# Reparameterising likelihood

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.

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 $$\log \sigma$$.