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Link Functions and Scale · Tutorial 105 of 981

Recognizing Canonical Links In Common Families

Match each response family’s natural parameter to its mean to identify its canonical link, and verify the result in worked examples.

Beginner 9 min read

What You'll Learn

  • Define a canonical link using the natural parameter in exponential-family form
  • Derive the Bernoulli canonical link from its probability mass function
  • Derive the Poisson canonical link from its probability mass function
  • Identify the normal canonical link under the GLM dispersion convention
  • Check each link by recovering the response mean from the natural parameter
  • Distinguish the normal GLM convention from a fixed-variance reparameterization

From a Response Family to Its Canonical Link

A link function maps a response mean to the scale on which a model specifies its linear predictor. Some links are especially closely tied to the response distribution: they turn the mean into the distribution’s natural parameter. These are the canonical links. Recognizing them for common response families is a matter of matching two expressions: the natural parameter in the family’s exponential form, and the mean of the response.

The three standard matches in this tutorial are Bernoulli with the logit link, Poisson with the log link, and normal with the identity link. The normal case requires care: the usual GLM convention represents the variance through a dispersion parameter, which determines how the canonical link is written.

Definition: Suppose a response distribution can be written in exponential-family form \(f(y;\theta,\phi)=\exp\{[y\theta-b(\theta)]/a(\phi)+c(y,\phi)\}\), where \(\theta\) is the natural parameter and \(\phi\) is a dispersion parameter. If the mean is \(\mu=b'(\theta)\), the canonical link is the function \(g\) satisfying \(g(\mu)=\theta\). Thus the canonical link maps the mean to the natural parameter, not to an arbitrary re-expression of it.

This definition uses the common GLM exponential-dispersion convention. The representation must be valid for the family under consideration, and the mean-to-natural-parameter relationship is considered on the interior range where it is one-to-one. We will obtain that relationship explicitly for each family below. No regression coefficients or fitted model are needed to identify the link.

Why the Mean Is the Derivative of the Log-Partition Function

The function \(b(\theta)\) in exponential-family form determines the mean. The next result justifies the matching procedure. For a discrete response, the argument uses a sum; for a continuous response, it uses an integral.

Result: Suppose \(Y\) has a density or mass function in the form above, its support does not depend on \(\theta\), and differentiation with respect to \(\theta\) can be interchanged with integration or summation in a neighborhood of the parameter value. Then \(E[Y]=b'(\theta)\).

Proof. Since \(f\) is a normalized density or mass function, its integral or sum over the support equals 1. Differentiate this identity with respect to \(\theta\). The stated interchange condition permits differentiation inside the integral or sum. From the exponential form, \(\partial f(y;\theta,\phi)/\partial\theta = f(y;\theta,\phi)[y-b'(\theta)]/a(\phi)\). Therefore,

$$ 0=\frac{\partial}{\partial\theta}\int f(y;\theta,\phi)\,dy =\int f(y;\theta,\phi)\frac{y-b'(\theta)}{a(\phi)}\,dy =\frac{E[Y]-b'(\theta)}{a(\phi)}. $$

For a discrete response, replace the integrals by sums. Since \(a(\phi)>0\), the last equality implies \(E[Y]=b'(\theta)\). \(\square\)

In the regular one-parameter cases here, \(b'(\theta)\) can be inverted to express \(\theta\) as a function of \(\mu\). That inverse is the canonical link. The next derivation gives all three matches directly from the family distributions.

The Three Common Family Matches

1
Bernoulli response.
Write the probability mass function in exponential form and read off the natural parameter as the log-odds.
2
Poisson response.
Write the probability mass function in exponential form and read off the natural parameter as the logarithm of the mean.
3
Normal response.
Use the GLM form that places the variance scale in the dispersion parameter; the natural parameter then equals the mean.

For a Bernoulli response \(Y\in\{0,1\}\) with \(P(Y=1)=p\), the mean is \(\mu=p\). Its mass function is \(f(y)=p^y(1-p)^{1-y}\). Taking logarithms and collecting terms gives \(\log f(y)=y\log\{p/(1-p)\}+\log(1-p)\). With \(\theta=\log\{p/(1-p)\}\), the second term is \(-\log(1+\exp(\theta))\). Thus \(b(\theta)=\log(1+\exp(\theta))\), and \(b'(\theta)=\exp(\theta)/[1+\exp(\theta)]=p\). Solving for \(\theta\) in terms of \(\mu=p\) gives the canonical link \(g(\mu)=\log\{\mu/(1-\mu)\}\), the logit link.

For a Poisson response with mean \(\mu>0\), the mass function is \(f(y)=\exp(-\mu)\mu^y/y!\), for nonnegative integer \(y\). Its log is \(y\log(\mu)-\mu-\log(y!)\). Set \(\theta=\log(\mu)\), so \(\mu=\exp(\theta)\). The mass function becomes \(\exp\{y\theta-\exp(\theta)-\log(y!)\}\). Hence \(b(\theta)=\exp(\theta)\), and \(b'(\theta)=\exp(\theta)=\mu\). The canonical link is therefore \(g(\mu)=\log(\mu)\), the log link.

