Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 211 (Advanced Financial Models)

An arbitrage is a finite-horizon previsible strategy with no positive initial cost, nonnegative cash flows at every date, and a strictly positive cash flow with positive probability at some date, after liquidation. Equivalently, one may require zero initial value and a nonnegative terminal gain that is positive with positive probability, after retaining intermediate cash flows in a cash account.

A martingale deflator is a strictly positive adapted process $Y$ such that every deflated cum-dividend asset gain has zero conditional drift:Using the definitions of ${\pi }^H$ and ${\xi }^H$,The holdings $H_t$ are ${\mathcal{F}}_{t-1}$-measurable, so the right side is a martingale transform of the deflated asset-gain local martingale. Hence $Z$ is a local martingale.

The fundamental theorem of asset pricing says, in this discrete-time formulation, that the market has no arbitrage if and only if it admits a strictly positive martingale deflator. Under a chosen positive numeraire this is equivalent to the existence of an equivalent martingale measure for numeraire-discounted gains.

Let $Y$ be a martingale deflator for the original arbitrage-free market. Part b shows thatis a local martingale. Thus $Y$ also deflates the gains of the added asset whose price and dividend are $({\pi }^H,{\xi }^H)$. It already deflates the original $n$ assets, so it is a martingale deflator for the enlarged market. The fundamental theorem of asset pricing implies that the enlarged market has no arbitrage. This expresses the fact that adding a dynamically replicated asset cannot create an arbitrage.

The strategy $\eta$ is a self-financing portfolio because ${\xi }_t^{\eta }=0$, and its price is known one period in advance. Suppose its price first became nonpositive. On the event, known immediately before that date, that the next price is nonpositive while the current price is positive, an investor can short or buy the self-financing portfolio with the sign that gives no downside, finance the position at the current date, and close it at the known next price. This gives a nonnegative gain and a strictly positive gain whenever the price changes sign or reaches zero from a positive value.

More formally, stopping and scaling $\eta$ on the first such predictable event constructs an arbitrage. Since the market has no arbitrage and ${\pi }_0^{\eta }>0$, induction over dates gives

Normalize the self-financing strategy $\eta$ by definingPart e makes this well-defined, andMoreover,Both $K$ and $L$ have constant unit price and predictable dividends. Their difference has zero price and predictable dividend ${\xi }^K-{\xi }^L$. If that dividend were nonzero with positive probability, taking its known sign would produce an arbitrage. Therefore ${\xi }^K={\xi }^L$, proving

Assume condition (1). Given an adapted cash-flow process $X_1,\dots ,X_T$, construct the holdings backwards. Set $H_{T+1}=0$. Once $H_{t+1}$ is known, the random variableis ${\mathcal{F}}_t$-measurable. Condition (1) supplies an ${\mathcal{F}}_{t-1}$-measurable $H_t$ satisfyingHence ${\xi }_t^H=X_t$ for every $t\le T$, and setting later holdings to zero proves condition (2).

Conversely, let $X_T$ be any ${\mathcal{F}}_T$-measurable random variable and apply condition (2) to the adapted process with cash flow $X_T$ at $T$ and zero cash flow earlier. Since $H_{T+1}=0$,where $H_T$ is ${\mathcal{F}}_{T-1}$-measurable. This is condition (1). The conditions are therefore equivalent and describe market completeness.

Let $Y$ and $\tilde{Y}$ be normalized martingale deflators. For any date $T$ and bounded ${\mathcal{F}}_T$-measurable $X_T$, condition g supplies a strategy whose only prescribed cash flow is $X_T$ at $T$. Applying the martingale identity from part b givesThus $\mathbb{E}[(Y_T-{\tilde{Y}}_T)X_T]=0$ for every bounded ${\mathcal{F}}_T$-measurable $X_T$. Taking indicators, or the sign of the difference, shows $Y_T={\tilde{Y}}_T$ almost surely. Since $T$ was arbitrary, the normalized martingale deflator is unique.

A T-forward measure is a probability measure $Q^T$, equivalent to the physical measure, under which prices expressed in units of the positive maturity-$T$ bond are martingales. Equivalently, every attainable payoff $X_T$ has time-$t$ priceThe zero-coupon bond is the numéraire.

