Unbiased estimator for truncated poisson This is supposed to be a classic problem but I could not seem to find a solution anywhere. 
Let $X$ be a random variable with zero-truncated Poisson distribution s.t. it's pmf is given by:
$f(x)=\frac{e^{-\theta}\theta^x}{(1-e^{-\theta})x!}$  $x=1,2...$
Find an unbiased estimator for $1-e^{-\theta}$
 A: The answer you have given in a comment is correct. I will show (I am lazy so I will let maple do the work). Let 
$$
   T(x)=\begin{cases}   0 &\text{if $x$ is odd}\\
                        2 &\text{if $x$ is even}
        \end{cases}
$$ be an estimator. We will show it is unbiased. The maple code is below:
f := x -> exp(-theta)*theta^x/( (1-exp(-theta))*factorial(x) )
ans := sum( f(2*k)*2, k=1..infinity ) assuming theta>0;
                          -2 + 2 cosh(theta)
                   ans := ------------------
                            exp(theta) - 1  
ans := convert(ans, exp)
                     -2 + exp(theta) + exp(-theta)
              ans := -----------------------------
                            exp(theta) - 1        

which gives the expectation as
$$
{\frac {-2+{{\rm e}^{\theta}}+{{\rm e}^{-\theta}}}{{{\rm e}^{\theta}}-
1}}
$$
and this you can now simplify yourself to see that it is $1-e^{-\theta}$.  So we have found an unbiased estimator. And it is really absurd, since it can be only 0 or 2 so is extremely unhelpful. Now the real work would be to show there is no other unbiased estimator!
