Strictly positive barycentre cone lemma ($\{\mathbb{E}[\nu X]:\nu >0,\mathbb{E}[\nu \Vert X\Vert ]<\infty \}=\mathrm{ri}K$)

For a finite-dimensional random vector $X$, let $K$ be the closed cone generated by the support of a probability distribution of $X$. Every integrable strictly positively weighted barycentre lies in the relative interior of $K$: every nonzero supporting functional on its span is positive on a set of positive probability. Conversely, put $w=(1+\Vert X\Vert {)}^{-1}$ and $D=\{\mathbb{E}[wfX]:0\le f\in L^{\infty }\}$. Its dual cone equals that of $K$, so $\overline{D}=K$ and $\mathrm{ri}K\subseteq D$. For $z\in \mathrm{ri}K$, subtract a sufficiently small positive multiple of $\mathbb{E}[wX]$ and represent the remainder using $D$; adding the constant positive weight back gives $\nu =w(f+\epsilon )>0$.

For an integrable finite-dimensional random vector $Y$, let $L=\{h:h\cdot Y=0\text{ almost surely}{\}}^{\perp }$. The displayed convex cone is open relative to $L$. Given a bounded strictly positive $Z$, the perturbation $Z_{\epsilon }=Z(1+\epsilon \cdot Y/(1+\Vert Y\Vert ))$, $\Vert \epsilon \Vert <1$, changes its weighted expectation by $A_Z\epsilon$, whereFor $0\ne v\in L$, $v^T{A}_{Z}v>0$: otherwise $v\cdot Y=0$ almost surely. Thus $A_Z$ is invertible on $L$, and these perturbations give a neighborhood of every point of the cone. The hyperplane separation theorem consequently gives the alternative: either zero is a strictly positively weighted expectation, or a nonzero direction in $L$ has a nonnegative dot product with $Y$ almost surely and a positive dot product with positive probability. To obtain the latter assertion, test the separating direction with $Z=1_{\{h\cdot Y<0\}}+\epsilon$ and let $\epsilon \downarrow 0$.