Why Outcome Support Is the First Check
Before fitting a generalized linear model, identify what values the response can take and how each observation was produced. A model for whether an event occurred has a different response structure from a model for how many events occurred. Both differ from a model for a positive dollar amount. A plausible response family must be able to represent the observed outcomes and must make sense for the process generating them.
This is a first screening step, not a complete model specification. Outcome support can rule out a family, but it cannot by itself establish that the remaining family is appropriate. The sampling structure, dependence, exposure, and mean-variance behavior must also be plausible. Choosing a response family does not, by itself, choose a link function or establish that predictors explain the response well.
Three Common Outcome Structures
For binary outcomes, write \(Y_i=1\) if the event of interest occurs for unit \(i\), and \(Y_i=0\) otherwise. If trials are independent and have the same event probability \(p\), then \(Y_i\) has a Bernoulli distribution with parameter \(p\). Its support is \(\{0,1\}\). A binary response records one yes-or-no outcome per unit.
When the response is the number of successes among a fixed, known number \(m\) of independent trials, each with the same success probability \(p\), the count has a binomial distribution. Its support is the integers from 0 to \(m\). Thus Bernoulli data are the one-trial case; binomial data often arise when multiple binary trials are summarized as one count.
A count of events over a specified time, area, or other exposure is different. A Poisson model is a candidate when events occur independently at a stable rate over comparable exposure, and the response counts events in that exposure. Its support is the nonnegative integers: 0, 1, 2, and so on. Unlike a binomial count, it does not have a fixed maximum number of events.
For the simplest Poisson model with mean \(\lambda\), the variance is also \(\lambda\). For a rate \(\lambda\) per unit of exposure \(t\), a constant-rate model gives mean \(\lambda t\) over exposure \(t\). Those properties are checks on plausibility, not guarantees that a Poisson model fits. Clustering, dependence, changing rates, or substantially greater variability than the mean can undermine the model.
A Gamma response is a candidate for a strictly positive continuous quantity, such as a positive payment amount. It has support \(y>0\), so it does not assign probability to zero or negative values. Under a common mean-dispersion parameterization, a Gamma response with mean \(\mu>0\) and dispersion \(\phi>0\) has variance \(\phi\mu^2\). This gives a useful rough check: variability that grows with the square of the mean may be plausible for Gamma outcomes. The Gamma family does not become appropriate merely because measurements are positive; the distributional shape and variability must also be reasonable.
| Response recorded | Candidate family | Support and key condition |
|---|---|---|
| One event indicator per unit | Bernoulli | 0 or 1; one binary trial per unit |
| Success count from a fixed number of trials | Binomial | Integers from 0 to the fixed trial count |
| Event count over exposure | Poisson | Nonnegative integers; a count process with a plausible stable rate |
| Positive continuous measurement | Gamma | Strictly positive values; positive, often right-skewed outcomes |
Why Bernoulli Trials Produce a Binomial Count
The connection between a Bernoulli response and a binomial count follows directly from the way the count is formed. Assume \(m\) independent trials, each with success probability \(p\), and let \(K\) be their total number of successes. Any particular arrangement with \(k\) successes and \(m-k\) failures has probability \(p^k(1-p)^{m-k}\). There are \(\binom{m}{k}\) such arrangements, so the probability of a total of \(k\) successes is:
This derivation also shows why the fixed trial count matters: \(K\) cannot exceed \(m\). For independent Bernoulli trials, expected values and variances add. Each trial has mean \(p\) and variance \(p(1-p)\), so:
If trials have different success probabilities, the simple binomial formula does not apply; the total can still be a sum of independent indicators, but it is not generally binomial. Dependence between trials is another reason not to assume the binomial model automatically.
Worked Example: Individual Indicators or a Grouped Binomial Count?
A hypothetical insurer observes whether each of 80 similar policies has at least one reported glass claim during a six-month period. Assume the indicators are independent and each policy has the same claim probability, \(p=0.15\). The analyst needs a response family for individual policy records and for a single response formed by summing the 80 indicators. Identify both families, then calculate the mean and variance of the grouped response.
Method and assumptions. Each policy contributes one binary outcome, with values 0 and 1. Under the stated independence and common-probability assumptions, each individual outcome is Bernoulli with parameter 0.15. The sum over the fixed 80 policies is binomial with \(m=80\) and \(p=0.15\).
