Resolving Family and Link Choices in a Risk Modeling Case

Why Family and Link Must Be Chosen Together

A generalized linear model (GLM) has two distinct choices: a family, which describes the conditional distribution of the response, and a link, which relates its conditional mean to predictors. The outcome scale is the first clue. A yes-or-no event, a nonnegative integer count, and a strictly positive continuous cost do not have the same support, and they do not generally have the same mean–variance relationship.

A plausible family-link combination should respect the response’s possible values and give a sensible relationship between predictors and its conditional mean. That choice is a modeling assumption, not a conclusion guaranteed by the data type alone. For instance, a count outcome may be overdispersed relative to a Poisson model, and a cost variable that includes zeros cannot be modeled directly by an ordinary Gamma response model.

Definition: For response $Y_i$ and predictors $x_i$, a GLM specifies a conditional distribution for $Y_i$ and a link $g$ such that $g(\mu_i)=\eta_i$, where $\mu_i=E[Y_i\mid x_i]$ and $\eta_i=\beta_0+\beta_1x_{i1}+\cdots+\beta_px_{ip}$ is the linear predictor. The family determines the response support and variance pattern; the link determines how the conditional mean changes with the predictors.

Three Common Outcome Scales

Response scale	Common family	Typical link	Conditional variance pattern
Binary outcome or successes out of a fixed number of trials	Binomial	Logit	$np(1-p)$
Nonnegative integer event count	Poisson	Log	$\lambda$
Strictly positive continuous outcome	Gamma	Log	$\phi\mu^2$

These are standard starting points, not automatic answers. The binomial model assumes a specified number of trials and a success probability for each trial; in its usual form, trials are conditionally independent with a common probability. The Poisson model describes counts with a conditional mean equal to the conditional variance. A Gamma model describes positive continuous measurements whose variance grows with the square of their mean under the common GLM parameterization. If these assumptions are implausible, the chosen family may not represent the outcome adequately.

Parameter and link conditions: A binomial probability parameter satisfies $0\le p\le1$; the endpoints give degenerate distributions. With a finite logit linear predictor, the fitted probability is strictly between 0 and 1. A Poisson mean $\lambda$ is nonnegative; with a finite log-link predictor, the fitted mean is positive. A Gamma response is strictly positive, and its mean is positive. A log link for a Gamma model also gives a positive fitted mean for every finite predictor.

Binary Outcomes: Binomial Family and Logit Link

For one binary observation, write $Y_i=1$ for a success and $Y_i=0$ otherwise. If $Y_i\mid x_i$ is Bernoulli with success probability $p_i$, then $E[Y_i\mid x_i]=p_i$ and $\operatorname{Var}(Y_i\mid x_i)=p_i(1-p_i)$. For a count $Y_i$ of successes in $n_i$ conditionally independent trials with common success probability $p_i$, the binomial model gives mean $n_ip_i$ and variance $n_ip_i(1-p_i)$.

The logit link models the log odds as a linear predictor. It is useful because the inverse link always returns a probability between zero and one for a finite predictor:

\operatorname{logit}(p_i)=\log\left(\frac{p_i}{1-p_i}\right)=\eta_i, \qquad p_i=\frac{e^{\eta_i}}{1+e^{\eta_i}}.

For grouped binomial data, the modeled probability is $p_i$, while the expected number of successes is $n_ip_i$. Do not confuse those two quantities. The binomial trial count affects the response mean and variance; it is not itself the probability.

Worked Example: Predicting a Binary Warranty Event

A hypothetical warranty study records whether each covered device requires a repair during its first year. Suppose the binary response has a binomial distribution with $n=1$, and a fitted logistic model is $\operatorname{logit}(p)=-1.2+0.4x+0.3z$. For a device with $x=2$ and $z=1$, calculate the fitted repair probability and the conditional variance of the binary response.

Method. Use the logit link to calculate the linear predictor, transform it to a probability, and then use the Bernoulli variance $p(1-p)$.

First, $\eta=-1.2+0.4(2)+0.3(1)=-1.2+0.8+0.3=-0.1$. Thus $e^\eta=e^{-0.1}\approx0.904837$, and $p=0.904837/(1+0.904837)=0.4750$, to four decimal places. The fitted probability is therefore 0.4750.

Since $n=1$, the response variance is $p(1-p)=0.4750(0.5250)=0.249375$, or 0.2494 to four decimal places. This is a variance for the zero-or-one response, not a variance of 0.2494 dollars or repairs.

