unbiased estimate of a geometric model

Question

I was reading a book where in one of the section it shows how to find the unbiased estimate of a geometric model. This is from the book:

Let x denote $n =1$ realisation from a geometric $Geom(\pi)$ distribution with pmf $f(x;\pi) = \pi(1-\pi)^{(x-1)}$

Now there is only one unbiased estimate of $\pi$ because the requirement $E(T(X)) = \sum_{x= 1}^\infty T(x)\pi(1-\pi)^{(x-1)} = \pi$ for all $\pi \in (0,1)$ leads to the solution of $T(x) = 1$ if $x= 1$ , else $T(x) = 0$.

My question is how do the author get $T(x) = 1$ if $x =1$ , and $T(x) = 0$ for everything else.

The $T(x)$ here is just a symbol for some test statistics.

thank you

Answer 1

There are a couple semantic issues with your question (we want an estimator of p, not an estimate of a geometric R.V and, more importantly, T(X) is not a test statistic in this case as we are not doing any hypothesis testing) but I think I understand what you mean.

We need our estimator to be unbiased which, with one sample observation, means we need $E [T (x)]=p \iff \sum_{x=1}^{n} T(x)p(1-p)^{x-1}=p $. This will certainly be the case when we take T to be 0 for all x not equal to 1 as it will 0 out all contributions from the 1-p term.