Show that for the binomial likelihood $y$ ~$Bin(n, \theta)$, $p(\theta) \propto \theta^{-1} (1-\theta)^{-1}$ is the uniform prior distribution for the natural parameter of the exponential family.
I am trying to simply understand the solution to the question given above. The solution goes as follows:
The binomial can be put in the form of an exponential family with (using the notation of Section 2.4) $f(y)$ = $ {n}\choose{k}$, $g(\theta) = (1-\theta)^n$ and $u(y) = y$ and natural parameter $\phi(\theta) = log(\theta/(1-\theta))$.
A uniform prior density on $\phi(\theta)$, $p(\phi) \propto 1$ on the entire real line, can be transformed to give the prior density for $\theta = \frac{e^{\phi}}{1+e^{\phi}}$:
(And here comes the part I do not understand in the solution)
$$q(\theta) = p(\frac{e^{\phi}}{1+e^{\phi}})|\frac{d}{d\theta}log(\frac{\theta}{1-\theta})| \propto \theta^{-1} (1-\theta)^{-1}$$
Could anybody pleas help me understand how they get top the last part?
self-studytag. Have you heard of the change of variable or Jacobian formula? The question is poorly worded (there is no such thing as a Uniform distribution on the real line) and so is the solution: $\phi$ should not appear in a formula involving $\theta$). $\endgroup$