0
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ you can pull it out, the expectation is with respect to $X$ and $g'(\theta)$ is constant with respect to $X$ $\endgroup$ – user365239 Oct 4 '17 at 4:18
  • $\begingroup$ of course... Thank you!!! $\endgroup$ – Hugh Entwistle Oct 4 '17 at 4:19
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.