I'm self-studying Bayesian Data Analysis by Gelman et al. and I'm struggling to understand the solution to exercise 2.7a. The question:

For the binomial likelihood, $y\sim Bin(n, \theta)$, show that $p(\theta) \propto \theta^{-1}(1-\theta)^{-1}$ is the uniform prior distribution for the natural parameter of the exponential family.

What I understand:

Per the question -- the natural parameter = $\phi = log(\frac{\theta}{1-\theta})$. So, $p_\phi(\phi)$ needs to be uniformly distributed. I need to show that setting $p_\theta(\theta) \propto \theta^{-1}(1-\theta)^{-1}$ results in a uniform distribution on $p_\phi$.

The solution per Gelman's website:

$q(\theta)=|\frac{d}{d\theta}log(\frac{\theta}{1-\theta})|p(\frac{e^{\phi}}{1+e^{\phi}}) \propto \theta^{-1}(1-\theta)^{-1}$

and $\frac{e^{\phi}}{1+e^{\phi}} = \theta$.

Earlier in the text we're given that $p_v(v)=|J|p_u(f^{-1}(v))$ is how one transforms a continuous R.V.

My reading of the above solution, transforming it to fit the formula above, we have $v = \phi, u = \theta$, because $f^{-1}$ here is $\frac{e^\phi}{1 + e^\phi}$, so the $p_u$ on the right takes values of $\theta$. The derivative is equal to the proposed prior, so we get the term on the right of the solution if $p(\theta) = 1$. But that doesn't answer the question, as I've understood it. Also, $q(\theta)$ on the left, takes $\theta$ as an argument, so we have two parts of this taking $\theta$ when it should be one on the right or the left. So the solution makes no sense to me. Is the answer as given on Gelman's website correct? Where am I misunderstanding what the answer is supposed to show?


1 Answer 1


You're almost there, but I would derive this in the other direction. You know the prior you want to specify on $\phi$, which is uniform (i.e. proportional to a constant) and want to derive the induced distribution on $\theta$ from this. So in the notation from BDA3 you have $p_v(v) = |J|p_u(f^{-1}(v)$, where $v = \theta$, $u = \phi$ and $J$ is the Jacobian of the transformation $u = f^{-1}(v)$, i.e. $\phi = f^{-1}(\theta)$. This Jacobian is (as you've already shown) equal to $\theta^{-1}(1 - \theta)^{-1}$.

Putting this all together, we find that $$ p_\theta(\theta) = |J| p_\phi(f^{-1}(\theta) = \theta^{-1}(1 - \theta)^{-1} p_\phi(f^{-1}(\theta) \propto \theta^{-1}(1 - \theta)^{-1} $$

Here the final step in the derivation follows from the fact that $p_\phi(\phi) \propto 1$.


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