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?

  • 2
    $\begingroup$ That summation should be actually $\sum_{j=1}^{\infty} T(j)(1-p)^{j-1}$ instead of $\sum_{j=1}^{\infty} T(X)(1-p)^{j-1}$. $\endgroup$ – Marcelo Ventura May 30 '16 at 18:07
  • 1
    $\begingroup$ Recall the definition of expected value (to correct your formulas) and use $q=1-p$. $\endgroup$ – Viktor May 30 '16 at 18:12
  • $\begingroup$ 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 $\frac{2^{-X_1}+2^{-X_2}}{2}$. $\endgroup$ – StubbornAtom Jun 1 at 14:55

Following your lines, for just one observation X, you can get $$ \sum_{x=1}^{\infty} T(x) p (1-p)^{x-1} = p (1+p)^{-1} \\ \sum_{x=1}^{\infty} T(x) (1-p)^{x-1} = (1+p)^{-1} \\ \sum_{x=1}^{\infty} T(x) (1-p)^{x-1} (1+p) = 1. $$

So, any $T(x)$ that satisfy that last equation will be enough, like, for instance, $t(x) := 2^{-x}(1-p)^{-x+1} (1+p)^{-1}$, since $$ \sum_{x=1}^{\infty} t(x) (1-p)^{x-1} (1+p) = \sum_{x=1}^{\infty} 2^{-x}(1-p)^{-x+1} (1+p)^{-1} (1-p)^{x-1} (1+p) = \sum_{x=1}^{\infty} 2^{-x} = 1 $$ and, once you have $t$, you can use it with both $X_1$ and $X_2$ and take their average $$ T(X_1,X_2) = \frac{t(X_1)+t(X_2)}{2}, $$ for $$ E(T(X_1,X_2)) = E\left(\frac{t(X_1)+t(X_2)}{2}\right) = \frac{E(t(X_1))+E(t(X_2))}{2} = \frac{\frac{p}{1-p}+\frac{p}{1-p}}{2} = \frac{p}{1-p}. $$

That way, you will not need to deal with joint pmf's.


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