Question

Suppose that\(X\) is a random variable for which the p.d.f. or the p.f. is\(f\left( {x|\theta } \right)\) where the value of the parameter \(\theta \) is unknown but must lie in an open interval \(\Omega \). Let \({I_0}\left( \theta \right)\) denote the Fisher information in \(X\) . Suppose now that the

parameter \(\theta \) is replaced by a new parameter \(\mu \), where\(\theta = \psi \left( \mu \right)\) and\(\psi \) is a differentiable function. Let \({I_1}\left( \mu \right)\)denote the Fisher information in X when the parameter is regarded as \(\mu \). Show thatShow that\({I_1}\left( \mu \right) = {\left( {{\psi ^{'}}\left( \mu \right)} \right)^{2}}{I_0}\left( {\psi \left( \mu \right)} \right)\)

Short answer

\({I_1}\left( \mu \right) = {\left({{\psi^{'}}\left( \mu \right)} \right)^{2}}{I_0} \left( {\psi \left( \mu \right)} \right)\)

Step-by-step solution

Given information

X be the random variable for which the pdf is \(f\left( {x|\theta } \right)\) where \(\theta \) is unknown but it must lie in an open interval\(\Omega \)

\({I_0}\left( \theta \right)\) be the fisher information in X.

The parameter\(\theta \)when replaced by the new parameter\(\mu \), where\(\theta = \psi \left( \mu \right)\)and

\(\psi \)is a differentiable function.

\({I_1}\left( \mu \right)\) denote the fisher information in X when the parameter is regarded as \(\mu \)

Fisher information

Fisher information \(I\left( \theta \right)\) in the random variable X is defined as

\(I\left( \theta \right) = {E_\theta }\left\{ {{{\left( {{\lambda ^{'}}\left( {x|\theta } \right)} \right)}^{2}}} \right\}\)

Or

\(I\left( \theta \right) = - {E_\theta }\left\{ {\left( {{\lambda ^{''}}\left( {x|\theta } \right)} \right)} \right\}\)

Verifying \({I_1}\left( \mu  \right) = {\left( {{\psi ^{'}}\left( \mu  \right)} \right)^{2}}{I_0}\left( {\psi \left( \mu  \right)} \right)\)

Let \(g\left( {x|\mu } \right)\) be the pdf of X when \(\mu \) is parameter.

Then \(g\left( {x|\mu } \right) = f\left( {x|\psi \left( \mu \right)} \right)\)

Hence taking log on both the sides

\(\begin{align}\log g\left( {x|\mu } \right) &= \log f\left( {x|\psi \left( \mu \right)} \right)\\ &= \lambda \left( {x|\psi \left( \mu \right)} \right)\end{align}\)

Differentiate \(\lambda \left( {x|\psi \left( \mu \right)} \right)\) with respect to \(\mu \)

\(\begin{align}{\lambda ^{'}}\left( {x|\psi \left( \mu \right)} \right) &= \frac{\partial }{{\partial u}}\lambda \left( {x|\psi \left( \mu \right)} \right)\\ &= {\lambda ^{'}}\left( {x|\psi \left( \mu \right)} \right){\psi ^{'}}\left( \mu \right)\end{align}\)

Then firsher information in x when the parameter is regarded as \(\mu \) is

\(\begin{align}{I_1}\left( \mu \right) &= {E_\mu }\left\{ {{{\left( {{\lambda ^{'}}\left( {x|\psi \left( \mu \right)} \right)} \right)}^{2}}} \right\}\\ &= {E_\mu }\left\{ {{{\left( {{\lambda ^{'}}\left( {x|\psi \left( \mu \right)} \right){\psi ^{'}}\left( \mu \right)} \right)}^{2}}} \right\}\\ &= {\left( {{\psi ^{'}}\left( \mu \right)} \right)^{2}}{E_\mu }\left( {{{\left\{ {{\lambda ^{'}}\left( {x|\psi \left( \mu \right)} \right)} \right\}}^{2}}} \right)\end{align}\)

\({I_1}\left( \mu \right) = {\left( {\psi \left( \mu \right)} \right)^2}{I_0}\left( {\psi \left( \mu \right)} \right)\)

hence proved.