# Reparameterisation Using Fisher Information

If $I(\theta)$ is the Fisher Information for $\theta$

Then how do I find $J=I(g(\theta ))$ ?

Currently my thoughts are that:

$$I(\theta) = E\left( \left(\frac{\partial}{\partial \theta}\log f(X;\theta)\right)^2 \right)$$ and if $u=g(\theta)$ where $g( \cdot )$ is differentiable, then $$\frac{\partial}{\partial u}=\frac{\partial\theta}{\partial u} \frac{\partial}{\partial \theta} =\frac{1}{g'(\theta)}\frac{\partial}{\partial \theta}$$ So $$I(u) = E\left( \left(\frac{\partial}{\partial u}\log f(X;u)\right)^2 \right) = E\left( \frac{1}{(g'(\theta))^2}\left(\frac{\partial}{\partial \theta}\log f(X;\theta)\right)^2 \right) \ \ \text{as u is a function of} \ \theta$$ but at this point I don't see the justification for taking the $\frac{1}{(g'(\theta))^2}$ out of the bracket...

I'm guessing the result should arrive at

$$I(g(\theta))=\frac{I(\theta)}{(g'(\theta))^2}$$

but I'm not sure how to get this...

• you can pull it out, the expectation is with respect to $X$ and $g'(\theta)$ is constant with respect to $X$ – user365239 Oct 4 '17 at 4:18
• of course... Thank you!!! – Hugh Entwistle Oct 4 '17 at 4:19

The expectation operator is with respect to $X$ and $g'(\theta)$ is constant with respect to $X$ so it can be taken out of the operator -- giving the required result.