TITLE: For the binomial distribution, why does no unbiased estimator exist for $1/p$?
QUESTION [12 upvotes]: Suppose that $X \sim \mathrm{Binomial}(n,p)$ for $0 < p < 1$
Why does no unbiased estimator exist for $1/p$?
My approach:
We try to find the structure of $E_p(U(x))$, where $U(x)$ is any estimator of $1/p$.
Now, we will have:
$$\sum{U(x)\binom{n}{x}\theta^x(1-\theta)^{n-x}}<\sum{U(x)\binom{n}{x}}<M(n)<\infty$$
so that the expectation is bounded above. So this is supposed to mean that if $\theta < 1/M(n)$, then the expectation cannot attain $1/\theta$ but I am not sure why the above argument even makes sense and what being bounded means for the expectation. 
Thanks!

REPLY [3 votes]: Suppose $X_1,...,X_n$ iid Bernoulli($\theta$). If we want to estimate $\tau(\theta)=\frac{1}{\theta}$, we then find that no unbiased estimator exists.
We can use MLE of $\tau(\theta)=\frac{1}{\theta}$, which is $\tau(\hat\theta)=\frac{1}{\bar x}$, by invariance principle. But this is biased.
Now I provide the proof to show that no unbiased estimator exists. We can use contradiction.
Suppose W(x) is an unbiased estimator of $\frac{1}{\theta}$. That means whatever estimator we choose (not only MLE), it cannot be unbiased. Let M=supW(x). Recognize that $M<\infty$, since M is from finite sample. Note that our sample x($X_1,...,X_n$) is either 0 or 1. We have $2^n$ possible data sets. It is finite, thus $M<\infty$. (In analysis, we can also say sup is just your maximum due to finiteness.)
Then $E_\theta$ W(x) $\leq M$ for all $\theta$. In other words, M is an upper bound.
By unbiased definition, $E_\theta$ W(x) = $\frac{1}{\theta}. $But it is$ \rightarrow \infty$ as $\theta \rightarrow 0$. Note that it is allowed to let $\theta \rightarrow 0$ because $\theta$ is in (0,1) and we can let $\theta$ be very small. And when $\theta$ is very small, then $ \frac{1}{\theta}\rightarrow \infty$.
But previously, we show that $E_\theta$ W(x) $\leq M$ for all $\theta$, including the case when $\theta$ is very small. Hence, this is contradiction.