A question about parameters of Gamma distribution in Bayesian econometrics The Wikipedia article on the Gamma distribution,
lists two different parameterisation methods, one of them frequently used in Bayesian econometrics with $\alpha>0$ and $\beta>0$, $\alpha$ is shape parameter, $\beta$ is rate parameter.
$$X\sim \mathrm{Gamma}(\alpha,\beta).$$
In a Bayesian econometrics textbook written by Gary Koop, the precision paramether $\frac{1}{\sigma^2}=h$ follows a Gamma distribution, which is a prior distribution
$$h\sim \mathrm{Gamma}(\underline{s}^{-2},\underline{\nu}),$$ 
where $\underline{s}^{-2}$ is mean and $\underline{\nu}$ is degrees of freedom according to his Appendix. Also $s^2$ is standard error with definition 
$$s^2=\frac{\sum(y_i-\hat{\beta}x_i)}{\nu}.$$
Thus for me, these two definition of the Gamma distribution are completely different, since the mean and variances will be different. If we follow the wikipedia definition, the mean will be $\alpha/\beta$, not $\underline{s}^{-2}$. 
I am highly confused here, would anyone help me to streighten the thoughts here? 
 A: For anyone still struggling with Koops terrible notation: The problem is that Koop uses neither the scale nor the rate parametrization, but rather a "mean,degrees of freedom" parametrization (see Appendix, Def. B. 22).
The distribution of $h$ in a proper parametrization (shape, rate) is thus
$$
h \sim \text{Gamma}(shape = \underline{\nu}/2 , rate = \underline{\nu s}^2 / 2)
$$
using Koops notation for the parameters.
A: I think that the Wikipedia article is referring to a specific form of the gamma distribution known as $\chi^2$.  Chi square is $\rm{Gamma}(\nu,1/2)$ and $s^2$ would be the constant that the $\chi^2$ random variable is multiplied by to get a random variable with the distribution of a variance estimate.  That is $\alpha=\nu$ and $\beta=1/2$. It is s that is the standard error and not $s^2$.  In the article you referred to the $\chi^2$ is listed under specials cases (second bullet).
A: It is customary to impose (as a prior) either the gamma distribution to $h=\frac{1}{\sigma^2}$ or the inverse gamma distribution to $\sigma^2$. Then, the posteior will have a beautiful looking. I believe you can assign a gamma distribution to $\sigma^2$, and still all calculation to derive the marginal by integrating out $\sigma^2$ will go through.
