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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?

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  • $\begingroup$ 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 $\theta$). $\endgroup$
    – Xi'an
    Commented Aug 30, 2020 at 15:08
  • $\begingroup$ I have simply copy pasted the solution manual for "Bayesian Data Analysis" by Gelman. @Xi'an $\endgroup$ Commented Aug 30, 2020 at 17:43

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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 $$q_2(\theta)=q_1(\phi(\theta)) \times \underbrace{\left|\frac{\text{d}\phi(\theta)}{\text{d}\theta}\right|}_{\text{Jacobian}}$$

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