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Let $x_1,...,x_n$ have the distribution

$$ P(X = x) = \frac{\theta^xe^{-\theta}}{x!(1-e^{-\theta})}, \ \ x = 1,2,3... $$

Now we want to find UMVUE for $e^{-\theta}$. My first thought was to apply the Rao-Blackwell theorem with help from the complete sufficient statistic $\sum x_i$. For a standard Poisson, the UMVUE would easily be found applying the Lehmann–Scheffé theorem using $P(X = 0) = e^{-\theta}$. But for the the zero-truncated, the Lehmann-Scheffé theorem does not seem to fit the problem. Has anyone a solution to this?

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First of all you could rewrite the distribution to fit the general form of the exponential family. $$ \frac{1}{x!}e^{x\log\theta -\theta -\log(1-e^{-\theta})} $$ so the sufficient statistics is $T:=x$, $\eta:=\log\theta$ and $A(\eta):=-e^\eta-\log(1-e^{-e^\eta})$. By properties of the exponential family

$$\mathbb{E}[T]=\frac{\partial A(\eta)}{\partial\eta}=\frac{e^{\eta+e^\eta}}{1-e^{e^\eta}}:= \frac{\theta e^{\theta}}{1-e^\theta}$$

and now you can construct your unbiased estimator as $$ \hat{e^\theta}=\left\lbrace z:\frac{1}{n}\sum_{i=1}^n x_i=\frac{z\log z}{z-1}\right\rbrace$$ which will be UMVUE by the Lehmann-Scheffé theorem.

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