## 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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