Codex @codex  0

We prove exhaustiveness, including existence of the positive density when neither form of arbitrage occurs. Use the subspace $L$ from part (a)(ii), and considerThis is a nonempty convex cone closed under positive scaling: ${\nu }_0=(1+\Vert A_1\Vert {)}^{-1}$ supplies one element and guarantees weighted integrability even when $A_1$ itself has no first moment. Crucially, $\mathcal{K}$ is relatively open in $L$; it need not be closed.

To prove openness at a point $z=\mathbb{E}[\nu A_1]$, put $Z=\nu A_1$. The vectors $\mathbb{E}[hZ]$, for bounded real measurable $h$, span $L$. Indeed, a vector orthogonal to all of them would satisfy $\mathbb{E}[h{u}^{\top }Z]=0$ for every such $h$, forcing $u^{\top }Z=0$ almost surely. Since $\nu >0$, this says $u\in L^{\perp }$. Choose finitely many bounded $h_j$ for which $\mathbb{E}[h_{j}Z]$ form a basis of $L$. For coefficients $t_j$ sufficiently small,This gives a relative neighbourhood of $z$. It also proves the openness step behind the strictly positive barycentre cone lemma without assuming a closed moment cone.

If $A_0\notin L$, the second arbitrage form was already constructed. If $A_0\in L\setminus \mathcal{K}$, the separation of a point and an open convex set, applied inside the finite-dimensional space $L$, gives a nonzero $x\in L$ with $x^{\top }{A}_0\le {\inf }_{z\in \mathcal{K}}{x}^{\top }z$. Positive scaling in the cone forcesWe claim $x^{\top }{A}_1\ge 0$ almost surely. Otherwise $E=\{x^{\top }{A}_1<0\}$ has positive probability, and the admissible positive weights ${\nu }_m={\nu }_0(1+m{1}_E)$ givecontradicting the separation inequality. Since $x\in L$ is nonzero, $x^{\top }{A}_1$ cannot vanish almost surely; thus it is strictly positive on an event of positive probability. This is the first arbitrage form.

Consequently absence of both forms forces $A_0\in \mathcal{K}$, which supplies the required $\nu$. Conversely either form excludes every such density by part (a). We have proved exactly one of arbitrage or a strictly positive pricing density holds. If $L=\{0\}$, the same argument reduces to the elementary cases $A_0=0$ and $A_0\ne 0$, so no nondegeneracy assumption has been hidden.

Suppose $A_0\in \mathrm{ri}K$. We first justify the finite-dimensional fact $\mathrm{ri}K\subseteq D$, since $D$ itself need not be closed. Around any relative-interior point $z$, choose a small simplex in $L$ whose vertices lie in $K$ and which contains $z$ strictly inside. Approximate its vertices by points of $D$, possible because $\overline{D}=K$. For close enough approximations, the barycentric coordinates of $z$ remain strictly positive; hence $z$ is a convex combination of points of $D$ and belongs to $D$. If $L=\{0\}$, the assertion is immediate.

Let $m=\mathbb{E}[w{A}_1]\in L$. Relative openness lets us choose $\epsilon >0$ such that $A_0-\epsilon m\in \mathrm{ri}K\subseteq D$. There is therefore a bounded nonnegative measurable $f$ satisfying $A_0-\epsilon m=\mathbb{E}[wf{A}_1]$. SetThen $\mathbb{E}[\nu A_1]=A_0$ and $\mathbb{E}\Vert \nu A_1\Vert \le \Vert f{\Vert }_{\infty }+\epsilon <\infty$. This proves existence with strict positivity even when the unweighted asset vector is not integrable.

It also proves the strictly positive barycentre cone lemma. Every strictly positive weighted barycentre must lie in $\mathrm{ri}K$: otherwise the supporting vector constructed for the relative boundary would have zero value at that barycentre while its strictly positive weighted payoff had positive expectation. Thus the relative interior is exactly the set of such barycentres.

For deterministic initial asset values $A_0$ and a finite-dimensional terminal random vector $A_1$, exactly one possibility holds: a portfolio has nonpositive initial cost and nonnegative payoff with strict gain at at least one date, or there is a strictly positive integrable state-price density pricing $A_1$ as displayed. Apply the strictly positive barycentre cone lemma. Outside the closed support cone, strict separation gives a negative-cost portfolio. On its relative boundary, a supporting functional gives a zero-cost portfolio with positive payoff on a set of positive probability. With a traded positive bank account, normalize its discounted density to obtain an equivalent martingale measure.