I wonder why my professor said that Gamma's canonical link is $\frac{1}{\mu}$. My thoughts are:

EDIT: $\theta$ is the canonical parameter.

Since $$\mathbb{E}_\theta(Y)=b^{'}(\theta)=-\frac{1}{\theta}=\mu$$ then $$\mathbb{E}_\theta(Y)=-\frac{1}{\theta}=\mu \iff \theta=g(\mu)=-\mu^{-1}$$ So $g(\cdot)=$ minus-inverse.

I clearly see a minus!! Shouldn't it be $-\frac{1}{\mu}$ instead of $\frac{1}{\mu}$?

  • $\begingroup$ Could you please tell us what $\theta$ represents? It's not any usual parameter, because it's obviously negative in your formulation (knowing that the Gamma mean $\mu$ is necessarily positive). $\endgroup$
    – whuber
    Dec 27, 2018 at 20:55
  • $\begingroup$ @whuber the canonical parameter $\endgroup$
    – alienflow
    Dec 27, 2018 at 21:20
  • $\begingroup$ Could you please add the tag self-study? $\endgroup$ Dec 27, 2018 at 21:22

1 Answer 1


Yes, you are quite right. When we write the gamma distribution as an exponential dispersion model $$f(y;\mu,\phi)=a(y,\phi)\exp\big\{\frac1{\phi}(y\theta-\kappa(\theta))\big\}$$ we do find that the canonical parameter $\theta$ is related to the mean parameter $\mu$ by $$\theta = - 1/\mu.$$ In generalized linear model theory, the canonical link is supposed to be equal to the canonical parameter, so the canonical link should in principle be the minus-reciprocal function instead of reciprocal, and indeed the minus-reciprocal would be an increasing function of $\mu$ instead of decreasing as the reciprocal is. However, people find it somewhat simpler to talk of $1/\mu$ rather than $-1/\mu$, so GLM code always uses the former as the canonical link for the gamma distribution instead of the latter. The only consequence is that all the linear model coefficients in the GLM are minus what they otherwise would have been. It makes no difference to the fitted values and has no negative consequences from an inferential point of view.

By the way, this question has essentially been answered before, although in the previous question the gamma shape parameter ($1/\phi$) was assumed to be known: Is the canonical parameter (and therefore the canonical link function) for a Gamma not unique?