# UMVUE of parameter from Zero Truncated Poission distribution

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?

## 1 Answer

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 $$\hat{e^\theta}=\left\lbrace z:\frac{1}{n}\sum_{i=1}^n x_i=\frac{z\log z}{1-z}\right\rbrace$$ which will be UMVUE by the Lehmann-Scheffé theorem.

• I am confused. How did you construct the unbiased estimator in your last step? Nov 18, 2023 at 4:45
• Note that the empirical expectation of $T$ is simply the mean of all $\{x_i\}_{i\in[n]}$. Then, we let $z:=e^\theta$ and use the expression of $\mathbb{E}[T]$ to determine this quantity. Nov 22, 2023 at 9:43
• I see how $\hat{z}$ can be derived from $\bar{ x}$. But it is not so clear why that derived $\hat{z}$ should also be unbiased. Nov 22, 2023 at 11:34
• So, if we have $E[T] = f(\theta)$ does that make the derived estimator $\hat\theta = f^{-1}(T)$ in general unbiased? Nov 22, 2023 at 11:47
• Refer to the theorem I invoke in the reply: en.wikipedia.org/wiki/Lehmann%E2%80%93Scheff%C3%A9_theorem Nov 22, 2023 at 20:03