(Wikipedia page for background info: https://en.wikipedia.org/wiki/Principle_of_transformation_groups )
I'm reading Edwin Jaynes's paper, "Prior Probabilities":
http://bayes.wustl.edu/etj/articles/prior.pdf
and on page 17 he describes the principle of transformation groups, wherein he provides a method to derive objective prior distributions for different parameters of a given probability density function.
Trouble is, I don't get how it works.
Jaynes seems to be trying to find a change of variables for a pdf such that the new pdf, multiplied by the Jacobian determinant of the new parameterization, results in the exact same pdf as before the transformation. Then, he tries to find a prior distribution for a given set of parameters such that the prior distribution, when run through the same change of variables and multiplied by the same Jacobian determinant, remains the same.
His notation is a little vague, so here's my interpretation of one of his examples. Let's say you have a reparameterization of $\frac{1}{\sigma}f(\frac{x-\mu}{\sigma})dx$, denoted $\Phi$, where $\Phi(x,\mu, \sigma)=(a(x+b),a(\mu+b),a\sigma)$, where $a$ and $b$ are arbitrary values.. Plugging these new variables into $\frac{1}{\sigma}f(\frac{x-\mu}{\sigma})dx$, we have $\frac{1}{a\sigma}f(\frac{a(x+b)-a(\mu+b)}{a\sigma})dx'\mathbf{J_{\Phi}}$, which simplifies to:
$\frac{1}{a\sigma}f(\frac{x-\mu}{\sigma})dx' \left| \left( \begin{array}{ccc} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array} \right) \right| = \frac{a^{2}}{\sigma}f(\frac{x-\mu}{\sigma})dx' $
... which isn't the same as the original, though Jaynes says that it is. He then proceeds to take the Jacobian and apply it to a prior, which I also don't understand how he's able to do. Aren't Jacobian determinants supposed to multidimensional versions of the coefficient of $du$ in $u$ substitution? Why is Jaynes able to multiply one seemingly unrelated function by the Jacobian determinant of another?
I'm positive I've done something wrong with my calculations, or made a fumble in my reasoning somewhere. If anyone could run me through how this method works, I'd be very grateful.