Suppose that $X$ has the Poisson distribution truncated on the right at $a$, so that it has the conditional distribution of $Y$ given $Y \leq a$, where $Y$ is distributed as $P(\lambda)$. Show that $\lambda$ does not have an unbiased estimator.
I'm trying to prove this using a proof by contradiction. Suppose $\delta(X)$ is an unbiased estimator for $g(\lambda) = \lambda$. Then:
$$\sum_{k=0}^{\infty} \delta(k)\frac{e^{-\lambda} \lambda^k}{k!} 1_{k \leq a} = \lambda $$
Rearranging I get:
$$\sum_{k=0}^{\infty} \delta(k)\frac{\lambda^{k-1}}{k!} 1_{k \leq a} = e^{\lambda} $$
Is this a contradiction because the finite sum will never equal $e^\lambda$? Or do I need another argument?