# Finding unbiased estimator of function of p from a geometric distribution

Let $X_1$ and $X_2$ be a random sample from the geometric distribution with $Pr(X_i=j)=p(1-p)^{j-1}$ for $i = 1, 2$, $j = 1, 2, \ldots$ and $0<p<1$.

Which statistics $T(X)$ could be an unbiased estimator for $\frac{p}{(1+p)}$?

My attempt: Let $T(X)$ be the required unbiased estimator. Then, $E(T(X))=\frac{p}{(1+p)}$ and $$\sum T(j)p(1-p)^{j-1}=p(1+p)^{-1}=p(1-p+p^{2}-p^{3}+...) \\ \sum T(j)(1-p)^{j-1}=(1-p+p^{2}-p^{3}+...)$$

Can we find the estimator by comparing the two sides of above equation? If not, how can we find the required unbiased estimator?

As it is said that $X$ is a random sample of size $2$, should we consider the joint pmf while finding the unbiased estimator?

• That summation should be actually $\sum_{j=1}^{\infty} T(j)(1-p)^{j-1}$ instead of $\sum_{j=1}^{\infty} T(X)(1-p)^{j-1}$. – Marcelo Ventura May 30 '16 at 18:07
• Recall the definition of expected value (to correct your formulas) and use $q=1-p$. – Viktor May 30 '16 at 18:12
• Since the pgf of this geometric pmf is $E(s^X)=\frac{ps}{1-(1-p)s}$, setting $\frac{ps}{1-(1-p)s}=\frac{p}{1+p}$ gives $s=\frac{1}{2}$. Hence an unbiased estimator based on two observations is simply $\frac{2^{-X_1}+2^{-X_2}}{2}$. – StubbornAtom Jun 1 '19 at 14:55
• @StubbornAtom , you should transform your comment into an answer. – Marcelo Ventura May 21 at 5:48

As is shown here in a similar question, one can simply work with the PGF or MGF to find an unbiased estimator based on a single observation.

For a geometric variable $$X$$ with given pmf, we have

\begin{align} E\left[t^X\right]&=\sum_{j=1}^\infty t^jp(1-p)^{j-1} \\&=pt\sum_{j=1}^\infty ((1-p)t)^{j-1} \\&=\frac{pt}{1-(1-p)t}\qquad\quad,\,\,\small|t|<\frac1{1-p} \end{align}

Solving $$\frac{pt}{1-(1-p)t}=\frac{p}{1+p}$$ for $$t$$ gives $$t=\frac12$$.

That is, for every $$p\in(0,1)$$,

$$E\left[2^{-X}\right]=\frac{p}{1+p}$$

So an unbiased estimator of $$\frac{p}{1+p}$$ based on two observations $$X_1$$ and $$X_2$$ is simply $$T(X_1,X_2)=\frac{2^{-X_1}+2^{-X_2}}2$$