Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 30L (Stochastic Financial Models)

In discounted terms, an arbitrage is a portfolio with zero initial value whose terminal payoff is nonnegative in every state and strictly positive with positive probability. A risk-neutral measure is a probability measure equivalent to the physical measure under which every discounted asset price is a martingale. The one-period fundamental theorem of asset pricing says that a finite one-period market is arbitrage-free exactly when it has a risk-neutral measure.

If $q\in \mathcal{Q}$ and $\varphi \in \mathcal{A}$, thenbut strict positivity of $q$ and the nonzero vector $P\varphi \ge 0$ make the left side positive. Thus the two sets cannot both be nonempty.

Conversely, suppose $\mathcal{A}=\varnothing$. The subspace $L=\mathrm{Im}P$ is disjoint from the compact simplexStrict separation gives a vector $q$ that vanishes on $L$ and is positive on all of $\Delta$. The first property says $P^{T}q=0$; testing the vertices of $\Delta$ shows every $q_i>0$. After normalization ${\sum }_i{q}_i=1$, this gives $q\in \mathcal{Q}$. This is Stiemke theorem in normalized form. Therefore

Let $P$ be the state-by-asset matrix of discounted gains and identify $Y$ with its state vector. Fix $q^0\in \mathcal{Q}$. If $q\ge 0$, $P^{T}q=0$, and $q\ne 0$, thenbelongs to $\mathcal{Q}$. The assumption and passage to the limit imply $q^{T}Y\ge 0$ for every nonnegative $q\in \ker P^T$.

The finite-state superhedging alternative now gives $\theta$ withThis residual cannot vanish identically, since then every $Q\in \mathcal{Q}$ would have $E_{Q}Y=q^{T}P\theta =0$. Hence it is strictly positive in at least one state, which has positive physical probability. This is the required inequality.

The same mixing argument with a fixed $q^0\in \mathcal{Q}$ shows thatfor every $q\ge 0$ in $\ker P^T$, including boundary vectors. By the finite-state superhedging alternative, there is $\varphi \in {\mathbb{R}}^d$ such thatwhich is the asserted almost-sure inequality. No strict residual is required here.