From what I understand, the exponential family is defined as
$$f(y;\theta,\phi) = \exp\left(\frac{y\theta – b(\theta)}{a(\phi)}+c(y,\phi)\right) $$
I've read (but not seen shown anywhere), that the t distribution is not a member of the exponential family. But I don't understand why.
For instance, suppose I set $\theta = 0$, $b(\theta)=0$, and set $$c(y,\phi) = \ln\left(\frac{\Gamma\left(\frac{\phi+1}{2}\right)}{\sqrt{\phi\pi}\Gamma(\frac{\phi}{2})} \left(1+\frac{x^2}{\phi} \right)^{-\frac{\phi+1}{2}}\right)$$,
wouldn't then the t distribution then appear because
you would have
$$\exp\left(0+\ln\left(\frac{\Gamma\left(\frac{\phi+1}{2}\right)}{\sqrt{\phi\pi}\Gamma(\frac{\phi}{2})} \left(1+\frac{x^2}{\phi} \right)^{-\frac{\phi+1}{2}}\right)\right)=\frac{\Gamma\left(\frac{\phi+1}{2}\right)}{\sqrt{\phi\pi}\Gamma(\frac{\phi}{2})} \left(1+\frac{x^2}{\phi} \right)^{-\frac{\phi+1}{2}}$$
which is the t-distribution.
Why doesn't this work?
EDIT:
Ideally, I'd like to know why also my logic above is wrong (which I'm certain it is). So if you can fit that into your answer, that would be great.
Best Answer
No, the t distribution is not an exponential family. Exponential family distributions do have existing moment generating functions, and the t distribution do not. See also Why doesn't the exponential family include all distributions?