These two textbooks appear to differ in their probability distribution of $s^2$ given $\sigma^2$, and I was wondering if anybody could please explain why? Specifically the exponent of ${(n-1) \over \sigma^2}$ is differing between these two texts.
In "Mathematical Statistics with Applications" by Wackerly, and everywhere else I can find on the internet, you'll find that ${(n-1)s^2 \over \sigma^2}$ is distributed $\chi^2$ with (n - 1) degrees of freedom. This means it's gamma distributed with $\beta = 2$ and $\alpha = {(n - 1) \over 2}$. So, correct me if I'm wrong, for the pdf we have:
$$f\left({(n-1)s^2 \over \sigma^2}\right) = {\left({(n-1)s^2 \over \sigma^2}\right)^{{(n-1) \over 2}-1}exp\left[{(n-1)s^2 \over 2\sigma^2}\right] \over 2^{(n-1) \over 2}\Gamma({n-1 \over 2})} $$
Where I have just plugged in the values for $Y = {(n-1)s^2 \over \sigma^2}$, $\alpha = {(n-1) \over 2}$, and $\beta = 2$ into the formula for a generalized gamma distribution. Which btw is:
$$ f(y) = {y^{\alpha-1}e^{-y \over \beta} \over \beta^\alpha \Gamma(\alpha)} $$
I am also reading "Bayesian Inference in Statistical Analysis" by Box and Tiao (https://www.pdfdrive.com/bayesian-inference-in-statistical-analysis-wiley-classics-library-e162123466.html). On page 63 they say:
"Normal distribution, both mean and variance unknown: In this case, the likelihood function $l(\theta, \sigma^2\ |\ y)$ based on the n independent observations y is that given in (1.4.2). Now it is well known that:
a) $\bar y$ is distributed as $N(\theta, \sigma^2/n)$
b) $(n - 1)s^2 = \sum_n (y_n - \bar y)^2$ is distributed as $\sigma^2\chi^2_{n-1}$
c) $\bar y$ and $s^2$ are statistically independent
Thus, $$ p(\bar y, s^2 | \theta, \sigma) = p(\bar y | \theta, \sigma^2)p(s^2 | \sigma^2)$$ where $p(\bar y| \theta, \sigma^2)$ is that in (1.4.9) and
$p(s^2 | \sigma^2) = {1 \over \Gamma[\frac {1}{2}(n-1)]}\left({n-1 \over 2\sigma^2} \right)^{\frac 12(n-1)}(s^2)^{\frac 12(n-1)-1}exp[- {(n-1)s^2 \over 2\sigma^2}] $"
[End quote]
Allow me to draw your attention to the last part there, $p(s^2 | \sigma^2)$, where the exponent of ${(n-1) \over \sigma^2}$ is ${(n-1) \over 2}$ instead of ${(n-1) \over 2} - 1$.
Does anyone know why this is? It looks like there is an extra ${(n-1) \over \sigma^2}$ coming from somewhere in the Box and Tiao version. And I know it's not a typo because they repeat this argument on page 86, with the same exponent. Once again discussing the distribution of $s^2$ given $\sigma^2$.
Also I assume $(n - 1)s^2 = \sum_n (y_n - \bar y)^2$ is distributed as $\sigma^2\chi^2_{n-1}$ means the same thing as ${(n-1)s^2 \over \sigma^2}$ is distributed $\chi^2$ with (n - 1) degrees of freedom? Any help would be very much appreciated!!!! :D