# Why is the canonical link of a GLM with Gamma distribution the reciprocal?

I'm fitting a generalized linear model to a theoretically gamma-distributed dataset, and I'm confused about the canonical link. The gamma distribution has PDF $$f(y;a,\lambda) = \frac{\lambda^a e^{-\lambda y}y^{a-1}}{\Gamma(a)}$$ and can be shown to belong to the exponential family: $$f = exp\{alog(\lambda)-\lambda y+(a-1)log(y)-log(\Gamma(a))\}$$ Or, with some small changes, $$f = exp\{\frac{y\cdot\frac{\lambda}{a}-log(\lambda)}{(-1/a)} + [(a-1)log(y)-log(\Gamma(a))]\}$$ By this representation, if the canonical link is used for a generalized linear model, $$\eta = \frac{\lambda}{a} = \mathbf{x}^T\beta$$. This is contrary to the canonical links in books (Introduction to Linear Regression Analysis by Montgomery, Peck, and Vining) of $$\eta = 1/\lambda$$. What is the problem here? Is it because of the alternative PDF of the Gamma distribution, such as in @GordonSmyth's answer in https://stats.stackexchange.com/a/474351/411578?

• The canonical link for a gamma distribution is $\eta = 1/\mu$, not $\eta = 1/\lambda$. What is $\mu$ for a Gamma distribution as you've parameterized it? Commented Apr 26 at 21:52
• @jbowman thanks for the clarification, the textbook I read used lambda which is a very confusing notation. Commented Apr 26 at 22:54

Your formulation of the Gamma distribution is:

$$f(y;a,\lambda) = \frac{\lambda^a e^{-\lambda y} y^{a-1}}{\Gamma(a)}$$ where $$a$$ (often referred to as $$k$$ in some texts) is the shape parameter, and $$\lambda$$ is the rate parameter.

To express this distribution in the exponential family form, we write it as:

$$f(y; \theta, \phi) = \exp\left(\frac{y\theta - b(\theta)}{a(\phi)} + c(y,\phi)\right)$$ For the Gamma distribution:

• $$\theta = -\lambda$$
• $$b(\theta) = -\log(-\theta) = -\log(\lambda)$$
• $$a(\phi) = \frac{1}{k}$$ (using $$k$$ instead of $$a$$ for clarity, where $$k$$ is the shape)
• $$c(y, \phi) = (k-1)\log y - \log \Gamma(k)$$

In this exponential family form:

$$\theta = -\lambda, \quad b(\theta) = -\log(-\theta)$$ In the context of exponential families, $$b(\theta)$$ is known as the cumulant generating function. The derivative $$b'(\theta)$$ provides the expected value (or mean) of the distribution. This relationship is important for linking the natural parameter $$\theta$$ to the mean $$\mu$$, which is often the parameter of interest in statistical modeling.

Thus, the mean of the distribution $$\mu$$ can be obtained from the derivative of $$b(\theta)$$:

$$b'(\theta) = \frac{d}{d\theta}(-\log(-\theta)) = \frac{d}{d\theta}(-\log(\lambda)) = -\frac{1}{\lambda}$$ Since $$\theta = -\lambda$$, and $$\mu = -\frac{1}{\theta} = 1/\lambda$$, we see that:

$$\mu = 1/\lambda$$ The canonical link function $$g(\mu)$$ is defined as the function that equates the linear predictor $$\eta$$ to the natural parameter $$\theta$$, i.e., $$\eta = g(\mu) = \theta$$. For the Gamma distribution, since $$\theta = -\lambda$$ and $$\mu = 1/\lambda$$, the canonical link function is the reciprocal link:

$$g(\mu) = -1/\mu$$ The confusion in your question arises from the parameters used and their interpretations. When you equate $$\eta = \frac{\lambda}{a}$$, it seems like a mix-up in the interpretation of parameters in your formulation. The canonical link function should straightforwardly be $$\eta = -1/\mu$$, which implies $$\eta = -\lambda$$, fitting with the notion that for the Gamma distribution with rate $$\lambda$$, the reciprocal link $$\eta = 1/\mu$$ is natural and correct. This aligns with $$\mu = 1/\lambda$$, thus $$\eta = -1/\mu$$ makes $$\eta = -\lambda$$, the correct canonical form for $$\theta$$.