# How to check if an exponential family is regular?

A strictly $$k$$-parameter exponential family $$f=\exp\left(\sum \eta_i(\theta)T_i(x)-B(\theta)\right)h(x)$$ is regular if the natural parameter space $$\eta(\theta)$$ contains a $$k$$-dimensional open set.

I was given this counterexample but couldn't figure out why it is not regular: $$\exp\left[\sum_{j=1}^{k-1} \left(\log \theta_j−\log\left(1−\sum_{l=1}^{k-1}\theta_l\right)\right)N_j+n\log\left(1−\sum_{l=1}^{k-1}\theta_l\right)\right].$$ For reference this is the likelihood of $$n$$ i.i.d. random variables $$x_i$$, taking value $$j\in \{1,...,k\}$$ with probability $$\theta_j$$, and $$N_j=\sum_i1_{x_i=j}$$.

And in general, how do I check if an exponential family is regular?

• Should that last $=j$ be part of the subscript? Jan 5 at 23:47
• yes I've fixed that, thanks Jan 6 at 0:16

The easy way to see this one isn't regular is that the $$\theta_j$$ need to add up to 1, so (given sufficient smoothness) they can only fill up at most a $$k-1$$-dimensional set.
Somewhat more generally, you could work out the information matrix and see that it is of rank less than $$k$$ everywhere in the interior of the parameter space. I don't know if that's a necessary condition to be non-regular, though.
(To make the calculations easier, you might start with $$j=2$$)
• The $\theta_j$ in the OP's counterexample actually don't need to add to one because there are $k-1$ of them instead of $k$. The trouble is that the representation in the counterexample is not the natural parameter space, for which your statement is correct. Jan 6 at 2:57