Question

Question:Suppose that a random variable X has the geometric distribution with an unknown parameter p. (See Sec. 5.5.).Find a statistic \({\bf{\delta }}\left( {\bf{X}} \right)\)that will be an unbiased estimator of\(\frac{{\bf{1}}}{{\bf{p}}}\).

Short answer

\(\delta \left( X \right) = \overline X \) is an unbiased estimator of \(\frac{1}{p}\)

Step-by-step solution

Given information

It is given that the random variable X has the geometric distribution with an unknown parameter p.

Define the pdf and find the expectation

Therefore, the pdf of X is \(P\left( {X = x} \right) = p{\left( {1 - p} \right)^{x - 1}},x > 0\)

The expectation is:

\(\begin{aligned}{}\sum x P\left( {X = x} \right) &= \sum\limits_{i = 1}^n {{x_i}} p{\left( {1 - p} \right)^{{x_i} - 1}}\\ &= \frac{1}{p}\end{aligned}\)

Therefore, the expectation is \(\frac{1}{p}\) .

Define the unbiased estimator

An estimator \(\delta \left( X \right)\,\,of\,\,g\left( \theta \right)\)is unbiased if \(E\left( {\delta \left( X \right)} \right) = g\left( \theta \right)\)for all possible values of \(\theta \) .

We also know that sample mean is an unbiased estimator of population mean.

Therefore,

\(\begin{aligned}{}E\left( {\delta \left( X \right)} \right) &= g\left( \theta \right)\\ \Rightarrow E\left( {\overline X } \right) &= \frac{1}{p}\end{aligned}\)

Therefore \(\delta \left( X \right) = \overline X \) is an unbiased estimator of \(\frac{1}{p}\).