For an individual policy, the response is \(Y_i\sim\operatorname{Bernoulli}(0.15)\). For the grouped count \(K\), the parameter values are \(K\sim\operatorname{Binomial}(80,0.15)\). Its expected count is:
Its variance is:
The standard deviation is \(\sqrt{10.2}\approx3.194\). For a check, the mean is 12 claims across 80 policies, which corresponds to a 0.15 expected claim probability per policy. The variance is below the mean, as expected for this binomial setup. Common mistake avoided: treating the grouped count as Poisson just because it is a count. The fixed maximum of 80 and the fixed number of binary trials support a binomial response.
Counts, Exposure, and the Poisson Candidate
Poisson is a candidate when the response is a count of events, not merely because the recorded values happen to be whole numbers. The events should be defined consistently, the exposure should be meaningful, and the event-generating process should plausibly have an approximately stable rate over that exposure. If the same event cannot occur more than once per unit, the response may instead be a binary indicator. If the number of trials is fixed and each trial can succeed at most once, a binomial count may be the better description.
For comparable units with equal exposure, a Poisson model with mean \(\lambda\) predicts both mean and variance \(\lambda\). If exposures differ, expected counts differ with exposure under a stable-rate assumption; comparing raw counts without accounting for exposure can therefore mislead. A clear mismatch between observed variability and the Poisson mean is a warning to investigate assumptions. It does not, by itself, identify the correct alternative family.
Worked Example: Checking a Poisson Count Model
A hypothetical service network records the number of equipment failures at each of 16 depots during one month. All depots have the same exposure. Assume failures occur independently and at a stable rate, and consider a Poisson model. The recorded counts are 0, 1, 2, 1, 0, 3, 1, 2, 1, 0, 2, 1, 3, 2, 1, and 4. Calculate the mean and population variance, assess whether they are broadly compatible with a Poisson response, and estimate the mean count per depot-month.
Method and assumptions. These are event counts over equal exposure, with no fixed trial maximum. Under the stated stable-rate and independence assumptions, a Poisson response is a candidate. The model implies equal theoretical mean and variance; the observed values are only a diagnostic check.
There are 16 observations. Adding them gives:
The sum of squared counts is \(56\). The population variance calculated from these 16 observations is the mean squared deviation from 1.5. Using the identity that the mean squared deviation equals the mean of the squares minus the square of the mean:
The observed mean is 1.5 and the population variance is 1.25. They are in the same general range, so this small summary does not reveal a large mean-variance discrepancy. It does not prove the Poisson assumptions; independence, stability, and the event definition still need consideration. The estimated mean count is 1.5 failures per depot-month. For 40 depot-months at the same rate, the expected total would be \(40(1.5)=60\) failures. Check: the stated total of 24 divided by 16 depots gives the same estimate of 1.5 per depot-month.
Positive Continuous Outcomes and Gamma Responses
A positive continuous response can be a Gamma candidate when its values are strictly above zero and its distribution is reasonably compatible with positive, often right-skewed measurements. A Gamma response is not a direct match for a binary outcome, a count, a negative value, or a dataset containing exact zeros. In actuarial applications, it is important to distinguish the amount paid on a positive claim from the total payment recorded for every policy: the latter may include many zeros.
The mean-dispersion parameterization is useful for a rough variability check. If a positive response has mean \(\mu\) and variance \(\phi\mu^2\), then the implied dispersion is \(\phi=\operatorname{Var}(Y)/\mu^2\). Equivalently, a moment-matched Gamma shape parameter is \(1/\phi\), and its scale parameter is \(\phi\mu\). These moment calculations summarize a candidate distribution; they do not establish that individual observations follow a Gamma distribution.
Worked Example: Positive Claim Amounts With Zero Payments
A hypothetical analyst records total payment per policy for seven policies in a period: \$0, \$0, \$100, \$200, \$300, \$400, and \$500. The analyst proposes a Gamma response for all seven amounts. Assess that proposal. Then, for the positive payments only, calculate the mean and population variance and obtain moment-matched Gamma parameters using shape and scale.
Method and assumptions. A Gamma response has support strictly above zero, so it cannot represent the two observed zero payments as ordinary Gamma outcomes. If the analysis separates whether a payment occurs from the amount conditional on a positive payment, the first component is a binary indicator and the second component can be assessed as a positive continuous response. For the moment calculation, use the five positive amounts as the conditional severity observations.
