The classic example given in class was when the support depends on the parameter $\theta$. If this is not the case, can the distribution always be written as a member of the exponential family?
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2 Answers
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No. Consider the Negative Binomial distribution for $x \in \{0, 1, \dots\}$:
$$p(x;r,p) = {r+x-1 \choose x}p^r(1-p)^x$$
This can be mildly rewritten as:
$$p(x;r,p) = {1 \over x!}p^r e^{x\ln(1-p)}\prod_{i=r}^{r+x-1}i$$
Compare to (one of the) ways of writing out a member of the exponential family:
$$p(x;\theta) = h(x)g(\theta)e^{T(x)\eta(\theta)}$$
We're close, but not quite there. You can't get from the second equation to the third because of that product term; it doesn't allow for separation of the expression into a term involving $r$ and another term involving $x$ either on the "regular" scale or on the log scale. So, the Negative Binomial is not a member of the exponential family.
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Another typical illustration is the $t$ distribution: when observing a sample $x_1,\ldots,x_n$ from a $t_\nu$ distribution with unknown location $\mu$, the likelihood is $$\prod_{i=1}^n \left[1+(x_i-\mu)^2/\nu\right]^{-(1+\nu)/2}$$ and cannot be written in an exponential form since it is of the same complexity as a polynomial of degree $2n$, hence cannot depend on a fixed number of functions of the sample $x_1,\ldots,x_n$ (Darmois-Pitman-Koopman lemma).
