# Convexity of exponential family

It is known that (given a $$\sigma$$-finite Borel reference measure $$\nu$$ on $$\mathbb{R}$$) the parameter space of an exponential family is convex in Euclidean space. However, my question is, for an the associated exponential manifold $$\sqrt{EM(c)}$$ a convex subset of $$L^2_{\nu}$$? Where:

• $$EM(c)\triangleq \left\{ p(\cdot,\theta):\, \theta \in \Theta,\, p(x,\theta)=exp(\theta^Tc(x)-\psi(\theta)) \right\},$$ for some fixed continuously-differentiable scalar functions $$c=\{c_1,\dots,c_n\}$$ on $$\mathbb{R}^d$$,
• We assume that $$EM(c)\subseteq L^2_{\nu}$$.

## 1 Answer

Assuming that $$\sqrt{EM(c)}$$ is defined as $$\{\sqrt p\colon p\in EM(c)\}$$, the answer is "of course, no". Indeed, write $$t$$ and $$T$$ for $$\theta$$ and $$\Theta$$, respectively, and $$p_t$$ for $$p(\cdot,t)$$. Then the convexity of $$\sqrt{EM(c)}$$ would imply that, for any $$s$$ and $$t$$ in $$T$$, the function $$\frac14\,p_s+\frac14\,p_t+\frac12\,\sqrt{p_s}\sqrt{p_t}=(\frac12 \sqrt{p_s}+\frac12 \sqrt{p_t})^2$$ is in $$EM(c)$$ and hence a pdf, which (in view of the Cauchy--Schwarz inequality) will be the case only when $$p_s=p_t$$ $$\nu$$-almost everywhere for all $$s$$ and $$t$$ in $$T$$, i.e., when the exponential family has (essentially) only one member.