For a normal response \(Y\sim N(\mu,\phi)\), where \(\phi>0\) is the variance, expand the log density: \(\log f(y)=-\tfrac12\log(2\pi\phi)-(y-\mu)^2/(2\phi)\). Rearranging,

$$ \log f(y) =\frac{y\mu-\mu^2/2}{\phi} -\frac{y^2}{2\phi} -\frac12\log(2\pi\phi). $$

This is exponential-dispersion form with \(\theta=\mu\), \(b(\theta)=\theta^2/2\), \(a(\phi)=\phi\), and \(c(y,\phi)=-y^2/(2\phi)-\tfrac12\log(2\pi\phi)\). Since \(b'(\theta)=\theta=\mu\), the canonical link is \(g(\mu)=\mu\), the identity link.

Response familyMeanNatural parameterCanonical link
Bernoulli\(\mu=p\)\(\theta=\log\{\mu/(1-\mu)\}\)Logit: \(g(\mu)=\log\{\mu/(1-\mu)\}\)
Poisson\(\mu\)\(\theta=\log(\mu)\)Log: \(g(\mu)=\log(\mu)\)
Normal, GLM dispersion form\(\mu\)\(\theta=\mu\)Identity: \(g(\mu)=\mu\)

Worked Checks

Worked Example: Bernoulli Mean of 0.8

Suppose a Bernoulli response has mean \(\mu=0.8\). Identify its canonical parameter and verify that it maps back to the stated mean.

For a Bernoulli response, \(\mu=p\), so \(p=0.8\) and \(1-p=0.2\). The odds are \(p/(1-p)=0.8/0.2=4\). The natural parameter, and hence the canonical link value, is \(\theta=g(\mu)=\log(4)\approx 1.3863\).

To check, apply the inverse relationship \(p=\exp(\theta)/[1+\exp(\theta)]\). Since \(\exp(\log 4)=4\), \(p=4/(1+4)=0.8\), as required. The canonical parameter is the log-odds, not the probability itself.

Worked Example: Poisson Mean of 3.2

For a Poisson response with mean \(\mu=3.2\), find the canonical parameter and check it.

The Poisson natural parameter is \(\theta=\log(\mu)\). Therefore \(\theta=\log(3.2)\approx 1.1632\). Its inverse relationship is \(\mu=\exp(\theta)\), so \(\exp(\log(3.2))=3.2\). This recovers the original mean exactly.

The mean need not be an integer: a Poisson response takes nonnegative integer values, while its distribution’s mean can be any positive real number. That does not change the canonical link.

Worked Example: Normal Mean and Variance

Let \(Y\sim N(2.5,9)\), so its mean is \(2.5\) and its variance is \(9\). Identify the canonical link value using the GLM dispersion form.

Here \(\phi=9\), \(\theta=\mu=2.5\), and \(b(\theta)=\theta^2/2\). The derivative is \(b'(\theta)=\theta=2.5\), confirming that this parameterization has mean \(2.5\). Thus \(g(\mu)=\mu=2.5\).

As a check on the density decomposition, the terms involving \(y\) and \(\mu\) are \((2.5y-2.5^2/2)/9=(2.5y-3.125)/9\). The remaining terms are \(-y^2/18-\tfrac12\log(18\pi)\), exactly the terms that do not involve the natural parameter in the GLM exponential form.

A Scaling Pitfall and Why the Match Matters

A common mistake is to write the normal natural parameter as \(\mu/\sigma^2\) and conclude that the normal canonical link must be \(g(\mu)=\mu/\sigma^2\). That expression can arise if a normal density with fixed variance \(\sigma^2\) is written in a different one-parameter exponential-family convention, with the variance factor absorbed into the natural parameter. It is not the canonical-link convention used here for GLMs.

In the GLM exponential-dispersion form, the normal variance is \(a(\phi)=\phi\), outside the term \(y\theta-b(\theta)\). The density expansion above therefore gives \(\theta=\mu\) and the identity canonical link. For example, with \(\mu=2.5\) and \(\phi=9\), the GLM natural parameter is \(2.5\), not \(2.5/9\). Always inspect where the scale factor appears in the chosen exponential-family representation before identifying the canonical parameter.

Canonical links are useful because they give a direct correspondence between the mean and the response family’s natural parameter. For recognition, the procedure is concrete: express the distribution in the stated exponential form, identify \(\theta\), and express it as a function of \(\mu\). The result is a family-specific link; it is not selected by the response support alone.

Key takeaway: Under the GLM exponential-dispersion convention, the Bernoulli, Poisson, and normal canonical links are respectively logit, log, and identity. Verify the parameterization first, especially when the normal variance is written explicitly.

Check Your Understanding

Use the family’s exponential form to answer each question.

  1. For a Bernoulli response with mean \(0.25\), what is the canonical parameter? Show how applying the inverse relationship recovers the mean.
  2. Why does the Poisson canonical link equal \(\log(\mu)\), rather than \(\mu\)? Identify the natural parameter in the mass function.
  3. For a normal response in GLM dispersion form, what are \(\theta\) and \(b(\theta)\)? Verify that \(b'(\theta)\) is the mean.
  4. Explain why the expression \(\mu/\sigma^2\) may appear in a fixed-variance normal exponential-family representation but is not the canonical-link value under the GLM dispersion convention used here.
  5. State the canonical links for Bernoulli, Poisson, and normal responses in that order.
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