# Equivalence of gamma and inverse gamma:Does choosing a prior lead to possibly different evaluations?

Suppose we had an expressions that was proportional to

$$\frac{1}{\sigma^{3}}\exp(\frac{-\beta}{\sigma^{2}})$$

My question is, can you choose different possible parametrization for a prior distribution?

you could choose a prior, $$\frac{1}{\sigma^{2}} \sim Gamma(1/2,\beta)$$

Or you could choose

$$\sigma^{2} \sim Inverse-Gamma(\frac{5}{2},\beta)$$

which correspond to different kernels..

It seems we could take either, but the kernel would then be different leading to differences in integration and such.

So what is it I am missing here? Could you really just choose either one even though they lead to different inferences( I suppose because $$1/x^{2}$$ is not invariant?

• Forgot Jacobian.
– guy
Commented Dec 11, 2018 at 3:48
• Thanks but that still wouldn't allow us to know which form to choose for a prior in the example I gave would it? Or is that just a result that choosing a different form of prior that may not be invariant can lead to different results in integration and such Commented Dec 11, 2018 at 3:59
• If you want the inference to correspond, you have to have your priors correspond. Jeffreys priors do this, for example Commented Dec 11, 2018 at 4:05
• Got it, thanks. So if not specified, either interpretation could be seen as valid, even if leading to different conclusions? Commented Dec 11, 2018 at 4:06
• In the same sense that inference based on two distinct priors are both valid, even if leading to different conclusions. Commented Dec 11, 2018 at 4:07

The question is meaningless in my opinion because it is too vague: the "expression" stems out of nowhere (is it a probability density? a likelihood? as a function of $$\sigma$$? $$\sigma^2$$? $$\sigma^{-2}$$?) and in particular it does not specify the dominating measure. Is it the Lebesgue measure? the left invariant Haar measure? the right invariant Haar measure? Another measure?
• (it results from a posterior likelihood of normals with a prior of the form $\frac{1}{\sigma^{2}}$ Commented Dec 11, 2018 at 19:25