Q:

MGF Of a Normal-Gamma Distribution

According to wikipedia, the normal-gamma distribution is actually, unconditionally, a nonstandardized t-distribution.  The MGF of t-distributions do not exist.  However, I can make the following argument to calculate the MGF of a normal-gamma, where does this fail? 
Suppose X|$\sigma^2$~Normal(0,$\sigma^2$) and $\sigma^2$~Gamma(a,b). 
$E(e^{tX})=E(E(e^{tX}|\sigma^2))=E(e^{\mu+\frac{1}{2}t^2\sigma^2})=e^\mu E(e^{\frac{1}{2}t^2\sigma^2})$ and then use the result for the MGF of a gamma to calculate the last expectation. 
Thanks! 

A:

This is not a normal gamma distribution as written in the original post.  A normal gamma has 
$1/\sigma^2 \sim Gamma(a,b)$. Thus, using the original conditioning argument, you would still have:
$E(e^{tX})=E(E(e^{tX}|\sigma^2))=E(e^{\mu+\frac{1}{2}t^2\sigma^2})=e^\mu E(e^{\frac{1}{2}t^2\sigma^2})$
but $\sigma^2$ now has an inverse gamma distribution.  The MGF for inverse gamma distributions does not exist as a real valued function.