Check and common mistake. The probability is inside the required interval from zero to one, and the variance is nonnegative. A common error is to report $\eta=-0.1$ as the fitted probability. It is the log odds; applying the inverse-logit gives the probability.

Count Outcomes: Poisson Family and Log Link

A Poisson response takes values $0,1,2,\ldots$. Conditional on predictors, a Poisson model with mean $\lambda_i$ has probability mass function $P(Y_i=y\mid x_i)=e^{-\lambda_i}\lambda_i^y/y!$, for nonnegative integers $y$. Its conditional mean and variance are both $\lambda_i$. The equality of mean and variance is a substantive restriction, not merely a consequence of choosing a convenient link.

The log link gives $\log(\lambda_i)=\eta_i$, so $\lambda_i=e^{\eta_i}>0$. A predictor change of $c$ changes the mean by a multiplicative factor $e^c$, holding other predictors fixed. The model permits an observed zero count even though its fitted mean is positive: a positive expectation does not require every realized count to be positive.

Worked Example: Predicting a Count of Service Calls

A hypothetical equipment study records the number of service calls per unit during a fixed observation period. Assume the conditional count follows a Poisson distribution, and a fitted log-link model is $\log(\lambda)=0.2+0.3x-0.1z$. For a unit with $x=2$ and $z=1$, find the expected count and variance.

Method. Evaluate the linear predictor, exponentiate to obtain the Poisson mean, and use the Poisson equality of mean and variance.

The linear predictor is $\eta=0.2+0.3(2)-0.1(1)=0.2+0.6-0.1=0.7$. Therefore $\lambda=e^{0.7}\approx2.0138$, to four decimal places. The expected number of calls is 2.0138 for a unit with these predictors.

Under the stated Poisson assumption, $\operatorname{Var}(Y\mid x,z)=\lambda\approx2.0138$. The model does not predict that an individual unit makes 2.0138 calls; it predicts a conditional mean, while an observed count must be a nonnegative integer.

Check and common mistake. Exponentiating the predictor yields a positive count mean, as required. A common error is to treat 0.7 as the expected count. It is the log of the expected count, so the inverse link is necessary.

Positive Continuous Outcomes: Gamma Family and Log Link

A Gamma response is appropriate for a strictly positive continuous measurement when its distribution is positively skewed and its variability tends to increase with its mean. In a common GLM parameterization, $\operatorname{Var}(Y_i\mid x_i)=\phi\mu_i^2$, where $\phi>0$ is a dispersion parameter. The mean is $\mu_i=E[Y_i\mid x_i]$. The Gamma family does not include zero or negative response values.

A log link, $\log(\mu_i)=\eta_i$, guarantees a positive fitted mean. Its coefficients act multiplicatively on the mean: increasing a predictor by one unit while holding other predictors fixed multiplies the fitted mean by the exponential of its coefficient. The variance formula also shows why the model allows greater absolute variability at larger means: if the mean doubles, the variance becomes four times as large when $\phi$ is unchanged.

Worked Example: Modeling Positive Repair Costs

A hypothetical study records strictly positive repair costs in dollars. Suppose a Gamma GLM uses a log link with $\log(\mu)=\log(200)+\log(1.25)x$, where $x=1$ denotes a specified repair category. Let the dispersion parameter be $\phi=0.25$. Calculate the fitted mean and conditional variance.

Method. Apply the inverse log link to the linear predictor, then use the Gamma variance function $\phi\mu^2$.

At $x=1$, the predictor is $\eta=\log(200)+\log(1.25)=\log(200\times1.25)=\log(250)$. Therefore $\mu=e^\eta=250$ dollars. The fitted conditional mean cost is $250.

The conditional variance is $\phi\mu^2=0.25(250)^2=0.25(62{,}500)=15{,}625$ dollars squared. As a check under a Gamma shape-scale parameterization, the shape is $1/\phi=4$, the scale is $\phi\mu=62.5$, and the variance is $4(62.5)^2=4(3{,}906.25)=15{,}625$.

Check and common mistake. The mean is positive and the variance has squared-dollar units. A common mistake is to multiply the mean by $\phi$ and report $62.50 as the variance. The variance function uses the square of the mean: $\phi\mu^2$.

Why These Mean–Variance Rules Matter

The variance patterns help explain why the same linear predictor should not be used indiscriminately across response types. For a sum of $n$ independent Bernoulli variables with common probability $p$, the mean is $np$, and the variance is the sum of the individual variances, $np(1-p)$. For a Poisson response, the defining distribution gives mean and variance $\lambda$. For a Gamma response in the stated parameterization, the variance is $\phi\mu^2$. These relationships imply different expected spread at different mean levels.

