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Gamma Response Models · Tutorial 576 of 981

Canonical Negative-Inverse Link

Derive the gamma canonical link from the exponential-family form and see precisely how its negative-reciprocal predictor corresponds to the positive reciprocal convention.

Intermediate 10 min read

What You'll Learn

  • Express the gamma density in exponential-dispersion form
  • Identify the natural parameter and cumulant function
  • Derive the canonical link and its domain
  • Verify the link using the mean and variance
  • Translate between negative- and positive-reciprocal predictors
  • Recognize invalid predictor values under each convention

From the Gamma Density to Its Natural Parameter

For a gamma response, the mean \(\mu\) must be positive. The canonical link is determined by the distribution’s exponential-family representation, not chosen simply because a reciprocal transformation seems convenient. With the density convention used here, the resulting canonical link is the negative reciprocal of the mean.

Write a gamma response with mean \(\mu>0\) and dispersion \(\phi>0\) as having shape \(1/\phi\) and scale \(\phi\mu\). Its density, for \(y>0\), is

$$ f(y;\mu,\phi) = \frac{y^{1/\phi-1}\exp\{-y/(\phi\mu)\}} {\Gamma(1/\phi)(\phi\mu)^{1/\phi}}. $$

The exponential-dispersion form is

$$ f(y;\theta,\phi) = \exp\left\{ \frac{y\theta-b(\theta)}{\phi} +c(y,\phi) \right\}. $$

Here the response enters through the statistic \(y\), \(\theta\) is the natural parameter, \(b\) is the cumulant function, and \(\phi\) is the dispersion. To identify these quantities, take the logarithm of the gamma density and collect the terms involving \(y\) and \(\mu\):

$$ \log f(y;\mu,\phi) = -\frac{y}{\phi\mu} -\frac{\log\mu}{\phi} +\left(\frac{1}{\phi}-1\right)\log y -\frac{\log\phi}{\phi} -\log\Gamma(1/\phi). $$

The term involving \(y\) linearly is \(-y/(\phi\mu)\), so matching it to \(y\theta/\phi\) gives \(\theta=-1/\mu\). The mean-dependent term is \(-\log\mu/\phi\). Since \(-\theta=1/\mu\), we have \(\log(-\theta)=-\log\mu\), and therefore this term equals \(\log(-\theta)/\phi\). In the exponential-dispersion form it is \(-b(\theta)/\phi\), which identifies \(b(\theta)=-\log(-\theta)\).

Definition: For the gamma exponential-dispersion representation above, the natural parameter and cumulant function are \(\theta=-1/\mu\) and \(b(\theta)=-\log(-\theta)\), with natural-parameter domain \(\theta<0\). The canonical link is the map from the mean to the natural parameter: \(g(\mu)=\theta=-1/\mu\), for \(\mu>0\).

The remaining terms give \(c(y,\phi)=(1/\phi-1)\log y-(\log\phi)/\phi-\log\Gamma(1/\phi)\). These terms do not depend on \(\mu\). The domain \(\theta<0\) is necessary: the logarithm in \(b(\theta)\) requires \(-\theta>0\), and the corresponding mean \(\mu=-1/\theta\) is positive.

Worked Example: Identifying the Canonical Parameters

Suppose \(\mu=4\) and \(\phi=1/4\). Find the natural parameter, cumulant value, and canonical predictor.

The natural parameter is \(\theta=-1/\mu=-1/4\), which is in the required domain because \(-1/4<0\). The cumulant value is \(b(\theta)=-\log(-\theta)=-\log(1/4)=\log 4\). Thus the canonical link evaluated at this mean is \(g(4)=-1/4\).

The signs can be checked directly: \(-\log\mu=-\log 4\), while \(\log(-\theta)=\log(1/4)=-\log 4\). Also \(b(\theta)=\log 4\), so \(-b(\theta)=-\log 4\), matching the mean-dependent term in the log density after multiplication by \(1/\phi\).

