Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 32 / 3 / Solution

On a fixed finite horizon, a self-financing strategy consists of predictable stock holdings $h^i$, integrable against the price processes, and a bank account holding $h^0$. Its wealth and gains satisfyAn admissible trading strategy has discounted wealth $X_t/B_t$ bounded below by a deterministic constant throughout the horizon. With the bounded short rate in this part, an undiscounted deterministic lower bound is equivalent after adjusting the constant. An arbitrage is an admissible zero-initial-cost strategy with $X_T\ge 0$ almost surely and $\mathbb{P}(X_T>0)>0$.

Put $1=(1,\dots ,1{)}^{\top }$ and define the market price of riskThe bounded coefficients and bounded inverse make $\lambda$ bounded. Define the stochastic exponentialThe Novikov condition states that $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T\Vert {\lambda }_s{\Vert }^{2}ds)<\infty$ makes this exponential a true martingale with expectation one. Boundedness verifies the condition. Therefore $dQ=L_{T}dP$ defines an equivalent probability measure. The Girsanov theorem states thatis a $d$-dimensional Brownian motion under $Q$. Substituting into the asset equations givesThus $Q$ is an equivalent local martingale measure. The sufficient direction of the fundamental theorem of asset pricing says that such a measure excludes arbitrage among admissible self-financing strategies. Here the reason can be seen directly: $X/B$ is a local martingale under $Q$, and its lower bound makes it a supermartingale. Consequently $X_0=0$ implies ${\mathbb{E}}_Q[X_T/B_T]\le 0$, incompatible with the nonnegative terminal payoff of an arbitrage. Therefore the bounded-inverse market has no arbitrage.

For the requested law of one price counterexample, take a broader continuous-time local-martingale model on $[0,T]$. Let $X$ be a Brownian motion, let $\tau =\inf \{u\ge 0:X_u=-1\}$, and set $q(t)=t/(T-t)$ for $t<T$. DefineThe given hitting-time fact gives $\tau <\infty$ almost surely. The corresponding physical hitting time is $T\tau /(1+\tau )<T$, so $S$ reaches one and then stays one. In particular it is continuous at maturity and $S_t\ge 1$.

For a rigorous closed-horizon localization, stop additionally when $S$ hits $n+2$, or at $T$ if it never does. The stopped process is Brownian motion stopped on first exiting $(-1,n)$ and subjected to the deterministic clock; it is a bounded martingale, including at $T$. The stopping times increase to $T$, since every Brownian path on the finite random interval $[0,\tau ]$ has a finite maximum. Thus $S$ is a positive local martingale on the whole horizon. With the time-changed Brownian filtration, the original probability measure is an equivalent local martingale measure. The admissible-wealth supermartingale argument above proves no admissible arbitrage.

Nevertheless $S_T=B_T=1$ while $S_0=2\ne B_0=1$, so the law of one price fails. Shorting one share and buying two units of cash has zero initial cost and terminal gain one, but its wealth is $2-S_t$. It has no deterministic lower bound: for every $n$, the probability of Brownian motion hitting $n$ before $-1$ is $1/(n+1)>0$, obtained by optional stopping of the bounded stopped Brownian motion. That apparently profitable strategy is therefore inadmissible. The counterexample deliberately uses a strict local martingale; the bounded-coefficient class in the first part has true discounted asset martingales and does satisfy the law of one price.