 Solution

For $\lambda >0$, write the Poisson distribution mass function in exponential dispersion family form:Thus the natural parameter of an exponential family is $\theta =\log \lambda$ and the dispersion parameter is fixed at $1$. The counting-measure support is $0,1,\dots$; there is no independent free dispersion parameter in an ordinary Poisson model. The degenerate boundary $\lambda =0$ is obtained as a limit, rather than a finite natural parameter.

The exponential dispersion family identities give $\mathbb{E}Y=b^{\prime }(\theta )$ and $\mathrm{Var}(Y)=a(\varphi )b^{\prime \prime }(\theta )$. Since both derivatives of $e^{\theta }$ equal $e^{\theta }$,