Why the Canonical Link Is Negative Reciprocal

In a generalized linear model, the linear predictor is \(\eta=x^\mathsf{T}\beta\), and the link specifies \(\eta=g(\mu)\). For a canonical link, the link maps the mean to the natural parameter in the exponential-family representation. Since the gamma natural parameter just derived is \(-1/\mu\), the canonical choice is \(g(\mu)=-1/\mu\), not \(1/\mu\).

Theorem: For the gamma exponential-dispersion density with fixed \(\phi>0\), the canonical link is \(g(\mu)=-1/\mu\), with link domain \(\mu>0\) and predictor range \(\eta<0\). Its inverse is \(\mu=-1/\eta\) on \(\eta<0\).

Derivation. The exponential-family form contains the response term \(y\theta/\phi\). The gamma log density contains \(-y/(\phi\mu)\). Matching these coefficients for every \(y>0\) forces \(\theta=-1/\mu\). By definition, a canonical link sets \(g(\mu)=\theta\), giving \(g(\mu)=-1/\mu\). Because \(\mu>0\), this link always yields a negative value. Conversely, if \(\eta<0\), solving \(\eta=-1/\mu\) gives \(\mu=-1/\eta>0\). Hence the link and inverse have the stated domains. \(\square\)

The cumulant function also confirms the mean and variance encoded by this representation. Differentiating on \(\theta<0\) gives \(b'(\theta)=-1/\theta\) and \(b''(\theta)=1/\theta^2\). Substituting \(\theta=-1/\mu\) yields \(b'(\theta)=\mu\) and \(b''(\theta)=\mu^2\).

Result: In this gamma exponential-dispersion representation, \(E[Y]=b'(\theta)=\mu\) and \(\operatorname{Var}(Y)=\phi b''(\theta)=\phi\mu^2\).

For completeness, the derivative identities follow by differentiating the normalizing integral. At fixed \(\phi>0\), the part of the density involving \(y\) is proportional to \(y^{1/\phi-1}\exp(\theta y/\phi)\), for \(y>0\). On any compact interval of negative \(\theta\) values, this integrand and its first two derivatives with respect to \(\theta\) are bounded by integrable functions: near zero they are bounded by constants times \(y^{1/\phi-1}\), and at infinity by a power of \(y\) times a decaying exponential. Differentiation under the integral is therefore valid locally. Differentiating the normalizing identity once gives \(E[Y]=b'(\theta)\); differentiating a second time gives \(\operatorname{Var}(Y)=\phi b''(\theta)\). The formulas for \(b'\) and \(b''\) then give the stated mean and variance.

Worked Example: Checking the Exponential-D 가족 Form

Take \(\mu=4\) and \(\phi=1/4\). The gamma distribution then has shape \(1/\phi=4\) and scale \(\phi\mu=1\). Its density is \(f(y)=y^3e^{-y}/\Gamma(4)=y^3e^{-y}/6\), for \(y>0\).

From the exponential-dispersion representation, \(\theta=-1/4\), \(b(\theta)=\log 4\), and \(c(y,1/4)=3\log y-4\log(1/4)-\log 6\). Therefore its log density is

$$ \frac{y(-1/4)-\log 4}{1/4}+3\log y-4\log(1/4)-\log 6 = -y-4\log 4+3\log y+4\log 4-\log 6 = 3\log y-y-\log 6. $$

Exponentiating gives \(y^3e^{-y}/6\), exactly the gamma density above. In particular, the term \(-b(\theta)\) contributes \(-\log 4\) before division by \(\phi\); it does not contribute \(+\log 4\).

Connection to the Positive Reciprocal Link

A positive reciprocal link is also used for gamma means: \(g_+(\mu)=1/\mu\). It is not the canonical link under the density convention used here. The two links differ by a minus sign, so their linear predictors and regression coefficients must be compared with the same sign change.

Result: If the positive reciprocal model has predictor \(x^\mathsf{T}\beta_+=1/\mu\), then the same mean function written with the canonical link has predictor \(x^\mathsf{T}\beta_{\mathrm{can}}=-1/\mu\), where \(\beta_{\mathrm{can}}=-\beta_+\). The positive reciprocal predictor must be positive; the canonical predictor must be negative.

