Is the Erlang Distribution member of the Exponential family? Given the Erlang distribution
$f(x)={\begin{cases}\displaystyle {\frac  {\lambda ^{n}x^{{n-1}}}{(n-1)!}}\,{\mathrm  {e}}^{{-\lambda x}}&x\geq 0\\0&x<0\end{cases}}$
I want to determine, whether it belongs to the exponential family for $\lambda$ or $n$.
Case $\lambda$:
$$f(x)=\frac{x^{n-1}}{(n-1)!}\exp\left(-\lambda x+n\log\lambda\right)$$
Then in the usual notation we have $\eta(\lambda)=-\lambda, T(x)=x, B(\lambda)=-n\log\lambda $
For a density I have to check if $\mu$ is continous, $T(x)\ne0$ and $h(x)$ continuous. Furthermore, the support must not depend on the paramter in question. These conditions are satisfied for this case.
Case $n$:
$$f(x)=\left(\lambda e^{-\lambda x}\right) \exp\left( (n-1)\log(x) +(n-1)\log(\lambda)-\log[ (n-1)!]\right)$$
Here, $\eta(n)=(n-1)$, $T(x)=\log(x)$, $B(n)=(n-1)\log(\lambda)-\log[ (n-1)!]$.
This is where it gets a little strange. Now since functions are always continous at isolated points, I conclude that $\eta$ is continous. But then again the concept of continuity is not really well defined for a discrete space (at least not on the level for which I am supposed to solve this exercise). 
Then $T(x)$ is non vanishing so this condition is satisfied and $h(x)$ is continuous, too. 
I have the following problems: 


*

*I can't decide whether the Erlang distribution is a member of the exponential family. 

*Also, I am asked to determine for which natural parameters this distribution is an exponential family. Does this mean I just have to reparametrize by $\eta$? 


The whole concept and use of the reparametrization of an exponential family is still pretty new to me, so I would appreciate some comments on this as well. What is the use of reparametrizing?
 A: Yes, but it is not "regular".
Let's begin by looking at the Gamma distribution, which is a generalization of the Erlang distribution with density function
$$f(x|n,\lambda) = \frac{\lambda^nx^{n-1}}{\Gamma(n)}e^{-\lambda x} I_{(0,\infty)}(x).$$
To see that this is in the exponential family with respect to $\theta = (n, \lambda)$, we can write
$$f(x|n, \lambda) = I_{(0,\infty)}(x)\exp\left((n-1)\log x - \lambda x - (\log \Gamma(n) - n\log\lambda) \right).$$
It is easy to verify that the natural parameter for the gamma distribution is
$$\eta = \begin{bmatrix}
n-1 \\
-\lambda
\end{bmatrix}$$
with the sufficient statistic
$$T = \begin{bmatrix}
\log x \\
x
\end{bmatrix}.$$
Moreover, the natural parameter space is given by $$(-1, \infty) \times (-\infty, 0).$$ Since this space is non-empty and open, the gamma distribution is a regular exponential family. By placing the restriction $n \in \mathbb N$, nothing above changes except for the natural parameter space, which becomes $$(\{0\} \cup \mathbb N) \times (-\infty, 0).$$
This space, while non-empty, no longer contains an open set. Therefore the Erlang distribution is a member of the exponential family, but it is not regular.
Depending on your goals, this may or may not matter. Here's an example where it could matter. If a random sample $X_1, X_2, \cdots X_n$ is drawn from a regular exponential family, then the sufficient statistic is also complete (i.e. see Thm 6.2.25 from Casella & Berger).  If the exponential family is not regular, completeness (or lack thereof) must be demonstrated another way.