The forward contract initiated at $t$ has payoff $S_T-F_t^T$ and zero value. Pricing under the T-forward measure givesBecause $F_t^T$ is ${\mathcal{F}}_t$-measurable and $B_t^T>0$,The tower property of conditional expectation therefore makes $(F_t^T{)}_{t<T}$ a $Q^T$-martingale.

If $K_1\le K_2$, then pointwiseThe positive pricing formula under the T-forward measure givesTherefore $C_t^{T,K_1}\ge C_t^{T,K_2}$, so the call price is non-increasing in strike.

Define the piecewise-linear functionOn $[K_j,K_{j+1})$, its slope is $g^{\prime }(K_j)$. Since $g$ is convex, $g^{\prime }$ is nondecreasing and hence $h^{\prime }(s)\le g^{\prime }(s)$ wherever the derivatives exist. As $h(0)=g(0)$, integration gives $h(s)\le g(s)$ for every $s\ge 0$.

Positive no-arbitrage pricing, the forward identity ${\mathbb{E}}_{Q^T}[S_T\mid {\mathcal{F}}_t]=F_t^T$, and the call-price formula now give

The cash-discounted stock is a positive continuous local martingale, because its dynamics contain no drift. Applying Itô formula to ${\pi }_t=U(t,v_t,S_t)$, the displayed partial differential equation cancels its drift exactly, leaving another local martingale. Since $U$ is bounded, $\pi$ is in fact a true martingale.

Thus the physical measure itself is an equivalent local martingale measure relative to cash for all three traded assets. The continuous-time fundamental theorem of asset pricing rules out arbitrage, more precisely no free lunch with vanishing risk, in the usual admissible class.

The bounded local martingale $\pi$ is a true martingale. Its terminal condition is ${\pi }_T=U(T,v_T,S_T)=g(S_T)$, soThis is also the Feynman-Kac formula for the displayed backward equation.

The explicit stochastic exponential solutions satisfyConditionally on the path generated by $W$, the last stochastic integral is a centered Gaussian random variable with variance $Y_T$, because $W^{\perp }$ is independent of $W$. The conditional expectation of $g(S_T)$ is thereforeTaking expectations and using part b proves the result.

Integrating the variance equation givesThe Doléans-Dade exponential solution of $d{\tilde{S}}_t=\rho {\tilde{S}}_t\sqrt{v_t}d{W}_t$ isSubstitution yields

The pair $(v_t,Y_t)$ is a Markov diffusion with infinitesimal generatorPart d shows that the terminal condition in the equation for $\tilde{U}$ is exactlywhen evaluated at $(v_T,Y_T)$. The Feynman-Kac formula applied to the displayed backward equation therefore givesPart c identifies the right side with ${\pi }_0$.

Definefor $t\ge 1$, with $A_0=M_0=0$. The supermartingale property makes $\Delta A_t\ge 0$, and it is ${\mathcal{F}}_{t-1}$-measurable, so $A$ is previsible and nondecreasing. The increments of $M$ have conditional mean zero, so $M$ is a martingale. Finally,and summation gives the Doob decomposition in discrete time

The recursion makes $U$ a supermartingale and gives $U_t\ge Z_t$ at every date. The optional sampling theorem for a supermartingale for the bounded stopping time $\tau$ therefore gives

Take the first entry into the stopping region:This set is nonempty because $U_T=Z_T$. Before ${\tau }^{\ast }$, the recursion hasso the stopped process $U_{t\wedge {\tau }^{\ast }}$ is a martingale. Optional sampling and $U_{{\tau }^{\ast }}=Z_{{\tau }^{\ast }}$ giveThus ${\tau }^{\ast }$ is an optimal stopping time.

For every stopping time $\tau$, the martingale case of the optional sampling theorem for a supermartingale gives $\mathbb{E}{X}_{\tau }=X_0=0$. HenceUse the optimal stopping time from part c and the pathwise inequalityTaking expectations gives

Let $U=U_0+M-A$ be its Doob decomposition in discrete time and choose the martingaleSince $Z_t\le U_t$ and $A_t\ge 0$,For the first optimal stopping time ${\tau }^{\ast }$, the complementarity for the Snell envelope compensator implies $A_{{\tau }^{\ast }}=0$: all compensator increments before ${\tau }^{\ast }$ vanish. Since $Z_{{\tau }^{\ast }}=U_{{\tau }^{\ast }}$,Thereforepathwise, and taking expectations proves the asserted equality.