Convexity of exponential family

Question

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 $EM(c)\subseteq L^2_{\nu}$.

Answer 1

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.