Question

Question: Suppose that a random variable X has the Poisson distribution with unknown mean \({\bf{\theta }}\) >0. Find the Fisher information \({\bf{I}}\left( {\bf{\theta }} \right)\) in X.

Short answer

The Fisher Information is \(\frac{n}{\theta }\)

Step-by-step solution

Given information

It is given that X is a random variable that follows Poisson distribution with unknown parameter \(\theta \). Therefore \({X_1} \ldots {X_n}\) are iid \(Poisson\left( \theta \right)\) .

Define the pdf anStep 2: Define the pdf and find the expectation and varianced find the expectation and variance

\(f\left( {x{\rm{ }}|{\rm{ }}\theta } \right) = \frac{{{\theta ^x}{e^{ - \theta }}}}{{x!}},x = 0,1, \ldots \)

By the properties of a Poisson distribution

\(\begin{array}{l}E\left( X \right) = \theta \\Var\left( X \right) = \theta \end{array}\)

Define fisher information

Assume \(X \sim f\left( {x{\rm{ }}|{\rm{ }}\theta } \right)\) (pdf or pmf) with\(\theta \in \Theta \subset R\) .

Then fisher information is defined by

\(\begin{array}{c}{I_x}\left( \theta \right) = {E_\theta }\left[ {{{\left( {\frac{\partial }{{\partial \theta }}\log f\left( {X|\theta } \right)} \right)}^2}} \right]\\ = {E_\theta }\left( { - \frac{{{\partial ^2}}}{{\partial {\theta ^2}}}\log f\left( {X|\theta } \right)} \right)\end{array}\)

And

\({I_x}\left( \theta \right) = n{I_x}_{_1}\left( \theta \right)\)

Calculating fisher information for Poisson distribution

Let \(X = {X_1}\)

From the definition,

\(\begin{array}{c}{I_x}\left( \theta \right) = {E_\theta }\left( {{{\left( {\frac{\partial }{{\partial \theta }}\log f\left( {X|\theta } \right)} \right)}^2}} \right)\\ = {E_\theta }\left( {{{\left( {\frac{X}{\theta } - 1} \right)}^2}} \right)\\ = Va{r_\theta }\left( {\frac{X}{\theta }} \right)\,\,\,\left( {\,{\rm{Since}}\,\,{\rm{E}}\left( {\frac{{\rm{X}}}{{\rm{\theta }}}} \right){\rm{ = }}\frac{{{\rm{E}}\left( {\rm{X}} \right)}}{{\rm{\theta }}}{\rm{ = 1}}} \right)\end{array}\)

\(\begin{array}{c} = \frac{{Va{r_\theta }\left( X \right)}}{{{\theta ^2}}}\\ = \frac{\theta }{{{\theta ^2}}}\\ = \frac{1}{\theta }\end{array}\)

We know that

\(\begin{array}{c}{I_x}\left( \theta \right) = n{I_x}_{_1}\left( \theta \right)\\ = \frac{n}{\theta }\end{array}\)

Therefore, the Fisher Information is \(\frac{n}{\theta }\)