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In the comments of this question it is mentioned that, when calculating the log posterior of a Normal distribution with a uniform prior on $(\mu, \log\sigma)$, we can write down the same likelihood form for the parameterisation $p(y \mid \mu , \sigma)$ as we would for $p(y \mid \mu , \log\sigma)$.

I'm a little confused by this i.e. why the likelihood doesn't change for the parameterisation $p(y \mid \mu , \log\sigma)$?

Any help or further explanation appreciated.

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It depends on what you mean by doesn't change.

Let's start with rewriting the likelihood from parameters $\mu,\sigma$ to parameters $\mu, \sigma^2$

The likelihood in terms of $\sigma$ is $$L(\mu,\sigma; y) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(y-\mu)^2}{2\sigma^2}}$$

The likelihood in terms of $\sigma^2$ is $$L(\mu,\sigma^2; y) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(y-\mu)^2}{2\sigma^2}}$$

The only difference is that I've written $\sigma$ as $\sqrt{\sigma^2}$ in the denominator of the second one. They are the same expressions, and you normally wouldn't think twice about either of these being correct.

Why can I do this? Because $\sigma$ isn't random in the context of the likelihood. It's just a number, 17 or something.

The need for transformation comes when you have distribution (a measure) on $y$ or $\sigma$, and you need to transform the density to take account of the transformation of the measure. If $y$ is measured in feet, a probability density in $y$ has units feet$^{-1}$; if you transform to $z=y^2$, you have $z$ in square feet and the probability density needs to be transformed to be in feet$^{-2}$.

If you do what you were probably told not to do in calculus and write $$f(y;\mu,\sigma^2)\,dy = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(y-\mu)^2}{2\sigma^2}}\,dy$$ you can (not entirely legitimately but clearly) say there are issues with transforming $y$ because you need to fix up $dy$, but not with transforming $\sigma$ or with replacing $2\pi$ with the equivalent mathematical constant $\tau$

The same thing happens with $\log \sigma$.

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