# Solved – 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.

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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 $$p(Y)=|frac{partial X}{partial Y}|p(log(Y))propto1/Y$$.