# How to prove the Poisson link function is a canonical link function?

So I'm a 3rd year undergraduate doing my thesisin football score models right now. In my thesis I want to include a proof of what the link function for the Poisson distribution is and why it relates the mean to our linear predictors. I'm almost there, but there is one part that most literature seems to gloss over.

So we have our linear predictors η =β0 + β1_xi1+⋯+βp_xip and our natural parameter θ. Now I know why if η = θ then our link function is canonical, but my question is how do we prove that we have η = θ in the case of the Poisson distribution? Most literature just state that the canonical link sets this equality, or we can assume this equality for distributions that are members of the exp. family but don't actually prove why. Can anyone help?

• Welcome to Cross Validated! What link function do you mean? A regression with a conditional Poisson can have any number of link functions.
– Dave
Mar 7 at 18:15
• Sorry, I'm talking about the log link function Mar 7 at 18:23

In general, if the conditional mass/density function for a single observation is of the form $$f_{Y_i|\mathbf{X}_i}(y_i|\mathbf{x}_i)=h(y_i, \phi) \exp\left\{\frac{\theta_iy_i-b(\theta_i)}{\tau(\phi)} \right\}$$ where $$\theta_i$$ depends on $$\mathbf{x}_i$$, we say that the distribution belongs to an exponential dispersion family. The parameters $$\theta_i$$ and $$\phi$$ are location and scale parameters.
We can then show that the mean function is $$\mu(\mathbf{x}_i):=\mathbb{E}(Y_i|\mathbf{X}_i=\mathbf{x}_i)=b'(\theta_i)$$ the derivative of the function $$b$$.
If we then take the location parameter to be a linear function of the explanatory variables, $$\theta_i = \beta_0 + \beta_1 x_{i1}+\ldots + \beta_p x_{ip} = \mathbf{x}_i \mathbf{\beta}$$, we have $$\mu(\mathbf{x}_i)=b'(\mathbf{x}_i \mathbf{\beta}) \Rightarrow b'^{-1}(\mu(\mathbf{x}_i)) = \mathbf{x}_i \mathbf{\beta}$$ Thus, $$b'^{-1}$$ is a natural choice of link function and is known as the canonical link.