 Solution

An exponential dispersion family has density or mass functionwith support independent of the natural parameter $\theta$. Usually $a(\varphi )=\varphi /w$ for a known weight $w$. For the Poisson distribution, identifyThe dispersion is fixed at one. Differentiating the normalizing identity gives $\mathbb{E}Y=b^{\prime }(\theta )$; differentiating again gives $\mathrm{Var}Y=a(\varphi )b^{\prime \prime }(\theta )$. Here both derivatives equal $e^{\theta }=\lambda$, soThe variance function expresses $b^{\prime \prime }(\theta )$ in terms of the mean $\mu$. The canonical link function maps the mean to the natural parameter, so the Poisson canonical link is $g(\mu )=\log \mu$.