Tests for determining membership of exponential family

Question

Provided an arbitrary random variable $X$ with probability density/mass function $f$, are there any tests to determine if $f$ forms an exponential family?

Certainly, if $f$ can be written in the form $$ \tag{1} f(x|\theta)=h(x)\exp(\eta(\theta)\cdot T(x)-A(\theta)), $$ then it must form an exponential family. However, failure to write $f$ in such a form does not prove the converse.

In my particular problem I have the density given by the sum of independent Poisson and zero-mean normal variables $$ f(x|\lambda,\sigma)=\sum_{k=0}^\infty\frac{e^{-\lambda}\lambda^k}{k!}\frac{1}{\sigma}\phi\left(\frac{x-k}{\sigma}\right). $$

I suspect $f$ does not form an exponential family and up to this point have been unable to write it in the form $(1)$. So what else can be done to determine if this density function forms an exponential family?

Answer 1

$\newcommand\si{\sigma}\newcommand\la{\lambda}\newcommand\th{\theta}\newcommand\Th{\Theta}$Your family $(f_\th)$, where $\th:=(\lambda,\sigma)\in\Th:=(0,\infty)^2$, is not exponential.

Indeed, suppose the contrary. Since $f_\th>0$ for all $\th\in\Th$, we have $h>0$. So, we can define $$g(\th,x):=\ln f_\th(x)=\eta(\th)T(x)-A(\th)+\ln h(x)$$ for real $x$.

Take any $\th_0\in\Th$ and let $$G_{\th_0}(\th,x):=g(\th,x)+g(\th_0,0)-g(\th,0)-g(\th_0,x) =\xi(\th)S(x),$$ where $\xi(\th):=\eta(\th)-\eta(\th_0)$ and $S(x):=T(x)-T(0)$. So, for any $\th_1$ and $\th_2$ in $\Th$ and all real $x_1$ and $x_2$ we will have $$G_{\th_0}(\th_1,x_2)G_{\th_0}(\th_2,x_1)=G_{\th_0}(\th_1,x_1)G_{\th_0}(\th_2,x_2). \tag{1}\label{1}$$

To show that your family $(f_\th)$ is not an exponential family, it remains to find $\th_0,\th_1,\th_2$ in $\Th$ and real $x_1$ and $x_2$ such that \eqref{1} fails to hold. This is very easy to do, say by taking $\th_0,\th_1,\th_2$ in $\Th$ and real $x_1$ and $x_2$ "at random". For instance, if $\th_0=(1,1)=\th_1$, $\th_2=(2,1)$, $x_1=1$, and $x_2=2$, then the left- and right-hand sides of \eqref{1} are $-0.102\ldots$ and $0.075\ldots$, respectively, so that \eqref{1} fails to hold.