# Derive the prior on variance scale if uniform prior placed on logarithm scale

In page 64 of Bayesian Data Analysis by Gelman et.al. they write

... sensible vague prior density for µ and σ, assuming prior independence of location and scale parameters, is uniform on ($$\mu$$, $$\log~\sigma$$) or, equivalently, $$p(\mu, \sigma^2) \propto 1/\sigma^2$$.

Also in page 87 of The BUGS Book (pdf download) they discuss the equivalence of the Jeffreys prior to the uniform on the scale:

... the Jeﬀreys prior is $$p_J(\sigma) \propto \sigma^{-1}$$, which in turn means that $$p_J(\sigma^k) \propto \sigma^{-k}$$ for any choice of power k. ... we note that the Jeﬀreys prior is equivalent to $$p_J(log \sigma^k) \propto constant$$.

I have understood this to mean that a uniform prior on $$\log \sigma^2$$ should be $$\propto 1/\sigma^2$$. I have been unable to derive this. This is my attempt (with a nod to this answer):

\begin{align} \text{Let}~ Y =& \log \sigma^2 \\ p(Y) \propto& 1 \\ \frac{dY}{d\sigma^2} =& 2/\sigma \end{align}

Then to get the distribution on the $$\sigma^2$$ scale:

\begin{align} \text{If}~ X =& \sigma^2 \\ \text{then}~ p(X) =& p(Y) |\frac{dY}{d\sigma^2}| \\ =& 1 \times 2/\sigma \\ \propto & 1/\sigma \end{align}

Where have I gone wrong please?

Your error is going from $$\text{Let}~ Y = \log \sigma^2$$ to $$\dfrac{dY}{d\sigma^2} = 2/\sigma$$
You should have: $$\dfrac{dY}{d\sigma^2} = 1/{\sigma^2}$$ (simple derivative of a logarithm)
though perhaps you tried $$\dfrac{dY}{d\sigma} = 2\sigma \frac{1}{\sigma^2}= 2/{\sigma}$$ (chain rule).
This gives you: $$\text{If}~ X = \sigma^2 \text{ then } p(X) \propto 1/{\sigma^2}$$ for the improper prior.
It is worth noting that $$\log(\sigma^k)=k\log(\sigma)$$ and this indicates why something similar happens for all powers of the standard deviation.