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 $e^\theta$.
 A: 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.
A: 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. 
