# How to show that $\frac{1}{\theta}$ is a flat prior for $\log\theta$? [duplicate]

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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.

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

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:

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