why do some use ν=n instead of ν=n-1 in the t distribution I have always accepted that ν=n-1, but then I come across multiple authors showing a derivation of the t distribution by marginalizing out the variance from the normal-inverse-gamma distribution, or by marginalizing out the inverse variance (precision) from the normal-gamma distribution.
The results of deriving the t distribution from the normal-inverse-gamma distribution require the use of ν=n and not ν=n-1.
For example in the book "Bayesian Data Analysis: Third edition" on page 69, it derives $p(μ|μ_0,σ_0)=t(μ|μ=μ_n,σ=σ_n^2/κ_n,ν=ν_n)$ where
$$ μ_n={κ_0\over κ_0+n}μ_0+{n\over κ_0+n}\bar y $$
$$ κ_n=κ_0+n $$
$$ ν_n=ν_0+ν $$
$$ ν_nσ_n=ν_0σ_0^2+(n-1)s^2+{κ_0n\over κ_0+n}(\bar y-μ_0)^2 $$
Since the least informative prior has $κ_0=ν_0=0$ therefore $ν_n=ν_0+n=n$.  For example when using a normal gamma distribution to derive the t distribution, the prior on the precision is $Γ(α=ν_0/2,β=ν_0σ_0^2/2)$ [1] where clearly the least informative prior is obtained in the limit as $ν_0$ approaches 0.  Yet on page 66 ν=n-1 is derived, and seems to be correct.
Even if $ν_n=ν_0+n$ is correct there is no way to obtain $ν=n-1$ without $ν_0=-1$, which I can not make sense of, with a gamma or inverse gamma distribution.
I clearly must have some fundamental misunderstanding. What is it?
I thought that perhaps $σ_n$ would cancel out the different value of ν, but no, instead I get two different results depending on which derivation of the t distribution I use.
How do I get the same results from the two different derivations of the t distribution?
 A: The difference is in the prior distributions.

*

*The book starts with
$$p(\mu,\sigma) \propto \sigma^{-2} \phantom{\cdot \sigma^{-1} e^{- \frac{(\mu-\mu_0)^2}{2\sigma^2}}}$$
From which the t-distribution with $n-1$ degrees of freedom is derived as the marginal posterior distribution.


*The second derivation uses*
$$p(\mu,\sigma) \propto \sigma^{-2} \cdot \sigma^{-1} e^{- \frac{(\mu-\mu_0)^2}{2\sigma^2}}$$
From which the t-distribution with $n$ degrees of freedom is derived as the marginal posterior distribution.
So the difference is in the prior distribution. The former is uniform in $\mu$. With the latter this is not the case and you get a multiplication of this $\sigma^{-2}$ with a term for the distribution of $\mu$ conditional on $\sigma$. This changes the power from $\sigma^{-2}$ into $\sigma^{-3}$ and that changes the end result when you integrate out the $\sigma$.

*I have simplified it with $\nu_0 =0$ such that the difference between the two situations is more clear. But, the second case is more general and with $\nu_0\neq 0$ the equation is more complex. $$p(\mu,\sigma) \propto \sigma^{-2} \sigma^{-\nu_0} \exp \left( - \frac{1}{2\sigma^2} [\nu_0 \sigma_0^2 + \kappa_0(\mu_0-\mu)^2] \right)$$ But this complexity is not needed to see the difference why you get $n-1$ versus $n$.
