Let $X$ follow a Poisson $(\lambda)$ distribution, where $\lambda>0 $ is unknown. If $\delta(X)$ is the unbiased estimator of $g(\lambda)=e^{-\lambda}(3\lambda^2 + 2\lambda+ 1)$, then what is the value of $\sum_{k=0}^\infty \delta(k)$ ?
I know that unbiased estimator implies $E(\delta(X))=e^{-\lambda}(3\lambda^2 + 2\lambda + 1)$. But I am not sure how to proceed.
$\newcommand{\e}{\operatorname{E}}$You have $$ \e(\delta(X)) = e^{-\lambda}(3\lambda^2 + 2\lambda+ 1) \text{ for ALL positive values of $\lambda.$} $$ Therefore \begin{align} & e^{-\lambda}(3\lambda^2 + 2\lambda+ 1) = \e(\delta(X)) = \sum_{x=0}^\infty \delta(x) \Pr(X=x) \\[10pt] = {} & \sum_{x=0}^\infty \delta(x) \frac{\lambda^x e^{-\lambda}}{x!} \\[10pt] = {} & e^{-\lambda} \sum_{x=0}^\infty \delta(x) \frac{\lambda^x}{x!} \text{ This works because $e^{-\lambda}$ does not change as $x$ goes from $0$ to $\infty.$} \\[10pt] & \text{Therefore } 3\lambda^2 + 2\lambda+1 = \sum_{x=0}^\infty \delta(x) \frac{\lambda^x}{x!} \text{ for ALL positive values of $\lambda$.} \end{align} A theorem that is not very hard to prove says that two power series cannot be everywhere equal unless all of their coefficients are equal. Thus we have $$ 1 + 2\lambda + 3\lambda^2 + 0 \lambda^3 + 0 \lambda^4 + 0\lambda^5 + \cdots = \delta(0) + \delta(1) \lambda + \frac{\delta(2)} 2 \lambda^2 + \frac{\delta(3)} 6 \lambda^3 + \frac{\delta(4)}{24} \lambda^4 + \cdots $$ and so \begin{align} 1 & = \delta(0) \\[10pt] 2 & = \delta(1) \\[10pt] 3 & = \delta(2)/2 \\[10pt] 0 & = \delta(n) \text{ for all larger values of $n.$} \end{align} So $$ \delta(0) + \delta(1) + \delta(2) + \delta(3) + \cdots = 1 + 2 + 6 + 0 + 0 + 0 + \cdots. $$