On the Wikipedia article for exponential families the density of a distribution on a measure space $(X, \xi)$ from an exponential family is written as $$f_{\theta} \colon X \to \mathbb{R}_{\ge 0}, \qquad x \mapsto h(x) g(\theta) \exp\left( \eta(\theta) \cdot T(x) \right)$$ for $\theta \in \Theta$. On two occasions the Wikipedia articles mentions that "even if $\eta$ is not one-to-one, then ...". In the table of exponential families however, every natural parameter mapping $\eta$ is one-to-one (i.e. bijective onto its image) such that the inverse parameter mapping $\eta^{-1}$ can be written down.

Question 1. Is there an exponential family with non-injective parameter mapping?

Furthermore, it seems that for every exponential family in the above mentioned table, $\eta(\Theta)$ is convex. The aforementioned Wikipedia article states that the natural parameter space is always convex.

Question 2. Does this mean that $\eta(\Theta)$ is always convex? If not, is there an exponential family such that $\eta(\Theta)$ is non-convex?

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    $\begingroup$ The general definition means that $\theta$ could essentially be anything, e.g.$$\eta(\theta)=\theta_1\exp(\theta_2^{\theta_3})$$which is not at all useful, but illustrates that the distribution can be overparameterised in $\theta$. $\endgroup$
    – Xi'an
    Commented Jun 21, 2023 at 15:07

1 Answer 1


Not a complete answer, but primarily a response to

Question 1. Is there an exponential family with non-injective parameter mapping?

The short answer is: yes, there is. It's called the curved exponential family. Now the longer answer.

Let $\mathcal{P} = \{P_\eta:\eta \in \Xi\}$ be a full rank $s$-parameter exponential family with complete sufficient statistic $T$ and consider a sub-model $\mathcal{P}_0$, parametrized by $\theta\in\Theta$, with $\tilde{\eta}(\theta)$ the value of the canonical parameter associated with $\theta$. Thus

$$\mathcal{P}_0 = \{P_{\tilde\eta(\theta)}:\theta\in\Theta\}.$$

Often $\tilde\eta(\theta)$ is a one-to-one mapping from $\Theta$ to $\Xi$ and in such a case, $\mathcal{P} = \mathcal{P}_0$. Curved exponential families may arise when $\mathcal{P}_0$ is a proper subset of $\mathcal{P}$, generally with $\Theta\subseteq\mathbb{R}^r$ and $r<s$. Here are two possibilities:

(a) Points $\eta$ in the range of $\tilde{\eta}$, $\tilde{\eta}(\Omega) = \{\tilde\eta(\theta) : \theta \in\Theta\}$, satisfy a nontrivial linear constraint. In this case, $\mathcal{P}_0$ will be a $q$-parameter exponential family for some $q < s$. The statistic $T$ will still be sufficient, but will not be minimal sufficient.

(b) The points $\eta$ in $\tilde{\eta}(\Omega)$ do not satisfy a linear constraint. In this case, $\mathcal{P}_0$ is called a curved exponential family. Here $T$ will be minimally sufficient, but may not be complete.

A famous example of curved exponential family is $N(\theta,\theta^2)$, $\theta\in\mathbb{R}\setminus\{0\}$. In this case, $$\tilde{\eta}(\theta) = \left(\frac{1}{\theta},-\frac{1}{2\theta^2}\right).$$

The range space $\tilde\eta(\Theta)$ is a parabola (with one point removed). Points in this range space do not satisfy a linear constraint, so in this case, we have a curved exponential family and $T$ is minimal sufficient.

  • $\begingroup$ Thank you for your answer. Does you answer show that for curved exponential families, $\eta(\Theta)$ (in my notation) might not be convex? $\endgroup$ Commented Jun 21, 2023 at 11:59
  • $\begingroup$ @ViktorStein that's an interesting question. In the example of $N(\theta,\theta^2)$ convexity holds, but in general multidimensional cases, my guess is that the parameter space may not always be a convex set. $\endgroup$
    – utobi
    Commented Jul 4, 2023 at 7:39

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