Let $X\sim Poisson(\lambda)$, 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 $\sum_{k=0}^\infty\delta(k)$ ?
As $\delta(X)$ is an unbiased estimator of $g(\lambda)$, I know that $E(\delta(X))=\sum_{k=0}^\infty \frac{\delta(k)e^{-\lambda}\lambda^k}{k!}=g(\lambda)=e^{-\lambda}(3\lambda^2+2\lambda+1)$. How do I proceed further, how do I separate $\sum_{k=0}^\infty\delta(k)$ from the product with poisson distribution in the equation?
We know that $$\sum_{k=0}^\infty \delta(k) \frac{\lambda^k}{k!} = 3\lambda^2 + 2\lambda + 1.$$ So we could pick any $\delta$ that makes this true; e.g., $$\delta(X) = \begin{cases} (X+1)!, & X \in \{0, 1, 2\}, \\ 0, & \text{otherwise}. \end{cases}$$ This would give us $\sum_{k=0}^\infty \delta(k) = 1 + 2 + 6 = 9$.