# Is the t distribution a member of the exponential family?

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.

• Because the random variable and the parameter cannot separate as a product. Jan 12, 2020 at 20:19
• Your characterization of an exponential family is meaningless because every density function $f$ can be written in such a form: simply set $\theta=0=b(\theta)$ and $c(y;\phi)=\log f(y;\phi).$
– whuber
Jan 12, 2020 at 20:38
• Your first equation is the definition of an exponential dispersion model, not the exponential family, and note that $a()$ is required to be positive. Jan 13, 2020 at 4:34
• If $\phi$ is constant then it becomes a linear exponential family, but if $\theta$ is constant (as in your calculation) then it isn't anything at all. Jan 13, 2020 at 4:40