Solved – Proving a distribution is a member of the simple exponential family

distributionsexponential-familygamma distributionproof

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 $\theta$.

Best Answer

In the case of the gamma:

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.

*(as one might in the case of a GLM; in that case it is straightforward)

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 $\theta y$).

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