Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 41 / 1

We prove the positive state-price density alternative using finite-dimensional convex cones. Put $w=(1+\Vert A_1\Vert {)}^{-1}$ and defineAll displayed expectations are finite, since $w\Vert A_1\Vert \le 1$. The set $D$ is a convex cone, so $K$ is a closed convex cone. A vector $x$ is nonnegative on $K$ exactly when $x^{\top }{A}_1\ge 0$ almost surely. Indeed, one direction follows by taking every nonnegative bounded test weight $f$; for the other direction, if $x^{\top }{A}_1<0$ on a set of positive probability, take its indicator as $f$ to obtain a strictly negative expectation. This also identifies $K$ as the closed cone generated by the support of a probability distribution of $A_1$.

The three cases $A_0\notin K$, $A_0\in K\setminus \mathrm{ri}K$, and $A_0\in \mathrm{ri}K$ are treated below. They give respectively a negative-cost separating portfolio, a supporting zero-cost portfolio with a strictly positive payoff on an event of positive probability, or the required positive pricing weight. Thus the alternatives are exhaustive.

They are mutually exclusive. If $A_0=\mathbb{E}[\nu A_1]$ with $\nu >0$ and a portfolio has nonnegative payoff, then $x^{\top }{A}_0=\mathbb{E}[\nu x^{\top }{A}_1]\ge 0$. A negative initial cost is impossible. If the payoff is strictly positive with positive probability, this expectation is strictly positive, so nonpositive initial cost is impossible too. The given weighted integrability permits taking the expectation of this scalar product.

In a one-period market, $x^{\top }{A}_0$ is the portfolio's initial cost and $x^{\top }{A}_1$ its terminal value. The first two cases are investment-consumption arbitrage: nonnegative wealth can be taken out at both dates, with a strict benefit at one of them. If a strictly positive bank account is traded, any negative initial cost can be invested in it, converting the opportunity into a zero-cost arbitrage with nonnegative terminal payoff and positive payoff with positive probability.

In the third case, $\nu$ is a strictly positive state-price density. If a traded numéraire has price $B_0>0$ and terminal value $B_1>0$, its pricing identity is $\mathbb{E}[\nu B_1]=B_0$. ConsequentlyThe density has mean one and is strictly positive, so $Q$ is equivalent to the physical measure. Thus discounted asset prices are martingales in the one-period sense. This is the fundamental theorem of asset pricing formulation of absence of arbitrage; normalization to a probability measure uses the positive traded numéraire, not an unsupported assumption that $\mathbb{E}\nu =1$.

Suppose $A_0$ lies on the relative boundary of $K$ in its span $L$. There is a nonzero $x\in L$ with $x^{\top }y\ge 0$ for every $y\in K$ and $x^{\top }{A}_0=0$. Here is a direct construction of the supporting vector. Choose points $z_n\in L\setminus K$ converging to $A_0$, let $p_n$ be their nearest points in $K$, and set $x_n=(p_n-z_n)/\Vert p_n-z_n\Vert$. The minimization inequality for a nearest point gives $x_n^{\top }y\ge 0$ on $K$ and $x_n^{\top }{p}_n=0$. A subsequence of the unit vectors converges to a unit vector $x\in L$. Since $0\le x_n^{\top }{A}_0\le \Vert A_0-z_n\Vert$, it has the asserted support properties.

The test-weight characterization of $K$ now gives $x^{\top }{A}_1\ge 0$ almost surely. This random payoff cannot vanish almost surely: otherwise $x$ would annihilate every element of $D$, hence all of $L$, contradicting $x\in L$ and $\Vert x\Vert =1$. ThereforeThis supplies the positive-terminal-payoff alternative. If $K$ has zero-dimensional span, its relative boundary is empty, so that case requires no supporting vector.

If $A_0\notin K$, let $p$ be a nearest point of the closed convex cone $K$ to $A_0$. Existence follows by minimizing the continuous squared distance on a sufficiently large compact ball. The first-order minimization inequality is $(A_0-p{)}^{\top }(y-p)\le 0$ for every $y\in K$. Set $x=p-A_0$. Taking $y=0$ and $y=2p$, both in the cone, gives $x^{\top }p=0$. ConsequentlyThe cone's test-weight characterization again yields $x^{\top }{A}_1\ge 0$ almost surely. This is separation from a closed convex cone with the sign needed forThe strict benefit is already available at the initial date, even if the terminal payoff happens to be zero.

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.