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

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.

## marked as duplicate by Glen_b♦ self-study StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 29 '18 at 2:52

• $1/\theta$ 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:

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