The Cramer Rao Lower Bound for The Geometric Distribution
Introduction
Suppose that we have a unbiased estimator of a parameter of a distribution. The Cramer-Rao inequality provides us with a lower bound for the variance of such an estimator. After we compute the Cramer-Rao Lower Bound (CRLB), we can check whether the unbiased estimator at hand reaches a minimum variance over all the unbiased estimators of the parameter.
The following is the Cramer-Rao Inequality.
Let
be an unbiased estimator of
which is a parameter of the distribution with probability density function
and log-likelihood
. Then the Cramer-Rao Inequality is given by:
![]()
where the function
known as Fisher Information is given by
, and
is the sample size over which the estimator
is estimated.
Thus the Cramer-Rao Lower Bound for the variance of
is given by:
, that is,
.
In this article, we will derive the Cramer-Rao Lower Bound of an estimator of the parameter
of the geometric distribution.
The Geometric Distribution
The probability density function of the geometric distribution (having parameter
) is given by:
![]()
Using standard probability theory, it follows that for the geometric distribution
.
The Cramer-Rao Lower Bound for the Geometric Distribution
Consider first the Fisher Information for the Geometric Distribution with parameter
.
![Rendered by QuickLaTeX.com \begin{equation*}\begin{align}I(p)&=-n\mathbb{E}\Big[\frac{\delta^2 l(x;p)}{\delta p^2}\Big]\\ &=-n\mathbb{E}\Big[\frac{\delta^2}{\delta p^2} [\ln ((1-p)^{x-1}p)]\Big]\\ &=-n\mathbb{E}\Big[\frac{\delta^2}{\delta p^2} [(x-1)\ln (1-p)+\ln p]\Big]\\ &=-n\mathbb{E}\Big[\frac{\delta}{\delta p} [-\frac{x-1}{1-p}+\frac{1}{p}]\Big]\\ &=n\mathbb{E}\Big[\frac{\delta}{\delta p} [\frac{x-1}{1-p}-\frac{1}{p}]\Big]\\ &=n\mathbb{E}\Big[\frac{x-1}{(1-p)^2}+\frac{1}{p^2}\Big]\\ &=n\Big(\frac{\mathbb{E}[x]-1}{(1-p)^2}+\frac{1}{p^2}\Big)\\ &=n\Big(\frac{\frac{1}{p}-1}{(1-p)^2}+\frac{1}{p^2}\Big)\text{ (since }\mathbb{E}[x]=\frac{1}{p})\\ &=n\Big(\frac{1-p}{p(1-p)^2}+\frac{1}{p^2}\Big)\\ &=n\Big(\frac{1}{p(1-p)}+\frac{1}{p^2}\Big)\\ &=n\Big(\frac{p+(1-p)}{p^2(1-p)} \Big)\\ &=\frac{n}{p^2(1-p)}. \end{align} \end{equation*}](https://datasciencegenie.com/wp-content/ql-cache/quicklatex.com-20f0a478ed24e56e84274add94ab11f8_l3.png)
Thus, the Cramer-Rao Lower Bound for the variance of an estimator of
of the geometric distribution is given by:
