Why Match the Response Family Before the Link?
A link function connects a response mean to a linear predictor, but the appropriate link is not chosen from the observed values alone. First identify the response’s sampling distribution: a binary indicator, a count of successes out of known trials, an event count, or a continuous measurement may require different probability models. Once the response family is specified, its exponential-family representation identifies a natural parameter; setting the linear predictor equal to that parameter gives the canonical link.
The distinction is useful because the family describes the possible observations and their variability, while the link describes how the mean is related to predictors. A canonical link is a standard match, not a requirement that every model must use. The earlier material on link functions and canonical links establishes the general mapping; here the focus is matching that mapping to four common response families.
In a generalized linear model, the linear predictor is \(\eta\), and the link specifies \(\eta=g(\mu)\). With a canonical link, \(\eta=\theta\). The identity \(E(Y)=b'(\theta)\) provides the key step: solve it for \(\theta\) in terms of the response mean, then use that function as \(g\). The corresponding variance is \(Var(Y)=a(\phi)b''(\theta)\); it too is determined by the family, rather than chosen independently of it.
Four Standard Family–Link Matches
The table summarizes the match. For Binomial data, \(m\) is the known number of trials and \(p\) is the success probability on each trial. The canonical link is most directly stated for \(p\), even when the recorded response is the number of successes \(Y\).
| Response family | Response and mean | Natural parameter | Canonical link |
|---|---|---|---|
| Bernoulli | \(Y\in\{0,1\}\), \(\mu=p\) | \(\theta=\log\{p/(1-p)\}\) | Logit: \(g(p)=\log\{p/(1-p)\}\) |
| Binomial | \(Y\in\{0,\ldots,m\}\), \(E(Y)=mp\) | \(\theta=\log\{p/(1-p)\}\) | Logit of \(p=\mu/m\) |
| Poisson | \(Y\in\{0,1,2,\ldots\}\), \(\mu=\lambda\) | \(\theta=\log(\lambda)\) | Log: \(g(\mu)=\log(\mu)\) |
| Gaussian | \(Y\) continuous, \(\mu=E(Y)\) | \(\theta=\mu\) | Identity: \(g(\mu)=\mu\) |
Deriving the Matches
These derivations use the exponential-family form directly. They also explain why a Binomial count needs a small but important qualification: its mean count is \(mp\), whereas its natural parameter describes \(p\).
For \(Y=0\) or \(1\), with \(P(Y=1)=p\), the log probability is \(y\log p+(1-y)\log(1-p)\). Set \(\theta=\log\{p/(1-p)\}\); the expression becomes \(y\theta-\log(1+e^\theta)\). Thus \(b(\theta)=\log(1+e^\theta)\), and \(b'(\theta)=e^\theta/(1+e^\theta)=p\). Therefore the canonical link is the logit.
For \(Y\sim Binomial(m,p)\), the log probability is \(\log {m\choose y}+y\log p+(m-y)\log(1-p)\). Rewriting gives \(y\theta-m\log(1+e^\theta)+\log {m\choose y}\), where \(\theta=\log\{p/(1-p)\}\). Here \(b(\theta)=m\log(1+e^\theta)\), so \(b'(\theta)=mp=E(Y)\). Hence \(\theta=\log\{(\mu/m)/(1-\mu/m)\}\): the logit of the success probability, not the logit of the unscaled mean count.
For \(Y\sim Poisson(\lambda)\), the log probability is \(y\log\lambda-\lambda-\log(y!)\). Set \(\theta=\log\lambda\), so the expression is \(y\theta-e^\theta-\log(y!)\). Then \(b(\theta)=e^\theta\), \(b'(\theta)=e^\theta=\lambda=\mu\), and \(\theta=\log\mu\). The canonical link is the log link.
For \(Y\sim N(\mu,\sigma^2)\) with known variance \(\sigma^2\), its density can be written as \(\exp\{(y\mu-\mu^2/2)/\sigma^2-y^2/(2\sigma^2)-\log(2\pi\sigma^2)/2\}\). Thus \(\theta=\mu\), \(b(\theta)=\theta^2/2\), and \(b'(\theta)=\theta=\mu\). The canonical link is the identity.
The Gaussian form shows why the identity link is canonical when the response is modeled as Gaussian with a common known variance. It does not mean that every continuous response is Gaussian, or that a Gaussian model is suitable whenever the data include decimals. The response distribution and its assumptions still need to match the measurement process.
Worked Example: Bernoulli or Grouped Binomial?
Suppose a hypothetical clinic records whether each of 12 independently observed patients experiences a specified event during follow-up. Nine have the event and three do not. Identify the family and canonical link, then calculate the observed event proportion and its logit.
Each patient contributes one binary outcome, so an individual observation is Bernoulli, with \(Y_i=1\) for an event and \(Y_i=0\) otherwise. The Bernoulli mean is \(p=P(Y_i=1)\), and its canonical link is \(g(p)=\log\{p/(1-p)\}\). The observed proportion is \(9/12=0.75\). Substituting that proportion into the link gives $$ \log\left(\frac{0.75}{1-0.75}\right) =\log\left(\frac{0.75}{0.25}\right) =\log(3) \approx 1.099. $$
If the same information is recorded only as “9 events out of 12 patients,” the total can instead be represented as one Binomial observation \(Y\sim Binomial(12,p)\). Its mean count is \(12p\), but the canonical link remains the logit of \(p\), here estimated descriptively by \(\logit(9/12)\). These are two representations of the same trial information; they are not evidence that the individual outcomes are independent if the study design makes them dependent.