The link controls the range and scale of the fitted mean. The logit inverse maps every finite real predictor into the interval $(0,1)$. The exponential inverse maps a finite real predictor to a positive mean, which suits a Poisson mean or a Gamma mean. The link does not, however, establish that the distributional assumptions are right. For example, a Poisson log-link model can produce sensible positive fitted means while still being a poor description if the conditional counts have substantially greater variability than their means.

\begin{aligned} \text{Binomial: } & E[Y\mid x]=np,\quad \operatorname{Var}(Y\mid x)=np(1-p),\\ \text{Poisson: } & E[Y\mid x]=\lambda,\quad \operatorname{Var}(Y\mid x)=\lambda,\\ \text{Gamma: } & E[Y\mid x]=\mu,\quad \operatorname{Var}(Y\mid x)=\phi\mu^2. \end{aligned}

A practical choice proceeds in stages. First identify what one response represents and list its possible values. Then choose a family whose support fits those values. Next assess whether its conditional variance behavior is credible, and choose a link that keeps fitted means within their required range while giving a useful interpretation. Finally, verify that the predictors and response have been defined consistently. This approach is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus because it requires defending the model rather than matching a label mechanically.

Identify the response scale.
Distinguish a binary result or success count, an event count, and a strictly positive continuous measurement.

Match a candidate family.
Consider binomial, Poisson, or Gamma according to support and the process producing the response.

Check mean and variance.
Ask whether the family’s conditional variance pattern is reasonable for the outcome.

Choose and apply the link.
Use an inverse link to obtain a valid fitted mean, and interpret the result on the response scale.

Common Traps in Family and Link Choices

Calling every numeric response Gaussian. Counts are integers and cannot be negative; a continuous cost and a binary indicator have different support and variance behavior.
Confusing a response with its mean. A Poisson mean of 2.0138 is an expectation, not a possible observed fractional count.
Using the linear predictor as the fitted mean. With a log link, exponentiate $\eta$; with a logit link, apply the inverse-logit.
Treating the family’s parameter range as the link’s fitted range. Binomial probabilities may equal 0 or 1 as distribution parameters, but a finite logit predictor yields a probability strictly between them.
Ignoring zeros in positive-cost data. An ordinary Gamma response is strictly positive; observations at zero require a modeling approach that accounts for them rather than silently treating zero as a positive Gamma value.
Assuming a plausible link proves a plausible family. A positive fitted mean does not confirm the Poisson or Gamma variance assumptions.

Key takeaway: Choose the family to represent the response’s support and conditional variability, and choose the link to relate predictors to a valid conditional mean. Then calculate on the response scale and check that the assumptions—not just the fitted values—make sense.

Check Your Understanding

For each problem, identify the appropriate family and link where requested, and check the mean or variance calculation.

A hypothetical screening study records whether each applicant has a specified condition. For one applicant, a logistic model gives $\eta=\log(3/2)$. Find the fitted probability and the variance of the binary response. (Answer: Binomial with logit link; $p=0.6000$, variance $0.6000(0.4000)=0.2400$.)
A hypothetical monitoring system records event counts with a Poisson log-link model $\log(\lambda)=-0.4+0.2x$. For $x=3$, calculate the mean and variance to four decimal places. (Answer: $\eta=0.2$, so $\lambda=e^{0.2}=1.2214$; mean and variance are both 1.2214.)
A hypothetical positive-cost study uses a Gamma model with log link $\log(\mu)=\log(80)+\log(1.5)x$, $\phi=0.20$, and $x=1$. Find the fitted mean and conditional variance. (Answer: $\mu=80(1.5)=120$; variance $=0.20(120)^2=2{,}880$ dollars squared.)
A grouped binary outcome records 4 successes among 10 trials. Assume independent trials with a common success probability $p=0.3$. State the family, expected number of successes, and variance. (Answer: Binomial; mean $10(0.3)=3$; variance $10(0.3)(0.7)=2.1$.)
A hypothetical cost response is continuous, strictly positive, and has a fitted conditional mean of 400 dollars. Under a Gamma GLM with $\phi=0.15$, calculate the conditional variance and explain whether a fitted mean of 400 is itself an observed-cost prediction. (Answer: Gamma is a candidate family; variance $=0.15(400)^2=24{,}000$ dollars squared. The mean is a conditional expectation, not a claim that an observation must equal $400.)

What You'll Learn