# 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:

1. I can't decide whether the Erlang distribution is a member of the exponential family.
2. 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?

• Your factorization isn't the best, for example, why factor $\lambda^n$ into $\text{exp}((n-1)\log(\lambda) + \log(\lambda))$ instead of just $\text{exp}(n\log(\lambda))$? Dec 20, 2019 at 17:11
• You are correct. Nevertheless, this does not alter the question in any way. Dec 20, 2019 at 17:35
• Well, sure, but that's why it's a comment. Writing clean(er) math is a good thing in its own right :) Dec 20, 2019 at 18:06
• Also, when you get into the topological definition of continuity, which collapses to the one we all learned in Calc I or wherever, the concept of continuity is perfectly well defined for a discrete space, as you seem to know, so I'm not sure why this is a problem. Dec 20, 2019 at 18:14
• I will make an edit, to make it cleaner! The problem is, that I don't know with respect to which topology it has to be continuous. My Script does not specify this. I am (most naturally) assuming that the natural numbers have to be seen as an embedding into the reals, as this course is rather for beginners. Would you say it is an exponential family wrt. $n$? Dec 20, 2019 at 18:26

## 1 Answer

### 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.

• Great answer! What definition of natural parameter space did you use? I looked it up and only found it defined as $\mathcal{E}=\{\eta\colon A(\eta)<\infty\}$ where $A(\eta)$ is the reparametrized $B(\theta)$ in the density. Dec 20, 2019 at 23:14
• @EpsilonDelta, that's correct. Note that $$A(\eta_1, \eta_2) = \log(\Gamma(\eta_1+1))-(\eta_1+1)\log(-\eta_2).$$ This will surely be finite whenever $\eta_1 > -1$ and $\eta_2 < 0$. It is either infinite or undefined otherwise (since there are restrictions on $n$ and $\lambda$ which lead to restrictions for $\eta_1$ and $\eta_2$. Dec 21, 2019 at 1:51