# 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 $1-e^{-\theta}$

• I tried a few things..couldn't get it (+1) Commented Aug 16, 2017 at 4:22
• The answer is supposed to be absurd but analytically accurate. Specifically, 0 if X is odd and 2 if it is even. I still can't figure out how to make that happen though. Commented Aug 16, 2017 at 8:50
• stats.stackexchange.com/q/544225/119261 Commented Nov 22, 2023 at 6:51

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!