Worked Example: Matching a Poisson Count
In a hypothetical monitoring exercise, a sensor records the number of alerts during each of six equal one-hour intervals. Suppose the counts are \(2, 4, 3, 1, 5,\) and \(3\). Assume the interval counts are independent and arise from the same Poisson distribution, with a constant mean count per interval. Find the sample mean and its log-link value.
The outcomes are nonnegative integer counts, and the assumed family is Poisson. The total is \(2+4+3+1+5+3=18\), so the sample mean is $$ \bar{x}=\frac{18}{6}=3\text{ alerts per one-hour interval}. $$
For a Poisson mean \(\mu\), the natural parameter is \(\theta=\log(\mu)\), which is the canonical log link. Evaluating it at the sample mean gives \(\log(3)\approx1.099\). This arithmetic illustrates the family-link match; it is not a claim that a full regression has been fitted. The constant-mean assumption is appropriate only if the intervals are comparable in exposure and relevant conditions, and the independent-count assumption must also be plausible.
Worked Example: Matching a Gaussian Response
A hypothetical laboratory measures a continuous outcome, in milligrams, on four independent specimens under one condition. The measurements are \(8, 10, 11,\) and \(7\) mg. Suppose the measurements are normally distributed with common known variance \(4\text{ mg}^2\). Identify the canonical link and calculate its value at the sample mean.
The total is \(8+10+11+7=36\) mg, giving \(\bar{x}=36/4=9\) mg. For a Gaussian response with variance \(\sigma^2=4\text{ mg}^2\), the natural parameter is the mean itself, \(\theta=\mu\). Its canonical link is therefore \(g(\mu)=\mu\), the identity. At the sample mean, the link-scale value is \(9\) mg. The normal-family assumption concerns the measurement distribution, and the common known variance is part of this example’s stated setup; continuous measurements alone do not establish either assumption.
How to Choose and Check a Match
A practical selection starts with the unit of observation and the mechanism that produces its response. Ask whether each row represents one binary trial, a number of successes among a known number of trials, an event count over a defined exposure, or a continuous measurement. Then check that the proposed family’s possible values and assumptions make sense. The link comes after this family decision; selecting a familiar link first can conceal a mismatch in the response model.
For a basic GLM, observations are treated as independent conditional on the modeled predictors, with the specified response distribution and mean structure. The examples also rely on their stated conditions: a Binomial total requires a known trial count and a common success probability within the modeled observation; a Poisson count model requires an appropriate count process and exposure definition; and a Gaussian model specifies a continuous normal response with its variance structure. These are substantive modeling assumptions, not conclusions guaranteed by the observed response values.
The canonical links are not the only possible links. A different link may be useful for a scientific or modeling reason, but it is then noncanonical for that family. The response family and link answer different questions: for example, a log link does not turn a non-count response into Poisson data, and a Gaussian family does not become appropriate merely because the identity link yields convenient predictions.
Diagnostics should be interpreted in light of the chosen family. A fitted mean outside the response’s allowable range, systematic disagreement between observed and predicted response patterns, or variability substantially inconsistent with the family can signal a problem with the family, the mean structure, or the independence assumptions. Such evidence does not by itself identify which assumption failed. In particular, a poor fit is not repaired simply by changing to the canonical link if the response distribution is unsuitable.
The main alternatives differ in how they represent the response. Bernoulli observations retain one outcome per trial, while a Binomial observation summarizes successes and failures with a known denominator. Poisson models counts without a fixed success denominator. Gaussian models continuous responses and allow values across the real line. The canonical links—logit, logit of the Binomial probability, log, and identity—follow from these families’ natural parameters. The match is a principled starting point, but its validity remains limited by the assumptions and data-generating process.
Check Your Understanding
Answer each question, then compare with the solutions below.
- A recorded outcome is 0 or 1 for each independently observed trial. Name the family, mean, and canonical link.
- For \(Y\sim Binomial(8,p)\), suppose \(E(Y)=6\). What is \(p\), and what is the canonical-link value?
- A count response has Poisson mean \(\mu=4\). State its natural parameter and canonical-link value.
- A Gaussian response has mean \(\mu=2.5\) and known variance \(9\). What is its canonical link and natural parameter?
Solutions. (1) Bernoulli, with mean \(p\), and canonical link \(\log\{p/(1-p)\}\). (2) \(p=E(Y)/8=6/8=0.75\); the canonical-link value is \(\log(0.75/0.25)=\log(3)\approx1.099\). (3) \(\theta=\log(\mu)=\log(4)\approx1.386\), and the canonical link is the log. (4) The natural parameter is \(\theta=\mu=2.5\), and the canonical link is identity; the variance affects dispersion, not this family’s canonical-link form.