# Poisson canonical form question - $b(\cdot)$

For the Poisson distribution in exponential canonical form, why is the $$b(\cdot)$$ part of the canonical form expressed as $$e^{\theta} \dots$$ since the exponential form is written as $$\exp(y\log(\lambda) - \lambda - \log(y!)$$. I could understand if the b($$\cdot$$) component would be $$b(\theta)$$ = $$\lambda$$ when $$\theta=\log(\lambda)$$, but not $$e^\theta$$.

To start with, not everyone uses the same notation for the terms in the exponential family form, (eg there's no $$b$$ in the Wikipedia version)
The Poisson pmf, as you note, can be written as $$\exp\left( y\log\lambda -\lambda -\log (y!)\right)$$ That's not the canonical form. In the canonical form, the first term (the one involving both parameter and data) is $$t(y)\theta$$, where $$t(y)$$ is a sufficient statistic and $$\theta$$ is the canonical parameter.
For the Poisson, this means $$\theta=\log\lambda$$ must be the canonical parameter, and the sufficient statistic $$t(y)$$ is just $$y$$, so you have the canonical form $$\exp\left( y\theta -e^\theta -\log (y!)\right)$$
I think it's easiest to understand this if you think of the construction of an exponential family. You start out with a density or pmf $$h(x)$$ and put in an additional parameter by exponential tilting $$f(x;\theta) \propto e^{x\theta}h(x)$$ What makes $$\theta$$ the canonical parameter for the Poisson is that the Poisson family is what you get by doing this with $$h(x)$$ being any single Poisson distribution.
I think $b(\theta)$ should be a function of $\theta$, so $b(\theta) = \exp(\theta)$. However, you are right that $b(\theta) = \exp(\theta) = \exp(\log(\lambda)) = \lambda$, too.