# Why is uniform prior on log(x) equal to 1/x prior on x?

I'm trying to understand Jeffreys prior. One application is for 'scale' variables like the standard deviation $$\sigma$$ (or its square, the variance $$\sigma^2$$) of Gaussian distributions. It is often said that using a uniform prior over $$\sigma$$ is not really non-informative and instead one should either:

1. Use instead $$\ln \sigma$$ as the free parameter, with a uniform prior (this is often called a log-uniform prior)

2. Or keep using $$\sigma$$ as the free parameter but use $$1/\sigma$$ as the prior (which is not uniform).

Why are the above two methods/priors equivalent? I feel it has something to do with the fact that the derivative of ln $$\sigma$$ is $$1/\sigma$$ but I can't take the next step.

Also, why does this even matter, in simple language with minimal jargon? I see all these complicated explanations online involving the Fisher information matrix but in the end all I see is that the above log-uniform or $$1/\sigma$$ priors preferentially weight lower values of $$\sigma$$ more highly. Why? If possible, a simple analytic example or python snippet would be very helpful.

When transforming a uniform distribution on $$\log(\sigma)$$ to a distribution on $$\sigma$$ you need to take into account the Jacobian of the transformation. This corresponds, as you correctly intuited, to $$1/\sigma$$.
Writing this a little more clearly, let $$X=\log(\sigma)$$ and the transformation we're after is $$T(X)=\sigma=e^{X}=Y$$, which has inverse transformation $$T^{-1}(Y)=\log(Y)$$. The jacobian is then $$|\frac{\partial X}{\partial Y}|=1/Y$$. So since $$p(X)\propto 1$$, we have that the induced density for $$\sigma$$ is the $$p(Y)=|\frac{\partial X}{\partial Y}|p(\log(Y))\propto1/Y$$.
• That last line should be $p(X)$ instead of $p(e^X)$. – Moss Murderer Jan 19 '18 at 1:47
• Sorry but why is the Jacobian relevant and why are you introducing this new variable Y, instead of just calling it $\sigma$? And why is $p(X)\propto1$? Sorry I'm confused – quantumflash Jan 19 '18 at 2:49
• @quantumflash I introduced new variables X and Y for clarity. We have $p(X)\propto 1$ since it's uniform (thus proportional to a constant), and since it's uniform on the positive reals it actually doesn't have a density, so we can only specify it up to proportionality. As for why the jacoban is relevant, see the change of variables portion of en.wikipedia.org/wiki/…. This is all introductory probability theory that you should read up on before learning about Bayesian statistics. – aleshing Jan 19 '18 at 3:24