk-time. Therefore, the count itself is not a fair measure of unusualness: compare it with its fitted mean, or compare claim frequency $N_i/e_i$ with the fitted rate. Exposure also affects how much information the observation supplies to estimate the rate.

Worked Example: A Large Count That Is Not the Most Unusual

A hypothetical frequency model uses a fixed fitted rate of $0.5$ claims per exposure unit. Policy A has exposure $e_A=1$ and observes $N_A=2$ claims. Policy B has exposure $e_B=8$ and observes $N_B=6$. Calculate the fitted means, Pearson residuals, and signed deviance residuals. Which observation is more unusual by these residual measures?

Method. Use the offset mean $\mu=e\lambda$, then calculate each residual relative to its own fitted mean. This avoids treating a larger raw count as automatically more surprising.

For A, $\mu_A=1(0.5)=0.5$. Its Pearson residual is $ (2-0.5)/\sqrt{0.5}=1.5/0.707107=2.1213$. Its deviance is $2[2\log(2/0.5)-(2-0.5)]=2[2\log(4)-1.5]$. Since $\log(4)=1.386294$, the deviance is $2[2.772589-1.5]=2.545177$, and the signed deviance residual is $+\sqrt{2.545177}=+1.5954$.

For B, $\mu_B=8(0.5)=4$. Its Pearson residual is $(6-4)/\sqrt{4}=1.0000$. Its deviance is $2[6\log(6/4)-(6-4)]=2[6\log(1.5)-2]$. Since $\log(1.5)=0.405465$, the deviance is $2[2.432791-2]=0.865581$, giving signed deviance residual $+\sqrt{0.865581}=+0.9304$.

Policy B has the larger observed count, but Policy A has the larger Pearson and deviance residual magnitudes. B's six claims are assessed against an expected count of four, while A's two claims are assessed against only one-half. Check: both Pearson residuals are positive because both observed counts exceed their fitted means; both deviance residuals have the same sign.

Common mistake avoided: ranking observations by raw claim count. The model assesses counts relative to exposure-adjusted means, not in isolation.

Why Exposure Changes the Fit

The distinction between a residual and a fitting contribution follows directly from the Poisson likelihood. For one observation, omitting terms that do not depend on the model parameters, the log-likelihood is $N_i\log(\mu_i)-\mu_i$. With $\log(\mu_i)=\log(e_i)+\boldsymbol{x}_i^\mathsf{T}\boldsymbol{\beta}$, differentiation with respect to $\boldsymbol{\beta}$ gives the score contribution below. The expected information contribution follows by taking the negative expected second derivative.

\frac{\partial \ell_i}{\partial\boldsymbol{\beta}} =\boldsymbol{x}_i(N_i-\mu_i), \qquad \mathcal{I}_i =\mu_i\boldsymbol{x}_i\boldsymbol{x}_i^\mathsf{T}.

Thus exposure affects the score through $\mu_i=e_i\lambda_i$, and it affects the information through that same fitted mean. It does not appear as an additional multiplier on the score residual. Holding the rate and rating vector fixed, increasing exposure increases the expected count and the information about the rate. Whether the observation has a large realized score also depends on the difference $N_i-\mu_i$.

For an intercept-only Poisson rate model, the score equation is $\sum_i(N_i-\mu_i)=0$. Because $\mu_i=e_i\lambda$, solving gives $\widehat{\lambda}=\sum_iN_i/\sum_ie_i$, the total count divided by total exposure. This is the exposure-weighted rate, not the unweighted average of the individual rates.

Worked Example: A High-Exposure Record and the Fitted Rate

A hypothetical intercept-only Poisson model has three aggregated records: record 1 has exposure $1$ and $1$ claim; record 2 has exposure $4$ and $8$ claims; record 3 has exposure $1$ and $0$ claims. Fit the rate, calculate the score and information contributions, and assess the effect of excluding record 2.

Method. Fit the common rate by total claims divided by total exposure. Then calculate each fitted mean and score contribution. Each row is an aggregate record, so excluding record 2 removes its full four exposure units.

The total claim count is $1+8+0=9$, and total exposure is $1+4+1=6$. Therefore, $\widehat{\lambda}=9/6=1.5$ claims per exposure unit. The fitted means are $\mu_1=1(1.5)=1.5$, $\mu_2=4(1.5)=6$, and $\mu_3=1(1.5)=1.5$.

The three score contributions for the intercept are $N_i-\mu_i$: record 1 contributes $1-1.5=-0.5$, record 2 contributes $8-6=2$, and record 3 contributes $0-1.5=-1.5$. Their sum is $-0.5+2-1.5=0$, as required at the fitted intercept. The information contributions are $\mu_i$: $1.5$, $6$, and $1.5$, for a total of $9$.

Record 2's Pearson residual is $(8-6)/\sqrt{6}=2/2.449490=0.8165$. Despite having the largest claim count and the largest positive score contribution, its Pearson residual is less than one because its expected count is also large. If record 2 is excluded, the remaining records have $1$ claim over $2$ exposure units, so the refitted rate is $1/2=0.5$. The rate changes from $1.5$ to $0.5$, showing that the high-exposure aggregate has substantial potential to affect the fit.

Check: the full-data estimate reproduces the totals, since $6(1.5)=9$. The reduced estimate also reproduces the remaining data, since $2(0.5)=1$. Common mistake avoided: interpreting a modest standardized residual as proof that a record cannot influence the fitted coefficients. Residual size and information leverage are related but are not the same quantity.

Severity Weights: What Is Being Multiplied?

Severity models require careful interpretation of weights. A frequency weight $w_i$ can represent $w_i$ replicated observations with the same response and covariates. In a weighted GLM, that replication multiplies the observation's score and deviance contribution by $w_i$. This is different from a frequency-model exposure offset: an offset changes the mean through a known exposure amount, while a frequency weight multiplies a likelihood contribution.

