# Find the prior distribution for the natural parameter of an exponential family

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?

• Please add the self-study tag. 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 $\theta$). – Xi'an Aug 30 at 15:08
• I have simply copy pasted the solution manual for "Bayesian Data Analysis" by Gelman. @Xi'an – xxtensionxx Aug 30 at 17:43

The Jacobian or change-of-variable formula lets one find the density of a bijective transform of a random variable. If $$\theta$$ is a random variable with density $$q_1(\cdot)$$ and if the density of $$\phi=\phi(\theta)$$ is denoted $$q_2(\cdot)$$ then $$q_2(\theta)=q_1(\phi(\theta)) \times \underbrace{\left|\frac{\text{d}\phi(\theta)}{\text{d}\theta}\right|}_{\text{Jacobian}}$$