Find the UMVUE of $\theta$ from Zero-truncated Poisson distribution

Question

Let $X_1,...,X_n$ be a random sample from the distribution $f_{\theta} (x) = C(\theta)e^{-\theta}\theta^x/x!, x=1,2,...,\infty$, for some $C(\theta)$. Find the UMVUE of $\theta$ when n=1 and the UMVUE of $\theta$ when n=2.

Answer 1

Suppose $g(X)$ is unbiased for $\theta$ when $n=1$.

That is,

$$E_{\theta}[g(X)]=\theta \quad,\,\forall\,\theta>0$$

This implies $$\sum_{j=1}^\infty g(j)\cdot\frac{\theta^j}{j!(e^\theta -1)}=\theta \quad,\,\forall\,\theta $$

Or, $$\sum_{j=1}^\infty \frac{g(j)}{j!}\cdot\theta^j=\theta(e^{\theta}-1) \quad,\,\forall\,\theta$$

Now recall the power series expansion of $e^\theta$ and rewrite the right hand side of the last equation as a series. Equate the coefficients of $\theta^j$ from both sides and solve for $g(\cdot)$. This results in a unique unbiased estimator, which also makes it the best unbiased estimator.