This relationship is algebraic. For any fixed covariate vector \(x\), if \(x^\mathsf{T}\beta_+=1/\mu>0\), then setting \(\beta_{\mathrm{can}}=-\beta_+\) gives \(x^\mathsf{T}\beta_{\mathrm{can}}=-x^\mathsf{T}\beta_+=-1/\mu\). Conversely, negating a canonical coefficient vector gives the positive reciprocal coefficient vector. Thus, when coefficient spaces and predictor constraints are transformed together, both forms describe the same set of positive fitted means. They do not use the same numerical coefficient signs or predictor domain.

Worked Example: Converting a Positive Reciprocal Predictor

Suppose a positive reciprocal model uses covariates \(x=(1,2)^\mathsf{T}\) and coefficients \(\beta_+=(0.2,0.1)^\mathsf{T}\). Its predictor is \(x^\mathsf{T}\beta_+=1(0.2)+2(0.1)=0.4\), so the fitted mean is \(\mu=1/0.4=2.5\).

For the canonical link, negate the coefficient vector: \(\beta_{\mathrm{can}}=(-0.2,-0.1)^\mathsf{T}\). Its predictor is \(x^\mathsf{T}\beta_{\mathrm{can}}=1(-0.2)+2(-0.1)=-0.4\). Inverting the canonical link gives \(\mu=-1/(-0.4)=2.5\), the same positive mean. The predictor signs differ, but the fitted mean agrees.

Domain Checks and a Common Mistake

The link domain is not a minor technicality. Under the canonical link, every fitted linear predictor must be negative if it is to correspond to a finite positive mean. Under the positive reciprocal convention, the predictor must instead be positive. A coefficient vector that produces a predictor outside the required range cannot be inverted into an admissible mean under that link.

Worked Example: Detecting an Invalid Canonical Predictor

Suppose a proposed canonical predictor at a covariate value is \(\eta=0.3\). The inverse canonical link would require \(\mu=-1/\eta=-1/0.3=-10/3\), which is not a positive mean. Therefore this predictor is inadmissible for the gamma canonical link.

Under the positive reciprocal link, the same numerical predictor \(0.3\) would give \(\mu=1/0.3=10/3\). That is valid for the positive reciprocal convention, but it is not the canonical predictor: the matching canonical predictor for this mean is \(-0.3\). This check prevents the two sign conventions from being silently interchanged.

Common mistake: It is tempting to say that the canonical link is \(1/\mu\), because a reciprocal appears when solving for the mean. That reasoning drops the sign in the natural parameter. The gamma log density has the linear response term \(-y/(\phi\mu)\), so its natural parameter is \(-1/\mu\), not \(1/\mu\). The positive reciprocal link is a separate convention related by negating the predictor and coefficients.

A second sign error can occur while identifying the cumulant function. Since \(-\theta=1/\mu\), the correct identity is \(\log(-\theta)=-\log\mu\). Thus \(-\log\mu=\log(-\theta)\), whereas \(b(\theta)=-\log(-\theta)=\log\mu\). Keeping the cumulant \(b\) distinct from the term \(-b\) in the exponent avoids reversing this sign.

Key takeaway: With the gamma density written using \(y\theta-b(\theta)\), the natural parameter is \(\theta=-1/\mu\), the cumulant is \(b(\theta)=-\log(-\theta)\), and the canonical link is the negative reciprocal. The positive reciprocal form describes the same means only after reversing the predictor and coefficient signs.

Check Your Understanding

Use the stated exponential-dispersion convention and check the predictor domain in each answer.

  1. For \(\mu=5\), find the natural parameter and cumulant value. Verify that the natural parameter is in its domain.
  2. Starting from the gamma log density, identify the term that determines the sign of the natural parameter.
  3. A positive reciprocal model has predictor \(x^\mathsf{T}\beta_+=0.25\). What are the mean and the corresponding canonical predictor?
  4. Explain why a canonical predictor equal to zero cannot correspond to a finite positive gamma mean.
  5. State the mean and variance obtained from \(b'(\theta)\) and \(b''(\theta)\) for dispersion \(\phi\).
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