# Proving a distribution is a member of the simple exponential family

Does anyone have any tips/ideas/method for proving that a distribution is a member of the simple exponential family (SEF)?

Or is the process unique to each distribution?

For example, I am trying to show firstly that the Gamma distribution belongs to the exponential family:

\begin{align*} \rho(y|v,\alpha)&=\frac{\alpha^v}{\Gamma(v)}y^{v-1}e^{-\alpha y} \\ &=\exp[v\ln(\alpha)-\ln(\Gamma(v))+(v-1)\ln(y)-\alpha y] \end{align*}

But then I am not sure where to go from here to get it into the form

$$\exp[\frac{y\theta-b(\theta)}{\phi}+c(y,\phi)]$$

The obvious things to do is set $\theta=-\alpha$ but then $b(\theta)=v\ln(-\theta)$ which is still a function of $v$.

I'm just not sure of a general technique to pick the right $\theta$.

• Have you already done anything trying to solve the problem? Have you tried manipulating the density formula of your distribution to make it match the general formula of the exponential family? – Richard Hardy Feb 4 '15 at 19:47
• @RichardHardy I have edited my question to go through the example which I am currently working on. However, my problem is more general in that I want a general procedure to be able to go through to help find the right $\theta$ and $\phi$. – denby47 Feb 4 '15 at 20:13
• The current version of your question is much more user-friendly. I hope someone can pick it up and help you proceed. – Richard Hardy Feb 4 '15 at 20:17

Unless you want to do it for some fixed shape parameter, $\nu$* (I assume you mean for your "$v$" to be a parameter, not a variable, so I am using $\nu$), you could consider whether a vector-valued $\theta$ might solve your problem.
In particular note that there are two terms involving y-and-a-parameter. To work that way, you may need to allow the exponential family to be more general (say to have $\eta(\theta)\cdot T(y)$ instead of just $\theta y$).