Double exponential distribution is not an exponential family 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?
 A: 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.    
A: Meanwhile I figured out another proof. But the one of Byron is clearly more elegant.
Following the idea of Shao we consider the quotients
$\frac{p_\theta(x)}{p_{-\theta}(x)}= \frac{f_\theta(x)}{f_{-\theta}(x)}$.
This allows us to get rid of $h(x)$ and yields
\begin{equation}
    |x+\theta| - |x-\theta| = \left( \eta(\theta) - \eta(-\theta) \right)T(x) - \left( \xi(\theta) - \xi(-\theta) \right)
\end{equation}
Since $\eta(\theta) - \eta(-\theta)$ must be non-zero for some $\theta$ we
define 
\begin{equation}
    A = \eta(\theta) - \eta(-\theta)
\end{equation}
\begin{equation}
    B =\xi(\theta) - \xi(-\theta)
\end{equation}
and get $ |x+\theta| - |x-\theta| = A T(x) + B $. Moreover $|x+\theta| -
|x-\theta|$ is an antisymmetric function and therefore $A T(x) + B = -(A T(-x) + B)$.
This yields $T(x) + \frac{B}{A} = - T(-x) - \frac{B}{A}$ and implies that,
$T(x)$ shifted by $\frac{B}{A}$ is antisymmetric.
On the other hand, if we consider $\frac{p_\theta(x)}{p_{0}(x)}= \frac{f_\theta(x)}{f_{0}(x)}$ we 
get
\begin{equation}
    |x| - |x-\theta| = \left( \eta(\theta) - \eta(0) \right)T(x) - \left( \xi(\theta) - \xi(0) \right)
\end{equation}
For positive $\theta$ this function is constant in $x$ for $x<0$. But $T(x)$ is also antisymmetric. This 
implies $T(x) \equiv T$ is constant for all $x\in\mathbb R$. A contradiction.
