1
$\begingroup$

This question already has an answer here:

I do not really understand what the statement even means.

To my understanding, a prior $p(\theta)$ is said to be flat if $p(\theta) =$ constant $\forall \theta,$ where $p(\theta)$ is the prior distribution.

How do I show that $\frac{1}{\theta} =$ constant $\forall\log\theta$?

I am quite confused.

$\endgroup$

marked as duplicate by Glen_b self-study Dec 29 '18 at 2:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ $1/\theta$ is the Jacobian on the transformation... the constant is all that's left $\endgroup$ – Glen_b Dec 29 '18 at 2:52
2
$\begingroup$

Let $\phi = \log \theta$ denote the log-transformed parameter, and differentiate to get $d\phi = d \theta / \theta$. Now, using the improper uniform prior $p(\phi) \propto 1$ for the log-transformed parameter, and applying the standard rule for transformations of random variables you get:

$$p(\theta) = p(\phi) \cdot \Bigg| \frac{d\phi}{d\theta} \Bigg| \propto 1 \cdot \frac{1}{\theta} = \frac{1}{\theta}.$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.