Question 11.10: The P(θ) p.d.f. is of the exponential type....

Here

f(x;\theta)=\frac{e^{-\theta}\theta^{x}}{x!}I_{A}(x),\quad A=\{0,1,\,\dots\}.

Hence

f(x;\theta)=e^{-\theta}\times e^{(\log\theta)x}\times{\frac{1}{x!}}I_{A}(x),

so that f(x;\theta) is of the form (4)
f(x;\theta)=C(\theta)e^{Q(\theta)T(x)}\times h(x),\ \ \ \ x\in\Re,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)
with C(\theta)=e^{-\theta},\,Q(\theta)=\log\theta strictly increasing, T(x) = x, and h(x)={\textstyle\frac{1}{x!}}I_{A}(x).