Sufficiency [BC]

Here is an email I received about The Bayesian Choice a few days ago:

I am an undergraduate student in Japan. I am self-studying your classical book The Bayesian Choice. The book is wonderful with many instructive examples. Although it is a little bit hard for me right now, I think it will be very useful for my future research.

There is one point that I do not understand in Example 1.3.2 (p.14-15). I know a standard result that the sample mean and sample variance are independent, with the sample mean follows

\mathcal{N}(\mu,(1/n)\sigma^2)

while s^2/\sigma^2 follows a chi-square of n-1 degree of freedom. In this example is it correct that one must factorize the likelihood function to g(T(x)|\theta) which must be the product of these two normal and chi-square densities, and h(x|T(x)) which is free of \theta ?

In the book I do not see why $g(T(x) | \theta)$ is the product of normal and chi-square densities. The first part correctly corresponds to the density of \mathcal{N}(\mu,(1/n)*\sigma^2) . But the second part is not the density of n-1 degree of freedom chi-square of s^2/\sigma^2.

The example, as often, skips a lot of details, meaning that when one starts from the likelihood

\sigma^{-n} e^{-(\bar x-\theta)^2 n/2\sigma^2} \, e^{-s^2/2\sigma^2} / (2\pi)^n,

this expression only depends on T(x). Furthermore, it involves the normal density on \bar x and part of the chi-square density on . One can then plug in the missing power of to make g(T(x)|\theta) appear. The extra terms are then canceled by a function we can call h(x|T(x))However, there is a typo in this example in that \sigma^n in the chi-square density should be \sigma^{n-1}!

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