The question goes like this: Use Jeffreys prior for Bernoulli distribution and find the prior for $\eta$ where: $$\eta(p) = \left(\frac{p}{1-p}\right) $$
So $\eta$ here is some kind of a transformation on the Bernoulli parameter $p$, and I know that Jeffreys prior $\pi(p)$, for Bernoulli distribution, is:
$$\pi(p) = \sqrt{I(p)} = \sqrt{\frac{1}{p(1-p)}}$$
The solution for the question is:
$$f_n(\eta) = f_p(g^{-1}(\eta))\cdot (g^{-1}(\eta))'$$ $$= \sqrt{\frac{1}{p(1-p)}} \cdot p(1-p)$$ $$= \sqrt{p(1-p)}$$
I'm trying to understand the solution:
$f_p(g^{-1}(\eta))$ is just $\pi(p)$, right? cause I see its equal to $\sqrt{\frac{1}{p(1-p)}}$.
And I didn't get what is $(g^{-1}(\eta))'$.. here its equal to $p(1-p)$ so it can't be the derevative of $\pi(p)$, so what it is then?
And basically they didn't use the fact that $\eta(p) = \left(\frac{p}{1-p}\right)$, they just used the prior for Bernoulli, right?