Define a one-parameter exponential family as a family of densities of the form
$$f_\theta(x)=\exp(\eta(\theta)T(x) + \xi(\theta))h(x)$$
where $T(x)$ and $h(x)$ are Borel functions, $\theta\in\Theta\subset\mathbb R$ and $\eta$ and $\xi$ are real-valued functions defined on $\Theta$.
Double exponential distribution is a distribution having the density
$$p_\theta(x)= \frac{1}{2}\exp(-|x – \theta|)$$
for $\theta\in\mathbb R$.
I am looking for a simple proof of the theorem in the title. I found a proof in the book of Shao "Mathematical Statistics. Exercises and Solutions." but it uses a more general definition of exponential families and doesn't show why the classes are not compatible. What is the special feature of $p_\theta(x)$ that makes the representation as exponential family impossible?
Best Answer
Assume that $\Theta$ has at least three distinct points $\theta_1$, $\theta_2,$ and $\theta_3$. Suppose that $$\exp(\eta(\theta)T(x)+\xi(\theta)) h(x)={1\over 2}\exp(-|x-\theta|).$$ Since $h(x)$ never takes the value zero, we can write it as $h(x)={1\over 2}\exp(w(x))$, and deduce that, for all $x\in\mathbb{R}$ and $\theta\in\Theta$, $$\eta(\theta)T(x)+\xi(\theta) +w(x)= -|x-\theta|.$$
Substitute $\theta_1, \theta_2$ and subtract the two equations to get $$[\eta(\theta_1)-\eta(\theta_2)]\ T(x)+\xi(\theta_1)- \xi(\theta_2) = |x-\theta_2|-|x-\theta_1|.$$ Since the right hand side is not a constant function of $x$, we find that $\eta(\theta_1)\neq\eta(\theta_2)$ and hence that $T$ is differentiable in $x$, except possibly at $\theta_1$ and $\theta_2$. The same argument using the pairs $\{\theta_1 ,\theta_3\}$ and $\{\theta_2 ,\theta_3\}$ shows that $T$ is, in fact, differentiable everywhere.
We conclude that $|x-\theta_2|-|x-\theta_1|$ is everywhere differentiable in $x$, which is a contradiction.