Question

Question:Suppose that X has the geometric distribution with parameter p. (See Sec. 5.5.) Find the Fisher information I (p) in X.

Short answer

The Fisher information is \(I\left( p \right) = \frac{1}{{{p^2}\left( {1 - p} \right)}}\)

Step-by-step solution

Given information

X has the geometric distribution with parameter p.

Finding the Fisher information

The probability mass function of X is,

\(f\left( {x\left| p \right.} \right) = p{\left( {1 - p} \right)^x},\,\,for\,\,x = 0,1,...\)

Taking log

\(\begin{array}{c}\log f\left( {x\left| p \right.} \right) = \log \left( {p{{\left( {1 - p} \right)}^x}} \right)\\ = \log \left( p \right) + x\log \left( {1 - p} \right)\end{array}\)

The first-orderderivative with respect to p is,

\(\begin{array}{c}\frac{\partial }{{\partial p}}\left[ {\log f\left( {x\left| p \right.} \right)} \right] = \frac{\partial }{{\partial p}}\left[ {\log \left( p \right) + x\log \left( {1 - p} \right)} \right]\\ = \frac{1}{p} - \frac{x}{{\left( {1 - p} \right)}}\end{array}\)

The variance of geometric distribution is,

\(Var\left( X \right) = \frac{{\left( {1 - p} \right)}}{{{p^2}}}\)

The Fisher information is,

\(\begin{array}{c}I\left( p \right) = Var\left[ {\frac{1}{p} - \frac{X}{{\left( {1 - p} \right)}}} \right]\\ = \frac{{Var\left( X \right)}}{{{{\left( {1 - p} \right)}^2}}}\\ = \frac{{\frac{{\left( {1 - p} \right)}}{{{p^2}}}}}{{{{\left( {1 - p} \right)}^2}}}\\ = \frac{1}{{{p^2}\left( {1 - p} \right)}}\end{array}\)

So, the Fisher information is \(I\left( p \right) = \frac{1}{{{p^2}\left( {1 - p} \right)}}\)