There are five positive payments among seven policies. The observed positive-payment proportion is \(5/7\), while the conditional severity observations are \$100, \$200, \$300, \$400, and \$500. Their sum is \$1,500, so their mean is:
The deviations from \$300 are \(-\$200,-\$100,\$0,\$100,\$200\). Their squared deviations sum to \(40{,}000+10{,}000+0+10{,}000+40{,}000=100{,}000\) dollars squared. Dividing by five gives the population variance of these five observations:
For a Gamma distribution with shape \(k\) and scale \(\theta\), the mean is \(k\theta\) and the variance is \(k\theta^2\). Matching these two moments gives:
The moment-matched conditional candidate is therefore Gamma with shape 4.5 and scale \(200/3\) dollars. Checking the parameters, its mean is \(4.5(200/3)=\$300\), and its variance is \(4.5(200/3)^2=20{,}000\) dollars squared. Common mistake avoided: fitting one Gamma response to all seven amounts. The zeros violate Gamma support; a separate occurrence component and positive-amount component preserve the distinction in the data.
A Practical Selection Sequence
Use the response definition and sampling unit, rather than the predictor types or the appearance of a software output, to choose a candidate family. The following sequence keeps support, data structure, and variability checks in the right order.
State whether one row records an event indicator, a count over exposure, a success count from known trials, or a measured amount.
Binary values suggest Bernoulli; counts from a fixed number of trials suggest binomial; event counts suggest Poisson; strictly positive continuous values can suggest Gamma.
Ask whether trials are independent, whether the trial count is fixed, and whether event rates and exposures are comparable. For positive measurements, check that zeros and negative values are absent from the response being modeled.
Compare observed variability with what the candidate makes plausible: binomial variability depends on the success probability, Poisson variance equals its mean, and Gamma variance grows with the square of its mean at fixed dispersion.
Name the family as a candidate and identify the assumption that could make it unsuitable. Support and a rough moment check are evidence, not proof of fit.
Common Exam Traps
- Calling every integer-valued outcome Poisson. A count of successes from a fixed number of trials has a maximum and may be binomial. Check how the count was formed.
- Treating a binary indicator as a general event count. If each unit can contribute at most one event, a Bernoulli response at the individual level is natural; a fixed group total can be binomial.
- Ignoring exposure. A count observed over a longer period can be larger simply because it has more time in which events can occur. Compare rates only when the exposure is defined and handled appropriately.
- Choosing Gamma because values are positive on average. Every modeled response value must be positive for an ordinary Gamma response; a positive mean does not make zero observations compatible.
- Overinterpreting the mean-variance check. Similar sample mean and variance do not prove a Poisson process, and a plausible Gamma moment match does not prove a Gamma shape.
- Confusing family choice with the whole model. Selecting a response family does not establish independence, a suitable predictor structure, or an appropriate link.
Check Your Understanding
For each situation, identify a defensible response-family candidate and complete the requested calculation or qualification.
- A hypothetical screening program records whether each of 50 independent participants tests positive. Assume the same positive probability \(p=0.08\) for every participant. Give the individual and grouped response families, and calculate the mean and variance of the grouped count. (Answer: Individual Bernoulli; grouped \(\operatorname{Binomial}(50,0.08)\); mean \(4\), variance \(3.68\).)
- Each of 12 devices is inspected once, and the response is the number of failed components among exactly 20 components on that device. The component failures are assumed independent with a common probability. Should the response be modeled as Poisson or binomial, and why? (Answer: Binomial, because there are a fixed 20 trials per device and the count cannot exceed 20.)
- A hypothetical monitoring station records 15 events over 10 station-months of comparable exposure. Under a stable-rate Poisson assumption, estimate the rate and give the expected count over four station-months. (Answer: \(15/10=1.5\) events per station-month; expected count \(4(1.5)=6\).)
- Four strictly positive measurements are \$40, \$60, \$80, and \$100. Calculate their mean and population variance, then give the moment-matched Gamma shape and scale. (Answer: Mean \$70; variance \(500\) dollars squared; shape \(9.8\); scale \(50/7\approx\$7.14\).)
- A payment response across policies contains exact zeros and positive continuous amounts. Is one ordinary Gamma response appropriate for the complete response? State a family structure that respects the support. (Answer: No; use a Bernoulli occurrence component for zero versus positive payment and assess a Gamma candidate for the positive amounts.)