For a Gamma GLM with log link, positive severity response $Y_i$, mean $\mu_i$, and variance $\operatorname{Var}(Y_i)=\phi\mu_i^2$, the unit deviance is $d_i=2[(Y_i-\mu_i)/\mu_i-\log(Y_i/\mu_i)]$. With a frequency weight $w_i$, the contribution is $w_id_i$, and the signed deviance residual is $\operatorname{sign}(Y_i-\mu_i)\sqrt{w_id_i}$. This interpretation assumes the row represents replicated independent Gamma observations with the same response and predictors. Other weight types can encode different variance assumptions and should not automatically be interpreted as replication.

Worked Example: Replicated Severity Claims and Weighted Deviance

A hypothetical Gamma severity model with log link has an intercept only. Four claim-level severities are $ \$1{,}000$, $ \$2{,}000$, $ \$2{,}000$, and $ \$3{,}000$. The two identical middle observations are stored as one row with frequency weight $2$; the other rows have weight $1$. Estimate the fitted mean and calculate each distinct row's weighted deviance contribution and signed deviance residual.

Method. For this Gamma log-link intercept model, the score equation is $\sum_iw_i(Y_i/\mu-1)=0$, so $\widehat{\mu}=\sum_iw_iY_i/\sum_iw_i$. Then apply the Gamma unit deviance and multiply by the frequency weight.

The weighted severity total is $1(1000)+2(2000)+1(3000)=8000$, and the total weight is $1+2+1=4$. Thus $\widehat{\mu}=8000/4=\$2{,}000$. The score contributions are $1(1000/2000-1)=-0.5$, $2(2000/2000-1)=0$, and $1(3000/2000-1)=0.5$; their sum is zero.

For the $ \$1{,}000$ row, the unit deviance is $2[(-1000/2000)-\log(1000/2000)] = 2[-0.5+\log(2)] = 0.386294$. Its weight is $1$, so its weighted contribution is $0.386294$, and its signed deviance residual is $-\sqrt{0.386294}=-0.6215$.

For the $ \$2{,}000$ row, the observed value equals the fitted mean. Its deviance contribution and residual are both zero, even though its frequency weight is $2$. For the $ \$3{,}000$ row, the unit deviance is $2[0.5-\log(1.5)]=2[0.5-0.405465]=0.189070$. Its weight is $1$, so the weighted contribution is $0.189070$, and its signed residual is $+\sqrt{0.189070}=+0.4348$.

The total deviance is $0.386294+0+0.189070=0.575364$. Check: the total weighted score is zero at the estimated mean, and the total deviance equals the sum of the three row contributions. Common mistake avoided: treating the weight as a change to the fitted severity itself. Here it increases the replicated row's contribution; the fitted mean is determined by the weighted data as a whole.

Interpreting Influence in Practice

A useful diagnostic interpretation separates three questions:

Is the outcome unusual? Compare the observation with its fitted mean using an appropriate residual. In frequency models, exposure is already reflected in the mean; a raw count alone is not a residual diagnostic.
How much information does it carry? In the Poisson model, information depends on the fitted mean and the rating vector. High exposure can produce substantial information even when the observed count is close to expectation.
What does its weight mean? Determine whether the weight represents replicated observations or a different variance convention. A frequency weight scales score and deviance contributions; it is not interchangeable with an offset.

Cook’s distance and leverage, covered in the earlier material on GLM influence, combine aspects of how an observation’s fitted values or coefficients may change. The present distinctions help explain why those diagnostics should be read alongside residuals and the data construction. A large claim can have a large residual but modest leverage; a high-exposure record can have a modest residual yet carry substantial information; and a weighted severity row can have a larger diagnostic contribution because it represents multiple claims.

These interpretations depend on the model being appropriate. For example, if counts are overdispersed relative to the Poisson assumption, Poisson Pearson residual scaling and information calculations may understate uncertainty. If a severity weight is not a replication weight, multiplying the deviance contribution by it may not have the intended statistical interpretation. Always identify the response unit, exposure unit, distribution, and weight definition before comparing diagnostics.

Key takeaway: Large claims, high exposure, and weights influence a GLM through different mechanisms. Use residuals to judge departure from fitted expectations, account for exposure in the mean and information, and interpret weights according to how the data were constructed.

Check Your Understanding

For each problem, distinguish the fitted mean or rate from the observation's residual or weighted contribution.

A Poisson frequency model predicts $0.4$ claims per exposure unit. A policy has exposure $5$ and observes $4$ claims. Find its fitted mean, Pearson residual, and signed deviance residual. (Answer: $\mu=2$, Pearson residual $=1.4142$, deviance residual $=+1.2430$)
An intercept-only Poisson model has records $(N,e)=(2,2),(9,3),(1,1)$. Find the fitted rate and the rate after excluding the middle record. (Answer: full rate $=2.000$; excluding the middle record $=1.000$ claims per exposure unit)
A Gamma log-link intercept model groups exact replicated claim severities: $ \$800$ with weight $1$, $ \$1{,}200$ with weight $2$, and $ \$2{,}000$ with weight $1$. Find the fitted mean severity. (Answer: $\$1{,}300$)
At a fixed Poisson rate of $0.8$, compare a record with $e=10,N=9$ and one with $e=1,N=2$ using Pearson residuals. Which has the larger count, and which is more unusual by absolute Pearson residual? (Answer: the first has the larger count, $9$ versus $2$; the second has the larger absolute Pearson residual, $1.3416$ versus $0.3536$)
In a Gamma severity model, a row has $Y=\mu$ and frequency weight $w=5$. What are its weighted deviance contribution and signed deviance residual? (Answer: both are $0$)

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