Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 211 / 1

For holdings $h\in {\mathbb{R}}^n$, let $V_0=h\cdot P_0$ and $V_1=h\cdot P_1$. Under the investment-consumption arbitrage convention used here, an arbitrage satisfiesThe initial consumption is $-V_0$, and the terminal consumption is $V_1$: there is a possible gain with no external funding and no negative consumption. A pure-investment arbitrage additionally requires $V_0=0$. This convention is important because the last part distinguishes initial consumption from pure investment.

A pricing kernel is a strictly positive random variable $R$ such that $R{P}_1^i$ is integrable andIt is a one-period state-price density, converting a terminal payoff into its initial price. Strict positivity means $R>0$ almost surely, not merely $R\ge 0$. When cash has price $1$ at both dates, this identity forces $\mathbb{E}R=1$, so $d\mathbb{Q}=Rd\mathbb{P}$ defines an equivalent martingale measure.

For any holdings $h$, the pricing kernel identity implies $V_0=\mathbb{E}[R{V}_1]$. If $V_1\ge 0$ almost surely, then $V_0\ge 0$, so an arbitrage cannot have $V_0<0$. If also $V_0=0$, then the nonnegative integrable random variable $R{V}_1$ has zero expectation and hence is zero almost surely. Since $R>0$ almost surely, $V_1=0$ almost surely, ruling out a positive terminal gain. Therefore a pricing kernel rules out arbitrage.

Write $q_{15},q_{12},q_9$ for the state probabilities under an equivalent martingale measure. Since cash is constant and the European put option pays only in the lowest state, the pricing equations areTheir unique solution is $q_9={\xi }_0/2$, $q_{15}={\xi }_0/2-2/3$, $q_{12}=5/3-{\xi }_0$. Every physical state has positive probability, so equivalence requires all three values to be strictly positive. ThusFor each price in this open interval the pricing kernel takes values $3q_{15},3q_{12},3q_9$ in the three equally likely states, proving absence of arbitrage.

The necessity, including the exclusion of endpoints, can also be checked directly. The three terminal payoff vectors of cash, the stock and the European put option form the invertible matrixEach unit state payoff therefore has a replicating strategy, whose initial cost is the corresponding $q$. A zero or negative $q$ supplies an arbitrage, so the endpoints are genuinely excluded.

Write $(a,b,c)$ for holdings of cash, the stock and the European put option, and let $\gamma =-V_0$ be initial consumption. At the given price, $a=-10b-c-\gamma$, and the three terminal payoffs, in descending order of the stock price, areNonnegative terminal payoffs require $c+\gamma \le 2b$ and $c-\gamma \ge b$. Together with $\gamma \ge 0$, these imply $b\ge 2\gamma \ge 0$. If $b=0$ they force $\gamma =c=a=0$, which is not an arbitrage. If $b>0$, the highest-state payoff is at least $3b>0$, so every such choice is an arbitrage. Consequently the complete set isThe pure-investment arbitrages are exactly the choices $\gamma =0$:For example, $(a,b,c)=(-11,1,1)$ costs zero and pays $(4,1,0)$.