TITLE: Fisher Information of a function of a parameter
QUESTION [14 upvotes]: Suppose that $X$ is a random variable for which the p.d.f. or the p.f. is $f(x|\theta)$, where the value of the parameter $\theta$ is unknown but must lie in an open interval $\Omega$.  Let $I_0(\theta)$ denote the Fisher information in $X.$  Suppose now that the parameter $\theta$ is replaced by a new parameter $\mu$, where $\theta = \psi(\mu)$ and $\psi$ is a differentiable function.  Let $I_1(\mu)$ denote the Fisher information in $X$ when the parameter is regarded as $\mu.$  Show that $$I_1(\mu) = [\psi'(\mu)]^2 I_0[\psi(\mu)].$$
How would I do this?  Do I need to use a Taylor expansion?  Regardless, I would appreciate a written proof.  This isn't for class but the above statement has been mentioned in texts without any detail whatsoever.
Thanks!

REPLY [10 votes]: By definition $I_{0}(\theta)=-\mathbb{E}\left[\frac{d^{2}\log f\left(X\vert\theta\right)}{d\theta^{2}}\right].$
So $I_{1}(\mu)=-\mathbb{E}\left[\frac{d^{2}\log f\left(X\vert\mu\right)}{d\mu^{2}}\right]$.By
the chain rule we have 
$$I_{1}\left(\mu\right) =  -\mathbb{E}\left[\frac{d^{2}\log f\left(X\vert\theta\right)}{d\theta^{2}}\left(\frac{d\theta}{d\mu}\right)^{2}+\frac{d\log f\left(X\vert\theta\right)}{d\theta}\frac{d^{2}\theta}{d\mu^{2}}\right]$$
      $$=  -\mathbb{E}\left[\frac{d^{2}\log f\left(X\vert\theta\right)}{d\theta^{2}}\right]\left(\frac{d\theta}{d\mu}\right)^{2}+\mathbb{E}\left[\frac{d\log f\left(X\vert\theta\right)}{d\theta}\right]\frac{d^{2}\theta}{d\mu^{2}}. $$
But $\mathbb{E}\left[\frac{d\log f\left(X\vert\theta\right)}{d\theta}\right]=0.$  So we get
$$
I_{1}\left(\mu\right)=I_{0}\left(\theta\right)\left(\frac{d\theta}{d\mu}\right)^{2}.$$